the *quantum of conductance*

$$ G_0 = \frac{e^2}{h} = \frac{1}{25\,812.807\,\Omega}, \qquad (3.45) $$

which is itself expressed in the universal constants *e* and *h*. Its inverse, which may be called the quantum of resistance, has become the unit of resistance.

The expression (3.44) turns out to be extremely fruitful in the understanding of quantum electronic transport in several respects. First, we immediately see that the mean free path is a lower bound for the validity of the formula. Indeed, the conductanc would diverge in the limit *L* ≪ *l*<sub>e</sub>, where we need to take into account a ballistic approach to electric transport rather than the diffusive one of the Einstein relation. Furthermore, one also sees the intimite relation between the width or the cross section of the conductor and the Fermi wavelength. Although we have not explicitly taken into account the wave nature of the electrons in the previous section, this result already anticipates that the conductor may also be viewed as a wave guide, with transverse modes quantised by multiples of half the (Fermi) wavelength. The ratio *W*/*λ*<sub>*F*</sub> thus determines the *number of transverse modes that are transmitted*.

### 3.3.1 What is electric conductance and what is a conductance measurement? A new paradigm

Before a more detailed discussion of ballistic *vs*. diffusive transport, let us step back a bit and ask ourselves the question: what is more fundamental? the conductivity or the conductance of a device? In order to aboard this question, we should thoroughly think about the nature of the transport measurement. And ask ourselves: do we *measure* a conductivity or a conductance? From the experimental point of view, the answer is evident. One naturally measures a conductance and not a conductivity, since the latter is then deduced from the measured conductance and the geometrical properties of the conductor.

Let us be more precise than that and consider the different results of a transport measurement through the device, which is depicted in figure 3.3 and that consists of a metallic ring (in a magnetic field) contacted by four contacts. The different setups of the measurement shown on the right-hand side of the figure may seem quite unusual since we are commonly used to a
<figure><img src="image_24.png" /><figcaption>Figure 3.3: Conductance through a ring as a function of the magnetic field threading the surface surrounded by the ring, for three different arrangements of the contacts. Two of the contacts (in red) serve as the source and drain by which one drives the current through the device, while the other two (in blue) allow one to measure the voltage drop. After A. Benoit et al., Phys. Rev. Lett. 57, 1765 (1986).</figcaption></figure>

two-contact measurement, where the source and drain (the contacts used to drive a current $I$ through the device) are also the contacts by which we measure the voltage drop $V$, which allows us to obtain the conductance $G = I/V$. However, this is not necessary and one is perfectly allowed to use two other contacts, such as the blue ones in the setups sketched in figure 3.3 to measure a voltage drop, and reserve the source and drain (red contacts in the figure) just for the current. One needs to emphasise two experimental facts. First, the three measured conductance curves as a function of the applied magnetic field are clearly different from one another although the conductor remains the same. Only the arrangement of the contacts is changed. Second, while they seem to show some random fluctuations, they are perfectly reproducible, i.e. another run for the conductance measurement after a certain time reproduces the conductance curve provided that one keeps the arrangement of current and voltage contacts. Furthermore, there exist some relations between the measurements. Indeed, the first and the last curve happen to be identical if one changes the orientation of the magnetic field $B \rightarrow -B$.

Based on these experimental facts, we must accept that the relation $G = I/V$ is too simple to account for the complex experimental situation, we have just encountered. Instead, we must admit that the conductances and
<figure><img src="image_25.png" /><figcaption>Figure 3.4: 2D conductor. (Upper panel) In the case of a diffusive conductor, of length $L \gg l_e$, we expect Ohm’s law to remain valid, in line with the conductance (3.44), which one may measure as the ratio between the current $I$ in the conductor and the voltage drop $V_1 - V_2$ in the contacts. (Lower panel) For a ballistic conductor, there is yet a current through the conductor, but the voltage difference between the contacts is not necessarily zero so that the conductance could take a finite value. However, if we are capable – via a non-invasive measurement – of measuring the voltage drop between the two positions indicated by the black arrows, there is no voltage drop due to the absence of relaxation processes, and the corresponding conductance should thus diverge.</figcaption></figure>

the resistances depend on the manner how the device is contacted to the outer world, via the contacts and their arrangement. Quite generally, the conductance is thus a tensor

$$G_{ij,kl} = (R_{ij,kl})^{-1} = \frac{I_{ij}}{V_k - V_l} = \frac{I}{V_k - V_l} \quad (3.46)$$

where the indices $i$ and $j$ denote the source and the drain, respectively, and $V_k$ and $V_l$ are the respective electrostatic potentials in the contacts $k$ and $l$. Usually, we limit ourselves to a single source and a single drain, i.e. only two contacts are used to inject or collect current in the device so that $I_i = I_j = -I_{ij}$. This means that we do not have other contacts or current “leakages” where electrons could disappear or emerge in the device. Formally, this means that all other contacts than the source and the drain have an infinite internal resistance and are used only to measure a voltage drop.

In order to further appreciate the central role of the conductance and to set the stage for the following discussions of this chapter, let us take a closer look at the two 2D conductors shown in figure 3.4, connected by the
source (at voltage $V_1$) and the drain (at $V_2$) – one is taken to be a usual diffusive conductor (upper panel) and the other one is ballistic (lower panel). For the diffusive conductor, with $L \gg l_e$, we may use Ohm’s law and the conductance should be $G = 2(e^2/h)\pi(W/\lambda_F)(l_e/L)$ for electrons with a two-fold spin degeneracy ($g = 2$). How about the ballistic conductor? There is also a current $I$ injected at the source that passes through the device and that is collected at the drain. Indeed, all electrons injected at the source make their path through the conductor and reach the drain since there are no scatterers that allow the electrons to be backscattered. Naively, one may expect a zero resistance for this situation and thus a divergent conductance. However, this would require that the source and the drain be at the same electrostatic potential, $V_1 = V_2$. As we show below, this constraint is too strong. One may indeed put one contact to the ground, while the other one is at a different potential, in which case the conductance takes a finite value $G = I/(V_1 - V_2)$.

Notice that this is the classic measurement setup where source and drain are also the contacts, where we measure the voltage drop and thus a conductance or resistance. We may imagine, nevertheless, that we are capable of measuring the voltage difference between two positions in the conductor (indicated by the black arrows in figure 3.4) in a non-invasive manner, i.e. without disturbing the electrons and their wave functions on their path through the conductor. What result would you expect for the voltage difference between these two positions? The inevitable answer to this question is: $\Delta V = 0$. The reason is the following. Consider an electron injected at the source. This electron must be at thermodynamic equilibrium with the source, i.e. its energy is given by the chemical $\epsilon_F - eV_1$ of an electron at the Fermi level $\epsilon_F$ plus the electrostatic potential $-eV_1$ imposed in the contact. Similarly an electron injected at the drain is at thermodynamic equilibrium with the latter and thus has an energy $\epsilon_F - eV_2$. The case of an electron injected at the drain may, at first sight, seem contradictory since the drain is used to collect the current. This is, however, only an average statement: there may be a difference in the density of electrons $n_1$ injected in the source and that $n_2$ injected in the drain so that there is a net current flowing between source and drain. If we now measure the potential of the electrons in the conductor at a fixed position $x$, we will obtain

$$V(x) = \frac{n_1(x)V_1 + n_2(x)V_2}{n_1 + n_2}, \qquad (3.47)$$
where $n_{1/2}(x)$ is the density of electrons at the position $x$ that are at the potential $V_{1/2}$. Naturally, we have $n_1(x=0) = n_1$ and $n_2(x=L) = n_2$. Since we consider ballistic transport, the density of electrons moving from the source to the right is independent of the position since there is no obstacle that could scatter an electron back to the source, while in a diffusive conductor $n_1(x)$ would decrease once the position $x$ is further away from the source. Similarly, the density of electrons injected at the drain turns out to be constant over the ballistic conductor so that $V(x) = V$ is constant over the length of the conductor. Therefore, we find $\Delta V = 0$ for the voltage difference between the two black arrows regardless of the distance between them! We revisit this somewhat sketchy argument in more quantitative detail below in the discussion of the chemical-potential profile of electrons in diffusive and ballistic conductors, in section 3.4.1.

From this qualitative discussion, we may retain two important things. First, as mentioned above, there is no means to measure *the* (single) conductance of a conductor, but the conductance depends on the arrangement and the nature of the contacts used to measure it. If we measure the conductance between the source and the drain, via the voltage drop between precisely these two contacts, we measure a finite conductance, while we would measure an infinite conductance between two additional (non-invasive) contacts situated at the black arrows in figure 3.4. Second, in contrast to diffusive transport, the result may be independent of the device geometry. This is clearly the case for the latter *four-terminal* conductance associated with a zero resistance, and we show below that this is also the case for the former ballistic *two-terminal* conductance.

### 3.3.2 Classical ballistic vs. diffusive transport – Sharvin conductance

In order to obtain a first idea about the ballistic (two-terminal) conductance measured between the source and the drain, we consider classical ballistic electrons for the moment. They can pass from one $d$-dimensional volume $A$ to a second one $B$ via a slit of surface $S = W^{d-1}$. The setup is sketched in figure 3.5(a). The electronic densities in the reservoirs are determined by the chemical potentials $\mu_A$ and $\mu_B$, respectively, and the latter can be tuned by the voltages $V_{A/B}$, $\mu_{A/B} = \epsilon_F - V_{A/B}$. One obtains for the electronic densities directly from the density of states $\rho(\epsilon)$ and the Fermi-Dirac occupation factor
<figure><img src="image_26.png" /><figcaption>Figure 3.5: (a) Classical ballistic transport between two reservoirs. The two reservoirs A and B are at the chemical potentials $\mu_A$ and $\mu_B$, respectively. They may exchange electrons through the slit in a wall separating the reservoirs. This slit plays the role of the conductor. The densities in the reservoirs are given by $n_A$ and $n_B$. (b) Case of a conducting channel of width $L$. The conduction properties do not change with respect to (a) because an electron that enters into the opening (slit) maintains the x-component of its velocity during its passage through the channel, i.e. it is not backscattered.</figcaption></figure>

$$f(\epsilon, \mu = \epsilon_F - eV_{A/B}),$$

$$n_{A/B} = \int d\epsilon \rho(\epsilon) f(\epsilon, \epsilon_F - eV_{A/B}) \approx n - eV_{A/B} \frac{\partial n}{\partial \mu}, \quad (3.48)$$

where we have made a Taylor expansion of the occupation factor to linear order in $|eV_{A/B}| \ll \epsilon_F$. Furthermore, we have defined the average density at the Fermi level

$$n = \int d\epsilon \rho(\epsilon) f(\epsilon, \epsilon_F), \quad (3.49)$$

and we obtain from the symmetry of the Fermi-Dirac under exchange of $\epsilon$ and $\mu$ in the argument (3.2),

$$\frac{\partial n}{\partial \mu} = \int d\epsilon \rho(\epsilon) \frac{\partial f}{\partial \mu} = \int d\epsilon \rho(\epsilon) \left(-\frac{\partial f}{\partial \epsilon}\right) \approx \rho(\epsilon_F), \quad (3.50)$$

where we have again used the zero-temperature expression (3.34) of the derivative of the occupation factor.

The above discussion of the densities in the two reservoirs and of their relation to their respective densities of states turns out to be fruitful in the analysis of the current. Let us first consider the particle current that passes through the slit from A to B. It is given by the density $n_A$ of particles in the reservoir A times the surface of the d-1-dimensional slit S times the average velocity in the x-direction $\langle v_x \rangle_+$. Indeed, the particles the particles
arrive at a random angle at the slit, but the x-component must evidently be positive in order to pass to the other reservoir B. This is indicated by the subscript + at the average velocity.<fnref n="5" /> Similarly, the particle current from B to A is given by $n_B S$ times the average velocity in the -x-direction, $\langle v_x \rangle_-$ which is up to a sign the same as that in the +x-direction, $\langle v_x \rangle_+ = -\langle v_x \rangle_-$.
The net particle current is therefore given by

$$I_p = (n_A - n_B)S\langle v_x \rangle_+ = -e(V_A - V_B)\frac{\partial n}{\partial \mu}S\langle v_x \rangle_+ \approx -e(V_A - V_B)\rho(\epsilon_F)S\langle v_x \rangle_+ \quad (3.51)$$

Now remember that we are interested in the charge current rather than in the particle current of electrons, but both are simply related by the electric charge $-e$ so that the net electric current at low temperatures finally becomes

$$I = e^2 \rho(\epsilon_F) S \langle v_x \rangle_+ (V_A - V_B), \quad (3.52)$$

and we may immediately read off the ballistic conductance

$$G_{\text{bal}} = \frac{I}{V_A - V_B} = e^2 \rho(\epsilon_F) S \langle v_x \rangle_+, \quad (3.53)$$

which is also called *Sharvin conductance*. This result shows clearly, as mentioned above, that the ballistic conductance (for the moment, in the classical regime) is finite and does not diverge in this two-terminal measurement! Furthermore we notice that the Sharvin conductance depends only on the (average) velocity of the electrons in the x-direction. This x-component remains

$$dN_{\theta} = n_{A} S v_{x} dt \frac{d\Omega}{S_{d}},$$

where $S_d$ is again the surface of the unit sphere in $d$ dimensions, which plays the role of the integrated solid angle, and $d\Omega(\theta)$ is an infinitesimal solid angle at an angle of incidence $\theta$. The particle current is then given by $I_p = dN/dt$ integrated over the solid angle, and the average velocity reads

$$\langle v_x \rangle_+ = \int_+ \frac{d\Omega(\theta)}{S_d} v \cos \theta,$$

where the subscript + at the integral indicates that we only consider positive values of $\cos \theta$, and $v$ is the modulus of the average velocity, which is given by the temperature in a classical gaz, $mv^2 = dk_B T$, or by the Fermi velocity $v_F$ for the electron gaz in the degenerate limit $T \ll T_F$.

<footnote n="5">⁵To be more quantitative, the number of particles $dN_\theta$ that passes from A to B through the slit at an angle $\theta$ in an infinitesimal time interval $dt$ is given by</footnote>
<figure><img src="image_27.png" /><figcaption>Figure 3.6: Comparison between ballistic Sharvin (a) and diffusive (b) transport. (a) In the ballistic case ($L \ll l_e$), there are no scatterers present in the conducting channel, and each electron that enters into the channel with $v_x > 0$ transits from the left to the right to the conservation of $v_x$ in the reflection processes at the channel walls. The conductance is thus independent of the channel length $L$. (b) For diffusive transport ($L \gg l_e$), the channel contains many scatterers, and electrons can thus be backscattered. The backscattering probability increases (linearly) with $L/l_e$.</figcaption></figure>

unaltered even if we consider a conducting channel of width $L$ instead of the slit as long as the length is smaller than the mean free path. This is sketched in Fig. 3.5(b). In this case, the electrons bounce back at the walls of the conducting channel (in the $y$-direction) while they are never backscattered due to the conservation of the $x$-component of their velocity. Backscattering would require scattering events in the form of static disorder that is absent in the ballistic limit $L \ll l_e$ we consider here.

## Sharvin and diffusive conductances in 2D conductors

Let us be a little bit more quantitative about this result and evaluate it for a 2D conductor, for which the density of states (3.6) reads

$$ \rho(\epsilon) = \frac{4\pi}{h} \frac{1}{\lambda_F v_F} \qquad (3.54) $$

and $S = W$. We still need to evaluate the average electronic velocity in the $x$-direction, the modulus of which is simply the Fermi velocity since we consider the zero-temperature limit of the electron gas here,

$$ \langle v_x \rangle_+ = \frac{1}{2\pi} \int_{-\pi/2}^{\pi/2} d\theta v_F \cos\theta = \frac{v_F}{\pi}, \qquad (3.55) $$
where we have used $S_2 = 2\pi$ and $d\Omega(\theta) = d\theta$ in $d = 2$ dimensions. The 2D Sharvin conductance may thus be written as

$$G_{\text{bal}} = 2 \frac{e^2 2W}{h \lambda_F}. \qquad (3.56)$$

We may want to compare this result to the corresponding diffusive conductance (3.44) in two dimensions,

$$G_{\text{diff}} = 2 \frac{e^2 W l_e}{h \lambda_F L}. \qquad (3.57)$$

One notices that, apart from a difference in the geometric prefactor (2 instead of $\pi$), the main difference is the factor $l_e/L$, which is absent in the Sharvin conductance of a ballistic conductor. This is indeed expected from the argument elaborated at the end of the last paragraph: each electron that enters into the conducting channel is transmitted from the left to the right (and vice versa) regardless of the length of the conductor, as long as $L$ is smaller than $l_e$. In other words, the mean free path ceases to be a relevant length scale once the system size is smaller than the mean free path [see also figure 3.6(a)]. As a thumb rule, one might this obtain the ballistic conductance from the diffusive one by omitting the *diffusion factor* $l_e/L$. As a consequence, we realise that the conductance is independent of the length of the conductor, in contrast to Ohm's law. The diffusive conductor is sketched in figure 3.5(b) as compared to the ballistic case in figure 3.6(a).

While the Sharvin conductance is a classical quantity, obtained by considering electrons as classical particles, it bears already a hint of quantum mechanics. Indeed, similarly to the derivation of the conductance (3.44) we have used some ingredients from quantum mechanics in the form of the Pauli exclusion principle for fermions, which has been taken into account in the form of the Fermi-Dirac occupation factor. For this reason we can write the Sharvin conductance (as well as the classical diffusive one) in terms of the Fermi wavelength. This Fermi wavelength is precisely the bridge to a quantum-mechanical interpretation of Sharvin's result since it is the wavelength associated with an electron at the Fermi level. One may therefore also interpret the ratio $2W/\lambda_F$ as the *number of transverse modes* of wavelength $\lambda_F$ that match the width $W$ of the slit. Indeed, the slit imposes boundary conditions to the wave passing through it, in analogy with an optical wave guide: the wave function must vanish at the edges of the slit, as it is the case
for waves with a transverse wavelength that is a multiple of twice the width [see figure 3.5(b)]. As we show below in a full quantum treatment of the quantum ballistic regime $L \ll l_e \ll L_\phi$, the expression (3.56) for the Sharvin conductance remains valid in this regime if we take the integer part of the ratio $2W/\lambda_F$, which indicates the number of modes below the Fermi level that contribute to the quantum transport.

## 3.4 Conductance through a 1D wire and the Landauer formula

Before presenting a full calculation of the conductance through a quantum wire, let us briefly evaluate the Sharvin conductance (3.53) in an arbitrary dimension $d$. The density of states at the Fermi level is then given by the general expression (3.6), with

$$ \rho(\epsilon) = \frac{2}{h} S_d \frac{1}{\lambda_F^{d-1} v_F}, \qquad (3.58) $$

and we still need to evaluate the average velocity

$$ \frac{\langle v_x \rangle_+}{v_F} = \frac{1}{S_d} \int_+ d\Omega(\theta) \cos\theta. \qquad (3.59) $$

The solid angle $d\Omega$ can indeed be evaluated from the following geometric consideration: in 3D, it would simply be a circumference of a disk of radius $\sin\theta$, while in a general dimension $d$ the circumference of the disk needs to be replaced by the surface $S_{d-1}$ of the unit sphere in $d-1$ dimensions, and its radius is given by $\sin^{d-2}\theta$,

$$ d\Omega(\theta) = d\theta S_{d-1} \sin^{d-2}\theta. \qquad (3.60) $$

We can thus perform the integration to obtain the average velocity, with the help of $\int_+ d\Omega(\theta) \cos\theta = S_{d-1} \int_0^{\pi/2} d\theta \cos\theta \sin^{d-2}\theta = S_{d-1}/(d-1)^6$ so that we have

$$ \frac{\langle v_x \rangle_+}{v_F} = \frac{S_{d-1}}{(d-1)S_d}, \qquad (3.61) $$

<footnote n="6">⁶Notice that we have to integrate the angle $\theta$ now only in an interval $[0, \pi/2]$ since we have taken into account its “negative values” already in the surface $S_{d-1}$. Otherwise, we would count twice the solid angle.</footnote>
and the explicit form of the Sharvin conductance for electrons in d dimensions thus reads

$$G_{\text{bal}} = 2 \frac{e^2}{h} \frac{S_{d-1}}{d-1} \left(\frac{W}{\lambda_F}\right)^{d-1} = 2 \frac{e^2}{h} A_{d-1} \left(\frac{W}{\lambda_F}\right)^{d-1}, \quad (3.62)$$

where $A_d = S_d/d$ is the volume of the unit sphere in d dimensions. For $d = 2$ (and $A_2 = \pi$), we retrieve the expression (3.56) for the 2D Sharvin conductance.

We finish this section on classical ballistic transport with a short notice on the apparent contradiction of this regime with respect to the length scales at work. We have seen that the ballistic regime is that of devices with a system size smaller than the mean free path $L \ll l_e$, while the classical regime requires the system size to be larger than the phase-coherence length $L \gg L_\phi$. However, we have also argued that the phase-coherence length is larger than the mean free path so that there is in principle not much space for the classical ballistic regime. Indeed, the reader may consider the above discussion of this regime as a purely academic exercise to get prepared for the study of transport in the quantum ballistic regime and to appreciate the disappearance of one length scale that comes along with the loss of a well-defined conductivity as a relevant microscopic quantity here. However, this is a little bit too harsh, and we will see in the discussion of a milestone experiment in section 3.6 that the regime of classical ballistic transport may be reached in the so-called *quantum point contact*, a lateral constriction of a 2D high-mobility electron gas, at high temperatures.

### 3.4.1 Single-channel conductance

In the preceding section, we have seen that, in order to define a conductance, we need not only to take into account the intrinsic properties of the conductor, but the manner how the latter is connected to the outer world is highly relevant. We therefore discuss now the electric transport through a 1D quantum wire, which may be viewed as the paradigmatic example of quantum transport, connected to two electric contacts that are the source and the drain. The setup is shown in figure 3.7, and let us be precise about the different elements. The system, which is here generally considered to be a diffusive quantum wire (grey part in the figure) is connected via ballistic leads to the source and the drain, by which the current $I$ is driven through
<figure><img src="image_28.png" /><figcaption>Figure 3.7: Quantum transport through a 1D quantum wire. The system, which may be diffusive or ballistic (region named “scatterer”), is connected to the source and drain (called *contacts*, *terminals* or *reservoirs*) via ballistic wires (leads).</figcaption></figure>

the system. Naturally, the source and drain are electric contacts that are also called terminals or reservoirs due to their thermodynamic role in the electronic transport. Indeed, we consider that an electron that is injected into the conductor at one of the contacts to be in thermodynamic equilibrium with the contact, i.e. the injected electron is at the temperature $T_i$ and the chemical potential $\mu_i = \epsilon_F - eV_i$ of electrons in the contact $i$, which thus plays the role of a thermodynamic particle reservoir. The chemical potential can to some extent be varied by the electrostatic voltage $V_i$ applied to this contact, and we generally consider that the associated energy $-eV_i$ is much smaller than the Fermi energy, $e|V_i| \ll \epsilon_F$.

Now that we have fixed the terminology, we make some further assumptions about the system and the contacts. First, the system is supposed to be either ballistic or diffusive, but we consider only elastic scattering. The electrons thus remain phase-coherent during the full time they remain in the conductor, leads included. On the other hand, we suppose that the electrons that enter into a contact lose their phase memory and get thermalised in the bath. As already mentioned above, an electron leaving the contact is at thermodynamic equilibrium with the contact, i.e. it is injected at the temperature $T$ and the chemical potential $\mu = \epsilon_F - eV$ of the contact. There is thus no coherence between an incoming and an outgoing electron in the contact, which we call *Ohmic contact*. We may consider in some occasions “non-invasive contacts”, in which case the electron does not lose its phase coherence during a passage in this contact, but we specify such situation explicitly. Otherwise, we always mean an Ohmic contact or a thermodynamic
reservoir when we speak about a contact.

As one sees from the setup shown in figure 3.7, we are now confronted with a problem of 1D quantum mechanics the bases of which are treated in any introductory class of quantum physics. The scatterer may be viewed as a potential barrier, and we may thus appeal to the phenomenon of quantum-mechanical tunneling through a barrier described by a tunneling amplitude $t$ that yields the transmission coefficient $\mathcal{T} = |t|^2$. Let us consider the current density through the barrier carried by an electron in the state $|k, s\rangle$, in a parabolic band, transmitted from the left contact to the right one,

$$j_k = -e \frac{\hbar}{m} \mathrm{Re} \left( \psi_R^* \frac{\partial}{i\partial x} \psi_R \right), \qquad (3.63)$$

where

$$\psi_R = \frac{t}{\sqrt{L}} e^{ikx} \qquad (3.64)$$

is the transmitted part of the wave function on the right-hand side of the scattering region. If we allow the transmission coefficient to depend on the energy $\epsilon_k$ of the state, as it is usually the case, this yields

$$j_k = -e \frac{\hbar k}{mL} |t|^2 = -e \frac{v_k}{L} \mathcal{T}(\epsilon_k) \qquad (3.65)$$

for the current density in the state $|k, s\rangle$ regardless of its spin orientation $s$.<fnref n="7" />

Since we are in one spatial dimension, the current is identical to the current density. We obtain the current from the left to the right contact by summing over all states with a positive wave vector, $k > 0$,

$$I \rightarrow -2 \frac{e^2}{L} \sum_{k>0} v_k \mathcal{T}(\epsilon_k) f(\epsilon_k, \epsilon_F - eV_1), \qquad (3.66)$$

$$j_{n,k} = -\frac{e}{L} \mathcal{R} \langle u_n(k) | v | u_n(k) \rangle_R = -\frac{e}{\hbar L} \frac{\partial E_n(k)}{\partial k} \mathcal{T}(\epsilon_k),$$

where $|u_n(k)\rangle_R = t|u_n(k)\rangle$ is the transmitted part of the Bloch state $|u_n(k)\rangle$, and we have made use of Eq. (3.26) for the average velocity in a Bloch state. For our illustration, however, we stick to a single-band model with a parabolic dispersion to simplify the discussions, keeping in mind that the generalisation is then straight-forward.

<footnote n="7">⁷While we have derived this expression for a 1D electron in a parabolic band, it is valid more generally if we use Eq. (3.24) for a single state in the band $n$,</footnote>
where the factor of two reflects the two-fold spin degeneracy, and the Fermi-Dirac occupation factor takes into account that the transmitted electrons are originally injected in the left contact with an electrostatic potential $V_1$. This current is partially compensated by the electrons injected in the right contact (at $V_2$) and transmitted through the scattering region to the left contact

$$I_{\leftarrow} = -2 \frac{e^2}{L} \sum_{k<0} v_k \mathcal{T}(\epsilon_k) f(\epsilon_k, \epsilon_F - eV_2), \quad (3.67)$$

i.e. they have a negative wave vector, $k < 0$. Since $v_{-k} = -v_k$, the total current is simply given by

$$I = I_{\rightarrow} + I_{\leftarrow} = -2 \frac{e^2}{L} \sum_{k>0} v_k \mathcal{T}(\epsilon_k) [f(\epsilon_k, \epsilon_F - eV_1) - f(\epsilon_k, \epsilon_F - eV_2)], \quad (3.68)$$

which may be rewritten in terms of the 1D density of states $\rho(\epsilon) = 2/\pi\hbar v(\epsilon)$,

$$ \begin{aligned} I &= -e \int d\epsilon \frac{\rho(\epsilon)}{2} v(\epsilon) \mathcal{T}(\epsilon) [f(\epsilon, \epsilon_F - eV_1) - f(\epsilon, \epsilon_F - eV_2)] \\ &= -2 \frac{e}{h} \int d\epsilon \mathcal{T}(\epsilon) [f(\epsilon, \epsilon_F - eV_1) - f(\epsilon, \epsilon_F - eV_2)]. \end{aligned} \quad (3.69) $$

Notice the different factors of two. The original factor for the spin is absorbed in the definition of the density of states, but we must only take half of the density of states since the original sum runs over the *positive* wave vectors only. The negative ones are discarded although they are taken into account in the definition of the density of states, which therefore needs to be divided by two. More importantly, we see that the velocity $v(\epsilon)$ has completely disappeared from the expression of the current. Indeed, the original velocity, which stems from the current density, is cancelled by that in the denominator of the density of states.

A series expansion of the Fermi-Dirac occupation factor allows us to obtain the standard form of the electric current, $I = G(V_1 - V_2)$, with the conductance

$$G = 2 \frac{e^2}{h} \int d\epsilon \left(-\frac{\partial f}{\partial \epsilon}\right) \mathcal{T}(\epsilon). \quad (3.70)$$

This is the Landauer formula for quantum conductance at an arbitrary temperature and it may be brought into its zero-temperature ($T \ll T_F$) form
with the usual replacement of the derivative of the Fermi-Dirac occupation factor by the Dirac $\delta$-function (3.34),

$$G = 2 \frac{e^2}{h} \mathcal{T}(\epsilon_F). \qquad (3.71)$$

This is a remarkable result! The Landauer conductance across a potential barrier that mimics a scatterer is given in terms of the universal quantum of conductance $G_0$, which we have already introduced in Eq. (3.45), and a pure transmission coefficient $\mathcal{T}$. The factor of two simply reflects the two-fold spin degeneracy.

If we consider the ballistic limit, $\mathcal{T} = 1$, we see that the quantum transport of electrons in a single band is entirely given by (twice) the quantum of conductance, regardless of the sample geometry, that is the length of the wire in the present example! Moreover, in line with Sharvin's classical ballistic treatment of the last section, the conductance is finite and does not diverge. This means that the two-terminal resistance does not vanish even in the ballistic regime. As we have mentioned above, this is somewhat at odds with our intuitions, and several questions should be raised naturally. First, if there is a finite resistance, there must be a potential drop somewhere since $V_1 \neq V_2$. Where does that take place? Second, related to the first question, there is some power dissipated, $P = I(V_1 - V_2)$. Again, where? Finally, one may ask the question how to measure the conductance/resistance of the scatterer itself. Can one define something like an intrinsic conductance? As for the last question, we have already had a glimpse at it. It is at the origin of the distinction between the two-terminal and the four-terminal resistance, as shown in figure 3.4, where the black arrows indicate additional, but non-invasive contacts used to measure the potential drop inside the conductor. The detailed answer to these questions is the object of the next paragraphs.

## Profile of the chemical potential in a ballistic conductor

In order to understand the position of the potential drop, we first consider a fully ballistic conductor in the two-terminal configuration, such as that shown in the lower panel of figure 3.4. As we know by now, the right-moving electrons are in thermodynamic equilibrium with the left contact, i.e. they have a chemical potential $\mu_1$, while left-moving electrons share the chemical potential $\mu_2$ with the right reservoir. What is the average chemical
<figure><img src="image_29.png" /><figcaption>Figure 3.8: Potential profile in a ballistic conductor. Electrons injected into the conductor from the left contact remain at their energy given by the chemical potential $\mu_1$ of the left reservoir. Similarly, electrons injected on the left contact remain at an energy $\mu_2$. If we want to define an average chemical potential (in red) in the conductor, it is given by $\mu_m = (\mu_1 + \mu_2)/2$.</figcaption></figure>

potential $\mu_m$ in the conductor? A first well-educated guess would be to take the arithmetic average of the chemical potentials

$$ \mu_m = \frac{\mu_1 + \mu_2}{2} \quad \text{and} \quad V_m = \frac{V_1 + V_2}{2} \qquad (3.72) $$

For the moment, let us just accept this result, which will be proven in detail in the next paragraph. Naturally, this is the average potential all over the conductor since it is ballistic. There is thus no potential drop and no resistance inside the conductor, and the inevitable consequence of this fact is to attribute a resistance to the two contacts, where the potential drops from $V_1$ to $V_m$ and from $V_m$ to $V_2$ [see figure 3.8]. The contact resistance is therefore given by

$$ R_c = \frac{V_1 - V_m}{I} = \frac{V_m - V_2}{I} = \frac{h}{4e^2} \qquad (3.73) $$

per contact, that is half of the ballistic two-terminal resistance. This is indeed in line with our intuition and Kirchhoff's laws that tell us that the total resistance is the sum of resistances in series, and here the two contact resistances may be viewed as being in series. Furthermore, the dissipated power is $P = I(V_1 - V_2) = (2e^2/h)T(\epsilon_F)(V_1 - V_2)^2$, and it is equally dissipated at the two contacts.

## Profile of the chemical potential in a diffusive conductor

Let us now investigate the more complex situation where there is a potential barrier mimicking a disordered region, yet connected to the reservoirs via
<figure><img src="image_30.png" /><figcaption>Figure 3.9: Potential profile in a diffusive conductor. Electrons injected into the conductor from the left contact remain at their energy given by the chemical potential $\mu_1$ of the left reservoir. Similarly, electrons injected on the left contact remain at an energy $\mu_2$. Due to backscattering by the diffusive central region, some left-moving electrons are now at the potential $\mu_1$ and some right-moving ones at $\mu_2$ (dashed blue lines). The average potential (in red) is therefore no longer constant over the whole length of the conductor, and the potential $V_A$ in the left lead is generally different from that $V_B$ in the right lead.</figcaption></figure>

ballistic leads. In contrast to the ballistic situation, the potential is no longer constant all along the conductor due to backscattering events represented by the dashed blue lines in figure 3.9. As a consequence, some right-moving electrons – the elastically backscattered ones – are at the potential $\mu_2$ instead of $\mu_1$. While the potential may change in the disordered part of the conductor, it remains then constant in the ballistic leads. In order to calculate the associated electrostatic potentials $V_A$ and $V_B$ in the leads A and B, respectively, as a function of $V_1$ and $V_2$ (the potentials in the source and the drain) and the transmission coefficient $T$, we first calculate the electronic densities $n_A$ and $n_B$ in the leads. The density in the lead A may be calculated in the usual manner, with the help of the density of states and the Fermi-Dirac occupation factor. It consists of three parts: (i) the density of electrons injected from the left contact at the potential $V_1$; (ii) that of electrons reflected by the scattering region – these electrons are at the same potential since the scattering is supposed to be elastic –; (iii) the density of electrons transmitted through the scattering region from the right contact at
the potential $V_2$,

$$ 
\begin{aligned}
n_A &= \frac{1}{2} \int d\epsilon \rho(\epsilon) (1 + \mathcal{R}) f(\epsilon, \epsilon_F - eV_1) + \frac{1}{2} \int d\epsilon \rho(\epsilon) \mathcal{T} f(\epsilon, \epsilon_F - eV_2) \\
&\approx n(\epsilon_F) - \frac{e}{2} \int d\epsilon \rho(\epsilon) \left(-\frac{\partial f}{\partial \epsilon}\right) [(1 + \mathcal{R})V_1 + \mathcal{T}V_2] \\
&\approx n(\epsilon_F) - \frac{e}{2} \rho(\epsilon_F) [(1 + \mathcal{R})V_1 + \mathcal{T}V_2],
\end{aligned} 
\quad (3.74) $$

where $\mathcal{R} = 1 - \mathcal{T}$ is the reflection coefficient and where we have used again the series expansion of the occupation factor under the assumptions $e|V_i| \ll \epsilon_F$ and $T \ll T_F$. Furthermore $n(\epsilon_F) = \int d\epsilon \rho(\epsilon) f(\epsilon, \epsilon_F)$ plays the role of an average density in the conductor, in the absence of a potential bias, $V_1 = V_2 = 0$. As in the calculation of the conductance through a quantum wire, the factors 1/2 are due to the fact that we must only consider electrons with a fixed sign of the wave vector, $k > 0$, and thus take again only half of the density of states. If we consider a thermodynamic equilibrium in the lead A, the density may also be written in terms of the chemical potential $\mu_A = \epsilon_F - eV_A$ in the lead A,

$$ n_A = \int d\epsilon \rho(\epsilon) f(\epsilon, \epsilon_F - eV_A) \approx n(\epsilon_F) - e\rho(\epsilon_F)V_A. \quad (3.75) $$

Notice that this average potential $V_A$ is that of all electrons in the lead A, regardless of its direction of propagation, so that the factor of 1/2 is now absent. This average potential may now be obtained by identifying the two expressions obtained above,

$$ V_A = \frac{1}{2} [(1 + \mathcal{R})V_1 + \mathcal{T}V_2], \quad (3.76) $$

and an analogous argument for the lead B yields

$$ V_B = \frac{1}{2} [\mathcal{T}V_1 + (1 + \mathcal{R})V_2]. \quad (3.77) $$

Notice first that, in the ballistic limit with $\mathcal{T} = 1$ and $\mathcal{R} = 0$, we obtain $V_A = V_B = (V_1 + V_2)/2$, in agreement with the average potential used above in the calculation of the potential in a ballistic quantum wire. Furthermore, we see from these two expressions that the potential difference between the leads is related to that between source and drain via the reflection coefficient,
$V_A - V_B = \mathcal{R}(V_1 - V_2)$. This allows us now to calculate the four-terminal conductance/resistance

$$ G_4 = \frac{I}{V_A - V_B} = 2 \frac{e^2 \mathcal{T}}{h \mathcal{R}} \quad \text{and} \quad R_4 = G_4^{-1} = \frac{h \mathcal{R}}{2e^2 \mathcal{T}}, \qquad (3.78) $$

while the two-terminal conductance is still given by the Landauer formula (3.71), $G_2 = (2e^2/h)\mathcal{T}$. In order to see that the four-terminal resistance can indeed be interpreted in terms of an intrinsic resistance of the central part of the conductor containing the scatterers (without the leads), we may investigate the resistances in series that are associated with the potential drops. As it is clear from figure 3.9, there are three regions where the potential changes: in addition to the central region (with $\mathcal{T} < 1$), the potential drops again at the contacts.

Quite generally, the additivity of the resistances “in series” is simply a consequence of the sum of potential differences over the entire conductor. Since the full resistance between source and drain is given by the two-terminal resistance $R_2 = (V_1 - V_2)/I$, we may cut the conductor into two parts at a position $x$, where we consider the potential to be $V_x$. There are now two potential drops, one between the source and $x$ and one between $x$ and the drain, with the associated resistances

$$ R_{1,x} = \frac{V_1 - V_x}{I} \quad \text{and} \quad R_{x,2} = \frac{V_x - V_2}{I}, \qquad (3.79) $$

where the current remains the same in both denominators because we do not have any electron leakage over the conductor, and one immediately finds the additivity of the resistances

$$ R_2 = R_{1,x} + R_{x,2}. \qquad (3.80) $$

As a consequence of the additivity of the resistances, we see that the two-terminal resistance $R_2$ can now be decomposed into three parts corresponding to the three regions where the potential changes: the two contact resistances $R_c$ and the intrinsic four-terminal resistance related to the potential drop in the diffusive part between the leads. We thus obtain the two-terminal resistance

$$ R_2 = 2R_c + R_4 = \frac{h}{2e^2} \left( 1 + \frac{\mathcal{R}}{\mathcal{T}} \right) = \frac{h}{2e^2} \frac{1}{\mathcal{T}}, \qquad (3.81) $$
<figure><img src="image_31.png" /><figcaption>Figure 3.10: Schematic view of the scattering matrix $S$, which relates the incoming waves $i$ and $i'$ (in blue) to the outgoing ones $o$ and $o'$ (in red).</figcaption></figure>

where the contact resistance is the same as that (3.73) obtained in the ballistic regime, $R_c = h/4e^2$ per contact. In summary, we thus need to subtract, in an experimental measurement of the two-terminal resistance, the contact resistances from the measured result in order to obtain the intrinsic resistance of a conductor, which we identify with the four-terminal resistance. This is noteworthy since the contact resistance is rarely discussed in electronic transport. This distinction between intrinsic four-terminal and two-terminal resistance is indeed often not necessary if the contact resistance becomes negligible with respect to the intrinsic one. In a macroscopic transport measurement, one may indeed neglect the contact resistance which is then not given by $R_c = h/4e^2$ but by $R_c = h/4Me^2$ where $M = (2W/\lambda_F)^{d-1} \gg 1$ is the number of channels in the conductor connected to the macroscopic contact (of width $W$), as we show in detail below in section 3.6.

## 3.5 Scattering and transfer matrices

In the preceding section, we have argued in detail that the description of electric transport in small conducting devices in the mesoscopic regime must be revisited with respect to Ohm's law, and we have seen that the conductance, in units of the quantum of conductance, is then essentially given by a transmission coefficient $T$. We may however ask ourselves the reasonable question under which circumstances one retreives Ohm's law from Landauer's approach. This is the aim of the present section. In order to make this connection, which is based on the study of a large number of individual scatterers
in a 1D quantum wire, it is useful to introduce the so-called transfer-matrix formalism, which is itself based on the scattering matrix $S$. The latter relates the incoming to the outcoming waves in a scattering process, as it is shown in figure 3.10. Consider the outgoing wave $o$ on the left-hand side. It has two contributions, one coming from the incoming wave $i$ on the left-hand side that is reflected by the scatterer with an amplitude $r$ and one from the incoming wave $i'$ on the right-hand side transmitted by the scatterer with an amplitude $t'$,

$$o = ri + t'i'. \qquad (3.82)$$

Similarly the outcoming wave $o'$ on the right-hand side has a transmitted component $ti$ from the incoming wave on the left-hand side and a reflected one $r'i'$ from that on the right-hand side,

$$o' = ti + r'i'. \qquad (3.83)$$

The scattering matrix thus reads

$$S = \begin{pmatrix} r & t' \\ t & r' \end{pmatrix}. \qquad (3.84)$$

There are some constraints imposed on the scattering matrix. Indeed, the current is conserved, i.e. $|i|^2 + |i'|^2 = |o|^2 + |o'|^2$, so that the scattering matrix must be unitary, $S^†S = 1$. This implies the intuitive result

$$\mathcal{R} + \mathcal{T} = \mathcal{R}' + \mathcal{T}' = 1, \qquad (3.85)$$

in terms of the reflection and transmission coefficients $\mathcal{R} = |r|^2$, $\mathcal{R}' = |r'|^2$, $\mathcal{T} = |t|^2$ and $\mathcal{T}' = |t'|^2$, and

$$|t| = |t'| \quad \text{and} \quad |r| = |r'|. \qquad (3.86)$$

Notice that we have already used the last property implicitly in the derivation of the Landauer formula, where we have used the same transmission coefficient for the left- and right-moving electrons in the diffusive region of the quantum wire. Another constraint is imposed by time-reversal symmetry that yields the so-called *reciprocity relations* for the elements of the scattering matrix in the presence of a magnetic field. This constraint is discussed in detail in the next chapter.

While the scattering matrix is useful in the description of a global scattering process, it is less convenient when we want to investigate several scattering processes in series. Indeed, in this case we prefer to relate the waves on
the right-hand side to those on the left-hand side because multiple scattering processes can then be viewed as a concatenation of individual processes that is represented by a simple matrix product. The transfer matrix is defined as

$$ \begin{pmatrix} o' \\ i' \end{pmatrix} = \mathcal{M} \begin{pmatrix} i \\ o \end{pmatrix} = \begin{pmatrix} t - \frac{rr'}{t'} & \frac{r'}{t'} \\ -\frac{r}{t'} & \frac{1}{t'} \end{pmatrix} \begin{pmatrix} i \\ o \end{pmatrix}, \quad (3.87) $$

where the matrix elements in the last expression are readily obtained from those of the scattering matrix. Two successive scattering events that are described each by a scattering matrix $\mathcal{M}_1$ and $\mathcal{M}_2$ yield a new scattering matrix

$$ \mathcal{M} = \mathcal{M}_1 \mathcal{M}_2. \quad (3.88) $$

If we consider the lower right element of the matrix $\mathcal{M}$, we obtain the transmission amplitudes

$$ t' = \frac{t'_1 t'_2}{1 - r'_1 r_2} \quad \text{and} \quad t = \frac{t_1 t_2}{1 - r'_1 r_2}, \quad (3.89) $$

in terms of the transmission and reflection amplitudes of the two individual scatterers.<fnref n="8" />

In the following paragraphs, we investigate these relations both in a full quantum-mechanical treatment and in the high-temperature regime, where we may neglect quantum coherence. While the second regime allows us to retrieve Ohm's law for classical diffusive transport, the former treatment allows us to understand the so-called *Anderson localisation*, which is a quantum phenomenon that yields insulating behaviour in 1D disordered systems.

### 3.5.1 From the Landauer formula to Anderson localisation in a 1D conductor

In a first step, we propose a slightly different perspective on the composition law (3.89) for transfer matrices, which may be viewed as the electronic analogue of the optical Fabry-Pérot interferometer. The transmission amplitude

<footnote n="8">⁸Notice that, in principle, we need to take into account the free propagation of the (electronic) waves between the two scattering events. Indeed, we do not have $o_1 = i_2$ for the outgoing wave $o_1$ after the first scattering event and the incoming one $i_2$ at the second scatterer, but rather $o_1 \exp(i\phi_{12}) = i_2$, where the phase $\phi_{12} = k_F l_{12}$ takes into account this free propagation in terms of the ballistic path $l_{12} \sim l_e$ separating the two scatterers. However, these phases can be absorbed into the phases of the different transmission and reflection amplitudes so that we do not write them down explicitly.</footnote>
<figure><img src="image_32.png" /><figcaption>Figure 3.11: Transmission through a system that consists of two scatterers. In addition to the direct transmission given by the amplitude $t_1t_2$, we may have multiple reflection in the region between the two scatterers. This is the electronic analogue of a Fabry-Pérot interferometer.</figcaption></figure>

may indeed be viewed as a sum of multiple reflection events between the two scatterers; instead of a direct transmission given by the product $t_1t_2$, an electron may also be backscattered by the second scatterer and then on the back side of the first scatterer. These events are associated with the amplitudes $r_2$ and $r'_1$, respectively. In addition to the direct transmission, we therefore need to take into account one round of the electron between the scatterers and a total amplitude of $t_1(r_2r'_1)t_2$ as well as, more generally, $n$ rounds with an amplitude $t_1(r_2r'_1)^n t_2$. This is sketched in figure 3.11, and the sum of all contributions yields the geometric series

$$t = \sum_{n=0}^{\infty} t_1 (r_2 r'_1)^n t_2 = \frac{t_1 t_2}{1 - r_2 r'_1}, \qquad (3.90)$$

which is precisely the result (3.89) obtained by matrix multiplication. The transmission through multiple scatterers is therefore similar to optical Fabry-Pérot interferometers in series.

In order to make a connection with the (two- and four-terminal) resistance, we are interested in the transmission coefficient rather than in the amplitude,

$$\mathcal{T} = |t|^2 = \frac{\mathcal{T}_1 \mathcal{T}_2}{1 + \mathcal{R}_1 \mathcal{R}_2 - 2\sqrt{\mathcal{R}_1 \mathcal{R}_2} \cos \theta}, \qquad (3.91)$$

where the angle $\theta$ takes into account the phases, which are acquired during the reflection processes, as well as the phase associated with the free propagation in the region between the scatterers. The intrinsic four-terminal
resistance (3.78) then reads

$$R = \frac{h(1-T)}{2e^2 T} = \frac{h \mathcal{R}_1 + \mathcal{R}_2 - 2\sqrt{\mathcal{R}_1 \mathcal{R}_2} \cos \theta}{\mathcal{T}_1 \mathcal{T}_2} \approx \frac{h \mathcal{R}_1 + \mathcal{R}_2}{2e^2 \mathcal{T}_1 \mathcal{T}_2} \quad (3.92)$$

where we have taken the average $\langle \cos \theta \rangle \simeq 0$ in the last step. This average is motivated by the fact that we are searching a composition law for the resistances associated with each of the scatterers, which are themselves randomly spaced. This yields a random phase $\theta$, so that it averages out in the cosine function. Notice that the last equation may be rewritten as

$$R = R_1 + R_2 + 2R_1 R_2 \frac{2e^2}{h} \quad \text{or} \quad 1 + 2 \frac{R}{R_0} = \left(1 + 2 \frac{R_1}{R_0}\right) \left(1 + 2 \frac{R_1}{R_0}\right), \quad (3.93)$$

in terms of the resistance quantum $R_0 = h/2e^2$. From this expression, we may already conclude that the resistance grows faster than that expected on the basis of Ohm's law, because $R > R_1 + R_2$. This growth is entirely due to the quantum interferences. The last expression is then easily generalised to $n$ scatterers,

$$1 + 2 \frac{R}{R_0} = \left(1 + 2 \frac{R_1}{R_0}\right)^n, \quad (3.94)$$

where $R_1$ is the (average) intrinsic resistance associated with a single scatterer. In order to make a connection with the length of the conductor, we may use the following argument in orders of magnitude: since the distance between the scatterers is given by the mean free path $l_e$, there are $n \simeq L/l_e$ scatterers in the diffusive conductor of length $L$. Introducing the localisation length $L_0 = l_e/(R_1/R_0)$ and using $(1+x)^n \simeq e^{xn}$ in the large-$n$ limit, we obtain

$$R(L) = \frac{h}{4e^2} (e^{2L/L_0} - 1) \quad (3.95)$$

for the resistance of a diffusive wire of length $L$. We notice that this resistance of the 1D conductor diverges as soon as its length exceeds the localisation length $L_0$. This localisation, first described by P. W. Anderson in 1958, is at the origin of an insulating behaviour of electrons in diffusive 1D systems. It has later been investigated in 2D and 3D systems as a function of the disorder strength, and it was shown that the resistance diverges logarithmically in 2D while in 3D disordered systems there is a metal-insulator transition at a critical strength of the disorder. However, a quantitative discussion of this phenomenon, which is central in random media, is far beyond the scope of these classes.
### 3.5.2 From the Landauer formula to Ohm's law

As we have seen in the last paragraph, the full quantum treatment of a diffusive 1D conductor does not allow us to retrieve Ohm's law because even if the electrons were originally free, i.e. there are available states at the Fermi level to allow for electronic transport, they get localised due to quantum interferences in multiple scattering events as soon as the system size exceeds the localisation length. One may nevertheless make a connection with Ohm's law if we consider the setup sketched in figure 3.11 in terms of transmission and reflection coefficients instead of their respective amplitudes. While we allow still for multiple reflections in the regions between the scatterers, we consider them to be classical now. This approach is justified as soon as we are at sufficiently high temperatures so that the electrons immediately lose their phase memory on the average distance $l_e$ between two scatterers. In this case, we obtain the transmission coefficient

$$ \mathcal{T} = \frac{\mathcal{T}_1 \mathcal{T}_2}{1 - \mathcal{R}_1 \mathcal{R}_2}, \qquad (3.96) $$

which may be rewritten as

$$ \frac{1 - \mathcal{T}}{\mathcal{T}} = \frac{1 - \mathcal{T}_1}{\mathcal{T}_2} + \frac{1 - \mathcal{T}_2}{\mathcal{T}_2} \quad \Leftrightarrow \quad \frac{\mathcal{R}}{\mathcal{T}} = \frac{\mathcal{R}_1}{\mathcal{T}_1} + \frac{\mathcal{R}_2}{\mathcal{T}_2}. \qquad (3.97) $$

The last expression indicates directly that the intrinsic resistances associated with each of the scatterers $R_i = (h/2e^2)R_i/\mathcal{T}_i$ add up to give the total resistance, and in the case of $n \sim L/l_e$ scatterers, one obtains

$$ R = \frac{h}{2e^2} \sum_i \frac{R_i}{\mathcal{T}_i} \approx \frac{h L}{2e^2 l_e} \rho, \qquad (3.98) $$

where we have defined an average dimensionless resistance $\rho = \langle R_i/\mathcal{T}_i \rangle$. We realise from this result that the resistance shows some similarity with Ohm's law in that the resistance grows linearly with the system length. In order to complete this link with Ohm's law, we still need to show that the resistance is also inversely proportional to the cross section $S = W^{d-1}$. This requires a generalisation of the present approach, illustrated in the framework of a 1D conductor, to higher dimensions. It is the object of the *multi-channel* conductance, which is treated in the following section.
<figure><img src="image_33.png" /><figcaption>Figure 3.12: (a) Sketch of a 2D conductor of width W connected in the two-terminal geometry. In the scattering region an electron that was originally in the mode b (in the left lead) can be scattered into another mode a in the right lead. The associated transmission amplitude is given by $t_{ab}$. (b) Family of 1D bands associated with the conductor. The modes correspond to the states of the different bands $n_y$, each of which contributes to the electric transport as long as its band bottom is situated below the Fermi energy $\epsilon_F$.</figcaption></figure>

## 3.6 Multi-channel conductance

In the preceding section, we have obtained the Landauer formula (3.70) for electronic transport in a single 1D band. Let us now consider the situation of a 2D conductor of width W, under the assumption that this width be much smaller than the length of the conductor, $W \ll L$ [see figure 3.12(a)]. We have already seen at the end of section 3.1 that the confinement of a 2D electron gas in the y-direction yields a family of 1D bands

$$ \epsilon_n(k_x) = \frac{\hbar^2 k_x^2}{2m} + \epsilon_{n_y}, \qquad (3.99) $$

where $\epsilon_{n_y}$ indicates the quantised energy due to the confined motion in the y-direction.<fnref n="9" /> These bands are sketched in figure 3.12(b). These bands are associated with different transverse modes, with $n_y - 1$ nodes in the wave function $\phi_{n_y}(y)$, which is the y-component of the full wave function

$$ \psi_{n_y, k_x}(x, y) = \frac{1}{\sqrt{L}} \phi_{n_y}(y) e^{ik_x x}. \qquad (3.100) $$

Let us now generalise the Landauer formula – or more precisely its expression in the zero-temperature limit – in order to account for the fact that in a diffusive quantum wire, there is now the possibility for an electron to

<footnote n="9">⁹Remember that for a simple square potential, we have $\epsilon_{n_y} = \hbar^2\pi^2n_y^2/2mW^2$.</footnote>
be scattered from a mode b (in the left lead) to a mode a (in the right lead). We still consider elastic scattering processes that are allowed as long as the bottom of the two bands associated with the two modes a and b remain below the Fermi level [see figure 3.12(b)] because then both bands cross the Fermi energy. The transmission of a mode b into a mode a may now be associated with a transmission amplitude $t_{ab}$ and a transmission coefficient $\mathcal{T}_{ab} = |t_{ab}|^2$, in analogy with the case of a 1D wire with a single mode discussed in section 3.4.1. Before proceeding, let us fix the terminology. We have already insisted in section 3.3 on the role of the contact, which we also call terminals or reservoirs. In addition to those, we now have different electronic modes that we also call channels inside the conductor due to the lateral confinement. The situation depicted in figure 3.12(a) thus corresponds to a multi-channel conductor in the two-terminal configuration. We may thus write down the Landauer formula for the multi-channel two-terminal conductance

$$I = G(V_1 - V_2) \quad \text{with} \quad G = 2 \frac{e^2}{h} \sum_{a,b} \mathcal{T}_{ab} = 2 \frac{e^2}{h} \text{tr}tt^\dagger, \qquad (3.101)$$

where the sum runs over all allowed modes (below the Fermi level). This generalised formula indicates that all scattering events from one channel b to another channel a contribute to the conductance as long as it is a forward-scattering process that is taken into account precisely by the transmission coefficient. In the last step, we have made use of a matrix description, where trtt$^\dagger$ indicates the trace of a product of matrices t whose elements are the transmission amplitudes $t_{ab}$.

If we consider the ballistic limit, there is naturally no inter-mode (or inter-channel) scattering, and an electron injected at the source into the left lead in the mode b remains in this channel all through the conductor. The transmission coefficients are therefore diagonal $\mathcal{T}_{ab} = \delta_{a,b}$, so that the ballistic two-terminal conductance becomes

$$G_{\text{bal}} = 2 \frac{e^2}{h} \sum_b = 2 \frac{e^2}{h} M, \qquad (3.102)$$

where M is simply the number of modes whose energy minima remain below the Fermi level. This number is indeed readily evaluated since we must simply require $\epsilon_{ny} \le \epsilon_F$. If we consider the lateral (square) confinement, this condition is equivalent to $k_y = (\pi/W)n_y \le k_F$, and the number of modes is
simply given by the integer part of the ratio $k_F W / \pi = 2W / \lambda_F$,

$$M = \mathrm{Int} \left[ \frac{k_F W}{\pi} \right] = \mathrm{Int} \left[ \frac{2W}{\lambda_F} \right]. \qquad (3.103)$$

As promised, we thus obtain the very same result (3.56) for Sharvin’s conductance as in the classical ballistic description if we replace the ratio $2W/\lambda_F$ by its integer part, which plays the role of the number of modes below the Fermi level! Notice that we have already implicitly made use of a multi-channel version of the conductance even in the context of a single-channel conductor, because of the spin. One may also consider the two spin states as two channels that are at the origin of the factor of two in our expressions for the conductance. One may therefore equally speak, in the present context, of 2M channels, or more generally gM channels if we are confronted with a g-fold internal degeneracy. In the following parts of these classes, we will rather speak about M two-fold degenerate channels unless the spin degeneracy is explicitly lifted, e.g. by a prominent Zeeman effect.

Again, this result can be generalised to higher dimensions in which case simply the mode counting changes. Consider, for example, three spatial dimensions, where the transverse motion is quantised in, say, the x- and y-direction while we keep plane-wave states in the z-direction. We then need to count the number of modes

$$\frac{\hbar^2 \pi^2}{2mW^2} (n_x^2 + n_y^2) \le \epsilon_F \Leftrightarrow \frac{\pi}{W} \sqrt{n_x^2 + n_y^2} \le k_F, \qquad (3.104)$$

and the last equation can be viewed as the number of states, which occupy each a 2D volume of $(W/\pi)^2$, inside the positive quadrant of a circle (2D sphere) of area $\pi k_F^2 / 4$. The number of modes in a 3D conductor of cross section $S = W^2$ is then given by

$$M = \mathrm{Int} \left[ \frac{\pi W^2}{\lambda_F^2} \right], \qquad (3.105)$$

and generally we have for a d-dimensional conductor

$$M = \mathrm{Int} \left[ \frac{\mathcal{A}_{d-1} k_F^{d-1} / 2^{d-1}}{(\pi / W)^{d-1}} \right] = \mathrm{Int} \left[ \frac{\mathcal{A}_{d-1} W^{d-1}}{\lambda_F^{d-1}} \right], \qquad (3.106)$$

in terms of the volume $\mathcal{A}_d$ of the d-dimensional unit sphere. This result agrees again perfectly with that obtained from the classical ballistic calculation of
<figure><img src="image_34.png" /></figure>

Figure 3.13: (a) Sketch of a quantum point contact. The high-mobility 2D electron gas at a GaAs/AlGaAs interface is ballistic between the two contacts but depleted below two electrodes called the *gates*. An application of a gate voltage allows one to define a constriction at the centre of the 2D electron gas of a width *W* that is controlled by the gate voltage *V*. (b) Measurement of the conductance through the device as a function of the gate voltage. The curves correspond to different temperatures and are offset for a better visibility. Adapted from B. J. van Wees *et al.*, Phys. Rev. Lett. **60**, 848 (1988).

the Sharvin conductance (3.62) in *d* spatial dimensions if we replace the
geometric ratio $A_{d-1}W^{d-1}/\lambda_F^{d-1}$ by its integer part, which represents the
number of modes.

**3.6.1 Conductance through a quantum point contact**

The above multi-channel Landauer conductance has found a remarkable ex-
perimental proof (in 1988 by B. J. van Wees *et al.*) in the two-terminal con-
ductance measured in a quantum point contact (QPC) in a high-mobility 2D
electron gas that shows ballistic transport over the full system size, *i.e.* be-
tween the two contacts [see figure 3.13(a) for the setup]. The QPC is defined
by gate electrodes fabricated on top of the GaAs/AlGaAs heterostructure,
and the application of a voltage (the *gate voltage*) allows one to modify the
density below the gates. These regions can be fully depleted so that one
creates a bottleneck in the 2D electron system at the centre between the
source and drain contacts. The gate voltage *V* allows then for a definition of
the width *W* of the constriction. The measured two-terminal conductance is
depicted in figure 3.13(b) in units of 2e²/h as a function of the applied gate
voltage. The largest (negative) gate voltages are found on the left-hand side
of the figure and correspond to low values of the QPC width, while the width is increased towards the right-hand side. The different curves, which are offset for a better visibility, correspond to different temperatures, between 0.3 K and 4.2 K.

At low temperatures, one clearly sees that the conductance increases in steps of $2e^2/h$ upon increase of the gate voltage and thus enhancement of the QPC width, as one would expect indeed on the basis of Eq. (3.102). Indeed, if one increases the width of the constriction, the number of modes $M$ increases because the bottom of the associated bands is shifted below the Fermi level. Furthermore, the discrete change in the number of modes upon increase of the QPC width is washed out at higher temperatures, where the system behaves more and more classically and where one thus retrieves the classical Sharvin conductance. We may even be more quantitative about this crossover to the classical result if we appeal to the finite-temperature version (3.70) of the Landauer formula, which in the multi-channel case reads

$$G_{\text{bal}} = 2 \frac{e^2}{h} \int d\epsilon M(\epsilon) \left(-\frac{\partial f}{\partial \epsilon}\right), \qquad (3.107)$$

where $M(\epsilon) = \sum_n \Theta(\epsilon - \epsilon_n)$ counts the number of modes below the energy $\epsilon$. In the zero-temperature limit, we naturally retrieve Eq. (3.101). We may rewrite the above expression by integration by parts, with $\partial M/\partial \epsilon = \sum_n \delta(\epsilon - \epsilon_n)$, so that one obtains

$$G_{\text{bal}} = 2 \frac{e^2}{h} \sum_n f(\epsilon_n, \epsilon_F - eV). \qquad (3.108)$$

The form of the measured steps is therefore given by the temperature-rounded form of the Fermi-Dirac occupation factor

$$f(\epsilon_n, \epsilon_F - eV) = \frac{1}{e^{(\epsilon_n - \epsilon_F)/k_B T} + 1}. \qquad (3.109)$$

As an order of magnitude, we may say that the crossover between the quantum and classical regimes occurs once the typical energy scale $\hbar^2/mW^2$ equals the thermal energy $k_B T$. In order to obtain this energy scale numerically, we may report it to the Rydberg, which is the typical (atomic) energy scale $\hbar^2/m_0 a_B^2 \sim 10$ eV at the Bohr radius $a_B \sim 10^{-10}$ m, where $m_0$ is the bare electron mass. The band mass for GaAs is $m/m_0 \simeq 0.068$ and the experimental estimate for the width is $\sim 100$ nm (even if the latter changes via the
<figure><img src="image_35.png" /><figcaption>Figure 3.14: Experimental evidence for the crossover from ballistic to diffusive transport as a function of the channel length in GaAs/AlGaAs quantum wires. The measured resistances are plotted as a function of the channel length for several values of the conductor width ($W = 2, 4$ and $8$ µm). For short systems, the resistance saturates at the ballistic (Sharvin) limit while it crosses over to a linear increase for larger values of $L$. Adapted from S. Tarucha et al., Phys. Rev. B **47**, 4064 (1993).</figcaption></figure>

gate voltage). This yields a rough estimate

$$ \frac{\hbar^2}{mW^2} \approx 10 \text{ eV} \times \frac{m_0}{m} \left(\frac{a_B}{W}\right)^2 \sim 0.1 \text{ meV}, \quad (3.110) $$

which corresponds to a temperature in the Kelvin range, in agreement with the observed quantum-to-clasical crossover in figure 3.13(b).

### 3.6.2 Experimental evidence for a crossover from ballistic to diffusive transport

The multichannel Landauer approach allows us to make a first connection between the conductance and geometric effects since the number of modes $M \sim W^{d-1}$ is determined by the typical size of the cross section. It is a bridge between the pure quantum limit, which the Landauer formalism was designed for, and the classical Sharvin conductance for a ballistic conductor. Interestingly, one may also retrieve the diffusive Drude limit in the large-$L$ limit. In order to see this point, let us rephrase the arguments we have elaborated above in section 3.5 for the single-channel limit. First, the contact
resistance needs to be modified, and one naturally has

$$R_c(M) = \frac{h}{4Me^2}, \qquad (3.111)$$

now that we have M channels. The argument is the same as for a single channel: the full two-terminal conductance is $2Me^2/h$ if we consider a ballistic wire, and the potential drop happens at the two contacts, whence the associated resistance. The total resistance is then given by the sum of the contact resistances and the intrinsic four-terminal resistance $R_{\text{diff}} \approx (h/2Me^2) \sum_i \langle R_i/T_i \rangle \approx (h/2Me^2)(L/l_e)\rho$

$$R = 2R_c(M) + R_{\text{diff}} = \frac{h}{2Me^2} \left( 1 + \frac{L}{l_e} \rho \right), \qquad (3.112)$$

in terms of some dimensionless average resistance $\rho = \langle R_i/T_i \rangle$ associated with a scatterer in the multi-channel case. We notice that Eq. (3.112) now interpolates between the ballistic and the diffusive regime. It saturates for $L \ll l_e$ at the value for the ballistic two-terminal conductance, while it crosses over to the Drude result $R \propto L/W^{d-1}$ in the large-channel limit $L \gg l_e$.

This crossover has indeed been measured experimentally by S. Tarucha et al. in 1993, who studied the evolution of the resistance in quantum channels etched in 2D GaAs/AlGaAs quantum gases for three values of the width and several values of their length. Each measured value in figure 3.14 thus corresponds to a different sample. The width is so large that there are several hundreds of channels below the Fermi level, in which case one may make no distinction between the Sharvin and the Landauer multi-channel conductances.<fnref n="10" /> The resistance saturates well at a value in agreement with the Sharvin resistance for ballistic transport in the small-length limit, while above some 10 µm it crosses over to a behaviour linear in the channel length, in agreement with the Drude formula and Ohm's law.

## Notions and concepts to retain

* conductance and conductivity, resistance and resistivity

<footnote n="10">¹⁰Indeed, the channel width is one order of magnitude larger than that in the previously discussed experiment of the QPC in figure 3.13, so that the characteristic temperature for the quantum-to-classical crossover is reduced by two orders of magnitude, i.e. in the 10 mK range, while the experiments were performed at 1.5 K.</footnote>
* Drude model and Einstein relation for the conductivity in the classical diffusive limit

* ballistic *vs*. diffusive transport

* Sharvin conductance/resistance

* transport in a 1D quantum wire, contacts/terminals/reservoirs, leads

* Landauer formula

* multi-channel (-mode) transport and multi-channel generalisation of the Landauer formula

* two-terminal and four-terminal resistance, contact resistance

* quantum point contact (QPC)
# Chapter 4

## Conduction in multi-terminal
quantum devices:
Landauer-Büttiker formalism

In the preceding chapter, we have discussed the basics of quantum electronic transport in the mesoscopic regime. We have seen that it is more appropriate, especially in the ballistic limit, to discuss this type of transport in terms of conductances and resistances rather than in terms of conductivity and resistivity. Most saliently, we have noticed that the outcome of a resistance measurement depends significantly on the manner how the system is connected to the outer world. While we have mainly discussed the so-called two-terminal geometry, with a quick glimpse at a specific four-terminal configuration, we present here a generalisation of Landauer’s approach to multi-terminal systems, within a formalism proposed by M. Büttiker. This is the so-called *Landauer-Büttiker formalism*. This formalism has proven to be extremely fruitful in the understanding of fundamental effects, such as the quantum Hall effect (Chap. 5) or the quantum spin Hall effect, which is the experimental manifestation of a 2D topological insulator (Chap. 7).

The general formalism is presented in the next section (section 4.1) along with some general properties, in the absence and in the presence of a magnetic field. The formalism is then illustrated in section 4.2 for three-terminal and in section 4.3 for four-terminal devices.
<figure><img src="image_36.png" /><figcaption>Figure 4.1: (a) Sketch of a four-terminal conductor. The four contacts $i = 1, 2, 3$ and 4 are characterised by the chemical potentials $\mu_i = \epsilon_F - eV_i$ and connected to the conductor by ballistic leads (white regions of the device). The conductor itself (shaded area) is taken to be diffusive and the flux $\phi$ through the conductor indicates the presence of a possible magnetic field that breaks time-reversal symmetry. In general, there may be currents $I_i$ injected into the conductor by each of the terminals. (b) Example of an experimental multi-terminal device.</figcaption></figure>

## 4.1 Landauer-Büttiker formalism for multi-terminal systems

A typical multi-terminal device is sketched in figure 4.1(a) along with an experimental example in figure 4.1(b) that consists of a quantum wire contacted by 12 contacts. As in the two-terminal example discussed in the previous chapter, we consider that the contacts are thermodynamic reservoirs at a chemical potential $\mu_i = \epsilon_F - eV_i$ which can be tuned to some extent by an electrostatic potential $V_i$, where we still take the limit $e|V_i| \ll \epsilon_F$. While one may generally use all contacts to inject a current into the device, we stick to only two contacts, the *source* and the *drain*, where there is current flow. The sign of the current is defined as positive if it flows into the conductor, and we thus have $I_S = I$ at the source and $I_D = -I$ at the drain if there is no current leakage that would violate current conservation.

In order to investigate the conduction properties of such multi-terminal systems, let us just rewrite the Landauer formula for a two-terminal device represented in figure 4.2(a) in a slightly different manner. For the current in the source, we can formally rewrite the Landauer formula (at $T \ll T_F$) as

$$I_1 = I = -2\frac{e}{h}\mathcal{T}(\mu_1 - \mu_2) = 2\frac{e^2}{h}\mathcal{T}(V_1 - V_2) = 2\frac{e^2}{h}[(1 - \mathcal{R}_{11})V_1 - \mathcal{T}_{12}V_2], \quad (4.1)$$

where we reinterpret $\mathcal{R}_{11} = \mathcal{R}$ as the reflection coefficient into the contact 1 and $\mathcal{T}_{12} = \mathcal{T}$ as the transmission coefficient from the contact 2 to 1. Similarly,
<figure><img src="image_37.png" /><figcaption>Figure 4.2: (a) Sketch of a two-terminal conductor. The currents flowing into the conductor are $I_1 = I$ and $I_2 = -I$. (b) Three-terminal conductor contacted by ballistic leads to the contacts 1,2, and 3, at the respective chemical potentials $\mu_j = \epsilon_F - eV_j$. In this case, we choose 1 and 2 to be the source and the drain, respectively, with $I_1 = I$ and $I_2 = -I$, while there is no current passing through contact 3.</figcaption></figure>

we can rewrite the Landauer formula in the form of a balance equation in the contact 2 (drain),

$$I_2 = -I = 2\frac{e^2}{h}\mathcal{T}(V_2 - V_1) = 2\frac{e^2}{h}[(1 - \mathcal{R}_{22})V_2 - \mathcal{T}_{21}V_1], \quad (4.2)$$

with the reflection coefficient $\mathcal{R}_{22} = \mathcal{R}$ into the contact 2 and the transmission coefficient $\mathcal{T}_{21} = \mathcal{T}$ from contact 1 to contact 2. For the moment, we have not gained anything by this rewriting of the Landauer formula in a more complicated manner, apart from unveiling an emergent matrix structure

$$\begin{pmatrix} I_1 \\ I_2 \end{pmatrix} = 2\frac{e^2}{h} \begin{pmatrix} (1 - \mathcal{R}_{11}) & -\mathcal{T}_{12} \\ -\mathcal{T}_{21} & (1 - \mathcal{R}_{22}) \end{pmatrix} \begin{pmatrix} V_1 \\ V_2 \end{pmatrix} \quad (4.3)$$

that allows us to generalise more easily the Landauer formula to a situation with more than two contacts. Notice furthermore, that the two balance equations in the two contacts bear exactly the same information. Mathematically, this means that the two linear equations are not independent so that the (conductance) matrix in the above equation has a vanishing determinant.

In order to generalise the Landauer formula, we first consider the three-terminal device shown in figure 4.2(b), where contact 1 is chosen to be the source and 2 to be the drain, while there is no current flowing in contact 3. In order to obtain the balance equation, we may count the contributions: there is first the injection into the conductor from the source lowered by the back-scattered part at the potential $V_1$, $(1 - \mathcal{R}_{11})V_1$, and second the parts transmitted from the contacts 2 and 3 at their respective potentials, $-\mathcal{T}_{12}V_2$ and $-\mathcal{T}_{13}V_3$. One thus obtains for the balance in contact 1

$$I_1 = I = 2\frac{e^2}{h}[(1 - \mathcal{R}_{11})V_1 - \mathcal{T}_{12}V_2 - \mathcal{T}_{13}V_3]. \quad (4.4)$$
Similar balance equations are obtained for the other contacts, and the general Landauer-Büttiker formula in the case of N contacts reads

$$I_i = 2 \frac{e^2}{h} \left[ (M_i - \mathcal{R}_{ii})V_i - \sum_{j \neq i} \mathcal{T}_{ij}V_j \right], \qquad (4.5)$$

for the contact i. Several comments need to be made. First, we have smugled in a factor $M_i$ that denotes the number of modes in the lead connected to contact i below the Fermi level, in order to have directly an expression for the multi-channel case. We therefore need to reconsider the range of the reflection coefficients $\mathcal{R}_{ii}$, which may now take values between zero and $M_i$. Furthermore, we have are confronted with several transmission coefficients $\mathcal{T}_{ij}$ that denote the transmission from the contact j to the contact i and that are generally different from one another. The role of the usual equation $\mathcal{T} = 1 - \mathcal{R}$ needs to be reconsidered in the multi-terminal case since there is not only one transmission and one reflection coefficient. Finally, we have already mentioned in the two-terminal case that the two balance equations in the two contacts are essentially the same. There is thus some redundancy, as we show in detail in the next paragraph.

Notice finally that the Landauer-Büttiker formula (4.5) is indeed a matrix equation that can be rewritten in the form

$$\vec{I} = \mathcal{G}\vec{V}, \qquad (4.6)$$

with the conductance matrix

$$\mathcal{G} = 2 \frac{e^2}{h} \begin{pmatrix} (M_1 - \mathcal{R}_{11}) & -\mathcal{T}_{12} & \dots & -\mathcal{T}_{1N} \\ -\mathcal{T}_{21} & (M_2 - \mathcal{R}_{22}) & \dots & -\mathcal{T}_{2N} \\ \vdots & -\mathcal{T}_{32} & \ddots & \vdots \\ -\mathcal{T}_{N1} & \dots & \dots & (M_N - \mathcal{R}_{NN}) \end{pmatrix} \qquad (4.7)$$

for a conductor connected by N contacts.

### 4.1.1 Reference potential and current conservation

The above-mentioned points about the transmission and reflection coefficients and the redundancy of the balance equations. Indeed, in the two-terminal case, Eqs. (4.1) and (4.2) are equivalent. This is due to the invariance of the equations under the change of the reference potential. If we shift
$V_j \to V_j + V_0$ by the same potential $V_0$ in all contacts, the conduction should not change, i.e. the currents should remain unaltered, $I_i \to I_i + \Delta I_i$ with $\Delta I_i = 0$. If we substitute this into Eq. (4.5), we obtain

$$ \Delta I_i = 0 = 2 \frac{e^2}{h} \left[ (M_i - \mathcal{R}_{ii}) - \sum_{j \neq i} \mathcal{T}_{ij} \right] V_0, \quad (4.8) $$

which is nothing other than the intuitive statement that there is no current flowing through the conductor if all contacts are at the same potential. This yields

$$ M_i - \mathcal{R}_{ii} = \sum_{j \neq i} \mathcal{T}_{ij}, \quad (4.9) $$

which generalises the equation $1 - \mathcal{R} = \mathcal{T}$ in the one-channel two-terminal setup discussed in the previous chapter. Equation (4.5) can thus be rewritten in the form

$$ I_i = 2 \frac{e^2}{h} \sum_{j \neq i} \mathcal{T}_{ij} (V_i - V_j). \quad (4.10) $$

This tells us that the sum over all elements of each row of the conductance matrix (4.7) is equal to zero.

In order to obtain the entire redundancy of the $N$ balance equations for a conductor connected to $N$ contacts, we also need to consider current conservation. If we omit any current leakage in the conductor apart from the contacts, we have $\sum_i I_i = 0$. Therefore we have

$$ \sum_i \sum_{j \neq i} \mathcal{T}_{ij} V_i = \sum_i \sum_{j \neq i} \mathcal{T}_{ij} V_j. \quad (4.11) $$

This double sum excludes just the terms where $i = j$, and we can thus rewrite the term on the left-hand side as $\sum_i \sum_{j \neq i} \mathcal{T}_{ji} V_j$ by a simple change of variable in the sum so that we obtain

$$ \sum_i \sum_{j \neq i} \mathcal{T}_{ji} V_j = \sum_i \sum_{j \neq i} \mathcal{T}_{ij} V_j. \quad (4.12) $$

This equation is valid for any choice of the electrostatic potentials and therefore also for the very special choice $V_j \neq 0$ and $V_i = 0$ if $i \neq j$. As a consequence the sum remains the same if we exchange the indices

$$ \sum_{j \neq i} \mathcal{T}_{ij} = \sum_{j \neq i} \mathcal{T}_{ji}, \quad (4.13) $$
and the sum over a column of the conductance matrix (4.7) therefore vanishes as much as the sum over any row of the matrix. While these two properties may seem similar from a mathematical point of view, they are physically quite different. In one case, it is the invariance of the system upon a change of the reference potential (and thus the choice of the zero in energy of the thermodynamic potentials), while in the other case we need to appeal to current conservation. As a consequence of these properties, which are due to the above-mentioned redundancy of the balance equations, the conductance matrix has a vanishing determinant

$$ \det(\mathcal{G}) = 0, \qquad (4.14) $$

in a agreement with the conclusions drawn from the two-terminal geometry. This is, at first sight, a drawback because we cannot simply invert the conductance matrix in order to obtain the multi-terminal resistances. We will see, however, that this problem can be circumvented in many situations, namely in the three-terminal and four-terminal geometries, discussed in sections 4.2 and 4.3, respectively, as well as in the Hall geometry with six terminals in the quantum Hall, anomalous quantum Hall and quantum spin Hall effects, presented in the following chapters.

### 4.1.2 The role of a magnetic field and Onsager reciprocity relations

Before we investigate the conductance matrix and the related multi-terminal resistances in some particular geometries, let us continue to examine some general properties of the conductance matrix (4.7). A particularly important property is revealed in the presence of a magnetic field. In order to approach this situation, it turns out to be convenient to appeal again to the scattering matrix and its behaviour under time-reversal symmetry. Indeed, the conduction through the conductor connected by $N$ contacts can be viewed as a scattering process described in terms of an $N \times N$ generalisation of the scattering matrix (3.84), schematically represented in figure 4.3. The diagonal elements $s_{ii}$ of the scattering matrix are nothing other than the reflection amplitudes, so that we have $R_{ii} = |s_{ii}|^2$ for the reflection coefficients, while the transmission coefficients $T_{ij} = |s_{ij}|^2$ are given by the off-diagonal elements of the scattering matrix, for $i \neq j$. We have already seen that the scattering matrix must be unitary, $S^\dagger S = 1$, to account for current conservation. We
<figure><img src="image_38.png" /><figcaption>Figure 4.3: Schematic view of the scattering matrix S, which relates the incoming waves $i_n$ (in blue) to the outgoing ones $o_n$ (in red) in a multi-terminal (here four-terminal) device.</figcaption></figure>

thus have,

$$ \begin{pmatrix} o_1 \\ o_2 \\ \vdots \\ o_N \end{pmatrix} = S(B) \begin{pmatrix} i_1 \\ i_2 \\ \vdots \\ i_N \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} i_1 \\ i_2 \\ \vdots \\ i_N \end{pmatrix} = S^{\dagger}(B) \begin{pmatrix} o_1 \\ o_2 \\ \vdots \\ o_N \end{pmatrix}, \quad (4.15) $$

where the argument in the scattering matrix denotes the magnetic field, which we take into account explicitly here. Notice that the latter equation can also be understood in a slightly different manner if we consider time-reversal symmetry that inverts the arrows in figure 4.3 and thus exchanges the incoming and outgoing waves in the scattering process. The scattering matrix can then be analysed in the light of the Schrödinger equation in the presence of a magnetic field,

$$ H\psi(\mathbf{r}, t; \mathbf{A}) = \left[ \frac{(-i\hbar\nabla - e\mathbf{A}(\mathbf{r}))^2}{2m} + V(\mathbf{r}) \right] \psi(\mathbf{r}, t, \mathbf{A}) = i\hbar \frac{\partial}{\partial t} \psi(\mathbf{r}, t, \mathbf{A}), \quad (4.16) $$

where we have explicitly taken into account the vector potential $\mathbf{A}$ in the argument of the wave function, with $\mathbf{B} = \nabla \times \mathbf{A}$. Let us now take the
complex conjugate of this equation,

$$ \left[ \frac{(-i\hbar\nabla + e\mathbf{A}(\mathbf{r}))^2}{2m} + V(\mathbf{r}) \right] \psi^*(\mathbf{r}, t, \mathbf{A}) = i\hbar \frac{\partial}{\partial t} \psi^*(\mathbf{r}, t, \mathbf{A}). \quad (4.17) $$

This equation indicates that the wave function $\psi^*(\mathbf{r}, t, \mathbf{A})$ is the solution of the same Hamiltonian if we change $t \to -t$ and invert the sign of the magnetic field, $\mathbf{B} \to -\mathbf{B}$ (and thus the vector potential, $\mathbf{A} \to -\mathbf{A}$), i.e.

$$ \psi^*(\mathbf{r}, -t, -\mathbf{A}) = \psi(\mathbf{r}, t, \mathbf{A}), \quad (4.18) $$

or else for the incoming and outgoing components of the wave function in the scattering process (4.15)

$$ o_j(t, B) = i_j^*(-t, -B). \quad (4.19) $$

This symmetry, which is called *time-reversal symmetry* and that is discussed in more detail in section 7.1 at the beginning of chapter 7, needs to be respected by the scattering matrix. If we now rewrite the equation (4.15) explicitly in terms of the wave functions,

$$ \vec{\sigma}(t, B) = S(B)\vec{i}(t, B) \quad \text{and} \quad S^\dagger(B)\vec{\sigma}(t, B) = \vec{i}^\dagger(t, B), \quad (4.20) $$

inversion of the sign in $t$ and $B$, along with complex conjugation, of the second equation yields

$$ S^{\dagger*}(-B)\vec{\sigma}^*(-t, -B) = \vec{i}^*(-t, -B). \quad (4.21) $$

In view of the time-reversal-symmetry relation (4.19), a comparison with the first equation of (4.20) tells us that $S^{\dagger*}(-B) = S(B)$, i.e.

$$ S^\dagger(B) = S^*(-B) \quad \text{or in components} \quad s_{ij}(B) = s_{ji}(-B). \quad (4.22) $$

If we change the orientation of the magnetic field, we therefore need to trans- pose the scattering matrix, and the transmission and reflection coefficients inherit this symmetry, i.e.

$$ \mathcal{T}_{ij}(B) = \mathcal{T}_{ji}(-B) \quad \text{and} \quad \mathcal{R}_{ii}(B) = \mathcal{R}_{ii}(-B). \quad (4.23) $$

These relations are also called *Onsager reciprocity relations*.
<figure><img src="image_39.png" /><figcaption>Figure 4.4: (a) Measured two-terminal conductance through a metallic ring in a magnetic field. After L. Angers et al., Phys. Rev. B 75, 115309 (2007). (b) same as figure 3.3, four-terminal conductances through a metallic ring for different arrangements of the contacts. After A. Benoit et al., Phys. Rev. Lett. 57, 1765 (1986).</figcaption></figure>

As a consequence of the reciprocity relations, the scattering and the conductance matrices are symmetric in the absence of a magnetic field,

$$s_{ij}(B=0) = s_{ji}(B=0) \quad \text{and} \quad \mathcal{T}_{ij}(B=0) = \mathcal{T}_{ji}(B=0). \quad (4.24)$$

Along with Eq. (4.9), which expresses the diagonal elements of the conductance matrix in terms of the off-diagonal ones, the conductance matrix thus has $N(N-1)/2$ independent elements for $B=0$, while it has $(N-1)^2$ independent elements at non-zero magnetic fields.

The fact that the transmission coefficients are not symmetric under the exchange of the contacts $i \leftrightarrow j$ in the presence of a magnetic field may seem somewhat disturbing. Indeed, in the elaboration of the Landauer formula in the preceding chapter, we have made use of the fact that the transmission coefficient from the source to the drain is the same as that from the drain to the source, i.e. $\mathcal{T}_{12} = \mathcal{T}_{21}$. Notice that this remains true because we considered the two-terminal configuration for which the conductance matrix is thus a $2 \times 2$ matrix. In this case the sums in Eq. (4.13) contain only one element, and we have $\mathcal{T}_{12} = \mathcal{T}_{21}$ regardless of the magnetic field. As a consequence, the two-terminal conductance

$$G_2(B) = 2 \frac{e^2}{h} \mathcal{T}_{12} = G_2(-B) \quad (4.25)$$

is symmetric under the change of orientation of the magnetic field. That this is indeed the case can be seen from figure 4.4(a), where the two-terminal
conductance through a metallic ring is shown as a function of the magnetic field. The measured conductance displays Aharonov-Bohm oscillations the origin of which is not the relevant issue here, but rather the fact that the curve is perfectly symmetric under the exchange $B \to -B$. This needs to be contrasted with the four-terminal conductance through the same type of device, which we have already encountered in figure 3.3 and that is reproduced in figure 4.4(b) for a direct comparison. Clearly, neither of the three curves has $G_{ij,kl}(-B) = G_{ij,kl}(B)$. However, comparison between the first and the third conductance curve indicates that $G_{14,23}(B) = G_{23,14}(-B)$, i.e. the curves are identical if the change in the orientation of the magnetic field is accompanied by an exchange of the pair of contacts that serve as drain and source with those used to measure the voltage drop. While, this result arises from a full calculation of the four-terminal resistance (see section 4.3), it can be understood physically in a simple manner. In the absence of a magnetic field, there should be no difference between the contact arrangements in the first and the third line so that $G_{14,23}(B=0) = G_{23,14}(B=0)$ because the system is rotation-symmetric around the axis that cuts the ring in two halves in the x-axis. A simple continuous rotation by 180° around this axis would thus change the arrangement of the first line into that of the third one. This symmetry is broken by the magnetic field in the direction perpendicular to the plane, and the rotation by 180° precisely changes the orientation of the magnetic field.

The general Onsager reciprocity relations usually involve the resistances $R_{ij,kl} = 1/G_{ij,kl}$ associated with the above-mentioned conductances and read

$$R_{ij,kl}(B) = R_{kl,ij}(-B). \qquad (4.26)$$

### 4.1.3 Some other properties of the conductance matrix

While the reciprocity relations are certainly the most relevant properties of the conductance matrix, there are two further aspects that may be of practical interest. The first one is concerned with the balance in a contact $i$ that is neither the source nor the drain and in which we have $I_i = 0$. In this case, one obtains directly from Eq. (4.10) Millman's theorem

$$V_i = \frac{\sum_{j \neq i} T_{ij} V_j}{\sum_{j \neq i} T_{ij}}, \qquad (4.27)$$
which states that the potential in a non-current-carrying contact is given by the weighted sum of the potentials in the contacts j connected by the transmission coefficients $T_{ij}$ to i.

Finally, let us consider the dissipated power in the system. Since all scattering events in the conductor are considered to be elastic, the electrons keep their energy during their passage through the conductor. The power is therefore dissipated only in the contacts,

$$P = \sum_{i} I_{i}V_{i}. \qquad (4.28)$$

One first notices that the dissipated power is independent of the reference potential as we should have expected. Indeed, if we change all potentials $V_i \to V_i + V_0$ by a constant value $V_0$, the power does not change because we have $\sum_i I_i = 0$ due to current conservation. We may now replace the current in the expression by the expression (4.10) so that one finds

$$P = 2 \frac{e^2}{h} \sum_{i} \sum_{j \neq i} \mathcal{T}_{ij}(V_i - V_j)V_i = 2 \frac{e^2}{h} \sum_{i} \sum_{j \neq i} \mathcal{T}_{ji}(V_i - V_j)V_i, \qquad (4.29)$$

where we have made use of $\sum_{j \neq i} \mathcal{T}_{ij} = \sum_{j \neq i} \mathcal{T}_{ji}$ in the first term and a simple change of indices $\sum_{i} \sum_{j \neq i} \mathcal{T}_{ij}V_iV_j = \sum_{i} \sum_{j \neq i} \mathcal{T}_{ji}V_iV_j$ in the second term.<fnref n="1" /> We may therefore add the two sides of the identical expressions in the last equation and divide it by two to obtain the power

$$P = \frac{e^2}{h} \sum_{i} \sum_{j \neq i} (\mathcal{T}_{ij} + \mathcal{T}_{ji})(V_i - V_j)V_i = -\frac{e^2}{h} \sum_{i} \sum_{j \neq i} (\mathcal{T}_{ij} + \mathcal{T}_{ji})(V_i - V_j)V_j, \qquad (4.30)$$

where the last term has again been obtained by an exchange of the indices i and j. Again, we may take the half sum of the latter two expressions, and we eventually obtain

$$P = \frac{e^2}{2h} \sum_{i} \sum_{j \neq i} (\mathcal{T}_{ij} + \mathcal{T}_{ji})(V_i - V_j)^2. \qquad (4.31)$$

First, notice that, contrary to the resistances, the dissipated power does not depend on the orientation of the magnetic field because it is the symmetric

<footnote n="1">¹Indeed, the expression is symmetric under the exchange of the indices $\leftrightarrow j$ because so is $V_iV_j$, and the double sum runs over all pairs of indices that are not identical.</footnote>
<figure><img src="image_40.png" /><figcaption>Figure 4.5: Sketch of a conductor in the three-terminal geometry. The current is driven through the device via the source $i = 1$ ($I_1 = I$) and the drain $i = 2$ ($I_2 = -I$), while there is now current flowing into the contact 3, $I_3 = 0$.</figcaption></figure>

combination of the transmission coefficients, $T_{ij} + T_{ji}$, that enters the expression (4.31). Second, it is apparent from this equation that the dissipated power is positive since the transmission coefficients $T_{ij} = |s_{ij}|^2$ are positive, and so is the square of the potential differences.

## 4.2 Conductance through a three-terminal device

After this discussion of the general properties of the conductance matrix, let us now consider in detail the three-terminal device shown in figure 4.5. The Landauer-Büttiker formula (4.5) is then given in terms of a 3 × 3 conductance matrix,

$$ \begin{pmatrix} I \\ -I \\ 0 \end{pmatrix} = 2 \frac{e^2}{h} \begin{pmatrix} (\mathcal{T}_{12} + \mathcal{T}_{13}) & -\mathcal{T}_{12} & -\mathcal{T}_{13} \\ -\mathcal{T}_{21} & (\mathcal{T}_{21} + \mathcal{T}_{23}) & -\mathcal{T}_{23} \\ -\mathcal{T}_{31} & -\mathcal{T}_{32} & (\mathcal{T}_{31} + \mathcal{T}_{32}) \end{pmatrix} \begin{pmatrix} V_1 \\ V_2 \\ V_3 \end{pmatrix} . \quad (4.32) $$

As we have already mentioned, there are not three independent equations, but only two, and we may thus omit one of the lines from the matrix equation. For computational convenience,<fnref n="2" /> we omit the first line. Furthermore, we have seen that we may choose one of the potentials to be zero, e.g. $V_3 = 0$, in which case one may also omit the last column of the conductance matrix. The remaining 2 × 2 matrix may then be easily inverted, in order to obtain

<footnote n="2">²Unfortunately, this convenience sometimes arises only later, after some trials.</footnote>
the related resistances, and one thus has

$$ \begin{pmatrix} V_1 \\ V_2 \end{pmatrix} = \frac{h}{2e^2} \frac{1}{D} \begin{pmatrix} -\mathcal{T}_{32} & -(\mathcal{T}_{21} + \mathcal{T}_{23}) \\ \mathcal{T}_{31} & -\mathcal{T}_{21} \end{pmatrix} \begin{pmatrix} -I \\ 0 \end{pmatrix}, \quad (4.33) $$

in terms of the determinant

$$ \begin{aligned} D &= \mathcal{T}_{21}\mathcal{T}_{32} + \mathcal{T}_{31}(\mathcal{T}_{21} + \mathcal{T}_{23}) \\ &= \mathcal{T}_{21}(\mathcal{T}_{31} + \mathcal{T}_{32}) + \mathcal{T}_{31}\mathcal{T}_{23} = (\mathcal{T}_{21} + \mathcal{T}_{31})(\mathcal{T}_{21} + \mathcal{T}_{23}) - \mathcal{T}_{21}\mathcal{T}_{12}, \end{aligned} \quad (4.34) $$

where we have rewritten the expression in two different manners that turn out to be useful in the following.

The above matrix expression allows us to easily obtain the various resistances

$$ R_{12,ij} = \frac{V_i - V_j}{I}, \quad (4.35) $$

such as

$$ R_{12,13} = \frac{V_1}{I} = \frac{h \mathcal{T}_{32}}{2e^2 D} \quad \text{and} \quad R_{12,32} = \frac{-V_2}{I} = \frac{h \mathcal{T}_{31}}{2e^2 D}. \quad (4.36) $$

Let us check how these resistances evolve under the change of the orientation of the magnetic field. The determinant D remains invariant,

$$ D(B) = D(-B), \quad (4.37) $$

as one may immediately see from its last expression in Eq. (4.34), with the help of (i) $\mathcal{T}_{ij}(-B) = \mathcal{T}_{ji}(B)$ and Eq. (4.13)

$$ \mathcal{T}_{21}(B) + \mathcal{T}_{31}(B) = \mathcal{T}_{12}(B) + \mathcal{T}_{13}(B) = \mathcal{T}_{21}(-B) + \mathcal{T}_{31}(-B). \quad (4.38) $$

Similarly, $\mathcal{T}_{21} + \mathcal{T}_{23}$ is invariant under $B \to -B$, and so is the symmetric term $T_{21}T_{12}$. However, since generally $\mathcal{T}_{32}(B) \neq \mathcal{T}_{32}(-B)$ and $\mathcal{T}_{31}(B) \neq \mathcal{T}_{31}(-B)$, the resistances are not invariant,

$$ R_{12,13}(B) \neq R_{12,13}(-B) \quad \text{and} \quad R_{12,32}(B) \neq R_{12,32}(-B). \quad (4.39) $$

However, we see from Eq. (4.36) that the resistances remain invariant if the change in the orientation of the magnetic field is accompanied by an exchange of the contact 2 with 3, and 1 with 3, respectively,

$$ R_{12,13}(B) = R_{13,12}(-B) \quad \text{and} \quad R_{12,32}(B) = R_{32,12}(-B), \quad (4.40) $$
in agreement with the general reciprocity relations (4.26).

In contrast to these resistances, we may also investigate the two-terminal resistance

$$R_{12,12} = \frac{V_1 - V_2}{I} = \frac{h \mathcal{T}_{32} + \mathcal{T}_{31}}{2e^2 D}. \qquad (4.41)$$

As we have already seen, the denominator in this resistance is invariant under $B \to -B$, and so is the numerator, so that we obtain the invariance of the two-terminal resistance

$$R_{12,12}(B) = R_{12,12}(-B), \qquad (4.42)$$

as one would have expected. Furthermore, we may obtain the two-terminal conductance from this expression,

$$G_3 = \frac{1}{R_{12,12}} = 2 \frac{e^2}{h} \frac{D}{\mathcal{T}_{32} + \mathcal{T}_{31}} = 2 \frac{e^2}{h} \left( \mathcal{T}_{21} + \frac{\mathcal{T}_{31}\mathcal{T}_{23}}{\mathcal{T}_{31} + \mathcal{T}_{32}} \right), \qquad (4.43)$$

where we have used the second-last expression of the determinant (4.34). Notice that this expression of the two-terminal conductance in the three-terminal geometry is different from that

$$G_2 = 2 \frac{e^2}{h} \mathcal{T}_{21} \qquad (4.44)$$

in the two-terminal geometry, which we have obtained in the last chapter! Even if the third contact is not used for any measurement in the present two-terminal conductance, it affects the conductance. Indeed, the result for $G_3$ may be interpreted as the addition of two contributions, one that stems from the direct transmission from contact 1 to 2 (given by $G_2$), while a second contribution represents the contribution from electrons that “visit” the contact 3 before being transmitted to contact 2. Both contributions add up as one would expect for conductances in series. Furthermore, the contribution from the passage in contact 3 may be viewed as an incoherent contribution due to the thermalisation of the electrons transmitted from 1 to 3 with the electrons in the reservoir 3.

Notice finally that we retrieve the two-terminal conductance in the two-terminal geometry if we reduce drastically the transmissions from 1 to 3 and from 3 to 2. For the sake of the argument, let us consider no magnetic field and a symmetric transmission, $\mathcal{T}_{13} = \mathcal{T}_{23} = \epsilon \ll 1$. In this case, the incoherent contribution from the passage through the contact 3 becomes

$$\frac{\mathcal{T}_{31}\mathcal{T}_{23}}{\mathcal{T}_{32} + \mathcal{T}_{31}} = \frac{\epsilon}{2}, \qquad (4.45)$$
and the two-terminal conductance thus reads

$$G_3 = 2 \frac{e^2}{h} \left( \mathcal{T}_{21} + \frac{\epsilon}{2} \right) \rightarrow G_2 \quad \text{for } \epsilon \rightarrow 0. \qquad (4.46)$$

This last argument in terms of an infinitesimal coupling of the third contact to the conductor (or a vanishing *invasiveness*) allows us also to clarify an aspect that might seem counterintuitive: equation (4.43) indicates that in addition to the direct transmission from the source and the drain, $T_{21}$, there is an incoherent contribution due to electrons that passes by the third contact. This should indeed astonish us: adding a third contact to the conductor *increases* the conductance even if we have added a source of decoherence? This is actually only an apparent contradiction since we must reconsider the meaning of the transmission $T_{21}$. In order to illustrate this point, let us consider $B = 0$ for simplicity so that the conduction matrix becomes symmetric. Furthermore, let us consider that there is no reflection into the source and the drain, i.e. $R_{11} = R_{22} = 0$. Since the coupling to the third contact is still given by the transmission coefficient $T_{13} = T_{23} = \epsilon$, it would be a mistake to associate with the vanishing reflection a perfect transmission $T_{21} = M$ because we need to respect the sum rules (4.9). In view of the conductance matrix

$$\mathcal{G} = 2 \frac{e^2}{h} \begin{pmatrix} M & -\mathcal{T}_{21} & -\epsilon \\ -\mathcal{T}_{21} & M & -\epsilon \\ -\epsilon & -\epsilon & M - \mathcal{R}_{33} \end{pmatrix}, \qquad (4.47)$$

we immediately see that the sum rule imposes a reduced value

$$\mathcal{T}_{21} = M - \epsilon, \qquad (4.48)$$

along with $\mathcal{R}_{33} = M - 2\epsilon$. The reduction of the transmission coefficient $T_{12}$ is therefore *larger* than the incoherent contribution (4.45), and we have

$$G_3 = 2 \frac{e^2}{h} \left( M - \frac{\epsilon}{2} \right). \qquad (4.49)$$

While this value still converges to the ballistic two-terminal case $G_2 = 2Me^2/h$ in the limit where $\epsilon \to 0$, we clearly see that $G_3$ is *smaller* than $G_2$.
<figure><img src="image_41.png" /><figcaption>Figure 4.6: Sketch of a conductor in the four-terminal geometry. The current is driven through the device via the source $i = 1$ ($I_1 = I$) and the drain $i = 2$ ($I_2 = -I$), while there is now current flowing into the contacts 3 and 4, $I_3 = I_4 = 0$.</figcaption></figure>

## 4.3 Conductance through a four-terminal device

The above treatment of the conductance matrix in the three-terminal geometry and the calculation of the resistances may be directly transferred to the four-terminal geometry. Indeed, the discussion of the three-terminal geometry may be viewed as a warm-up for this more interesting situation, where we can revisit some aspects already mentioned in the previous chapter on the conductance through a quantum wire and where we may be confronted with *negative* resistances. The setup is depicted in figure 4.6. As before, the contacts 1 and 2 are the source and the drain, respectively, while there is no current flowing in the *floating* contacts 3 and 4. The Landauer-Büttiker formula is now given in terms of a 4 × 4 matrix,

$$ \begin{pmatrix} I \\ -I \\ 0 \\ 0 \end{pmatrix} = 2 \frac{e^2}{h} \begin{pmatrix} (M_1 - \mathcal{R}_{11}) & -\mathcal{T}_{12} & -\mathcal{T}_{13} & -\mathcal{T}_{14} \\ -\mathcal{T}_{21} & (M_2 - \mathcal{R}_{22}) & -\mathcal{T}_{23} & -\mathcal{T}_{24} \\ -\mathcal{T}_{31} & -\mathcal{T}_{32} & (M_3 - \mathcal{R}_{33}) & -\mathcal{T}_{34} \\ -\mathcal{T}_{41} & -\mathcal{T}_{42} & -\mathcal{T}_{43} & (M_4 - \mathcal{R}_{44}) \end{pmatrix} \begin{pmatrix} V_1 \\ V_2 \\ V_3 \\ V_4 \end{pmatrix}, \quad (4.50) $$

i.e. in terms of four equations, one of which is again redundant. As in the case of the three-terminal geometry, we omit the first line of the matrix equation as well as the last column (by choosing $V_4 = 0$) so that we need to invert a 3 × 3 matrix in order to obtain the (six) different resistances
$R_{12,ij} = (V_i - V_j)/I$. The reduced system of equation therefore reads

$$ \begin{pmatrix} -I \\ 0 \\ 0 \end{pmatrix} = 2 \frac{e^2}{h} \begin{pmatrix} -\mathcal{T}_{21} & (\mathcal{T}_{21} + \mathcal{T}_{23} + \mathcal{T}_{24}) & -\mathcal{T}_{23} \\ -\mathcal{T}_{31} & -\mathcal{T}_{32} & (\mathcal{T}_{31} + \mathcal{T}_{32} + \mathcal{T}_{34}) \\ -\mathcal{T}_{41} & -\mathcal{T}_{42} & -\mathcal{T}_{43} \end{pmatrix} \begin{pmatrix} V_1 \\ V_2 \\ V_3 \end{pmatrix}, \quad (4.51) $$

and its inversion yields the voltages

$$ \begin{pmatrix} V_1 \\ V_2 \\ V_3 \end{pmatrix} = \frac{h}{2e^2 D} \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \begin{pmatrix} -I \\ 0 \\ 0 \end{pmatrix}, \quad (4.52) $$

in terms of the determinant

$$ D = -\mathcal{T}_{21}\mathcal{T}_{32}\mathcal{T}_{43} - \mathcal{T}_{41}\sum_j \mathcal{T}_{2j}\sum_i \mathcal{T}_{3i} - \mathcal{T}_{23}\mathcal{T}_{31}\mathcal{T}_{42} - \mathcal{T}_{21}\mathcal{T}_{42}\sum_j \mathcal{T}_{31} \\ -\mathcal{T}_{31}\mathcal{T}_{43}\sum_j \mathcal{T}_{2j} + \mathcal{T}_{23}\mathcal{T}_{32}\mathcal{T}_{41}, \quad (4.53) $$

and the matrix elements $a_{ij}$ are the adjoint elements of $\tilde{T}_{ji}$, where $\tilde{T}_{ji}$ is the element in line $j$ and row $i$ of the reduced matrix in Eq. (4.51). Naturally, the expressions we need to handle are more cumbersome, but the structure is relatively simple. In a first step, let us investigate the sign of the determinant. Just by looking at its expression and knowing that the transmission coefficients are all positive, you may bet that the determinant is negative: the only positive term is the last one. You have then bet on the right horse. The determinant is indeed negative because this last term cancels an identical one that occurs in the sum of the second term of the determinant. We may therefore use the modulus of the determinant and simply change $-I \to I$ on the right-hand side of Eq. (4.52) to avoid sign problems in the calculations.

Out of the six different resistances, there are two that are of special interest here: the two-terminal resistance $R_{12,12}$, which is related to the voltage drop in the source and drain, and the four-terminal resistance $R_{12,34}$, which measures the voltage drop between the floating contacts 3 and 4, i.e. between the contacts where no current is flowing into or out of the conductor. The four-terminal resistance $R_{12,34}$ is also sometimes called a non-local resistance. In order to obtain this resistance, we simply need to evaluate

$$ R_{12,34} = \frac{V_3 - V_4}{I} = \frac{V_3}{I} = \frac{h}{2e^2} \frac{a_{31}}{|D|}. \quad (4.54) $$
As mentioned above, $a_{31}$ is the adjoint element of the upper right element $-\mathcal{T}_{23}$ of the reduced conduction matrix (without the unit $2e^2/h$), and it is thus given by the determinant

$$a_{31} = \mathcal{T}_{31}\mathcal{T}_{42} - \mathcal{T}_{32}\mathcal{T}_{41}, \qquad (4.55)$$

so that

$$R_{12,34} = \frac{h \mathcal{T}_{31} \mathcal{T}_{42} - \mathcal{T}_{32} \mathcal{T}_{41}}{2e^2 |D|}. \qquad (4.56)$$

Similarly, we obtain for the two-terminal resistance

$$R_{12,12} = \frac{V_1 - V_4}{I} = \frac{h a_{11} - a_{21}}{2e^2 |D|} = \frac{h \sum_{j \neq 3} \mathcal{T}_{3j} \sum_{i \neq 4} \mathcal{T}_{4i} - \mathcal{T}_{34} \mathcal{T}_{43}}{2e^2 |D|}. \qquad (4.57)$$

Here, the patient reader may start to be a bit frustrated. Matrix inversion is a cumbersome bookkeeping task that yields lengthy expressions, and we yet do not know anything about the transmission coefficients. Indeed, the transmission coefficients would need to be calculated within a microscopic quantum treatment of the scatterer that is not the object of the present classes. However, as we show below, these expressions bear already relevant information about the system, such as the possibility to obtain negative resistances, and we can furthermore make a connection with the quantum transport through a quantum wire, where we have first introduced the distinction between the two- and four-terminal resistances, albeit in an approach, where we considered the contacts 3 and 4 (connecting the ballistic leads) to be non-invasive (see section 3.4.1).

### 4.3.1 Test of the Onsager reciprocity relations

Let us first investigate the two resistances (4.56) and (4.57) in the light of the reciprocity relations according to which the two-terminal resistance must be symmetric if we change the orientation of the magnetic field, while this is generally not the case for the four-terminal resistance. Indeed, the determinant is invariant, $D(-B) = D(B)$, as one may check directly from its expression (4.53). Notice that this property is inherited from the full conductance matrix, where we have $G_{ij}(-B) = G_{ji}(B)$, i.e. the conduction matrix simply needs to be transposed if we change B to $-B$. However, the determinant is invariant under matrix transposition. Furthermore, the
numerator in Eq. (4.56) is invariant because the first term contains the sum over rows of the conductance matrix, for which we have $\sum_{j \neq i} T_{ij} = \sum_{j \neq i} T_{ji}$ and the second term in the numerator is symmetric under the exchange of the indices. We thus retrieve

$$R_{12,12}(B) = R_{12,12}(-B), \qquad (4.58)$$

as in the three-terminal geometry and as it is expected on general grounds, while this is not the case for the four-terminal resistance,

$$R_{12,34}(B) \neq R_{12,34}(-B), \qquad (4.59)$$

since generally $T_{ij} \neq T_{ji}$. However, we have

$$R_{12,34}(-B) = R_{34,12}(B), \qquad (4.60)$$

in agreement with the property discussed above that the change in the orientation of the magnetic field must be accompanied by an exchange of the current and voltage contacts. This property can be explicitly checked if we change 1 ↔ 2 and 2 ↔ 4, in line with the four-terminal measurement depicted in Figs. 3.3 and 4.4(b).

### 4.3.2 Negative four-terminal resistances

A particular situation may arise in which the four-terminal resistance, in contrast to the two-terminal one, can become negative. That the two-terminal resistance is always positive is apparent when looking at the expression (4.57), where the negative last term is compensated by one of the terms in the (positive) sum in the numerator. From a mathematical point of view, a negative four-terminal resistance requires simply that the transmission between 1 and 3 as well as that between 2 and 4 are reduced with respect to the transmission between 1 and 4 and that between 2 and 3, $T_{31}T_{42} < T_{32}T_{41}$. What does this mean from a physical point of view? For the sake of the argument, we consider the $B = 0$ case here so that the transmission from i to j is the same as that from j to i. In a usual situation, with the naming of the contacts as in figure 4.6, the voltage contact 3 is spatially closer to the source 1 than to the drain 2, while the voltage contact 4 is closer to the drain than to the source. If we consider a normal diffusive conductor, the transmission between two contacts is then inversely proportional to the distance, and the voltage drop
<figure><img src="image_42.png" /><figcaption>Figure 4.7: Four-terminal measurement of a (single-wall) carbon nanotube that shows negative four-terminal resistances. (a) Image of the device. (b) Sketch of the setup. The source and the drain are metallic (gold) contacts while the contacts 3 and 4 used to measure the voltage drop consist of metallic (multi-wall) nanotubes. (c) Sketch of the potentials in the contacts for a usual diffusive conductor that yield a positive four-terminal resistance $R_{4pt} = R_{12,34} > 0$. (d) Sketch of the inverted situation, where the potential in contact 4 is higher than in contact 3, such that one measures a negative four-terminal resistance $R_{4pt} < 0$. (e) Measured four-terminal resistance (devided by the two-terminal resistance) of a device at different temperatures, showing negative resistances at low temperatures. (f) Low-temperature measurement of another device. After Gao *et al.*, Phys. Rev. Lett. **95**, 196802 (2002).</figcaption></figure>
is expected to be monotonic over the conductor. This situation is depicted in figure 4.7(c), in which case one would measure a positive four-terminal resistance. However, let us assume that for some reason the transmission from the source to the contact 4 and that from the drain to contact 3 is larger than that from the source to contact 3 (and that from the drain to contact 4), in contrast to our naïve expectations. In this case, sketched in figure 4.7(d), the chemical potential (and thus the voltage) in contact 4 is better equilibrated with that of the source, while the voltage in contact 3 is closer to its value in the drain. Since we then have $V_3 < V_4$, while $V_1 > V_2$, one naturally obtains a negative resistance. While this situation is at odds with our intuitions that stem from a classical diffusive conductor, it may arise in a quantum conductor due to quantum-mechanical interference effects if the latter suppress the transmission $T_{13}$ and $T_{24}$ with respect to $T_{14}$ and $T_{23}$.

Such negative four-terminal resistances have indeed been measured experimentally, as shown in figure 4.7. The represented setup consists of a metallic (single-wall) carbon nanotube that is connected to two gold contacts which serve as the source and the drain. The voltage contacts consist of other metallic (multi-wall) carbon nanotubes. The measured four-terminal resistances (devided by the two-terminal resistances) at different temperatures are shown in figure 4.7(e), as a function of the gate voltage that allows one to change the electronic density and thus the Fermi level in the conductor. One clearly notices that the four-terminal resistance becomes negative at certain values of the gate voltage in the low-temperature limit, while the effect disappears at high temperatures. Figure 4.7(f) shows the low-temperature resistance in another device. The fact that the four-terminal resistance is no longer negative at higher temperatures indicates the quantum origin of the effect, i.e. quantum interferences are responsible for a better transmission over larger distances, as mentioned above.

### 4.3.3 Non-invasive four-terminal device

In this final section, we revisit, in view of the Landauer-Büttiker formalism, the potential landscape of a diffusive conductor in the four-terminal geometry, which we have discussed in section 3.4.1). The setup is sketched in figure 4.8 and coincides with the situation shown in figure 3.9. We suppose that there is no magnetic field present so that $T_{ij} = T_{ji}$. As before, the contacts 1 and 2 are the source and drain, respectively, and they are connected to the diffusive conductor by ballistic leads. The transmission through the diffusive region
<figure><img src="image_43.png" /><figcaption>Figure 4.8: Sketch of a conductor in the four-terminal geometry, connected by ballistic leads to the source and the drain. The contacts 3 and 4 are connected to the leads on both sides of the diffusive conductor, and their transmission is controlled by the value $\epsilon$ which describes the “invasiveness”. When $\epsilon \to 0$, these contacts are non-invasive, while they are fully Ohmic for $\epsilon = 1$.</figcaption></figure>

is characterised by the coefficient $\mathcal{T}$, while $\epsilon$ describes the “invasiveness” of the contacts 3 and 4. It can be viewed as the transmission to the ballistic leads, but it should not be mixed up with the transmission coefficient to the source or the drain. In order to see this, let us consider the transmission coefficient from the source 1 to contact 3. An electron can naturally be transmitted directly, and that yields a contribution $\epsilon$. However, it can also indirectly reach the contact 3 if it remains in the lead (with a probability $1 - \epsilon$) before being reflected by the scatterer (the diffusive part of the wire) with a probability $\mathcal{R} = 1 - \mathcal{T}$ and then reach the contact (again with the probability $\epsilon$). The transmission coefficient is then given by

$$ \mathcal{T}_{13} = \epsilon + (1 - \epsilon)\mathcal{R}\epsilon = \mathcal{T}_{24}, \qquad (4.61) $$

if we neglect multiple backscattering processes between the contact 3 and the diffusive part. By symmetry, this is also the transmission coefficient from the drain to contact 4. Furthermore, the direct transmission from the source to the drain is then given by

$$ \mathcal{T}_{12} = (1 - \epsilon)\mathcal{T}(1 - \epsilon) \qquad (4.62) $$

and that from the source to contact 4 (and from the drain to contact 3) by

$$ \mathcal{T}_{14} = \mathcal{T}_{23} = (1 - \epsilon)\mathcal{T}\epsilon, \qquad (4.63) $$

by the same probability multiplication as explicit for the coefficient $\mathcal{T}_{13}$. If we inject these expressions in the general expressions (4.57) and (4.56) we
obtain, after a cumbersome but straight-forward calculation

$$R_{12,12} = \frac{h}{2e^2} \left( \frac{1}{\mathcal{T}} + \frac{2\epsilon}{2 - \epsilon} \right) \qquad (4.64)$$

for the two-terminal resistance and

$$R_{12,34} = \frac{h}{2e^2} \left( \frac{\mathcal{R}}{\mathcal{T}} + \frac{\epsilon}{2 - \epsilon} \right) \qquad (4.65)$$

for the four-terminal resistance.

Let us interpret these two results in the different limits, starting with the limit where $\epsilon \to 0$. This is the limit, where the contacts 2 and 4 are barely coupled to the leads so that they can be considered as non-invasive. We discussed this limit in the framework of the quantum wire in section 3.4.1, and we retrieve the Landauer formula

$$R_{12,12} = \frac{h}{2e^2} \frac{1}{\mathcal{T}} \quad \text{and} \quad R_{12,34} = \frac{h}{2e^2} \frac{\mathcal{R}}{\mathcal{T}}, \qquad (4.66)$$

in agreement with Eqs. (3.78) and (3.81).

In the opposite limit, with $\epsilon \to 1$, where the contacts are perfectly coupled to the leads, we find

$$R_{12,12} = \frac{h}{2e^2} \left( \frac{1}{\mathcal{T}} + 2 \right) \quad \text{and} \quad R_{12,34} = \frac{h}{2e^2} \frac{1}{\mathcal{T}}, \qquad (4.67)$$

in agreement with the sum of resistances in series and Ohm's law. To see this point, we may decompose the two-terminal resistance into three parts

$$R_{12,12} = \frac{V_1 - V_3}{I} + \frac{V_3 - V_4}{I} + \frac{V_4 - V_2}{I} = R_{13}^{\text{bal}} + R_{34} + R_{42}^{\text{bal}}. \qquad (4.68)$$

Indeed, the fact that $\epsilon = 1$ indicates that all electrons that arrive in the diffusive part from the source inevitably pass through the contact 3 (at the potential $V_3$, which they keep until they arrive in the scattering region). Similarly all electrons are indirectly transmitted from the drain to the diffusive part, passing by the potential $V_4$. The transmission from contact 3 to 4 is therefore effectively that of the conductor in the two-terminal geometry $R_{12,12} = h/2e^2\mathcal{T}$ (in the limit $\epsilon = 0$), while that from 1 to 3 (and 4 to 2) is
<figure><img src="image_44.png" /><figcaption>Figure 4.9: Resistance measurement of a ballistic GaAs quantum wire in the four-terminal geometry. The setup is depicted on the right-hand side, and the lower curve on the left-hand side shows the two-terminal resistance $R_{12,34}$ while the upper curve shows the four-terminal resistance $R_{12,12}$, as a function of the gate voltage. The inset represents the ratio between the four-terminal resistance as a function of the invasiveness and the two-terminal resistance at $\epsilon = 0$. After De Picciotto et al., Nature 411, 51 (2001).</figcaption></figure>

that of a ballistic conductor in the two-terminal geometry, with the resistance $R_{13}^{\text{bal}} = R_{42}^{\text{bal}} = h/2e^2$. This yields the ratio

$$ \frac{R_{12,34}}{R_{12,12}} = \frac{1}{1 + 2\mathcal{T}} \qquad (4.69) $$

for the two resistances.

The latter statement is also in line with Millman's theorem (4.27) for the contacts 3 and 4. The potentials are given by

$$ V_3 = \frac{V_1 + \mathcal{T}V_4}{1 + \mathcal{T}} \quad \text{and} \quad V_4 = \frac{V_2 + \mathcal{T}V_3}{1 + \mathcal{T}}, \qquad (4.70) $$

because $\epsilon = 1$ yields $\mathcal{T}_{12} = 0$, $\mathcal{T}_{13} = 1$, $\mathcal{T}_{14} = 0$, $\mathcal{T}_{23} = 0$, $\mathcal{T}_{24} = 1$ and $\mathcal{T}_{34} = \mathcal{T}$, as one may see directly from the general expressions. We therefore obtain the relation between the potential differences

$$ V_3 - V_4 = \frac{1}{1 + 2\mathcal{T}}(V_1 - V_2) \quad \leftrightarrow \quad \frac{R_{12,34}}{R_{12,12}} = \frac{V_3 - V_4}{V_1 - V_2} = \frac{1}{1 + 2\mathcal{T}}. \qquad (4.71) $$

To finish this section on the four-terminal resistance, we present the experiment depicted in figure 4.9, where the four-terminal resistance of a ballistic
conductor (a GaAs quantum wire) has been measured as a function of the invasiveness $\epsilon$ of the contacts 3 and 4 (A and B) in the figure. The particularity of these contacts is that they are also formed of GaAs quantum wires, whose transmission can be controlled by side gates. The inset of the figure shows the ratio between the four-terminal resistance $R_{12,34}(\epsilon)$ as a function of $\epsilon$ and the two-terminal resistance in the absence of the contacts 3 and 4, $R_{12,12}(\epsilon = 0)$. From the general expressions (4.64) and (4.65) in the limit $T = 1$,

$$R_{12,34}(\epsilon) = \frac{h \epsilon}{2e^2 (2 - \epsilon)} \quad \text{and} \quad R_{12,12}(\epsilon) = \frac{h (2 + \epsilon)}{2e^2 (2 - \epsilon)}, \qquad (4.72)$$

we obtain the ratio

$$\frac{R_{12,34}(\epsilon)}{R_{12,12}(\epsilon = 0)} = \frac{\epsilon}{2 - \epsilon}, \qquad (4.73)$$

in agreement with the measured values. Notice that our expressions represent the single-channel case, while the experiment has been performed on a quantum wire of finite width. Both resistances therefore need to be divided by the number of channels $M(V)$ (i.e. the number of electronic bands below the Fermi level), which can be controlled by the gate voltage, as we have seen in section 3.6. This is at the origin of the step-like increase of the two-terminal resistance as a function of the (negative) gate voltage (see figure 4.9), but it does not affect the measured ratio between the two resistances.

## Notions and concepts to retain

* multi-terminal devices and Landauer-Büttiker formalism

* conductance matrix

* relations between transmission and reflection coefficients (current conservation and reference potential)

* Onsager reciprocity relations in the presence of a magnetic field

* Millman's theorem

* dissipated power in the contacts

* negative four-terminal resistances

* invasiveness of contacts
# Chapter 5

## Quantum Hall effects

One of the key experiments in quantum mesoscopic physics is definitely the quantum Hall effect, which is observed in high-mobility 2D electron systems in a strong perpendicular magnetic field. The effect, which was discovered by K. v. Klitzing in 1980 (Nobel Prize in 1985), consists of the appearance of plateaus in the Hall resistance, at precisely quantised values $R_H = h/Me^2$, where *M* is an integer. A typical Hall-resistance curve is shown in figure 5.1. In view of the terminology introduced in the preceding chapters, the Hall resistance is a *transverse* four-terminal resistance $R_H = R_{14,35}$, measured between, e.g., the contacts 3 and 5, while the contacts 1 and 4 are the source and the drain, respectively. The Hall resistance is accompanied by another *longitudinal* resistance, which is also a four-terminal resistance $R_L = R_{14,56}$ and that vanishes whenever the Hall resistance shows a plateau. On the basis of the knowledge acquired in the preceding chapters, we can already see that the transport must be ballistic through the device and that the integer *M* counts the relevant channels even if we do not know yet where the current flows through the device. The answer to this question and an understanding of the origin of this ballistic transport is the aim of this chapter. Notice that the quantisation of the Hall resistance is so precise that the effect is used nowadays as the resistance standard. Furthermore, this so-called integer quantum Hall effect (IQHE), which denotes the effect when *M* is an integer number, is accompanied in high-mobility samples by its fractional brother, the fractional quantum Hall effect (FQHE), which occurs at rational values of *M*. While this effect merits a discussion of its own – it is the prototype of topological phases that occur due to strong electronic correlations – it will not be covered in these lecture notes, where we concentrate on non-interacting
<figure><img src="image_45.png" /><figcaption>Figure 5.1: Experimental signature of the quantum Hall effect. Instead of a linear dependence on the magnetic field, as expected from the classical Hall effect (dashed gray line), the Hall resistance measured by the potential drop $V_y$ between the contacts 3 and 5 in the device (schematically represented in the inset) shows plateaus at quantised values $R_h = h/e^2M$. If $M$ is an integer, the effect is called integer quantum Hall effect (IQHE), while one speaks of the fractional quantum Hall effect (FQHE) if $M = p/q$ is a rational number, in terms of the integers $p$ and $q$. The plateaus in the Hall resistance are accompanied by a vanishing longitudinal resistance, measured by the potential drop $V_x$ between the contacts 5 and 6. The current is driven through the device by the contacts 1 (source) and 4 (drain).</figcaption></figure>

electronic systems, as mentioned in the introduction.

The present chapter aims at an understanding of the quantum Hall ef-
fect in the framework of the Landauer-Büttiker formalism introduced in the
preceding chapter. It is furthermore based on *Landau quantisation*, that is
the quantisation of the electronic energy in the presence of a magnetic field.
Indeed, the cyclotron motion of 2D electrons is reminiscent of a harmonic
oscillator, and Landau quantisation similarly yields discrete albeit highly
degenerate energy levels. Their degeneracy is given by the number of flux
quanta threading the 2D plane, as we show explicitly below. Landau quanti-
sation is the object of section 5.2, after a short reminder of the classical Hall
effect treated within the Drude model (section 5.1). Once an integer number of Landau levels is completely filled, the Fermi level resides between two adjacent Landau levels and one expects the system to be insulating since adding an electron comes with a cost in energy that is given by the Landau-level separation. Naturally, this is a very particular type of insulator since we need to interpret the Landau levels as infinitely flat bands. Moreover, this is only half of the truth. As we show in section 5.3, the Landau levels are bent upwards in energy upon approaching the edges of the system and thus cross the Fermi level there. These crossings are at the origin of metallic edge states that allow the electrons to travel from one contact to the other. Most saliently, these edge states happen to be chiral, i.e. the direction of propagation is imposed by the orientation of the magnetic field so that backscattering is suppressed. Indeed, backscattering would require that an electron could travel across the insulating bulk from one edge to the other, and such processes are exponentially suppressed in the distance between the edges. As a consequence, the edge transport is ballistic, as one may have already inspected from the precise resistance quantisation mentioned above. Such ballistic transport via the edge channels is then discussed in detail within the Landauer-Büttiker formalism in section 5.4.

This chapter on the quantum Hall effect is central since it also serves as a bridge to the physics of topological insulators that is discussed in the following chapters. Indeed, the dichotomy between an insulating bulk and metallic edges is at the heart of the so-called bulk-edge correspondance, which characterises topological materials. This central role played by the quantum Hall effect justifies the length of the present chapter of which the reader should be averted. Furthermore, we have chosen, in contrast to many textbooks, a gauge-independent approach to Landau quantisation. While this approach is perhaps a little bit lengthier than a direct solution of the Landau-level problem by fixing a particular gauge for the vector potential, it unveils the physical structure of the effect. First, it allows us to treat the edge states and the localised states in the bulk on an equal footing with the help of smoothly varying electrostatic potentials. Second, this approach happens to be useful in a completely different context, namely that of edge states in topological materials discussed in chapter 7.
<figure><img src="image_46.png" /><figcaption>Figure 5.2: Sketch of a 2D electron system in a perpendicular magnetic field **B**. Due to the deviation of electrons by the magnetic field, one obtains a charge gradient between opposite edges when one drives a current through the system by the contacts C1 and C4. This gradient yields a voltage drop between the two edges and thus a Hall resistance that one may measure between the contacts C3 and C5. In the classical Hall effect, the Hall resistance varies linearly with the magnetic field.</figcaption></figure>

## 5.1 Classical Hall effect

Let us first briefly remember the classical Hall effect, which E. Hall discovered in 1879. When one drives a current *I* through a metallic film, e.g. through the contacts C1 and C4 in figure 5.2, the electrons are deflected by the magnetic field due to the Lorentz force. This yields a charge difference between the upper and the lower sample edge and thus a voltage drop *V*<sub>H</sub> that can be measured between the contacts C3 and C5, for example. One thus obtains the Hall resistance *R*<sub>H</sub> = *V*<sub>H</sub>/<i>I</i> = *B*/*en*<sub>el</sub>, which is proportional to the applied magnetic field *B* and inversely proportional to the 2D charge density *en*<sub>el</sub>. Notice that the classical Hall effect is still used today to measure the carrier density and charge in metals.

The effect can be understood in the framework of the Drude model, which we have already encountered in the introduction and that is based on Newton's equation of motion. The latter needs to account for the Lorentz force in addition to the force due to the electric field,

$$ \frac{d\mathbf{p}}{dt} = -e \left( \mathbf{E} + \frac{\mathbf{p}}{m} \times \mathbf{B} \right) - \frac{\mathbf{p}}{\tau}, \qquad (5.1) $$

in terms of the band mass *m* and the relaxation time *τ*. As before, one obtains a linear relation between the current density and the electric field if
we search for the stationary solution $d\mathbf{p}/dt = 0$,

$$ 
\begin{aligned}
\sigma_0 E_x &= \frac{en_{el}}{m} [-p_x - (\omega_C \tau) p_y] \\
\sigma_0 E_y &= \frac{en_{el}}{m} [(\omega_C \tau) p_x - p_y],
\end{aligned}
\qquad (5.2) $$

where we have introduced the cyclotron frequency

$$ \omega_C = \frac{eB}{m} \qquad (5.3) $$

and the same Drude conductivity $\sigma_0 = n_{el}e^2\tau/m$ as in the introduction. The cyclotron frequency turns out to be an essential quantity. In the case of free electrons with a parabolic dispersion, it is the frequency associated with the harmonic cyclotron motion, and the reminiscence of the present system of electrons in a strong magnetic field with the harmonic oscillator indicates that the energy levels obtained within a quantum treatment depends on $\hbar\omega_C$, as we show in detail below.

In order to obtain the Drude model for electrons in a magnetic field, we bring again the above equations of motion into the standard form

$$ \mathbf{E} = \rho \mathbf{j}, \qquad (5.4) $$

in terms of the current density $\mathbf{j} = -en_{el}\mathbf{v} = -en_{el}\mathbf{p}/m$. The resistivity is now a matrix quantity, as much as the conductivity,

$$ \rho = \frac{1}{\sigma_0} \begin{pmatrix} 1 & \omega_C \tau \\ -\omega_C \tau & 1 \end{pmatrix} \quad \text{and} \quad \sigma = \rho^{-1} = \frac{\sigma_0}{1 + (\omega_C \tau)^2} \begin{pmatrix} 1 & -\omega_C \tau \\ \omega_C \tau & 1 \end{pmatrix}. \qquad (5.5) $$

The off-diagonal element of the resistivity matrix represents the Hall resistivity,

$$ \rho_H = \frac{\omega_C \tau}{\sigma_0} = \frac{B}{en_{el}}. \qquad (5.6) $$

It is noteworthy that the Hall resistivity coincides with the Hall resistance, the latter being related to the former by a geometric aspect ratio, $R_H = (L/W^{d-1})\rho_H$, for a conductor of length $L$ and width $W$ in $d$ dimensions. In the present case of a 2D conductor ($d = 2$), resistivity and resistance are dimensionally equivalent $[\rho] = [R]$, as well as the conductivity and the conductance, $[\sigma] = [G]$. Furthermore, we are interested in the Hall resistance, which is a transverse resistance, and the width of the conductor is thus given by the length itself.
A hint of quantum mechanics

Notice that we may also write the conductivity and the resistivity (5.5) in terms of the mobility $\mu = e\tau/m$ if we replace

$$ \omega_C \tau = \mu B. \qquad (5.7) $$

This equation turns out to be a useful relation both for the following parts and experimentally in the experimental determination of the mobility in diffusive metals. Indeed, once the cyclotron frequency outcasts the inverse scattering time, an electron can make a full circle without being scattered. From a quantum-mechanical point of view one may then expect novel effects because the associated wave function can connect with itself, and this is at the origin of *quantum oscillations* that occur when $\omega_C \tau \simeq 1$.<fnref n="1" /> In order to understand this point, without appealing to a full quantum treatment that may be found in the following section, let us consider this point within a semiclassical framework of Bohr-Sommerfeld quantisation. As for the Bohr model for atoms, a closed trajectory<fnref n="2" /> in phase space needs to satisfy the quantisation condition

$$ \oint d\mathbf{r} \cdot \mathbf{p} = h(n + \gamma), \qquad (5.8) $$

in terms of the integer number $n$. This ensures the single-valuedness of the wave function everywhere on the trajectory. Here, $\gamma$ is a *geometric* phase, and $h = 2\pi\hbar$ is the Planck constant. As a consequence of the linear relation between the (cyclotron) radius $R_C$ of the circular trajectory and the velocity, the Bohr-Sommerfeld quantisation rule yields a quantised radius<fnref n="3" />

$$ R_C = l_B \sqrt{2(n + \gamma)}, \qquad (5.9) $$

$$ \mathbf{p} = m\mathbf{v} - e\mathbf{A}(\mathbf{r}). $$

<footnote n="1">¹To simplify the discussion we consider $B > 0$ here so that the cyclotron frequency is a positive quantity. Otherwise we would need to appeal to the modulus of the frequency that renders the notations a bit cumbersome.</footnote>

<footnote n="2">²This is a semiclassical notion since you know that trajectories cease to be relevant in canonical quantum mechanics.</footnote>

<footnote n="3">³Even if this point is not essential for the remainder of these classes, we provide in this footnote the explicit calculation of this radius. In a first step, we need to express the momentum $\mathbf{p}$ in the Bohr-Sommerfeld formula in terms of the (tangential) electron velocity $\mathbf{v}$ since both are no longer directly proportional to each other (even for electrons with a parabolic dispersion relation) because of the presence of the vector potential,</footnote>
in terms of the magnetic length

$$l_B = \sqrt{\frac{\hbar}{eB}}. \qquad (5.10)$$

Not only the velocity, but also the (semiclassical) wave vector depends on the cyclotron radius, $\omega_C R_C = v = \hbar k/m$, the wave vector (at the Fermi level) inherits from the cyclotron radius its quantisation,

$$k_F \approx k_n = l_B^{-2} R_C = l_B^{-1} \sqrt{2(n + \gamma)}. \qquad (5.11)$$

Furthermore, this yields the energy quantisation $E_n = E(k_n)$. As an example, we may take a look at a parabolic isotropic band, e.g. in GaAs in the vicinity of the Γ-point, with $E(\mathbf{k}) = \hbar^2(k_x^2 + k_y^2)/2m$. In this case, the semiclassical quantisation scheme yields, with $k = \sqrt{k_x^2 + k_y^2} \rightarrow k_n$, the discrete energy spectrum

$$E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{\hbar^2}{ml_B^2} (n + \gamma) = \hbar\omega_C(n + \gamma). \qquad (5.12)$$

Similarly, we obtain for the graphene dispersion $E = \pm \hbar vk$

$$E_n = \pm \hbar \frac{v}{l_B} \sqrt{2(n + \gamma)}. \qquad (5.13)$$

While this semiclassical treatment yields indeed the correct form of the energy spectra, only the full canonical quantum treatment, presented in section 5.2, provides naturally the correct values for the geometric phases, which happen to be $\gamma = 1/2$ for a parabolic band and $\gamma = 0$ for Dirac fermions.

The semiclassical approach allows us to already understand the appearance of quantised energy levels that are called Landau levels and an intriguing effect discovered by Shubnikov and de Haas in 1930. They observed that, at large magnetic fields with $\omega_C \tau \gtrsim 1$ or $\mu B \gtrsim 1$, the longitudinal resistance starts to oscillate, as it is schematically shown in figure 5.3(a), while the Hall

$$\oint d\mathbf{r} \cdot \mathbf{p} = 2\pi e B R_C^2 - e \oint d\mathbf{r} \cdot \mathbf{A}(\mathbf{r}) = 2\pi e B R_C^2 - e \int d\sigma B = 2\pi e B R_C^2 - e \pi R_C^2 B = \pi e B R_C^2,$$

<footnote>The tangential velocity may then be rewritten as $\mathbf{v} = \omega_C R_C \mathbf{u}_\theta$, in terms of the tangential unit vector $\mathbf{u}_\theta$, and one thus finds</footnote>

<footnote>where we have made use of Stoke's theorem to replace the line integral over the vector potential by a surface integral over the enclosed (homogeneous) magnetic field. The surface being $\pi R_C$, substitution into Eq. (5.8) and division by $2\pi e B$ yield the result.</footnote>
<figure><img src="image_47.png" /><figcaption>Figure 5.3: (a) Sketch of the Hall and the longitudinal resistances, the latter of which starts to oscillate above a characteristic magnetic field $B_c$. (b) Sketch of the density of states, where the energy levels are broadened by disorder (continuous line: weak disorder; dashed line: stronger disorder).</figcaption></figure>

resistance remains linear.<fnref n="4" /> The effect is due to the above-mentioned energy quantisation and can be qualitatively understood on the basis of the Einstein relation (3.37) for diffusive transport discussed in section 3.2,

$$ \sigma = e^2 D \rho(E_F). \qquad (5.14) $$

The longitudinal conductivity is proportional to the density of states, which we have sketched in figure 5.3(b) for our case. Since this density of states consists of more or less large peaks centred around the Landau levels $E_n$, the longitudinal conductivity starts to oscillate as a function of the magnetic field. Every time that a Landau level crosses the Fermi level, there is a maximum in the density of states and thus in the conductivity. Notice that the onset of the Shubnikov-de-Haas oscillations at $\omega_C\tau = \mu B_c \simeq 1$ is commonly used to determine experimentally the mobility of metals.

<footnote n="4">⁴Notice that these Shubnikov-de-Haas oscillations are also visible in figure 5.1 below $B \sim 1$ T.</footnote>
## 5.2 Landau quantisation

The quantisation of the energy of a 2D electron in a perpendicular magnetic field is nowadays known under the name Landau quantisation after the Soviet physicist L. Landau who published his calculations in 1930. However, names do often not reflect the original calculation, and it is the case here, as Landau pointed out himself in his publication. Indeed, the more complicated calculation of the energy spectrum of a relativistic (Dirac) particle in a strong magnetic field had been published before by the Italian physicist I. Rabi (in 1928), who himself mentioned in his publication the “well-known” result of energy levels for non-relativistic electrons in a magnetic field. Landau thus repeated these calculations, not to understand this level quantisation in itself, but in the perspective of understanding the diamagnetism of metals, i.e. the tendency of a metal to form an orbital magnetisation that is oriented opposite to the magnetic field. Landau diamagnetism is indeed one of the most central manifestations of Landau quantisation, along with the quantum Hall effect and its precursor, Shubnikov-de-Haas oscillations, which were discovered just some months after Landau’s work.

As mentioned above, Landau chose free non-relativistic particles with a parabolic dispersion in his original calculation. Here, we choose a more general approach based on a Bloch Hamiltonian $H(k)$ in the low-energy limit such as to treat electrons with a parabolic band and those in graphene (and yet others) on an equal footing. Notice that we cannot *a priori* use the full Bloch Hamiltonian due to the presence of the vector potential $A(r)$. It necessarily depends linearly on the position $r$ in order to provide a homogeneous magnetic field $B = \nabla \times A(r)$ and thus violates Bloch’s theorem that requires periodic potentials. We have a certain freedom in the choice of the vector potential, called *gauge choice*.<fnref n="5" /> The most popular gauges are the Landau gauge

$$A_L(r) = B(-y, 0, 0) \qquad (5.15)$$

$$A(r) \rightarrow A'(r) = A(r) + \nabla\Lambda(r),$$

does not change the magnetic field since $\nabla \times \nabla\Lambda(r) = 0$.

<footnote n="5">⁵Remember from your classes on electromagnetism that adding the gradient of a function $\Lambda(r)$ to the vector potential</footnote>
and the symmetric gauge

$$A_{\text{sym}}(\mathbf{r}) = \frac{B}{2}(-y, x, 0), \qquad (5.16)$$

both of which respect the *Coulomb gauge* $\nabla \cdot \mathbf{A} = 0$ of electromagnetism.

We have already discussed, in chapter 2, that the low-energy limit involves an expansion of the Bloch Hamiltonian in $|q|a \ll$ around some particular points in the first Brillouin zone, e.g. the $\Gamma$-point in the case of GaAs or Si and around the K- and K'-points in graphene or boron nitride. The hypothesis we need to make to account for the magnetic field is that it does not invalidate the full band structure but quantises the energy only in a tiny window around the Fermi level. This is equivalent to requiring that the Heisenberg uncertainty in the wave-vector components in the presence of the magnetic field be small on the scale of the inverse lattice spacing $a$. As we have seen above, the magnetic field provides us with a novel length scale that is the magnetic length (5.10)

$$l_B = \sqrt{\frac{\hbar}{e|B|}} \approx 26 \text{ nm}/\sqrt{|B|[T]}, \qquad (5.17)$$

which will accompany us throughout this chapter. We show below explicitly that its inverse is the order of magnitude for the uncertainty in the wave vector $|q| \sim l_B^{-1}$ so that the above low-energy (or small wave-vector) condition reads $|q|a \sim a/l_B \ll 1$. Since in most materials, we have $a \lesssim 1$ nm, this condition is satisfied in most materials,<fnref n="6" /> and, as we discussed in sections 2.2.1 and 2.3, the condition $|q|a \ll a$ validates the low-energy expansion of the Bloch Hamiltonian.

Landau quantisation is now achieved in three steps.

1. Start with the Bloch Hamiltonian in the low-energy limit. Here, we

<footnote n="6">⁶Indeed, experimentally accessible fields are in the 45 T range for continuous fields in hybrid magnets and ~ 100 T for pulsed fields, while fields up to roughly 200 T may be achieved in semi-destructive experiments where the coils used to produce the magnetic field are blown up (but usually not the probe). From a theoretical point of view, a magnetic length that is on the order of the lattice spacing is interesting and yields a fractal spectrum that is called *Hofstadter butterfly* and that can be probed to some extent in superstructures. However, as interesting this situation may seem, it is beyond the scope of these introductory notes on the quantum Hall effect.</footnote>
concentrate on 2D electrons with an approximately parabolic dispersion

$$H(\mathbf{q}) = \frac{\hbar^2 q^2}{2m}, \qquad (5.18)$$

with $q^2 = q_x^2 + q_y^2$, such as in semiconductor heterostructures, and on 2D Dirac fermions (2.71)

$$H_{\xi}(\mathbf{q}) = \begin{pmatrix} \Delta & \hbar v(\xi q_x - i q_y) \\ \hbar v(\xi q_x + i q_y) & -\Delta \end{pmatrix} = \hbar v(\xi q_x \sigma_x + q_y \sigma_y) + \Delta \sigma_z. \qquad (5.19)$$

that we may encounter in graphene (with $\Delta = 0$) and boron nitride ($\Delta \neq 0$).

2. In order to take into account the magnetic field, we make use of the Peierls substitution, $\hbar\mathbf{q} \rightarrow \hbar\mathbf{q} + e\mathbf{A}(\mathbf{r})$, in the low-energy Hamiltonian,

$$H_B = H(\mathbf{q} + e\mathbf{A}/\hbar). \qquad (5.20)$$

3. Remember that the wave vector is the lattice momentum $\mathbf{p} = \hbar\mathbf{q}$ that does not commute with the argument $\mathbf{r}$ of the vector potential so that we need to take into account quantum mechanics seriously via the canonical commutation relations,

$$[x_{\mu}, p_{\nu}] = i\hbar\delta_{\mu,\nu}, \quad \text{et} \quad [x_{\mu}, x_{\nu}] = 0 = [p_{\mu}, p_{\nu}] = 0. \qquad (5.21)$$

### 5.2.1 Peierls substitution and quantisation

Let us spell out the above recipe in more detail, namely the second and third steps. First, the Peierls substitution is the lattice equivalent of the usual minimal coupling of a free charge carrier to the electromagnetic field, which you have encountered in your quantum-mechanics classes,

$$\mathbf{p} \rightarrow \mathbf{\Pi} = \mathbf{p} + e\mathbf{A}(\mathbf{r}) = m\mathbf{v}, \qquad (5.22)$$

in terms of the velocity $\mathbf{v}$ of the free particle. In the case of a lattice electron, the situation is more envolved since we should, in principle, take into account the vector potential before adding the periodic potential. However, this is a formidable task that has not yet found an analytic solution because we can no longer use Bloch's theorem, as we have mentioned above. It must
therefore be viewed as an approximate scheme that consists of two steps. First, we interpret the wave vector as a quantum mechanical operator that is the conjugate of the position operator which intervenes in the argument of the vector potential. Second, we couple this operator minimally to the vector potential,

$$ \hbar \mathbf{k} \rightarrow \mathbf{p} \rightarrow \mathbf{\Pi} = \hbar \mathbf{k} + e \mathbf{A}(\mathbf{r}), \quad (5.23) $$

which is justified in the above-mentioned limit of small flux through the unit cell that is equivalent to $a/l_B \ll 1$.

Let us now consider quantum mechanics in the form of the commutation relations (5.21), which induce the following commutation relations for the components of the operator $\mathbf{\Pi} = (\Pi_x, \Pi_y)$,

$$ \begin{aligned} [\Pi_x, \Pi_y] &= [p_x + eA_x(\mathbf{r}), p_y + eA_y(\mathbf{r})] = e ([p_x, A_y(\mathbf{r})] - [p_y, A_x(\mathbf{r})]) \\ &= e \left( \frac{\partial A_y}{\partial x} [p_x, x] + \frac{\partial A_y}{\partial y} [p_x, y] - \frac{\partial A_x}{\partial x} [p_y, x] - \frac{\partial A_x}{\partial y} [p_y, y] \right) \\ &= -i\hbar e \left( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) = -i\hbar e B_z = -i \frac{\hbar^2}{l_B^2} \operatorname{sgn}(B), \end{aligned} \quad (5.24) $$

where we consider the magnetic field to be parallel to the z-axis, and the sign $\operatorname{sgn}(B) = B/|B|$ its orientation: we have $\operatorname{sgn}(B) = +$ for an orientation along the positive z-axis and $\operatorname{sgn}(B) = -$ if it points towards the negative z-axis. Here, we have made use, in the second line, of the operator relation

$$ [C, f(D)] = \frac{\partial f}{\partial D}[C, D] \quad \text{if} \quad [C, [C, D]] = [D, [C, D]] = 0 \quad (5.25) $$

between an operator C and a (derivable) function of another operator $f(D)$.<fnref n="7" /> It is noteworthy that the relation

$$ [\Pi_x, \Pi_y] = -i \frac{\hbar^2}{l_B^2} \operatorname{sgn}(B) \quad (5.26) $$

<footnote n="7">⁷More precisely, we have made use of a generalisation to gradients of this relation,</footnote>

<footnote>$[C, f(D_1, \dots, D_n)] = \sum_{j=1}^{n} \frac{\partial f}{\partial D_j} [C, D_j]$ if $[C, [C, D_j]] = [D_k, [C, D_j]] = 0$ for all $j, k$.</footnote>
has been obtained without specifying a gauge for the vector potential. It is indeed *gauge-invariant* since the operator $\Pi$, which is proportional to the physical velocity, is gauge-invariant itself. Furthermore, we notice that the two components $\Pi_x$ and $\Pi_y$ form a pair of conjugate variables as do the pairs $x$ and $p_x$ or $y$ and $p_y$. This allows us to introduce ladder operators, in close analogy with the quantum-mechanical treatment of the harmonic oscillator

$$ a = \frac{l_B}{\sqrt{2\hbar}}(\Pi_x - i\Pi_y) \quad \text{et} \quad a^\dagger = \frac{l_B}{\sqrt{2\hbar}}(\Pi_x + i\Pi_y) \qquad (5.27) $$

or

$$ \Pi_x = \frac{\hbar}{\sqrt{2}l_B}(a^\dagger + a) \quad \text{et} \quad \Pi_y = \frac{\hbar}{i\sqrt{2}l_B}(a^\dagger - a). \qquad (5.28) $$

With this definition of the ladder operators at hand, we obtain the slightly unusual commutation relations

$$ [a, a^\dagger] = \operatorname{sgn}(B), \qquad (5.29) $$

which depend on the orientation of the magnetic field in the $z$-direction. In order to obtain the more conventional commutation relations, one therefore needs to *exchange* $a \leftrightarrow a^\dagger$ if one changes the orientation of the magnetic field.

Let us emphasise an essential point about Landau quantisation that may seem quite unusual at first sight: while we started with a *two-dimensional* quantum-mechanical problem that is described in terms of *two pairs of conjugate variables* – $x$ and $p_x$ on the one-hand side and $y$ and $p_y$ on the other side – the Hamiltonian $H_B(\Pi_x, \Pi_y)$, after the Peierls substitution, depends only on one pair of conjugate variables, $\Pi_x$ and $\Pi_y$ or, equivalently, $a$ and $a^\dagger$. The system has somewhat mysteriously become *one-dimensional* as if we had lost information. This cannot be the case, and we will see in section 5.2.4 that there exists a second pair of conjugate variables that commutes with the Hamiltonian and can thus be seen as a symmetry.<fnref n="8" /> However, before discussing this symmetry and the resulting degeneracy of the Landau levels, let us first explicitly calculate some Landau-level spectra.

<footnote n="8">⁸As we discuss below, this symmetry can be viewed as an effective translation symmetry. Even if, strictly speaking, translation symmetry is broken by the position dependence of the vector potential in the Hamiltonian, the magnetic field is spatially homogeneous. The effective translation symmetry is related to the homogeneity of the magnetic field as we show in detail in section 5.2.4.</footnote>
### 5.2.2 Landau levels of electrons in a parabolic band

We start with the standard case of Landau levels for electrons with a parabolic band that we consider to be isotropic. Upon Peierls substitution, the Hamiltonian (5.18) becomes

$$H_B = \frac{1}{2m} (\Pi_x^2 + \Pi_y^2) = \hbar\omega_C \left(a^\dagger a + \frac{1}{2}\right), \qquad (5.30)$$

where we have used the expression (5.28) for the operator $\Pi$ in terms of ladder operators as well as the commutation relations $[a, a^\dagger] = 1$.<fnref n="9" /> Furthermore, we have used the expression

$$\omega_C = \frac{e|B|}{m} = \frac{\hbar}{ml_B^2} \qquad (5.31)$$

for the cyclotron frequency in terms of the magnetic length.<fnref n="10" /> One thus obtains the quantum-mechanical spectrum of a 1D harmonic oscillator

$$E_n = \hbar\omega_C \left(n + \frac{1}{2}\right), \qquad (5.32)$$

in terms of the integer $n$, which is the eigenvalue of the number operator $a^\dagger a$,

$$a^\dagger a|n\rangle = n|n\rangle. \qquad (5.33)$$

In contrast to the spectrum, the states $|n\rangle$ and their associated wave functions depend explicitly on the gauge of the vector potential. Their expressions for the Landau gauge are derived in section 5.2.5, but for the moment, we are only interested in the spectrum, which is depicted in figure 5.4

Notice that the quantum-mechanical result for the Landau-level spectrum agrees with the semiclassical Bohr-Sommerfeld quantisation, which gave the result (5.13), $E_n = \hbar\omega_C(n + \gamma)$. However, the full quantum-mechanical treatment allows us also to determine the geometric phase $\gamma = 1/2$.

<footnote n="9">⁹In order to simplify the notations, we consider here a magnetic field oriented in the positive z-direction.</footnote>

<footnote n="10">¹⁰Contrary to the classical case, we consider the cyclotron frequency to be positive in order to avoid the cumbersome modulus in the expression when the cyclotron frequency is related to an energy.</footnote>
<figure><img src="image_48.png" /><figcaption>Figure 5.4: Landau-level spectra (a) for a parabolic band, $E_n = \hbar\omega_C(n + 1/2)$, and (b) for graphene electrons (2D massless Dirac fermions), $E_{\lambda,n} = \lambda\hbar(v/l_B)\sqrt{2n}$. In the absence of a mass term, the Landau levels for Dirac fermions are twofold valley-degenerate.</figcaption></figure>

### 5.2.3 Landau levels of 2D Dirac fermions

Let us now calculate, with the same recipe, the Landau levels for the Dirac Hamiltonian (5.19). With the help of the Peierls substitution, one finds

$$H_B^\xi = \begin{pmatrix} \Delta & \xi v(\Pi_x - i\xi\Pi_y) \\ \xi v(\Pi_x + i\xi\Pi_y) & -\Delta \end{pmatrix}, \quad (5.34)$$

where we remind the reader that the index $\xi = \pm$ denotes the valleys $K$ and $K'$ around which the Bloch Hamiltonian has been expanded. In terms of ladder operators, the Hamiltonian may be rewritten as

$$H_B^K = \begin{pmatrix} \Delta & \sqrt{2\frac{\hbar v}{l_B}}\hat{a} \\ \sqrt{2\frac{\hbar v}{l_B}}\hat{a}^\dagger & -\Delta \end{pmatrix} \quad \text{and} \quad H_{B'}^{K'} = \begin{pmatrix} \Delta & -\sqrt{2\frac{\hbar v}{l_B}}\hat{a}^\dagger \\ -\sqrt{2\frac{\hbar v}{l_B}}\hat{a} & -\Delta \end{pmatrix} \quad (5.35)$$

in the valleys $K$ and $K'$, respectively.

Before diagonalising the Hamiltonians to obtain their spectra, it should be emphasised that the Hamiltonian in the $K'$ valley is related to that in $K$ by the simple interchange $a \leftrightarrow a^\dagger$ (apart from an unimportant sign in the off-diagonal terms that does not affect the spectra). This is reminiscent to the change in the orientation of the magnetic field and can be related
more fundamentally to time-reversal symmetry that will be discussed in more detail in chapter 7: upon time reversal, the sign of the wave vector is changed $\mathbf{k} \to -\mathbf{k}$, i.e. the valleys $K$ and $K'$ are exchanged. Similarly, the magnetic field is odd under time-reversal symmetry and its orientation must equally change.

The spectrum is obtained by solving the eigenvalue equation

$$H_B^\xi \psi_{\lambda,n;\xi} = E_{\lambda,n}^\xi \psi_{\lambda,n;\xi}, \qquad (5.36)$$

in terms of spinorial eigenstates

$$\psi_{\lambda,n;\xi} = \begin{pmatrix} u_{\lambda,n;\xi} \\ v_{\lambda,n;\xi} \end{pmatrix}, \qquad (5.37)$$

dictated by the $2 \times 2$ matrix form of the Hamiltonian (5.35). In the $K$ valley ($\xi = +$), the eigenvalue equation thus yields two coupled equations

$$\begin{aligned} (\Delta - E_{\lambda,n}^K) u_{\lambda,n;K} + \sqrt{2} \frac{\hbar v}{l_B} av_{\lambda,n;K} &= 0 \\ \sqrt{2} \frac{\hbar v}{l_B} a^\dagger u_{\lambda,n;K} - (\Delta + E_{\lambda,n}^K) v_{\lambda,n;K} &= 0 \end{aligned} \qquad (5.38)$$

for the two spinor components. Substituting the first equation into the second one tells us that

$$2 \left( \frac{\hbar v}{l_B} \right)^2 a^\dagger a v_{\lambda,n;K} = (E_{\lambda,n}^{K2} - \Delta^2) v_{\lambda,n;K}, \qquad (5.39)$$

i.e. the second spinor component $v_{\lambda,n;K}$ is an eigenstate of the number operator $a^\dagger a$, that is proportional to $|n\rangle$. This yields the spectrum

$$E_{\lambda,n}^K = \lambda \sqrt{\Delta^2 + 2\hbar^2 \frac{v^2}{l_B^2} n}, \qquad (5.40)$$

for the integers $n \neq 0$. The case $n = 0$ needs a special treatment. Indeed, the upper component $u_{\lambda,n;K}$ of the spinor (5.37) is related to the lower one by

$$u_{\lambda,n;K} = - \frac{\sqrt{2\hbar v}}{l_B(\Delta - E_{\lambda,n}^K)} a v_{\lambda,n;K} \propto |n-1\rangle, \qquad (5.41)$$

as one may directly see from the first line of equation (5.38). Thanks to the ladder operator $a$, this component is thus proportional to the eigenstate $|n-1\rangle$ of $a^\dagger a$ because we have $v_{\lambda,n;K} \propto |n\rangle$. The upper spinor component must therefore vanish in the case $n = 0$.
Landau levels with $n \neq 0$

Before calculating the $n = 0$ Landau level explicitly, let us finish the discussion of the eigenstates $\psi_{\lambda,n;K}$ for $n \neq 0$. We have just seen that they are of the form

$$ \psi_{\lambda,n;K} = \begin{pmatrix} c_{\lambda,n;K} |n-1\rangle \\ d_{\lambda,n;K} |n\rangle \end{pmatrix}, \qquad (5.42) $$

and we thus have to determine the relative amplitudes which respect the normalisation $|c_{\lambda,n;K}|^2 + |d_{\lambda,n;K}|^2 = 1$. However, this is the same calculation as that of the spin-1/2 particle in a magnetic field treated in basic quantum mechanics (see e.g. the book by Jean-Louis Basdevant and Jean Dalibard). With the help of the trigonometric functions

$$ \sin \theta_{\lambda,n} = \frac{\frac{\hbar v}{l_B} \sqrt{2n}}{\sqrt{\Delta^2 + 2(\hbar v / l_B)^2 n}} \quad \text{and} \quad \cos \theta_{\lambda,n} = \frac{\Delta}{\sqrt{\Delta^2 + 2(\hbar v / l_B)^2 n}}, \qquad (5.43) $$

we thus find

$$ \psi_{\lambda=+,n;K} = \begin{pmatrix} \cos \frac{\theta_{\lambda,n}}{2} |n-1\rangle \\ \sin \frac{\theta_{\lambda,n}}{2} |n\rangle \end{pmatrix} \quad \text{and} \quad \psi_{\lambda=-,n;K} = \begin{pmatrix} -\sin \frac{\theta_{\lambda,n}}{2} |n-1\rangle \\ \cos \frac{\theta_{\lambda,n}}{2} |n\rangle \end{pmatrix}. \qquad (5.44) $$

Until now, we have only discussed the states in the $K$ valley, but we do not need to redo explicitly the calculations for the other valley. Indeed, we have seen that the Hamiltonian in the other valley $K'$ is obtained by exchanging the ladder operators $a$ and $a^\dagger$. Just by inspecting the corresponding system of equations

$$ \begin{aligned} \left(\Delta - E_{\lambda,n}^{K'}\right) u_{\lambda,n;K'} - \sqrt{2} \frac{\hbar v}{l_B} a^\dagger v_{\lambda,n;K'} &= 0 \\ -\sqrt{2} \frac{\hbar v}{l_B} a u_{\lambda,n;K'} - \left(\Delta + E_{\lambda,n}^{K'}\right) v_{\lambda,n;K'} &= 0 \end{aligned} \qquad (5.45) $$

in the $K'$ valley, one notices that this is the same system of equations as that for the $K$ valley upon exchange of the components $u_{\lambda,n}$ and $v_{\lambda,n}$ accompanied by a slight change of the parameters

$$ u_{\lambda,n} \leftrightarrow v_{\lambda,n}, \quad \Delta \leftrightarrow -\Delta \quad \text{et} \quad v \leftrightarrow -v. \qquad (5.46) $$

This amounts to adding $\pi$ to the angle $\theta_{\lambda,n;K}$,

$$ \theta_{\lambda,n;K'} = \theta_{\lambda,n;K} + \pi, \qquad (5.47) $$
in the expressions of the spinors (5.45). While the Landau-level spectrum is not affected by this change, so that we can write in a compact manner

$$E_{\lambda,n}^{\xi} = \lambda \sqrt{\Delta^2 + 2\hbar^2 \frac{v^2}{l_B^2} n} \quad \text{for} \quad n \neq 0, \qquad (5.48)$$

evidencing clearly the twofold valley degeneracy of the “relativistic” Landau levels, the spinors in the K' valley read

$$\psi_{\lambda=+,n;K'} = \begin{pmatrix} \cos \frac{\theta_{\lambda,n}}{2} |n\rangle \\ -\sin \frac{\theta_{\lambda,n}}{2} |n-1\rangle \end{pmatrix} \quad \text{and} \quad \psi_{\lambda=-,n;K'} = \begin{pmatrix} -\sin \frac{\theta_{\lambda,n}}{2} |n\rangle \\ -\cos \frac{\theta_{\lambda,n}}{2} |n-1\rangle \end{pmatrix}. \qquad (5.49)$$

Notice that, for graphene with $\Delta = 0$, we have $\theta_{\lambda,n} = \pi/2$ for each energy level with $n \neq 0$ thanks to the equations (5.43) and consequently $\cos(\theta_{\lambda,n}/2) = \sin(\theta_{\lambda,n}/2) = 1/\sqrt{2}$. In the absence of a gap (for massless Dirac fermions such as in graphene), the Bloch wave functions have therefore the same weight on the two components, i.e. on the two sublattices of the graphene lattice, a property that the spinors in a magnetic field inherit from the $B=0$ case.

### Landau levels with $n=0$

Let us now concentrate on the fate of the $n=0$ Landau levels. If we take a closer look at the form of the spinors (5.44) and (5.49), we notice that the upper component vanishes for an electron in the K valley, as a consequence of the relation (5.41) for the ladder operators, $a|n=0\rangle = 0$, while for electrons in the K' valley, it is the lower component that is zero. This means that K-valley electrons only occupy the B sublattice while K' electrons reside solely on the A sites, as long as we consider electrons in the $n=0$ Landau level, and the associated spinors thus become

$$\psi_{n=0;K} = \begin{pmatrix} 0 \\ |n=0\rangle \end{pmatrix} \quad \text{and} \quad \psi_{n=0;K'} = \begin{pmatrix} |n=0\rangle \\ 0 \end{pmatrix}. \qquad (5.50)$$

In contrast to the other Landau levels with $n \neq 0$, the spectrum now depends explicitly on the valley index $\xi$, as one may directly inspect from equations (5.38) and (5.45), the $n=0$ level sticks to the top of the valence band in the K valley, $E_{n=0,K} = -\Delta$, and it is glued to the conduction-band bottom in
<figure><img src="image_49.png" /><figcaption>Figure 5.5: Landau levels for massive Dirac fermions (gapped graphene), on the left for the valley K and on the right for the valley K'.</figcaption></figure>

the K' valley, $E_{n=0,K'} = \Delta$. One may thus rewrite the energy of these levels as

$$E_{n=0}^{\xi} = -\xi\Delta, \qquad (5.51)$$

in terms of the valley index $\xi$.

This result should astonish us greatly! While the spectrum, in the vicinity of each of the valleys, is perfectly particle-hole symmetric in the absence of a magnetic field, particle-hole symmetry is broken at $B \neq 0$ if we consider any of the valleys separately. This highly unusual situation, which is depicted in figure 5.5 for both valleys, is the prototype of an anomaly, and one speaks of the parity anomaly in the present case. Notice that the spectrum retrieves its particle-hole symmetry if we superpose the levels of both valleys. Moreover, the anomaly also disappears if the gap vanishes, $\Delta = 0$, such as in graphene, where then $E_{n=0}^{\xi} = 0$. The most important points to retain from this discussion of the $n = 0$ Landau levels are

(i) the valley degeneracy: while all Landau levels are twofold valley degenerate for $n \neq 0$, this is not the case for $n = 0$, where the valley degeneracy is lifted by a non-zero gap parameter, $\Delta \neq 0$;

(ii) their dispersion with the magnetic field: the energy (5.51) of the $n = 0$
Landau levels does not depend on the value of the magnetic field, while that of the other levels with $n \neq 0$ grows monotonically with the B field according to equation (5.48).

Experimental evidence for graphene Landau levels

From the above discussion, it is clear that the electrons with a parabolic band dispersion yield a different Landau-level spectrum as compared to relativisitic electrons in graphene. In the former case, equation (5.32) indicates that the energy of the Landau levels grows linearly with the magnetic field, while the levels for electrons in graphene (with $\Delta = 0$) disperse as the square root of the magnetic field, $E_{\lambda,n} \propto \lambda\sqrt{Bn}$, according to equation (5.48). This difference is indeed measurable in so-called magneto-optical spectroscopy that is sensitive to inter-Landau-level transitions; each time the incoming photon is in resonance with a transition from an occupied level $n$ to an unoccupied one $n \pm 1$, it can be absorbed by exciting an electron.<fnref n="11" />

In the case of a single parabolic band, there is only one possible absorption line observable at the cyclotron energy $\hbar\omega_c$. Indeed, only one transition is possible from $n$ to $n+1$, if the Fermi level lies in between these two Landau levels. This phenomenon is known under the name cyclotron resonance, and its linear-in-$B$ behaviour has been observed in many systems. The magneto-optical absorption properties of graphene are remarkably different. First, due to the presence of a conduction and a valence band, there are many transitions from an occupied Landau level $n$ to an unoccupied one $n \pm 1$, e.g. if the former resides in the valence and the second in the conduction band and if the system is charge neutral. Second, the absorption lines inherit their $\sqrt{B}$ dispersion from the Landau levels, and one thus expects absorption peaks at the energies

$$ \Delta_n = \sqrt{2} \frac{\hbar v}{l_B} (\sqrt{n} + \sqrt{n+1}) \propto \sqrt{B}. \qquad (5.52) $$

These expectations have been experimentally verified in infrared magneto-optical spectroscopy at the Grenoble High-Magnetic-Field Laboratory, in 2007. The results are shown in figure 5.6, where one notices the particular evolution of the absorption peaks as a function of the magnetic field, in

<footnote n="11">¹¹We do not prove the optical selection rules $n \to n \pm 1$, here, because this is beyond the scope of the present lecture notes even if it is not complicated to show. Just take it for granted.</footnote>
<figure><img src="image_50.png" /><figcaption>Figure 5.6: Magneto-optical transmission spectra. The relative transmission is normalised with respect to the transmission in the absence of a magnetic field. (a) Transmission lines at a fixed value of $B = 0.4$ T, as a function of the photon energy. The minima in the transmission indicate absorption peaks where a photon with the according energy is absorbed while exciting an electron within an optically active inter-Landau-level transitions ($n \rightarrow n \pm 1$, see inset). (b) Evolution of the absorption peaks upon variation of the magnetic field, for the B and C transitions. (c) Evolution of the energy of the absorption peaks (measured transmission minima) with the square root of the magnetic field for the transitions A, B and C. From Sadowski et al., Phys. Rev. Lett. **97**, 266405 (2007).</figcaption></figure>
agreement with equation (5.52). When the energy of the absorption peaks is plotted as a function of the square root of the magnetic field, one obtains a perfectly linear behaviour [see figure 5.6(c)]. The only fitting parameter is the slope, which is given by the velocity parameter $v$. On the basis of these experimental results, this parameter has been found to be $1.03 \times 10^6$ m/s, in excellent agreement with theoretical estimates [Sadowski et al., Phys. Rev. Lett. 97, 266405 (2007)].

### 5.2.4 Degeneracy of Landau levels

Until now, we have only been interested in the Landau-level spectrum, which has been obtained with the help of the Peierls substitution (5.22). As mentioned above, these levels are in a certain sense one-dimensional because the Hamiltonian is determined by only one pair of conjugate variables, $\Pi_x$ and $\Pi_y$. However, since we know that the system is two-dimensional, there must exist a second pair of conjugate variables that needs to be taken into account to complete the quantum-mechanical problem of electrons in a strong magnetic field even if the Hamiltonian does not depend on these variables. This is the origin of a strong degeneracy of the Landau levels that must not be mixed up with the usual internal degeneracies (such as the spin, or perhaps also the valley degrees of freedom), but it is more correctly to be viewed as an orbital degeneracy.

In order to understand this orbital degeneracy from a physical point of view, let us adopt again the semiclassical perspective according to which the energy of the electron is determined by the (discrete) radius of the classical circular trajectory. In contrast to its radius, the energy of the trajectory does not depend on the position $\mathbf{R}$ of the centre of the cyclotron motion, due to the homogeneity of the magnetic field. Let us therefore decompose the position of the electron as the sum

$$ \mathbf{r} = \mathbf{R} + \eta \qquad (5.53) $$

of the centre of the cyclotron orbit $\mathbf{R}$ (also called guiding centre and its position $\eta$ on the circular trajectory (cyclotron variable)). The latter cyclotron variable is perpendicular to the tangential velocity $\mathbf{v}$ of the electron, $\eta = \mathbf{v} \times \mathbf{u}_z / \omega_C$, where $\mathbf{u}_z$ is the unit vector in the $z$-direction. Furthermore, the velocity is nothing other (apart from the mass $m$) than the gauge-invariant
momentum, $\Pi = mv$, so that we have

$$ \eta = \frac{1}{eB} \Pi \times u_z \Leftrightarrow \eta_x = \frac{\Pi_y}{eB} \quad \text{and} \quad \eta_y = -\frac{\Pi_x}{eB}, \qquad (5.54) $$

and one obtains the equivalent commutation relations

$$ [\eta_x, \eta_y] = \frac{1}{e^2 B^2} [\Pi_x, \Pi_y] = -il_B^2. \qquad (5.55) $$

We notice that it is equivalent to write the Hamiltonian in terms of the components of $\Pi$ or of $\eta$, and in the case of a parabolic band we have

$$ H = \frac{\Pi_x^2 + \Pi_y^2}{2m} \Leftrightarrow H = \frac{1}{2} m \omega_C^2 (\eta_x^2 + \eta_y^2). \qquad (5.56) $$

From a classical point of view, the guiding centre is an integration constant obtained from the integration of the classical equations of motion or, equivalently, a constant of motion. Translated to quantum mechanics, a constant of motion is an observable that commutes with the Hamiltonian, so that we would expect

$$ [X, H_B(\Pi)] = [Y, H_B(\Pi)] = 0. \qquad (5.57) $$

This is indeed the case, since the above commutation is induced by the commutation relations

$$ [X, \Pi_x] = [X, \Pi_y] = [Y, \Pi_x] = [Y, \Pi_y] = 0 \qquad (5.58) $$

or, equivalently,

$$ [X, \eta_x] = [X, \eta_y] = [Y, \eta_x] = [Y, \eta_y] = 0, \qquad (5.59) $$

as one may verify by straight-forward calculation.<fnref n="12" />

$$ [X, \Pi_x] = [x - \eta_x, \Pi_x] = [x, p_x + eA_x(\mathbf{r})] - \frac{1}{eB} [\Pi_y, \Pi_x] = [x, p_x] - i \frac{\hbar eB}{eB} = 0, $$

where we have used equation (5.54) in the second and the commutation relation (5.26 in the third step. Similarly, one may show with the help of equation (5.54) that

$$ [Y, \Pi_x] = [y - \eta_y, \Pi_x] = [y, \Pi_x] + \frac{1}{eB} [\Pi_x, \Pi_x] = [y, p_x + eA_x(\mathbf{r})] = 0, $$

as well as the other two relations.

<footnote n="12">¹²To see that the components of the guiding centre commute indeed with those of the gauge-invariant momentum, one may calculate, e.g.,</footnote>
We thus see that the guiding centres, which are constants of motion, represent a symmetry from a quantum-mechanical viewpoint. As we have already anticipated above, this symmetry can be interpreted as a particular form of translation symmetry, which *stricto sensu* is broken by the position dependence of the vector potential in the Hamiltonian. However, since the magnetic field is homogeneous, one would expect the restoration of translation symmetry in some manner, and this is precisely given by the guiding-centre operators. Most saliently, the two components of the guiding-centre operator do not commute, and one finds<fnref n="13" />

$$[X, Y] = il_B^2 \operatorname{sgn}(B). \qquad (5.60)$$

As simple as it is, this last equation has extremely rich consequences for quantum-Hall systems. First, we see that it constitutes indeed the second pair of conjugate variables required in a 2D quantum system. Second, one may understand the effective translation symmetry inside a Landau level from equation (5.60), which indicates that one may interpret the component X of the guiding-centre operator as the generator of translations in the y-direction and Y as the generator in the x-direction.<fnref n="14" /> These translation properties turn out to be responsible also for the so-called *Hall drift* in the presence of an additional electrostatic potential, which is discussed in detail in section 5.3 below. Third, the commutation relations (5.60) are responsible for the macroscopic degeneracy of the Landau levels.

In order to understand this degeneracy quantitatively, remember that any commutation relation between observables induces a Heisenberg uncertainty relation. In our case, the commutation relation (5.60) indicates that

$$ \begin{aligned} 0 &= [x, y] = [X + \eta_x, Y + \eta_y] = [X, Y] + [\eta_x, \eta_y] + [X, \eta_y] + [\eta_x, Y] \\ &= [X, Y] - i \frac{\hbar}{e|B|} \operatorname{sgn}(B). \end{aligned} $$

<footnote n="13">¹³One obtains this relation directly from $[x, y] = 0$ and the commutators (5.55) and (5.59),</footnote>

<footnote n="14">¹⁴The associated translation operators would be $\exp(-iy_0 X/l_B^2)$ for a translation by $y_0$ and $\exp(ix_0 Y/l_B^2)$ for a translation by $x_0$. Notice, however, that here the translations in orthogonal directions do not commute. This unusual behaviour, as compared to translations in the absence of a magnetic field, is the hallmark of the (non-Abelian) magnetic translation group. A detailed discussion of this group is, however, far beyond the scope of the present lecture notes.</footnote>
$\Delta X \Delta Y \sim l_B^2$. This means that the centre of the cyclotron motion is blurred over a minimal surface $\Sigma$ on the order of $l_B^2$. Actually, it is even possible to show that this surface is given precisely by

$$ \Delta X \Delta Y = \Sigma = 2\pi l_B^2, \qquad (5.61) $$

as a consequence of the Gaussian character of the harmonic-oscillator wave functions. Instead of providing a full proof of this statement, which turns out to be quite involved, we verify it below, in section 5.2.5, by calculating the wave functions for a particularly chosen gauge (the Landau gauge) for the vector potential. In order to obtain the number of quantum states per Landau level, it is sufficient to divide the total surface $S$ occupied by the 2D sample by this minimal surface $\Sigma$ occupied by single quantum state,<fnref n="15" />

$$ N_B = \frac{S}{\Sigma} = \frac{S}{2\pi l_B^2} = S \times \frac{|B|}{h/e}. \qquad (5.62) $$

This last expression is particularly transparent: it indicates that the number of quantum states per Landau level is given by the total number of flux quanta threading the 2D surface $S$ of the sample. Indeed, the magnetic field $B$ is nothing other than a flux density, and we may also define the useful quantity

$$ n_B = \frac{1}{2\pi l_B^2} = \frac{|B|}{h/e} \qquad (5.63) $$

as the flux density in units of the flux quantum $\phi_0 = h/e$ that characterises the orbital degeneracy of the Landau levels. Notice that this degeneracy has been obtained by considering only the guiding-centre operator $\mathbf{R}$ and the non-commutativity of its components, independently of its energy spectrum. Indeed this is an essential property of Landau levels: the orbital degeneracy $N_B$ of the Landau levels is independent of the $B$-field dispersion of the latter. Apart from their internal degeneracy (spin and/or valley degeneracy), graphene Landau levels therefore have the same orbital degeneracy as electrons in GaAs heterostructures with a parabolic dispersion relation, and they depend only on the strength of the magnetic field.

<footnote n="15">¹⁵Notice that this argument is reminiscent of that usually invoked in statistical physics in order to count the number of (quantum) states in phase space. The latter is generated by the positions and momenta of the $N$ particles and divided into (hyper-)volumes of size $h^{3N}$. This partition of phase space is equally a consequence of the commutation relations (5.21) for the associated position and momentum operators.</footnote>
The orbital degeneracy turns out to be an essential ingredient in the understanding of the quantum Hall effect, that is of the formation of plateaus in the Hall resistance at special values of the magnetic field. As we discuss in detail below, these values correspond to situations where a certain number of Landau levels are completely filled. In order to characterise this situation, it turns out to be fruitful to introduce a quantity that defines the filling of the Landau levels. Since electrons are fermions, this quantity is given by the ratio between the total number of electrons $N_{el}$ and the number of flux quanta (or alternatively by the ratio between their respective densities),

$$ \nu = \frac{N_{el}}{N_B} = \frac{n_{el}}{n_B} = 2\pi l_B^2 n_{el} = \frac{hn_{el}}{e|B|}, \qquad (5.64) $$

called the *filling factor*. If the filling factor is an integer, $\nu = n$, the n lowest Landau levels are indeed completely filled<fnref n="16" /> while all other levels are empty.

Notice that the filling factor is defined without appealing to the internal degeneracies but only on the basis of the orbital degrees of freedom associated with the guiding centre. If e.g. spin and/or valley degrees of freedom need to be taken into account in the form of a g-fold internal degeneracy, all Landau levels occur as g copies of n. In this case, the condition of having the n lowest Landau levels completely filled, with a Fermi level between the Landau levels n − 1 and n, must reflect the number of copies so that

$$ \nu = gn \qquad (5.65) $$

then. In GaAs heterostructures, where one only needs to consider the spin degree of freedom, we have g = 2, while we have g = 4 in the case of graphene, where the twofold spin degeneracy is accompanied by the twofold valley degeneracy.<fnref n="17" />

$$ \nu_G = 4n + 2, $$

<footnote n="16">¹⁶Attention! the index of the last fully occupied Landau level is n − 1 due to the convention according to which the lowest Landau level has the index n = 0.</footnote>

<footnote n="17">¹⁷We shall see that $\nu = 4n$ does actually not correspond to the filling factor sequence at which the quantum Hall effect is observed in graphene. While the jumps in the filling factor indeed follow this law, as multiples of four, the measured series reads</footnote>

<footnote>i.e. the jumps are displaced by a factor of two. The origin of this displacement is discussed below in section 5.4.4.</footnote>
Now it should be great time for the reader to protest. We all know that the spin degeneracy is naturally lifted in the presence of the magnetic field due to the Zeeman effect, which has not at all been taken into account in the discussion of the Landau-level spectrum until now. Indeed, the Zeeman effect lifts the spin degeneracy of the Landau levels by the energy

$$ \Delta_Z = g\mu_B B, \qquad (5.66) $$

where $\mu_B = e\hbar/2m_0$ is the *Bohr magneton*. Notice first that the mass that characterises the Zeeman effect is the *bare electron mass* $m_0$, while the Landau-level spectrum (in the case of GaAs) depends on the *band mass*. Furthermore, the $g$-factor depends on the material environment; while it is $\approx 2$ in vacuum as well as in graphene, its value and sign can change for electrons in materials such as GaAs, where we have $g = -0.44$. If we consider the ratio between the band and the bare electron mass $m/m_0 = 0.068$ in GaAs, we find that the Zeeman effect in this material is suppressed by a factor $\Delta_Z/\hbar\omega_C = gm/m_0 \approx 1/70$ with respect to the Landau-level spacing. It is in this sense that the Landau levels can be taken as almost spin-degenerate.

The relative role of the Zeeman effect is yet smaller in graphene than in GaAs heterostructures. Even if the Landau-level spacing – here between the Landau levels $n = 0$ and $n = \pm 1$ – scales as $\sqrt{B}$, while the Zeeman effect depends linearly on the magnetic field, $\sim B$, the latter is roughly two orders of magnitude smaller than the energy scale $\sqrt{2\hbar v/l_B}$ for fields up to 10 T.

Let us discuss a last consequence of the presence of this second pair of conjugate variables that is closely related to the orbital degeneracy: it is essential in the complete description of the quantum states and their associated wave functions. For the moment, these states are characterised only by the Landau-level quantum number $n$ (and possible indices due to the internal degrees of freedom), $|n\rangle$ in the case of electrons in a parabolic band and $\psi_{\lambda,n;\xi}$ in the case of Dirac fermions. However, they need to be completed in a tensor product by a quantum number which is associated with the components of the guiding-centre operator and that we generically call $\mu$ here. In order to obtain this quantum number, one may formally use the commutation relation (5.60), $[X, Y] = il_B^2 \text{sgn}(B)$, in the same manner as for the components $\Pi_x$ and $\Pi_y$ to define a pair of ladder operators $b$ and $b^\dagger$, with $[b, b^\dagger] = 1$ and $b^\dagger b|\mu\rangle = \mu|\mu\rangle$. This yields the completed quantum states

$$ |n\rangle \otimes |\mu\rangle = |n, \mu\rangle \qquad (5.67) $$
in the case of Landau levels for a parabolic band dispersion and $\psi_{\lambda,n;\mu;\xi} = \psi_{\lambda,n;\mu;\xi} \otimes |\mu\rangle$, i.e.

$$ \psi_{\lambda,n;\mu;K} = \begin{pmatrix} c_{\lambda,n;K} |n-1,\mu\rangle \\ d_{\lambda,n;K} |n,\mu\rangle \end{pmatrix} \quad \text{in the valley } K \\ \text{et } \psi_{\lambda,n;\mu;K'} = \begin{pmatrix} c_{\lambda,n;K'} |n,\mu\rangle \\ d_{\lambda,n;K'} |n-1,\mu\rangle \end{pmatrix} \quad \text{in the valley } K' $$

for Dirac fermions. Notice that, in practice, the meaning of this second quantum number $\mu$ and of the associated states and wave functions depends on the gauge chosen for the vector potential. The above-mentioned ladder operators emerge naturally in the symmetric gauge, while in the Landau gauge the second quantum number is naturally associated with the wave vector in a certain spatial direction, which remains a constant of motion.

### 5.2.5 Hamiltonian and wave functions in the Landau gauge

The above results have been obtained in a rather modern approach to the Landau-level spectrum that is based on canonical quantisation. This approach has been chosen here because it is apparently gauge-invariant, and consequently all results obtained are gauge-invariant. This is, however, not the traditional approach chosen in many textbooks that is based on fixing a particular gauge, usually the Landau gauge, and then solving the quantum-mechanical problem. Not only for completeness but also because it provides a special illustration of the results obtained above, the present section revisits this approach in which we fix the vector potential in the Landau gauge (5.15), $A = B(-y, 0, 0)$. In view of our previous approach, the components of the gauge-invariant momenta thus read

$$ \Pi_x = p_x - eBy \quad \text{et} \quad \Pi_y = p_y, \qquad (5.68) $$

and, similarly, the components of the guiding-centre operator are

$$ X = x - \frac{p_y}{eB} \quad \text{et} \quad Y = \frac{p_x}{eB}. \qquad (5.69) $$

Remember that the latter are constants of motion, and furthermore none of the components explicitly depends on the $x$-component of the position
operator r. Its conjugate variable $p_x$ is thus itself a constant of motion, as one immediately sees also from the last equation, $Y = p_x/eB$, or else yields a good quantum number since also the Hamiltonian naturally commutes with it, $[p_x, H_B] = 0$. The eigenstates of $p_x$ – these are nothing other than the plane-wave states in the x-direction – are therefore also eigenstates of the Hamiltonian, which reads

$$H_B = \frac{\Pi_x^2 + \Pi_y^2}{2m} = \frac{(p_x - eBy)^2}{2m} + \frac{p_y^2}{2m} = \frac{1}{2}m\omega_C^2(y - y_0)^2 + \frac{p_y^2}{2m}, \quad (5.70)$$

where we have defined $y_0 = p_x/eB$, which is precisely the eigenvalue of the guiding-centre component Y and that can be used itself as a good quantum number. It is equivalent to $p_x = \hbar k_x$, and it is our second quantum number which we have called $\mu$ (in its generic form) in the last section. This is itself a remarkable result, in line with what we have already understood: the Hamiltonian is that of a 1D harmonic oscillator, but here we explicitly see that it is a harmonic oscillator in the y-direction centred around the position $y_0$ that is determined by the momentum $p_x = \hbar k_x$ in its perpendicular direction. The fact that $y_0$ is a constant of motion is apparent from the fact that the energy of a harmonic oscillator does not depend on the position of its origin. The eigenstates are thus given by the tensor product between the 1D harmonic-oscillator states $|n\rangle$ and the plane-wave states $|k_x\rangle$ in the x-direction, $|n, k_x\rangle$, and the associated wave functions can thus be written in terms of the Hermite polynomials $H_n(y - y_0)$,

$$\psi_{n,k_x}(y,x) = \langle x, y | n, k_x \rangle \propto \frac{e^{ik_x x}}{\sqrt{L}} H_n(y - k_x l_B^2) e^{-\frac{(y-k_x l_B^2)^2}{2l_B^2}}, \quad (5.71)$$

where we have omitted the normalisation constant which is of no interest here.

Let us now check whether we retrieve the same degeneracy of the Landau levels as that in equation (5.62), obtained on the basis of our general approach. In order to do so, we consider a rectangular sample<fnref n="18" /> with the extensions $L_x$ and $L_y$ in the x- and y-directions, respectively, so that the total surface is given by $S = L_x L_y$. The finite extension in the x-direction can be taken into account by choosing periodic boundary conditions for the plane waves in this direction so that we retrieve the usual quantisation

$$k_x = \frac{2\pi}{L_x} m_x \quad (5.72)$$

<footnote n="18">¹⁸This is the geometry for which the Landau gauge is the natural gauge choice.</footnote>
for the associated wave vector, in terms of the integer $m_x$, which can be used to count our states. What are its possible values? We have already seen that the $x$-component of the wave vector is linearly related to the (average) position $y_0$ of our harmonic-oscillator wave function in the $y$-direction. In order to account for the chosen sample geometry, this position must be restricted to the sample width $L_y$, e.g. by $0 \le y_0 \le L_y$. This automatically restricts the possible values of the wave vector to the interval $0 \le k_x \le L_y eB/\hbar$ so that its maximal value is given by

$$k_{x,max}l_B^2 = \frac{2\pi}{L_x}m_{x,max}l_B^2 = L_y. \qquad (5.73)$$

Naturally, the maximal value $m_{x,max}$ of the integer $m_x$ coincides with the number of states per Landau level in the thermodynamic limit, and one thus finds

$$m_{x,max} = \frac{L_x L_y}{2\pi l_B^2} = N_B, \qquad (5.74)$$

i.e. one retrieves the total number of flux quanta threading the 2D sample, in agreement with equation (5.62) obtained on the basis of our general arguments.

## 5.3 Extended and localised states in an electrostatic potential

Until now, we have extensively discussed Landau quantisation, i.e. the quantisation of the 2D electronic energy into highly degenerate Landau levels in the presence of a strong magnetic field. We have seen that the large degeneracy is a consequence of a symmetry reminiscent of translation symmetry if one appeals to the position of the centres of the electrons' circular motion, the so-called guiding centres $R = (X, Y)$ which commute with the Hamiltonian. However, there is yet a link to establish between this Landau quantisation (in terms of energy) and conductance quantisation, which gives rise to the observed plateaus in the Hall resistance. As a guideline through the remaining chapter, let us spoil the result: the (integer) quantum Hall effect manifests itself when an integer number of Landau levels is completely filled, i.e. when the filling factor $\nu = n$ is an integer. For the moment, we forget about internal degeneracies to simplify the already complex discussion and set $g = 1$.
Towards the end of this chapter, we lift this simplification and discuss explicitly the role of the spin and valley degeneracies. As a consequence of the relation (5.64), this condition is encountered at particular values of the magnetic field if we fix the electronic density $n_{el}$.<fnref n="19" /> In this case the Fermi level is situated between two Landau levels. Consider for the moment that these Landau levels are energy bands. Indeed, this assumption is not that far-fetched because we have seen in the previous section, where we studied the Landau-level problem in the Landau gauge, that the levels $\epsilon_n = \epsilon_n(k_x)$ can be viewed as bands as a function of the wave vector $k_x$, which constitutes the second quantum number in addition to the level index $n$. In this sense, *Landau levels are perfectly flat 1D energy bands as a function of the wave vector k$_x$*.

Since the Fermi level lies between two levels, say $n$ and $n + 1$ in the “band gap” [see figure 5.7(a)], the system is supposed to be an insulator. However, as we show below, the energy of the Landau levels is bent upwards when we approach the sample edges, at $y \sim 0$ and $y \sim L_y$, which may be modelled in terms of a confinement potential $V(y)$ that is zero inside the sample and increases strongly at the sample edges. In our analogy between Landau levels and 1D electronic bands, the latter acquire a dispersion because the wave vector (in the $x$-direction) is linearly related to the position in the $y$-direction, $V(y \approx k_x l_B^2)$ as we have seen above. The 1D bands thus inevitably cross the Fermi level at the edges, where one thus finds *metallic 1D edge channels*. Moreover, we shall see below that the sense of propagation in these metallic edge channels is imposed by the orientation of the magnetic field perpendicular to the plane. The edge channels are therefore *chiral*, and as a consequence the electrons cannot be backscattered at the same edge. For an electron to be backscattered would require that it be transferred from one edge to the opposite one via the bulk of the sample, but such transfers are strongly suppressed unless the edges are locally brought together, *e.g.* by a constriction such as a quantum point contact. In the absence of backscattering, the electronic transport along the sample edges is therefore *ballistic* so that the associated conductance is perfectly quantised. It is $e^2/h$ for each of the $M = n + 1$ channels formed by the $n + 1$ occupied Landau levels that successively cross the Fermi level at the sample edges [see

<footnote n="19">¹⁹Naturally, one may also fix the magnetic field and vary the electronic density. However, from an experimental point of view, this is more complicated to achieve in general, apart from graphene where the electronic density can be easily changed with the help of a back gate due to the capacitive charging effect.</footnote>
<figure><img src="image_51.png" /><figcaption>Figure 5.7: Evolution of Landau levels at the sample edge modelled by a confinement potential $V(y)$ in the $y$-direction. (a) Landau-level spectrum as a function of the position $y \simeq y_0 = k_x l_B^2$. The spectrum may be viewed as a family of 1D bands (as a function of the wave vector in the $x$-direction) that are flat in the bulk of the sample and acquire dispersion at the sample edges, where the levels are bent upwards. Consequently, all Landau levels successively cross the Fermi level $E_F$ when approaching the edge. At the crossing points $y_{max}^n$, they form conducting edge channels. (b) Sketch of the filling: upon approaching the sample edge, the Landau levels, which are filled up to the $n$-th level in the bulk ($\nu = n + 1$), are successively emptied. This yields steps in the electronic density (in the absence of interactions that have a tendency to smoothen the abruptness of the steps). Due to the bending of the Landau levels $\partial V/\partial y_0 \propto \partial V/\partial k_x$, all edge states have a drift velocity of the same sign, i.e. the same chriality.</figcaption></figure>
<figure><img src="image_52.png" /></figure>

Figure 5.8: Sketch of the potential landscape in a quantum-Hall sample. Charged impurities give rise to an electrostatic potential in the bulk of the sample that is represented by equipotential lines. Edge confinement can be modelled by (open) equipotential lines that follow the sample edges and that connect the left contact (at a chemical potential $\mu_L$) to the right one (at a chemical potential $\mu_R$). The left curly arrows indicate the sense of propagation along the equipotential lines, which is imposed by the orientation of the magnetic field (here in the +z-direction.

figure 5.7(b)], similarly to the ballistic quantum wire studied in section 3.4.1 of chapter 3. The measured Hall resistance turns out to be precisely that associated with this ballistic conductance $Me^2/h$, as we show below with the help of the Landauer-Büttiker formalism introduced in the last chapter.

Finally, in order to observe a plateau in the Hall resistance, i.e. a resistance that does not vary over a substantial range of the magnetic field in the vicinity of integer values of $v$, we need to ensure that the additional electrons promoted to a higher, originally unoccupied, Landau level upon decrease of $B$ (or the additional holes if one increases $B$) do not contribute to the electric transport. Indeed, we show in the remainder of this chapter that these additional charges get trapped on closed equipotential lines that characterise the electrostatic potential formed by underlying (charged) disorder. Let us now put together, step by step, these different ingredients.

### 5.3.1 Guiding-centre dynamics

As we have seen in the above sketch of the explanation of the quantum Hall effect, the basic additional ingredient, which we need to consider from now
on, is an electrostatic potential $V(\mathbf{r})$. This potential is used to model both the sample edges, which are described by a strong increase of the potential there (confinement potential), as well as to represent a disorder potential (see sketch in figure 5.8). How does this position-dependent potential affect Landau quantisation described in the previous section? One first notices that it breaks translation invariance explicitly, and the loss of this symmetry is expected to lift the Landau-level degeneracy. Indeed, the components of the guiding-centre operators $X$ and $Y$, which are constants of motion in the absence of a position-dependent potential, do no longer commute with the full Hamiltonian $H = H_B(\Pi_x, \Pi_y) + V(\mathbf{r} = \mathbf{R} + \mathbf{\eta})$ because

$$[X, H] = [X, V(\mathbf{R} + \mathbf{\eta})] = [X, Y] \frac{\partial V}{\partial Y} = i l_B^2 \operatorname{sgn}(B) \frac{\partial V}{\partial Y}$$

$$\text{and similarly} \quad [Y, H] = -i l_B^2 \operatorname{sgn}(B) \frac{\partial V}{\partial X}, \qquad (5.75)$$

where we have used the commutation relation (5.25) between an operator and the function of another operator, as well as the commutator (5.60) between the components $X$ and $Y$. One notices a second consequence of the potential, related to the lifted orbital degeneracy: the guiding centres acquire themselves dynamical properties since they are no longer constants of motion. Indeed, if we appeal to the Ehrenfest theorem,<fnref n="20" /> we see that the average velocities associated with the guiding-centre components are given by

$$\langle \dot{X} \rangle = \frac{1}{i\hbar} \langle [X, H] \rangle \approx \frac{l_B^2}{\hbar} \operatorname{sgn}(B) \langle \frac{\partial V}{\partial Y} \rangle \\ \text{and} \\ \langle \dot{Y} \rangle = \frac{1}{i\hbar} \langle [Y, H] \rangle \approx -\frac{l_B^2}{\hbar} \operatorname{sgn}(B) \langle \frac{\partial V}{\partial X} \rangle. \qquad (5.76)$$

These equations have a transparent interpretation: on the average, the guiding centre follows in time the equipotential lines in a direction that is imposed by the orientation of the magnetic field (Hall drift),

$$\langle \dot{\mathbf{R}} \rangle \perp \nabla V. \qquad (5.77)$$

$$i\hbar\langle\dot{A}\rangle = \langle[A, H]\rangle.$$

<footnote n="20">²⁰Remember your basic quantum-mechanics classes, where it was shown that the average time evolution of an observable $A$ is given by its commutator with the Hamiltonian,</footnote>
This is represented in figure 5.8) by the curly lines which indicate that the semiclassical circular motion evolves into cycloids due to the Hall drift experienced by the guiding centres.

As simple as this last result may seem, it is the cornerstone in understanding the quantum Hall effect. The Hall drift allows us to understand that an electron in the sample bulk is *trapped (or localised) on equipotential lines around a summit or a valley of the potential landscape and thus does not contribute to the electric transport probed by the contacts at the sample edges*.<fnref n="21" /> We may illustrate this trapping with a hike in the mountains. Imagine that we are lazy hikers and do not allow ourselves to change altitude during our walk. Certainly, we then only turn in circles either around a summit or a valley. The length of this closed path can be substantial and it can take us quite long to get back to the original position, but we cannot cross the mountains. Actually, this is not quite true. It is true for an overwhelming amount of paths, but there exists at least one path at a certain altitude that we do not specify here that allows us to pass the mountains (or that allows an electron to pass the bulk of the sample from one edge to the opposite one) without turning in circles. This phenomenon is called *percolation*, and it turns out to be most relevant in understanding the transitions between the plateaus in the Hall resistance.

While these extended percolating states are rare in the bulk of the sample, they are naturally found at the sample edges where they connect the different electric contacts. *These edge states are responsible for the quantised ballistic transport in the quantum Hall effect.* Furthermore, we see from equation (5.76) that the sense of propagation of the guiding centres, *i.e.* the Hall drift, is imposed by the orientation of the magnetic field. While this chirality is not so important for the sense of propagation for the localised states in the bulk,<fnref n="22" /> which do not contribute to the electric transport, it is essential for the edge transport.

<footnote n="21">²¹Notice that these equipotential lines can be much longer than the typical magnetic length so that the term *localised states* may lead to confusion. Even if they are *extended* on length scales much larger than the magnetic length, they are considered as *localised states* as long as they are closed (semiclassical) trajectories for the guiding centres that do not connect opposite sample edges.</footnote>

<footnote n="22">²²If one changes the orientation of the magnetic field, one simply changes the sense of the drift around a summit or a valley.</footnote>