This equation is no longer exactly equal to zero, as in the case of a ring without magnetic flux. A little later we will transform the obtained equation so that the periodic dependence on the magnetic flux is obvious. Now we will show that the obtained current is actually the thermodynamic equilibrium response of the system to the magnetic flux.

### Thermodynamic approach

To calculate the specified response, we recall from thermodynamics that a sample placed in an external magnetic field acquires a magnetic moment **M**. The magnitude of this magnetic moment is defined as the minus derivative of the thermodynamic potential Ω of the sample with respect to the magnetic field **B**:

$$ \mathbf{M} = -\frac{\partial \Omega}{\partial \mathbf{B}}. \qquad (9.30) $$

In the case we are considering, when the sample is a one-dimensional ring, the magnetic moment is the product of the current and the area of the ring $M = IS$. Dividing the left and right sides of equation (9.30) by the area of the ring $S$ and taking into account that the product of the magnetic field and the area is the magnetic flux $BS = \Phi$, we obtain the desired relation for the current $I$ flowing in a ring permeated by a magnetic flux $\Phi$:

$$ I = -\frac{\partial \Omega(\Phi)}{\partial \Phi}. \qquad (9.31) $$

The above equation is quite general and valid both for particles obeying the classical laws of motion and for quantum particles. The whole difference is concentrated in the dependence of the thermodynamic potential on the magnetic field (magnetic flux). Thus, charged quantum particles are sensitive to the Aharonov-Bohm effect, therefore the thermodynamic potential for a system of quantum particles, for example, electrons, in a one-dimensional ring depends on the magnetic flux. And the persistent current $I$ (9.31) is different from zero.
At the same time, classical particles, for example, Brownian particles diffusing along the ring, react only to the force action of the magnetic field penetrating the ring. Therefore, in weak fields $B \ll B_c$, the thermodynamic potential for such particles does not depend on the magnetic field and the current $I$ (9.31) carried by such particles is equal to zero.

Let us now show that for electrons in the ring, equation (9.31) is identical to equation (9.29). To do this, we write an equation for the thermodynamic potential $\Omega$ of a system of electrons with a discrete spectrum $E_n$ in the case when an exchange of energy and particles occurs with a reservoir having a temperature $T$ and a chemical potential $\mu$:

$$ \Omega = -k_B T \sum_n \ln \left( 1 + e^{\frac{\mu - E_n}{k_B T}} \right). \qquad (9.32) $$

Substituting here the electron spectrum $E_n$ (9.22), we obtain the dependence $\Omega(\Phi)$. Differentiating with respect to the magnetic flux according to equation (9.31), we obtain exactly equation (9.29):

$$ \begin{aligned} I(\Phi) = -\frac{\partial \Omega(\Phi)}{\partial \Phi} &= k_B T \sum_n \frac{\partial}{\partial \Phi} \ln \left( 1 + e^{\frac{\mu - E_n}{k_B T}} \right) = \\ &= k_B T \sum_n \left( -\frac{\partial E_n}{\partial \Phi} \right) \frac{1}{k_B T} \frac{e^{\frac{\mu - E_n}{k_B T}}}{1 + e^{\frac{\mu - E_n}{k_B T}}} = \\ &= \sum_n \left( -\frac{\partial E_n}{\partial \Phi} \right) \frac{1}{1 + e^{\frac{E_n - \mu}{k_B T}}} = \sum_n \left( -\frac{\partial E_n}{\partial \Phi} \right) f(E_n). \end{aligned} $$

Let us emphasize that the equation (9.31) connects the current in the ring with the spectrum of electrons. Whereas the equation (9.29) is actually obtained using wave functions.

Despite the fact that the formula (9.31) is widely used in the literature, in fact, it is applicable only for simple geometry. For example, for a single ring. However, when two rings are connected to each other at one point and the magnetic flux penetrates only one of the rings, the current calculated by the formula (9.25) differs from the current calculated by the formula (9.28).
The difference in the answers is due to the fact that the use of thermodynamic relations (9.28), (9.30) and (9.31) gives the correct answer for the magnetic moment of the sample observed from large (macroscopic) distances. However, a detailed distribution of microscopic currents responsible for the magnetic moment can be obtained only with a sequential quantum-mechanical consideration based on knowledge of the wave functions. Therefore, in the case of samples with complex geometry, for the correct calculation of the persistent current, it is necessary to find the wave functions and calculate the quantum-mechanical current using equation (9.25).

In conclusion, we note that Hund in his 1938 work [16] pointed out that the properties of an electron system should oscillate with the magnetic flux. In addition, Dingle in his 1952 work [17] wrote that the equilibrium properties of an electron system should depend on the magnetic flux. However, it was only in the 1959 work of Aharonov and Bohm [10] that the physical causes of such oscillations were revealed.

## Self-test questions

1. What is a persistent current?

2. Determine the spectrum of electrons in a ballistic ring without magnetic flux.

3. Calculate the current in a ring without magnetic flux.

4. Determine the spectrum of electrons in a ring with a magnetic flux.

5. What is the reason for the appearance of a persistent current?

6. Calculate the current carried by an electron in a certain quantum state in a ring with a magnetic flux.
## 10. Properties of a persistent current

From the equation (9.22) it is clear that the value of the energy $E_n(\Phi)$ varies monotonically with a change in the magnitude of the magnetic flux $\Phi$. However, the properties of the entire sample (the entire electronic system of the sample), as we expect, should be periodic with respect to the magnetic flux with a period of $\Phi_0$.

To resolve this apparent contradiction, consider the total energy $E$ of the electrons in the sample at zero temperature $T = 0$ as a function of the magnetic flux $\Phi$. To calculate the energy $E$, it is necessary to sum the energies of all electrons in the ring. It should be noted that the electrons occupy only such states whose energy is less than the chemical potential $\mu$:

$$E(\Phi) = \sum_{n=-\infty}^{\infty} E_n(\Phi)\theta[\mu - E_n(\Phi)],$$

where $\theta(x)$ is the Heaviside theta function, which equals unity for $x > 0$ and equals zero for $x < 0$.

It is possible to directly perform the summation in the above equation. However, we will do this later, when we calculate the magnitude of the persistent current. For now, we will limit ourselves to a qualitative analysis of the above equation and show that, indeed, the total energy $E$ is periodic in the magnetic flux.

Fig. 10.1 shows the single-electron energy levels $E_n$ as a function of the magnetic flux. The horizontal line in the figure indicates the level of the chemical potential $\mu$. We know that at zero temperature all levels located below the Fermi level are filled, and all levels located above are free. From Fig. 10.1 it is clear that when the magnetic flux changes, the energy of a certain level becomes greater than the Fermi energy. The electron that occupied such a level will pass from the sample to the reservoir, and the corresponding energy level in the sample will be freed. On the other hand, there is a level whose energy
<figure><img src="image_51.png" /><figcaption>Fig. 10.1. The thick line shows the periodic change in the energy of the upper filled level when the magnetic flux magnitude $\Phi$ changes. This periodicity is the reason why the total energy of the electron system in the ring changes periodically when the magnetic flux $\Phi$ changes, despite the fact that the energy of each individual level changes monotonically with $\Phi$. $\mu$ is the Fermi energy in the reservoir</figcaption></figure>

becomes less than the Fermi energy. Such a level will be filled by an electron that has passed from the reservoir to the sample. At the same time, if we leave aside the absolute value of the numbers of filled and free levels, then the filling pattern will be repeated periodically when the magnetic flux changes by the quantum value $\Phi_0$. That is, indeed, the energy of the entire system, which equals the sum of the energies of the filled states, is periodic in the magnetic flux with period $\Phi_0$.

Thus, the total energy $E$ of the system is a periodic function of the magnetic flux:

$$E(\Phi) = E(\Phi + \Phi_0),$$

in the case when only those one-electron energy levels $E_n$ that satisfy the conditions are filled in the sample

$$E_n(\Phi) < \mu. \qquad (10.1)$$
This condition is natural for the equilibrium state. And it is this condition that leads to the fact that the internal energy of the electron system in equilibrium is periodic with respect to the magnetic flux, despite the fact that the energy of each individual single-electron level changes monotonically with a change in the magnetic flux.

It should be said that every time when the energy level crosses the Fermi level with a change in the magnetic flux $\Phi$, it is necessary to wait for some time $\tau$ for the equilibrium filling of the levels, determined by the condition (10.1), to be established. For a ballistic ring connected to a reservoir by a point contact, characterized by a low probability of tunneling, $T \ll 1$, under the condition $E_n = \mu$, we can write $\tau \sim L/(v_F T)$. For typical values of $L \sim 10^{-6}$ m, $v_F \sim 10^5$ m/s, and $T = 0, 01$, we obtain $\tau \sim 10^{-9}$ s.

At the same time, with a sufficiently rapid change in the magnetic flux, the filling of the levels may not be equilibrium and the periodicity under discussion will be violated.

## 10.1. Analytical calculations

So, let's calculate the current $I$, (9.29). Since, as we expect, the persistent current is periodic with respect to the magnetic flux $\Phi$ with period $\Phi_0$, it is convenient to represent this dependence in the form of a Fourier series. The current must become zero at $\Phi = 0$. In addition, this current must change sign when the direction of the magnetic flux changes to the opposite, since when the sign of the magnetic flux changes, the sign of the magnetic moment must change. From the above, it follows that the persistent current must be an odd function of the magnetic flux, so we can write:

$$I(\Phi) = \sum_{q=1}^{\infty} i_q \sin \left( 2\pi q \frac{\Phi}{\Phi_0} \right), \qquad (10.2)$$

where $i_q$ is the amplitude of the corresponding current harmonic.

In order to represent the equation (9.29) in the form of a Fourier series (10.2), we use the Poisson summation formula:
$$ \sum_{n=-\infty}^{\infty} F(n) = \int_{-\infty}^{\infty} dn F(n) + 2 \operatorname{Re} \left( \sum_{q=1}^{\infty} \int_{-\infty}^{\infty} dn F(n) e^{i2\pi qn} \right), \quad (10.3) $$

where $F(n)$ is an arbitrary function of an integer $n$. We especially emphasize that in the left-hand side $n$ takes only integer values (discretely) while in the right-hand side $n$ takes any real values (continuously).

As $F(n)$ we will use the product $I_n f(E_n)$ in the equation (9.29):

$$ \begin{aligned} F(n) &= I_n f(E_n) \equiv \frac{eh}{m_e L^2} \frac{n - \Phi/\Phi_0}{1 + \exp\left(\frac{E_n - \mu}{k_B T}\right)}, \\ E_n &= \frac{h^2}{2m_e L^2} \left(n - \frac{\Phi}{\Phi_0}\right)^2. \end{aligned} \quad (10.4) $$

In this case, the equation for the current (9.29) takes the following form:

$$ I(\Phi) = \int_{-\infty}^{\infty} dn F(n) + 2 \operatorname{Re} \left( \sum_{q=1}^{\infty} \int_{-\infty}^{\infty} dn F(n) e^{i2\pi qn} \right). \quad (10.5) $$

First of all, let us analyze the two terms on the right side of the equation (10.5). Both of them are functions of a continuous variable $n$, which is integrated along the entire real axis. It can be seen from the equation (10.4) for $F(n)$ that the variable $n$ always enters the combination $n - \Phi/\Phi_0$, so it is convenient to make the following change of variables (shift):

$$ m = n - \frac{\Phi}{\Phi_0}. $$

After this replacement, the limits of integration will not change, since they are infinite, and the equation $F(m)$ will no longer depend on the magnetic flux $\Phi$:
$$F(m) = \frac{eh}{m_e L^2} \frac{m}{1 + \exp\left(\frac{E_m - \mu}{k_B T}\right)}, \qquad (10.6)$$

$$E_m = \frac{h^2 m^2}{2m_e L^2}.$$

The magnetic flux remains only in the second term of the right-hand side in the exponential factor:

$$I(\Phi) = \int_{-\infty}^{\infty} dm F(m) + 2 \operatorname{Re} \left( \sum_{q=1}^{\infty} \int_{-\infty}^{\infty} dm F(m) e^{i2\pi q m} e^{i2\pi q \frac{\Phi}{\Phi_0}} \right). \quad (10.7)$$

Since $F(m)$ (10.6) is an odd function of $m$, the first integral vanishes identically. Finally, we can write

$$I(\Phi) = 2 \operatorname{Re} \left[ \sum_{q=1}^{\infty} J(q) e^{i2\pi q \frac{\Phi}{\Phi_0}} \right], \qquad (10.8)$$

where the value of $J(q)$ does not depend on the magnetic flux:

$$J(q) = \int_{-\infty}^{\infty} dm F(m) e^{i2\pi q m}. \qquad (10.9)$$

From the equation (10.8) it is clear that, indeed, the current is periodic with respect to the magnetic flux with period $\Phi_0$:

$$e^{i2\pi q \frac{\Phi+\Phi_0}{\Phi_0}} = e^{i2\pi q \frac{\Phi}{\Phi_0}} e^{i2\pi q} = e^{i2\pi q \frac{\Phi}{\Phi_0}} \times 1.$$

We further note that $J(q)$ is an imaginary quantity. To verify this, we consider that $F(m)$ (10.6) is an odd function of the variable $m$, therefore:
$$ \int_{-\infty}^{\infty} dm F(m) \cos 2\pi qm \equiv 0, $$

$$ \int_{-\infty}^{\infty} dm F(m) \sin 2\pi qm \neq 0. $$

As a result, the equation (10.8) for the current $I$ will become:

$$ I(\Phi) = -2 \sum_{q=1}^{\infty} \mathrm{Im}[J(q)] \sin \left( 2\pi q \frac{\Phi}{\Phi_0} \right). \qquad (10.10) $$

So, we have presented the equation for the current in the form of a Fourier series. The amplitudes of the current harmonics are as follows:

$$ i_q = -2 \mathrm{Im}[J(q)]. $$

We see that the current is indeed an odd and periodic function of the magnetic flux $\Phi$ with period $\Phi_0$, as we assumed.

Next, we need to calculate the value of $J(q)$. Let's take into account that only the even part of the integrand affects the value of the integral. In this case, the value of the integral over the positive half-axis of the variable $m$ is equal to the value of the integral over the negative half-axis, so we get:

$$ J(q) = \frac{2eh}{m_e L^2} \int_0^\infty dm \frac{m e^{i2\pi qm}}{1 + \exp\left(\frac{E_m - \mu}{k_B T}\right)}, \qquad (10.11) $$

$$ E_m = \frac{h^2 m^2}{2m_e L^2}. $$
The integrand contains an oscillating function of the variable integration. Therefore, if the remaining factors change slowly with m, the integral will be close to zero. It follows that the main contribution to the integral comes from the integration region near the Fermi energy $E_m \sim \mu$, where the Fermi distribution function changes from unity for $E_m < \mu$ to zero for $E_m > \mu$. In this region, we can use the following expansion:

$$ 
\begin{aligned}
E_m - \mu &= \frac{h^2 m^2}{2m_e L^2} - \frac{h^2 m_F^2}{2m_e L^2} = \\
&= \frac{h^2 (m + m_F) (m - m_F)}{2m_e L^2} = \Delta_F (m - m_F),
\end{aligned}
$$

where we assume that $m \approx m_F$ for $E_m \sim \mu$. Here $\Delta_F$ is the distance between quantum levels near the Fermi energy, and $m_F$ is the quantity m corresponding to the Fermi energy $E_m = \mu$:

$$ 
\begin{aligned}
\Delta_F \equiv E_{m+1} - E_m &= \frac{m_F h^2}{m_e L^2} = \frac{hv_F}{L}, \\
m_F &= \sqrt{\frac{2m_e L^2 \mu}{h^2}} = \frac{L}{\lambda_F}.
\end{aligned}
$$

Let us introduce a new integration variable $x = \Delta_F(m - m_F)/(k_B T)$ into the equation (10.11) and write:

$$ J(q) = \frac{2ehm_F k_B T}{m_e L^2} e^{i2\pi qm_F} \int_{-\infty}^{\infty} dx \frac{e^{i2\pi qx \frac{k_B T}{\Delta_F}}}{1+e^x} \left(1+x \frac{k_B T}{\Delta_F} \frac{\lambda_F}{L}\right). \quad (10.12) $$

In the integrand we substitute $1/m_F = \lambda_F/L$. For a not very small ring $L \gg \lambda_F$, we can consider the equation in parentheses to be equal to unity. Using the value of the integral
$$ \int_{-\infty}^{\infty} dx \frac{e^{i\alpha x}}{1+e^x} = (-1) \frac{i\pi}{\sinh(\alpha\pi)} $$

and considering that in our case the parameter $\alpha$ is:

$$ \alpha = 2\pi k \frac{k_B T}{\Delta_F}, $$

we get that the equation (10.12) takes the following form:

$$ \mathrm{Im}[J(q)] = -\frac{2\pi ehm_F k_B T}{m_e L^2 \Delta_F} \frac{\cos(2\pi q m_F)}{\sinh\left(2\pi^2 q \frac{T}{\Delta_F}\right)}. $$

Substituting $J(q)$ into (10.10), we find the final equation for the current:

$$ \begin{align} I(\Phi) &= \sum_{q=1}^{\infty} i_q \sin \left( 2\pi q \frac{\Phi}{\Phi_0} \right), \tag{10.13} \\ i_q &= I_0 \frac{2T}{\pi T^*} \frac{\cos \left( 2\pi q \frac{L}{\lambda_F} \right)}{\sinh \left( q \frac{T}{T^*} \right)}. \nonumber \end{align} $$

We have introduced a notation for the characteristic value of persistent current:

$$ I_0 = \frac{ev_F}{L} $$

and for the crossover temperature:

$$ T^* = \frac{\Delta_F}{2\pi^2 k_B}, $$

above which the current is exponentially small.
Macroscopic limit

The magnitude of the current $I$ decreases with increasing ring size $L$, so the persistent current is absent in the macroscopic limit, $L \to \infty$.

At zero temperature, the current decreases as $1/L$. However, at finite temperatures, the current decreases with increasing ring size much faster. This is due to the fact that the crossover temperature $T^*(L) \sim 1/L$ decreases with increasing $L$. Therefore, at a fixed temperature $T$, the persistent current decreases exponentially with increasing ring length, if $L \gg L^*(T)$:

$$I \sim e^{-\frac{T}{T^*}} \sim e^{-\frac{L}{L^*(T)}},$$

where the critical size is:

$$L^*(T) = \frac{hv_F}{2\pi^2 k_B T}.$$

Below we will estimate the values of the quantities characterizing the persistent current.

## 10.2. Numerical extimates

### 10.2.1. Current amplitude

Let us estimate the current amplitude:

$$I_0 = \frac{ev_F}{L}.$$

For a ring with size $L \sim 10^{-6}$ m and for the Fermi velocity of electrons in a 2D gas $v_F \sim 10^4$ m/s, we obtain:

$$I_0 \sim 10^{-9} A.$$
<figure><img src="image_52.png" /><figcaption>Fig. 10.2. The signs “+” and “-” indicate the direction of the currents carried by electrons at successive energy levels in a one-dimensional ballistic ring, which is penetrated by a magnetic flux $\Phi/\Phi_0 \neq 2\pi n$, $2\pi(n + 0, 5)$. The total current is determined by the current carried by the upper occupied level, therefore, when the Fermi energy changes from $\mu_1$ to $\mu_2$, the current in the ring changes direction to the opposite</figcaption></figure>

From the equation for the current amplitude $I_0 = ev_F/L$ it follows that, in fact, the persistent current is created by a single electron with Fermi velocity, which moves along the ring. This happens because in the presence of a magnetic flux, the energy levels corresponding to the electron's clockwise and counterclockwise motion in the ring alternate, Fig. 10.2. In addition, with an increase in the energy of the level, the contribution to the current from this level increases, since the electron velocity increases. Therefore, if we consider the contribution to the current from several successive levels, we find that the contribution of each subsequent level compensates for the contribution of the previous level and adds a small amount over and above it. As a result, firstly, the current changes sign when the number of filled levels changes, that is, when the number of particles in the ring changes. This is the so-called parity effect. And, secondly, the magnitude of the total current, in fact, coincides with the magnitude of the current created by the electron in the upper filled level.

We can say that the persistent current is created by an electron moving in a circular orbit in an artificially created atom (ring). Let us compare the
magnitude of this current with the current that would be created by an electron moving in a flat circular orbit in the ground state of the hydrogen atom. However, it should be said that this example is used only for numerical comparisons, since in the hydrogen atom the electron moves in a spherical orbit and does not create an orbital magnetic moment. Thus, for the first Bohr orbit we have $R \sim 5 \times 10^{-11}$ m and $v_1 \sim 5 \times 10^6$ m/s. As a result, we find the current

$$I_a \sim 10^{-3} \text{A}.$$

Thus, the current in the ring is much smaller than the current in the atom:

$$\frac{I_0}{I_a} \sim 10^{-6}.$$

However, it is impossible to measure the electron current in an experiment. Usually, the magnetic moment created by the current $M \sim IS$ is measured, where $S$ is the area covered by the current. Let us compare the corresponding magnetic moments:

$$M \sim IS = \frac{e v L^2}{L 4\pi} = \frac{e v}{4\pi} L.$$

We see that the value of $M$ increases with increasing orbital size $L$, so the magnetic moment of the persistent current in the artificial atom (ring) turns out to be larger than the orbital magnetic moment of an electron in a regular atom:

$$\frac{M_0}{M_a} \sim \frac{I_0}{I_a} \left(\frac{L}{2\pi R}\right)^2 \sim 10.$$

### 10.2.2. Crossover temperature

Let us estimate the value $T^*$, which separates the region of low temperatures, $T \ll T^*$, where the persistent current does not depend on temperature,
from the region of high temperatures, $T \gg T^*$, where the persistent current is exponentially reduced. Considering that

$$ \Delta_F = \frac{hv_F}{L}, $$

for $L \sim 10^{-6}$ m and $v_F \sim 10^4$ m/s we obtain:

$$ T^* = \frac{\Delta_F}{2\pi^2 k_B} \sim 25 \times 10^{-3} K. $$

The crossover temperature is quite low. Note that this temperature is actually an order of magnitude smaller than the energy level separation $\Delta_F$.

## 10.3. Temperature dependence of a current

Let us analyze the dependence of the persistent current (10.13) on temperature. We will assume that the temperature remains low enough so that the state of electrons in the ring remains coherent and that the consideration based on the solution of the Schrödinger equation remains valid. That is, we will assume that one of the main conditions characteristic of mesoscopics is fulfilled, namely, that the phase failure length $L_\phi(T)$ is much larger than the dimensions of the ring $L_\phi(T) \gg L$. However, even under this condition, the dependence on temperature is significant, Fig. 10.3.

### 10.3.1. Low temperatures

At low temperatures

$$ T \ll T^* $$

we can put in the equation for current (10.13)

$$ \sinh \left( q \frac{T}{T^*} \right) \approx q \frac{T}{T^*}. $$
<figure><img src="image_53.png" /><figcaption>Fig. 10.3. The dependence of the persistent current on temperature $T$. Below the crossover temperature, $T < T^*$, the current is independent of temperature. Above the crossover temperature, $T > T^*$, the current decreases exponentially</figcaption></figure>

Substituting this into (10.13), we get:

$$I(\Phi) = I_0 \frac{2}{\pi} \sum_{q=1}^{\infty} \frac{\cos\left(2\pi q \frac{L}{\lambda_F}\right)}{q} \sin\left(2\pi q \frac{\Phi}{\Phi_0}\right). \quad (10.14)$$

Thus, within the limits of low temperatures, the dependence $I(\Phi)$ contains various harmonics with periods $\Phi_0/q$. The magnitude of the persistent current does not depend on temperature, but, in addition to the dependence on the magnetic flux, it also depends on the Fermi energy. The latter dependence is contained in the factor $\cos(2\pi qL/\lambda_F)$.

In the general case, at low temperatures, the dependence $I(\Phi)$ has a saw-tooth shape with a period of $\Phi_0$ (see below, Fig. 10.4 and Fig. 10.5). The jump-like change in the magnitude of the current is due to the fact that when the magnetic flux changes, one of the energy levels $E_n(\Phi)$ in the ring crosses the Fermi level, Fig. 10.1. At each such intersection, as we have already discussed earlier, the number of electrons in the ring changes by one. At the same time, the direction of the electron movement in the upper filled level changes, which changes the sign of the current to the opposite. Such a change corresponds to a jump in the dependence of the current on the magnetic flux.
At the same time, for some selected values of the chemical potential $\mu$, the number of particles in the ring does not change when the magnetic flux changes. This is the case depicted in Fig. 10.1. The position of the chemical potential relative to the energy levels in a zero magnetic field is important. If the Fermi energy coincides with one of the energy levels or is located exactly in the middle between neighbouring doubly degenerate levels, then when the magnetic flux changes, the chemical potential level will be crossed simultaneously by two energy levels: one of them will be released, and the other will be filled with an electron. In this case, the total number of particles in the ring remains unchanged. However, since the currents carried by electrons in the intersecting levels are different, the direction of the current in the ring will also change to the opposite. In this case, the equation for the current takes the simplest form. Here, two cases are possible.

### Even number of particles in a ring

If at $\Phi = 0$ the chemical potential $\mu$ coincides with one of the doubly degenerate energy levels $E_{n_0}$ in the ring, then the total number of electrons $N_e$ in the ring is even and does not change when the magnetic flux changes:

$$N_e = 2n_0.$$

In this case, the length of the ring is a multiple of the Fermi wavelength:

$$L = n_0 \lambda_F.$$

The multiplier in the equation (10.14), which contains the ratio $L/\lambda_F$, is the same for all harmonics $q$:

$$\cos\left(2\pi q \frac{L}{\lambda_F}\right) = \cos(2\pi n_0 q) = 1.$$

The equation for current (10.14) takes the following form:
<figure><img src="image_54.png" /><figcaption>Fig. 10.4. The dependence of the persistent current $I$ on the magnetic flux $\Phi$ for a ring with an even number of electrons $N_e = 2k$ is shown at zero temperature $T = 0$ (solid line) and at a temperature higher than the crossover temperature: $T > T^*$ (dashed line). $I_0 = ev_F/L$ is the current amplitude; $\Phi_0 = h/e$ is the magnetic flux quantum</figcaption></figure>

$$I_{2N}(\Phi) = I_0 \frac{2}{\pi} \sum_{q=1}^{\infty} \frac{1}{q} \sin \left( 2\pi q \frac{\Phi}{\Phi_0} \right).$$

Applying the inverse Fourier transform to this equation, i.e., summing the series over $q$, we find:

$$ \begin{align} I_{2N}(\Phi) &= -I_0 \left(2 \frac{\Phi}{\Phi_0} - 1\right), \quad 0 < \Phi < \Phi_0, \tag{10.15} \\ I_{2N}(\Phi) &= I_{2N}(\Phi + \Phi_0). \notag \end{align} $$

The graph of this dependence is shown in Fig. 10.4, solid curve.

Odd number of particles in the ring

If at $\Phi = 0$ the chemical potential $\mu$ is exactly in the middle between the doubly degenerate energy levels $E_{n_0}$ and $E_{n_0+1}$, then the total number of
electrons $N_e$ in the ring is odd and does not change when the magnetic flux changes:

$$N_e = 2n_0 + 1.$$

Note that the level with $n = 0$ is nondegenerate and contains one electron, and the remaining levels are doubly degenerate. At zero temperature (the case we are considering) the levels with $n \le n_0$ contain two electrons, and the levels with $n > n_0$ are not filled.

In this case, the ratio of the ring length to the Fermi wavelength is:

$$\frac{L}{\lambda_F} = n_0 + \frac{1}{2}.$$

In this case, the factor in the equation (10.14) containing this relation is:

$$\cos \left( 2\pi q \frac{L}{\lambda_F} \right) = \cos \left[ 2\pi (n_0 + 0, 5) q \right] = \cos(\pi q) = (-1)^q.$$

The equation for current (10.14) takes the following form:

$$I_{2N+1}(\Phi) = I_0 \frac{2}{\pi} \sum_{q=1}^{\infty} \frac{(-1)^q}{q} \sin \left( 2\pi q \frac{\Phi}{\Phi_0} \right).$$

The inverse Fourier transform results in the following:

$$\begin{align}
I_{2N+1}(\Phi) &= -2I_0 \frac{\Phi}{\Phi_0}, \quad -\frac{\Phi_0}{2} < \Phi < \frac{\Phi_0}{2}, \tag{10.16} \\
I_{2N+1}(\Phi) &= I_{2N+1}(\Phi + \Phi_0). \nonumber
\end{align}$$

The graph of this dependence is shown in Fig. 10.5.

It should be noted that the dependence (10.16) can be obtained from the
<figure><img src="image_55.png" /><figcaption>Fig. 10.5. The dependence of the persistent current $I$ on the magnetic flux $\Phi$ for a ring with an odd number of electrons $N_e = 2k+1$. The notation is the same as in Fig. 10.4.</figcaption></figure>

dependence (10.15) by shifting it by half a period:

$$I_{2N}(\Phi) = I_{2N+1}(\Phi + \Phi_0/2).$$

This can be interpreted as follows. Each electron brings with it into the ring an effective magnetic flux equal to half a quantum $\Phi_0/2$, which must be added to the external magnetic flux. Two electrons make a total contribution equal to one quantum of flux, which can be ignored due to the periodicity.

### 10.3.2. High temperatures

Тепер розглянемо високі температури:

$$T \gg T^*.$$

The denominator in the equation for the amplitude of current harmonics (10.13) becomes large and increases with increasing harmonic number $q$:

$$\sinh\left(q\frac{T}{T^*}\right) \sim \frac{1}{2} e^{q\frac{T}{T^*}} \gg 1.$$

Therefore, the amplitude of the first harmonic of the current $i_1$ turns out to
be dominant. Therefore, the magnitude of the current decreases exponentially with increasing temperature, and the shape of the dependence of the current on the magnetic flux becomes sinusoidal:

$$I \approx I_0 \frac{4}{\pi} \frac{T}{T^*} e^{-T/T^*} \cos \left( 2\pi \frac{L}{\lambda_F} \right) \sin \left( 2\pi \frac{\Phi}{\Phi_0} \right). \quad (10.17)$$

The exponential dependence on temperature and the smooth dependence on magnetic flux are a consequence of the temperature blurring of the edge of the Fermi distribution function. At high temperatures, a change in the magnetic flux does not lead to a change in the number of particles in the ring, so there are no sharp breaks in the dependence of the current on the magnetic flux. Compensation of contributions from partially filled levels leads to an exponential decrease in the current value with increasing temperature.

## 10.4. Parity effect

From equations (10.15) and (10.16) it is clear that the nature of the dependence of the persistent current on the magnetic flux strongly depends on the number of particles in the ring. If the number of particles in the ring is odd, then the response of the electronic system in a zero field is diamagnetic, i.e., the induced magnetic moment of the ring is opposite in sign to the applied magnetic flux. The magnitude of the diamagnetic moment is proportional to the magnetic flux and, therefore, vanishes at $\Phi \to 0$. The response of a ring with an even number of particles is paramagnetic, i.e., the induced magnetic moment of the ring coincides in sign with the applied magnetic flux. Unlike the diamagnetic moment, the magnitude of the paramagnetic moment is finite at $\Phi \to 0$.

This effect is called the parity effect. The parity effect, i.e., a significant, and moreover, periodic, change in the properties of a sample when the number of particles in it changes, is quite common in mesoscopics. A persistent current exhibits the parity effect because neighboring energy levels carry a current in the opposite direction. It should be noted that this is true only for simple geometry. In samples with a more complex shape, for example, in a ring with
a small conductor attached to it, entire groups of consecutive quantum levels carry a current in the same direction. In this case, the parity effect is modified.

In addition, we did not take into account the electron spin. If we take it into account, each energy level will be additionally doubly degenerate. In this case, for the persistent current, the number of electrons in the ring modulo 4 turns out to be significant. Only when the number of electrons in the ring changes by four does the dependence $I(\Phi)$ remain unchanged.

## Self-test questions

2. How does the amplitude of the persistent current change with temperature?

3. How does the amplitude of the persistent current change with increasing ring size?

4. What is the parity effect for the persistent current?

5. What is the crossover temperature for a persistent current and what does it depend on?

6. How does the persistent current depend on the magnetic flux at low temperatures? How does the parity effect manifest itself in this case?

7. What is the peculiarity of the persistent current at high temperatures? Does the parity effect manifest itself in this case?
# 11. Coulomb blockade effect

Using the example of a persistent current, we have seen that the motion of a single electron can determine a completely measurable effect in mesoscopic samples. In this regard, it is natural to assume that the discreteness of the electric charge is also essential for mesoscopics. Indeed, there are conditions under which the discreteness of the charge significantly affects the properties of mesoscopic samples [31, 32, 33].

It should be said that the question of the influence of the discreteness of the charge on the properties of solids has a long history. Actually, the measurement of the electron charge itself became possible due to the fact that such an influence is significant and can be studied experimentally. Thus, as early as 1911, Milliken measured the charge of an electron by observing the motion of a suspension of microscopic oil droplets. Later, the effect of single electron tunneling on the properties of granular films was experimentally studied in the works of Giaever and Zeller (1968) [34], Lambe and Jaklevic (1969) [35], and others. It should be emphasised that, unlike the above-mentioned works, in which, in fact, a collection of small granules was studied, the mesoscopic stage of development is associated with the possibility of studying a separately taken sample, a single granule.

## 11.1. Charge quantisation in a small island

In order for the effect of the charge of one electron to be noticeable, it is necessary, firstly, that the energy scale associated with charging the sample with a charge of 1e be comparable to other energy scales characteristic of the given sample, for example, temperature. And, secondly, it is necessary that the sample retain its charge for a sufficiently long time. For this, the sample must be well insulated from the environment. In particular, electrical contacts must be connected to such a sample using tunnel contacts with low transparency.

Let us estimate the scale of the energy associated with a change in the sam-
<figure><img src="image_56.png" /><figcaption>Fig. 11.1. The transfer of one electron from the reservoir to the granule changes its charge by $1e$, which is accompanied by a change in the energy of the entire system by $E_c = e^2/2C$, where $C$ is the electric capacitance of the granule.</figcaption></figure>

ple charge by $1e$, Fig. 11.1. This energy is nothing other than the electrostatic energy $E_c$, due to the sample's own capacitance $C$:

$$E_c = \frac{e^2}{2C}. \qquad (11.1)$$

The capacitance of a spherical sample with radius $R$ is $C = 4\pi\epsilon_0\epsilon R$, and for a sample in the form of a disk $C = 8\epsilon_0\epsilon R$. Here $\epsilon$ is the dielectric permittivity of the medium. The constant $\epsilon_0$ is as follows: $\epsilon_0 = 8,85 \times 10^{-12} C^2/(Nm^2)$. For characteristics values $R \sim 10^{-6}$ m and $\epsilon \sim 10$, we obtain $C \sim 10^{-15}$ F. The corresponding electrostatic energy is:

$$E_c = \frac{(1,6 \times 10^{-19})^2}{2 \times 10^{-15} \times 38 \times 10^{-23}} C^2/(FJ/K) \sim 1 K.$$

This means that when the ambient temperature is less than $1 K$, the charge $Q$ of such a sample will not fluctuate, but will be fixed. If a voltage is applied to the pellet, no current will flow, since for the current to flow, electrons must pass to the pellet. However, such a transition is impossible, since it would lead
<figure><img src="image_57.png" /><figcaption>Fig. 11.2. The change in the energy of the system by $E_c$ during the transition of the electron from the reservoir to the granule leads to the fact that the current $I \neq 0$ appears in the system only when $V > V^*$. The threshold voltage $|eV^*| \sim E_c$</figcaption></figure>

to a change in the charge of the pellet. Only if a considerable voltage is applied

$$V > V^* \sim \frac{E_c}{|e|} \sim 0,8 \times 10^{-3} \text{ V},$$

then the current will flow through the granule, since the source will be able to perform work against the electrostatic forces that prevent the electron from passing to the granule, Fig. 11.2. This is actually the effect of Coulomb blockade. Namely, at low temperatures the granule becomes conductive only starting from a certain threshold voltage

$$|eV^*| \sim E_c.$$

Next, we will analyze the requirement that the charge on the granule remains constant for a sufficiently long time. Why is this necessary? Above, we analyzed the possibility of an electron's transition to the granule, considering the electron as a classical, not a quantum particle. However, from the point of view of quantum mechanics, an electron can tunnel to the granule, stay there for some time $\delta t$, and then leave the granule. The electron can either return to the contact from which it initially tunnelled to the granule, or move to the other contact and thereby take part in the transfer of current through the granule. These are the so-called quantum fluctuations of the charge on the granule. For such virtual processes, there is no need to overcome the electrostatic barrier,
since the law of conservation of energy must be fulfilled only for the energies in the initial and final states, when the charge on the granule is absent and, accordingly, there is no electrostatic energy due to this charge. For intermediate, virtual states that exist for a short time $\delta t$, when the electron is on the granule and, as a consequence, electrostatic energy appears, the law of conservation of energy must not be fulfilled. More precisely, the law of conservation of energy is fulfilled with an accuracy of the order of $\delta E$. This quantity is related to $\delta t$ by the Heisenberg uncertainty relation for energy-time:

$$ \delta E \delta t \geq h, \qquad (11.2) $$

where $h$ is Planck's constant. If the energy uncertainty $\delta E$ is less than the electrostatic energy $E_c$ associated with the appearance of an electron charge on the granule,

$$ \delta E \sim \frac{h}{\delta t} < E_c, \qquad (11.3) $$

then the process of electron tunneling through the granule, in other words, the process of tunneling through a virtual electrostatic potential barrier, turns out to be impossible. If

$$ \delta E \gg E_c, $$

then tunneling is possible.

For the occurrence of Coulomb blockade, it is necessary that the uncertainty of the energy $\delta E$ be small compared to $E_c$. For this, it is necessary that the time $\delta t$, during which the electron is on the granule, be large. This time is in the order of magnitude equal to the time of charging or discharging the granule. It (time) is determined by the electrical characteristics - the capacitance $C$ and the resistance of the tunnel barrier $R_t$ between the contact (conductor) and the granule:

$$ \delta t \sim R_t C. $$
Substituting this estimate into equation (11.3), we obtain a constraint on the resistance of the tunneling contact separating the granule from the environment, at which tunneling is impossible (i.e., at which $\delta E \ll E_c$):

$$R_t > \frac{h}{e^2} \approx 25,8 \times 10^3 \Omega. \qquad (11.4)$$

Thus, in order for the effects associated with the discreteness of the electron charge to manifest themselves in the conductance of the granule, first, low temperatures are necessary,

$$T \ll E_c,$$

to avoid classical (thermal) over-barrier jumps and, secondly, considerable tunnel contact resistances to avoid jumps due to quantum tunneling,

$$\delta E \ll E_c.$$

Below, we will assume that both of these conditions are met, which will allow us to conduct a qualitative, semi-classical analysis of the influence of charge discreteness on the electrical conductance of the granule.

## 11.2. Coulomb blockade of an electron transport

Consider a granule connected by tunnel barriers to the left and right contacts. The conductance of the granule is proportional to the probability of an electron passing from the left contact to the right. In this case, this is the probability of tunneling through a double potential barrier. One potential barrier separates the granule from the left contact and the second potential barrier separates the granule from the right contact. We will assume that the granule is so small that it is necessary to take into account the quantization of the electron energy levels in the granule. In this case, we will assume that the temperature is zero.
11.2.1. Transport at resonance tunneling

First, we consider the conductance of the granule without taking into ac-
count the influence of the electrostatic energy $E_c$. Recall that the conductance,
defined as the ratio of the current to the applied voltage at $V \to 0$, is due
to the transfer of electrons with the Fermi energy through the sample. For
such electrons, the tunneling probability will be maximum in the case when
the Fermi level coincides with one of the quantum levels of the electrons in the
granule. This corresponds to the condition of resonant tunneling. In the case
of identical left and right tunneling barriers, the tunneling probability $T$ near
resonance can be represented by the following Breit-Wigner formula [7]:

$$
\mathcal{T} = \frac{\Gamma^2}{\Gamma^2 + (\mu - E_0)^2}, \qquad (11.5)
$$

where $\Gamma$ is the half-width of the resonance; $E_0$ is the value of the energy level
in the granule, which is closest to the chemical potential $\mu$ of the electrons in
the reservoir. The value of $\Gamma$ is proportional to the probability of tunneling
through a unit potential barrier.

From the equation (11.5) it is clear that the tunnel transparency $\mathcal{T}$, and
with it the conductance of the granule $G = G_0 \mathcal{T}$, depend on the difference
$\mu - E_0$. In the general case, this difference can be arbitrary and vary from
sample to sample. However, modern experimental techniques allow not only
to bring the contacts to an individual granule of micron dimensions, but also
to change the parameters of the granule. In particular, by using an additional
metal contact that forms a capacitor with the granule, it is possible to change
the potential of the granule $V_g$ with respect to the contacts. In this case, the
energy of each level in the granule will be shifted by the value $eV_g$, so the
conductance of the granule will depend on $V_g$:

$$
G(V_g) = G_0 \sum_n \frac{\Gamma_n^2}{\Gamma_n^2 + (\mu - E_n - eV_g)^2}. \qquad (11.6)
$$

Here, the summation is performed over all quantum levels in the granule. For
each level, we have written down its $\Gamma_n$, since the tunneling probability depends
on the energy and can be different for different energy levels.
<figure><img src="image_58.png" /><figcaption>Fig. 11.3. Without taking into account the Coulomb blockade effect, the dependence of the conductance G of a sample with a discrete spectrum on the sample potential V_g at zero temperature has a resonant character. The distance between the peaks is determined by the distance between the energy levels in the sample.</figcaption></figure>

Let us consider the dependence $G(V_g)$. This dependence will have the form of a set of peaks located at the following voltages:

$$eV_{g,n} = \mu - E_n. \qquad (11.7)$$

The distance between the conductance peaks will be determined by the distance between the energy levels in the granule, Fig. 11.3.

Thus, by studying the conductance, it is possible to determine the energies of the quantum levels in the granule, i.e., to perform quantum level spectroscopy. In this case, the difference between the voltages on the screen $V_g$, which correspond to the peaks of the dependence $G(V_g)$, is equal to the distance between the quantum levels in the granule:

$$-(eV_{g,n+1} - eV_{g,n}) = E_{n+1} - E_n \equiv \Delta E_n. \qquad (11.8)$$
It should also be said that to use this method, it is necessary that the temperature and voltage between the contacts are significantly smaller than the distance between the peaks $\Delta E_n$:

$$ \Delta E_n \gg k_B T |eV| . $$

We have described the picture of resonant tunneling without taking into account the fact that the electron charge is finite. Now we will consider how the effect of Coulomb blockade modifies the dependence of $G(V_g)$.

### 11.2.2. Coulomb blockade of transport

When we take into account the energy associated with the charge of the granule, we are actually taking into account the energy due to the Coulomb interaction of electrons between themselves and with other charged particles. An accurate account of these interactions requires consideration of the many-particle Schrödinger equation and is a rather difficult task. At least, it can be argued that in this case the single-particle energy levels $E_n$ calculated in the model of non-interacting electrons that we have used so far cannot be used to adequately characterise the transport properties.

However, there is an approximation that has proven itself well. It consists in the fact that the influence of electrostatic interaction is taken into account by introducing additional energy, which depends on the total number of electrons in the granule. In this model, the energies of single-particle levels depend on the number of electrons $N_e$ in the granule. This dependence is defined as follows. Let $E_n(N_e)$ be the single-particle energy levels in a granule with $N_e$ excess electrons. Excess electrons are those electrons whose charge is not compensated by the charge of the granule ions. Then the energy levels in a granule with $N_e + 1$ electron will be defined by the following equation:

$$ E_n(N_e + 1) = E_n(N_e) + E_c(N_e + 1) - E_c(N_e). \qquad (11.9) $$

Here, the electrostatic energy $E_c$ of the system depends on the charge of the
granule, i.e., on the number of electrons Ne in the granule, and on the potential $V_g$ of the granule, which can be changed using a metal screen:

$$E_c(N_e) = \frac{(eN_e)^2}{2C} + eN_e V_g. \qquad (11.10)$$

Substituting this equation into the equation (11.9), we obtain

$$E_n(N_e + 1) = E_n(N_e) + \frac{e^2}{C}(N_e + 1/2) + eV_g. \qquad (11.11)$$

Let us consider how the presence of electrostatic energy changes the conditions for the resonant transfer of electrons through the granule. Let the voltage on the screen $V_g$ correspond to the resonance condition, (11.7), without taking into account the electrostatic energy $eV_{g,n} = \mu - E_n$. What will happen in this case? It seems that the electron should move to the granule and fill the level $E_n$. But this does not happen. Why? Because when the electron moves to the granule, the electrostatic energy $E_c \sim e^2/(2C)$ will arise, which will shift the position of the energy levels in the granule relative to the chemical potential and, thereby, violate the resonance condition. It follows that the correct resonance condition must take into account the change in the electrostatic energy of the system when the number of electrons on the granule changes. Considering the above, the condition for resonant tunneling in the presence of the Coulomb blockade effect is the following

$$E_n(N_e + 1) = \mu. \qquad (11.12)$$

Using the equations (11.9) and (11.11), we obtain:

$$E_n(N_e) + \frac{e^2}{C}(N_e + 1/2) + eV_{g,n} = \mu. \qquad (11.13)$$

Thus, the value of the voltage on the screen at which the conductance of the granule has a resonant peak will be as follows:
$$ -eV_{g,n} = E_n(N_e) - \mu + \frac{e^2}{C}(N_e + 1/2). \qquad (11.14) $$

The following resonance will occur if the voltage across the screen is:

$$ -eV_{g,n+1} = E_{n+1}(N_e + 1) - \mu + \frac{e^2}{C}(N_e + 3/2). \qquad (11.15) $$

Note that the number of electrons in the granule has changed after the previous resonance and is equal to $N_e + 1$. The distance between the resonance peaks in the $G(V_g)$ dependence will be as follows:

$$ -(eV_{g,n+1} - eV_{g,n}) = E_{n+1}(N_e + 1) - E_n(N_e) + \frac{e^2}{C}. \qquad (11.16) $$

Comparing the equation (11.16) with the equation (11.8), we see that taking into account the electrostatic energy changes the distance between the peaks. If the electric capacitance $C$ is sufficiently small and, accordingly, the electrostatic capacitance $e^2/C$ is sufficiently large, then the distance between the conductance peaks does not depend on the location of the energy levels $E_n$ of the electrons in the granule and is determined only by the electrostatic energy $E_c$, Fig. 11.4.

Thus, depending on the ratio between $\Delta E_n$ and $E_c = e^2/(2C)$, the study of resonant tunneling allows either to investigate the quantum energy levels of the electrons in the granule, or to measure the electric capacitance of the granule:

$$ -(eV_{g,n+1} - eV_{g,n}) = \begin{cases} E_{n+1} - E_n & C \to \infty, \\ \frac{e^2}{C}, & C \to 0. \end{cases} \qquad (11.17) $$

Let us compare $\Delta E_n$ (when $E_n \sim \mu$) and $E_c$ for a two-dimensional disk-shaped granule with diameter $L$:
<figure><img src="image_59.png" /><figcaption>Fig. 11.4. For samples with a small size and, accordingly, a small electric capacitance $C \to 0$, the change in the electrostatic energy $E_c \sim e^2/C$, which is associated with the change in the sample charge during the tunneling of a single electron, becomes significant and, in fact, replaces the energy due to spatial quantization. For such samples, the effect of Coulomb blockade leads to the fact that the distance between the positions of the resonances in the dependence $G(V_g)$ is determined by the energy $E_c$</figcaption></figure>

$$ \Delta E_n = \frac{h^2}{4\pi m^* L^2}, $$

$$ E_c = \frac{e^2}{C} = \frac{e^2}{4\epsilon\epsilon_0 L}. $$

We see that as the granule size decreases, the distance between quantum levels $\Delta E_n$ grows faster than the electrostatic energy $E_c$. The energies $\Delta E_n$ and $E_c$ become of the same order for such a length:

$$ L_c \sim \frac{h^2 \epsilon \epsilon_0}{\pi m^* e^2} = 7,9 \times 10^{-9} \text{ m}, $$

where we set $\epsilon = 10 \text{ m}^* = 0,067 \text{ m}_e$.
In sufficiently small granules, for which $L < L_c$, the electrostatic energy $E_c$ can be ignored. Note that the value of $L_c$ is still larger than atomic dimensions ~$10^{-10}$ m, at which this approximation loses its meaning.

In granules of larger dimensions, for which $L > L_c$, the electrostatic energy exceeds the spatial quantization energy, and therefore it is the value of $E_c$ that determines the experimentally measured dependence of the conductance $G$ on the voltage on the barrier $V_g$. However, even in this case, the position of the quantum levels $E_n$ can be determined from the experimental data. For this purpose, as follows from the equation (11.16), it is necessary to study the violation of equidistance in the position of the peaks on the graph of the dependence of the sample conductance on the potential $V_g$.

## Self-test questions

1. What is the Coulomb blockade effect?

2. How does the magnitude of the electrostatic energy change when the size of the sample decreases?

3. What conditions are necessary to create the Coulomb blockade effect?

4. Describe the mechanism of current flow through a sample with a small electrical capacitance.

5. What is the difference between current flow during resonant tunneling and when the Coulomb blockade effect exists?
[1] Imry Y. Physics of mesoscopic systems / Y. Imry. In: Directions in Condensed Matter Physics, G. Grinstein and G. Mazenco (Eds.).– Singapore: World Scientific, 1986.– P. 101–163. 6, 35, 58

[2] Imry Y. Introduction to Mesoscopic Physics / Y. Imry.– New York Oxford: Oxford University Press, 2002. 6

[3] van Kampen N. G. In: Statistical Physics, Proceedings of the IUPAP International Conference, L. Pel and Szepflanszy (Eds.).– Amsterdam: North-Holland, 1976. 9

[4] Yanson I.K. Atlas of microcontact spectra of electron-phonon interaction in metals / I.K. Yanson, A.V. Khotkevich. – K.: Naukova Dumka, 1986.– 144 p. 17

[5] Omelyanchuk A. N. On the theory of nonlinear effects in the electrical conductivity of metal bridges / A. N. Omelyanchuk, I. O. Kulik, R. I. Shekhter // Letters to JETP.– 1977.– V. 25, N 10.– p 465–469. 17

[6] Landau L. D. Statistical Physics. Part 1 / L. D. Landau, E. M. Lifshitz. – M.: Nauka, GRFML, 1976.– 583 p. 33, 35

[7] Landau L. D. Quantum Mechanics. Non-relativistic Theory / L. D. Landau, E. M. Lifshitz. – M.: GIFML, 1963.– 702 p. 74, 89, 133, 166

[8] Aprikosov A. A. Fundamentals of the theory of metals / A. A. Aprikosov. – M.: Nauka, GRFML, 1987.– 520 p. 40, 45, 122

[9] Landauer R. Electrical resistance of disordered one-dimensionaled lattices / R. Landauer // Phil. Mag.– 1970.– V. 21.– P. 863. 47

[10] Aharonov Y. Significance of electromagnetic potentials in the quantum theory / Y. Aharonov, D. Bohm // Phys. Rev.– 1959.–V.115.–P. 485–491. 59, 141
[11] Chambers R. G. Shift of an electron interference pattern by enclosed magnetic flux / R.G. Chambers // Phys.Rev.Lett. - 1960. - V. 5. - P. 3-5. 62

[12] Little W. A. Observation of quantum periodicity in the transition temperature of a superconducting cylinder / W. A. Little, R. D. Parks // Phys. Rev. Lett. - 1962. - V. 9. - P. 9-12. 66

[13] London F. Superfluids. Volume 1 / F. London. - N.Y.: John Wiley and Sons, 1950. 66

[14] Altshuler B. L. Aharonov-Bohm effect in disordered wires days / B. L. Altshuler, A. G. Aronov, B. Z. Spivak // Letters to JETP. - 1981. - V. 33, N 2. - p. 101-103. 66

[15] Sharvin D. Yu. Quantization of magnetic flux in a cylindrical film of normal metal / D. Yu. Sharvin, Yu. V. Sharvin // Letters to JETP. - 1981. - V. 34, N 5. - p. 285-288. 66

[16] Hund F. Rechnungen über das magnetische Verhalten von kleinen Metallstücken bei tiefen Temperaturen / F. Hund // Annalen der Physik (Leipzig). - 1938. - V. 424, N 1-2. - P. 102-114. 66, 141

[17] Dingle R. B. Some Magnetic Properties of Metals. IV. Properties of Small Systems of Electrons / R. B. Dingle // Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences. - 1952. - V. 212, N 1108. - P. 47. 66, 141

[18] Büttiker M. Quantum oscillations in one-dimensional normal-metal rings / M. Büttiker, Y. Imry, M. Ya. Azbel // Phys. Rev. A. - 1984. - V. 30. - P. 1982-1989. 66

[19] Webb R. A. Observation of h/e Aharonov-Bohm Oscillations in Normal-Metal Rings / R. A. Webb, S. Washburn, C. P. Umbach, R. B. Laibowitz // Phys. Rev. Lett. - 1985. - V. 54, N 25. - P. 2696-2699. 67

[20] Chandrasekhar V. Observation of Aharonov-Bohm Electron Interference Effects with Periods h/e and h/2e in Individual Micron-Size, Normal-Metal
Rings / V. Chandrasekhar, M. J. Rook, S. Wind, D. E. Prober // Phys. Rev. Lett. – 1985. – V. 55, N 15. – P. 1610–1613. 67

[21] Griffith J. S. A free-electron theory of conjugated molecules. Part 1. Polycyclic hydrocarbons, // Trans. Faraday. Soc. – 1953. – V. 49. – P. 345–351. 77

[22] Ford C. J. B. Gated asymmetric rings as tunable electron interferometers / C. J. B. Ford, A. B. Fowler, J. M. Hong, C. M. Knoedler, S. I. Laux, J. J. Wainer, S. Washburn // Surf. Science. – 1990. – V. 229. – P. 307–317. 83

[23] Büttiker M. Generalized many-channel conductance formula with application to small rings / M. Büttiker, Y. Imry, R. Landauer, S. Pinhas // Phys. Rev. B. – 1985. – V. 31, N 10. – P. 6207–6215. 108

[24] Wharam D. A. One-dimensional transport and the quantization of the ballistic resistance / D. A. Wharam, T. J. Thornton, R. Newbury, M. Pepper, H. Ahmed, J. E. F. Frost, D. G. Hasko, D. C. Peacock, D. A. Ritchie, G. A. C. Jones // J. Phys. C.: Solid State Phys. – 1988. – V. 21, N 8. – P. L209–L214. 121

[25] van Wees B. J. Quantized Conductance of Point Contacts in a Two-Dimensional Electron Gas / B. J. van Wees, H. van Houten, C. W. J. Beenakker, J. G. Williamson, L. P. Kouwenhoven, D. van der Marel, C. T. Foxon // Phys. Rev. Lett. – 1988. – V. 60, N 9. – P. 848–850. 121

[26] Kulik I. O. Quantization of flux in normal metal / I. O. Kulik // Letters to JETP. – 1970. – V. 11, N 8. – p. 407–410. 124

[27] Büttiker M. Josephson behavior in small normal one-dimensional ring / M. Büttiker, Y. Imry, R. Landauer // Phys. Lett. A. – 1983. – V. 96, N 7. – P. 365–367. 124

[28] Levy L. P. Magnetization of mesoscopic cooper rings: evidence for persistent currents / L. P. Levy, G. Dolan, J. Dunsmuir, H. Bouchiat // Phys. Rev. Lett. – 1990. – V. 64, N 17. – P. 2074–2077. 124
[29] Chandrasekhar V. Magnetic response of a single, isolated gold loop / V. Chandrasekhar, R. A. Webb, M. J. Brady, M. B. Ketchen, W. J. Gallagher, A. Kleinsasser // Phys. Rev. Lett. – 1991. – V. 67, N 25. – P. 3578–3581. 124

[30] Mailly D. Experimental observation of persistent currents in a GaAs–AlGaAs single loop / D. Mailly, C. Chapelier, A. Benoit // Phys. Rev. Lett. – 1993. – V. 70, N 13. – P. 2020–2023. 124

[31] Kulik I. O. Kinetic phenomena and effects of charge discreteness in a granular medium / I. O. Kulik, R. I. Shekhter // JETP. – 1975. – V. 68, N 2. – p. 623–640. 161

[32] Averin D. V. Single electronics: a correlated transfer of single electrons and Cooper pairs in system of small tunnel junctions / D. V. Averin, K. K. Likharev // in Mesoscopic Phenomena in Solids, B. Altshuler, P. A. Lee, R. A. Webb (Eds.), Elsevier, Amsterdam. – 1991. – P. 173–272. 161

[33] Kastner M. A. The single-electron transistor / M. A. Kastner // Rev. Mod. Phys. – 1992. – V. 64, N 3. – P. 849–858. 161

[34] Giaever I. Superconductivity of Small Tin Particles Measured by Tunneled / I. Giaever, H. R. Zeller // Phys. Rev. Lett. – 1968. – V. 20, N 26. – P. 1504–1507. 161

[35] Lambe J. Charge-Quantization Studies Using a Tunnel Capacitor / J. Lambe, R. C. Jaclevic // Phys. Rev. Lett. – 1969. – V. 22, N 25. – P. 1371–1375. 161