Subfield: Modern Physics

Particles and Waves

Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons.

 Part A. Quantum particle in a box

Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by

$$
V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases}
\tag{1}
$$

While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls.
Context question:
A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$.
Context answer:
\boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$}


Extra Supplementary Reading Materials:

The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state.
Context question:
A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ).
Context answer:
\boxed{$E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$}


Context question:
A.3 Particle can undergo instantaneous transition from one state to another only by emitting or absorbing a photon of the corresponding energy difference. Find the wavelength $\lambda_{21}$ of the photon emitted during the transition of the particle from the first excited state $\left(E_{2}\right)$ to the ground state $\left(E_{1}\right)$.
Context answer:
\boxed{$\lambda_{21}=\frac{8 m c L^{2}}{3 h}$}


Extra Supplementary Reading Materials:

Part C. Bose-Einstein condensation

This part is not directly related to Parts A and B. Here, we will study the collective behaviour of bosonic particles. Bosons do not respect the Pauli exclusion principle, and-at low temperatures or high densitiesexperience a dramatic phenomenon known as the Bose-Einstein condensation (BEC). This is a phase transition to an intriguing collective quantum state: a large number of identical particles 'condense' into a single quantum state and start behaving as a single wave. The transition is typically reached by cooling a fixed number of particles below the critical temperature. In principle, it can also be induced by keeping the temperature fixed and driving the particle density past its critical value.

We begin by exploring the relation between the temperature and the particle density at the transition. As it turns out, estimates of their critical values can be deduced from a simple observation: Bose-Einstein condensation takes place when the de Broglie wavelength corresponding to the mean square speed of the particles is equal to the characteristic distance between the particles in a gas.
Context question:
C.1 Given a non-interacting gas of ${ }^{87} \mathrm{Rb}$ atoms in thermal equilibrium, write the expressions for their typical linear momentum $p$ and the typical de Broglie wavelength $\lambda_{\mathrm{dB}}$ as a function of atom's mass $m$, temperature $T$ and physical constants.
Context answer:
\boxed{$p=\sqrt{3 m k_{\mathrm{B}} T}$ , $\lambda_{\mathrm{dB}}=\frac{h}{\sqrt{3 m k_{\mathrm{B}} T}}$}


Context question:
C.2 Calculate the typical distance between the particles in a gas, $\ell$, as a function of particle density $n$. Hence deduce the critical temperature $T_{c}$ as a function of atom's mass, their density and physical constants.
Context answer:
\boxed{$\ell=n^{-1 / 3}$ , $T_{c}=\frac{h^{2} n^{2 / 3}}{3 m k_{\mathrm{B}}}$}


Extra Supplementary Reading Materials:

To realize BEC in the lab, the experimentalists have to cool gases to temperatures as low as $T_{c}=100 \mathrm{nK}$.
Context question:
C.3 What is the particle density of the Rb gas $n_{c}$ if the transition takes place at such= a temperature? For the sake of comparison, calculate also the 'ordinary' particle density $n_{0}$ of an ideal gas at the standard temperature and pressure (STP), i.e. $T_{0}=300 \mathrm{~K}$ and $p_{0}=10^{5} \mathrm{~Pa}$. How many times is the 'ordinary' gas denser? You may assume that the mass of the atoms is equal to 87 atomic mass units $\left(m_{\mathrm{amu}}\right)$.
Context answer:
Expression: $n_{c}=\frac{\left(3 \cdot 87 m_{\mathrm{amu}} k_{\mathrm{B}} T_{c}\right)^{3 / 2}}{h^{3}}$.

Numerical value: $n_{c} \approx 1.59 \cdot 10^{18} \mathrm{~m}^{-3}$.

Expression: $n_{0}=p /\left(k_{\mathrm{B}} T\right)$.

Numerical value: $n_{0} / n_{c} \approx 1.5 \cdot 10^{7}$.


Extra Supplementary Reading Materials:

Part D. Three-beam optical lattices

The first Bose-Einstein condensates were produced back in 1995, and since then the experimental work has branched out in diverse directions. In this part, you will investigate one particularly fruitful idea to load the condensate into spatially periodic potentials created by interfering a number of coherent laser beams. Due to the periodic nature of the resulting interference patterns, they are referred to as optical lattices. The potential energy $V(\vec{r})$ of an atom moving in an optical lattice is proportional to the local intensity of the light, and in your calculations you may assume that

$$
V(\vec{r})=-\alpha\left\langle|\vec{E}(\vec{r}, t)|^{2}\right\rangle
\tag{2}
$$

Here, $\alpha$ is a positive constant, and the angle brackets indicate time-averaging which eliminates the rapidly oscillating terms. The electric field produced by the $i$-th laser is described by

$$
\vec{E}_{i}=E_{0, i} \vec{\varepsilon}_{i} \cos \left(\vec{k}_{i} \cdot \vec{r}-\omega t\right)
\tag{3}
$$

with the amplitude $E_{0, i}$, the wave vector $\vec{k}_{i}$, and the unit-length polarization vector $\vec{\varepsilon}_{i}$.

<img_4443>

(a)

<img_4554>

(b)

<img_4345>

(c)

Figure 2. (a) Three-beam optical lattice: three plane waves with wave vectors $\vec{k}_{1,2,3}$ intersect and interfere in the area indicated by the grey circle. (b) Symmetries of a regular hexagon: solid and dashed lines show two sets of symmetry axes. (c) Saddle point: a point on a surface where the slopes in orthogonal directions are all zero, but which is not a local extremum of the plotted function. Travelling along the trajectory marked by the full line one encounters an apparent minimum. Additional analysis of the perpendicular direction (dashed line) is needed to distinguish a true minimum from a saddle point (shown).

Your task is to study triangular optical lattices that are produced by interfering three coherent laser beams of equal intensity. A typical setup is shown in Fig. 2a. Here, all three beams are polarized in the $z$ direction, propagate in the $x y$ plane and intersect at equal angles of $120^{\circ}$. Choose the direction of the $x$ axis parallel to the wave vector $\vec{k}_{1}$.

Question: D.1 Using Eqs. 2 and 3 obtain the expression for the potential energy $V(\vec{r})$ as a function of $\vec{r}=(x, y)$ in the plane of the beams.

Hint: the result can be neatly expressed as a constant term plus a sum of three cosine functions of arguments $\vec{b}_{i} \cdot \vec{r}$. Please write your result in this form and identify the vectors $\vec{b}_{i}$.

Solution:
We sum the three electric fields ( $z$ components)

$$
E(\vec{r}, t)=E_{0} \sum_{i=1}^{3} \cos \left(\vec{k}_{i} \cdot \vec{r}-\omega t\right)
\tag{4}
$$

and square the result

$$
\begin{aligned}
E^{2}(\vec{r}, t) & =E_{0}^{2} \sum_{i=1}^{3} \sum_{j=1}^{3} \cos \left(\vec{k}_{i} \cdot \vec{r}-\omega t\right) \cos \left(\vec{k}_{j} \cdot \vec{r}-\omega t\right) \\
& =\frac{E_{0}^{2}}{2} \sum_{i=1}^{3} \sum_{j=1}^{3}\left\{\cos \left[\left(\vec{k}_{i}-\vec{k}_{j}\right) \cdot \vec{r}\right]+\cos \left[\left(\vec{k}_{i}+\vec{k}_{j}\right) \cdot \vec{r}-2 \omega t\right]\right\} .
\end{aligned}
\tag{5}
$$

Time averaging gives

$$
\left\langle E^{2}(\vec{r}, t)\right\rangle=\frac{E_{0}^{2}}{2} \sum_{i=1}^{3} \sum_{j=1}^{3} \cos \left[\left(\vec{k}_{i}-\vec{k}_{j}\right) \cdot \vec{r}\right]
\tag{6}
$$

we analyse the 9 terms and simplify to

$$
\left\langle E^{2}(\vec{r}, t)\right\rangle=E_{0}^{2}\left(\frac{3}{2}+\sum_{j=1}^{3} \cos \vec{b}_{j} \cdot \vec{r}\right)
\tag{7}
$$

Here $\vec{b}_{1,2,3}=\left(\vec{k}_{2}-\vec{k}_{3}\right),\left(\vec{k}_{3}-\vec{k}_{1}\right),\left(\vec{k}_{1}-\vec{k}_{2}\right)$ or in terms of the Levi-Civita symbol, $\vec{b}_{k}=\varepsilon_{i j k}\left(\vec{k}_{i}-\vec{k}_{j}\right)$. Incidentally, they are known as the reciprocal lattice vectors.

Final answer: $V(\vec{r})=-\alpha E_{0}^{2}\left(\frac{3}{2}+\sum_{j=1}^{3} \cos \vec{b}_{j} \cdot \vec{r}\right)$,$\vec{b}_{1}=\vec{k}_{2}-\vec{k}_{3}, \quad \vec{b}_{2}=\vec{k}_{3}-\vec{k}_{1}, \quad \vec{b}_{3}=\vec{k}_{1}-\vec{k}_{2}$