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.

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.

Solution:
At temperature $T$, the average kinetic energy of translational motion is $\frac{3}{2} k_{\mathrm{B}} T$. Equating this result to $p^{2} /(2 m)$, we obtain typical momentum $p=\sqrt{3 m k_{\mathrm{B}} T}$ and the de Broglie wavelength

$$
\lambda_{\mathrm{dB}}=\frac{h}{p}=\frac{h}{\sqrt{3 m k_{\mathrm{B}} T}}
$$

Final answer: $p=\sqrt{3 m k_{\mathrm{B}} T}$ , $\lambda_{\mathrm{dB}}=\frac{h}{\sqrt{3 m k_{\mathrm{B}} T}}$