# Particle in a one-dimensional box

### Question

A point particle of mass $m$ moves in the region $0 \le x \le l$ and is reflected elastically at the walls at $x=0$ and $x=l$.

1. Calculate the volume $\Gamma_0(E)$ of the classical phase space with an energy smaller than $E$.
2. Assume that a particle initially has an energy $E_0$. Demonstrate that the phase-space volume $\Gamma_0(E)$ of this particle remains constant when the wall at $x=l$ is moved slowly (adiabatic invariance).
3. Calculate the number of states $\Omega_0(E)$ with energy smaller than $E$ for the corresponding quantum mechanical system. Compare the result with $\Gamma_0(E)$ for large $E$.

### Solution

1. The classical phase-space volume for this situation can be found from:

\Gamma_0 = \int_{0}^{l}dx\int_{-\sqrt{2mE}}^{\sqrt{2mE}}dp = 2l \sqrt{2mE}
\label{eq:gamma0}

since the energy of the particle is given by $E=\frac{p^2}{2m}$.

2. Since the expansion is adiabatic, $\delta Q=0$, where $\delta$ denotes an inexact differential.

Thus, from the second law of thermodynamics, $$\frac{\delta Q}{Τ}=\delta S=0$$

We know entropy is given by $S=k_B\ln \Omega$. Therefore $$\Delta S=k_B\ln \Omega_0 – k_B \ln \Omega_1 = 0$$Which necessitates $$\Omega_0=\Omega_1$$

And since $\Gamma = \Omega$, $\Gamma_0(E)$ remains constant.

3. The energies which correspond with each of the permitted wavenumbers for a particle in a one dimensional box may be written as $$E_n=\frac{h^2 n^2}{8ml^2}$$

The number of states up to energy $E$ may be found from summing over all of the accessible states of the system like so: $$\Omega_0(E) = \sum_{E_0}^{E}1 \stackrel {\text{large E}}{\approx} \sqrt{\frac{8ml^2}{h^2}E}$$ $$=\frac{2l \sqrt{2mE}}{h}=\frac{\Gamma_0(E)}{h}$$

So the difference between the classical and the quantum treatment is a factor of $h^{-1}$.