### Question

A point particle of mass moves in the region and is reflected elastically at the walls at and .

- Calculate the volume of the classical phase space with an energy smaller than .
- Assume that a particle initially has an energy . Demonstrate that the phase-space volume of this particle remains constant when the wall at is moved slowly (adiabatic invariance).
- Calculate the number of states with energy smaller than for the corresponding quantum mechanical system. Compare the result with for large .

### Solution

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

\begin{equation}

\Gamma_0 = \int_{0}^{l}dx\int_{-\sqrt{2mE}}^{\sqrt{2mE}}dp = 2l \sqrt{2mE}

\label{eq:gamma0}

\end{equation}

since the energy of the particle is given by .

2. Since the expansion is adiabatic, , where denotes an inexact differential.

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

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

And since , 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 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 .