## 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: \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 $$E=\frac{p^2}{2m}$$.

2. Since the expansion is adiabatic, $$\delta Q=0$$, where $$\delta$$ denotes an inexact differential. Inexact differentials are used in thermodynamics to express the path dependence of the 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 $$h^{-1}$$.