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 \(x=l\).
 Calculate the volume \(\Gamma_0(E)\) of the classical phase space with an energy smaller than \(E\).
 Assume that a particle initially has an energy \(E_0\). Demonstrate that the phasespace volume \(\Gamma_0(E)\) of this particle remains constant when the wall at \(x=l\) is moved slowly (adiabatic invariance).
 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

The classical phasespace 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 \(E=\frac{p^2}{2m}\).

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.

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 \(h^{1}\).