Particle in a one-dimensional box

Question

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

  1. Calculate the volume of the classical phase space with an energy smaller than .
  2. 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).
  3. 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 .