The classical and quantum mechanical phase volume

2.8 The classical and quantum mechanical phase volume

We investigate as an example the Hamiltonian of a free particle in \(s\)-dimensional space: (\(s\) is the number of degrees of freedom)

\begin{equation*} H = \sum_{j=1}^{s} \frac{p_j^2}{2m} \end{equation*}

(2.50)

The classical phase volume, i.e. the classical partition function, is

\begin{equation*} \Phi_{cl}(\epsilon) = \int_{V_s} d^s q \int_{H \leq \epsilon} d^s p = V_s K_s(\sqrt{2 m \epsilon}) = V_s \frac{\pi^{\frac{s}{2}}}{\left(\frac{s}{2}\right)!} (2 m \epsilon)^{\frac{s}{2}} \qquad . \label{psi_cl} \end{equation*}

(2.51)

In quantum mechanics we get for periodic boundary conditions the solution of the Schrödinger equation

\begin{equation*} \psi_k(q) = a \exp\left( i \sum_{j=1}^s k_j q_j\right) \qquad , \end{equation*}

(2.52)

with \(k_j = \frac{2\pi n_j}{l}\), respectively \(p_j = \hbar k_j = \frac{h n_j}{l}\), \(n_j = 0, \pm 1, \pm 2, ...\)
In an \(s\)-dimensional momentum space the Eigenvalues have a lattice distance of \(h/l\).
The number of momentum Eigenvalues with \(0 \leq \epsilon_k \leq \epsilon\) equals the number of Eigenvalues within the sphere \(K_s(\sqrt{2 m \epsilon})\). This number we can determine just by counting. As an approximation we substitute the counting by calculating the number of volume elements \(\left( \frac{h}{l}\right)^s\) within the sphere \(K_s(\sqrt{2 m \epsilon})\):

\begin{equation*} \Phi_{qm}(\epsilon) = \frac{V_s}{h^s} K_s(\sqrt{2 m \epsilon}) = \frac{V_s}{h^s} \frac{\pi^{\frac{s}{2}}}{\left(\frac{s}{2}\right)!} (2 m \epsilon)^{\frac{s}{2}} \qquad . \label{psi_qm} \end{equation*}

(2.53)

By comparison of Eq. (2.51) and Eq. (2.53) we find

\begin{equation*} \Phi_{qm}(\epsilon) = \frac{\Phi_{cl}(\epsilon)}{h^s} \qquad . \end{equation*}

(2.54)

For a many particle system we have to add some factors, since the quantum mechanical particles are not distinguishable

\begin{equation*} \Phi_{qm}(\epsilon,V_s,N) = \frac{\Phi_{cl}(\epsilon, V_s,N)}{N! h^{Ns}} \qquad . \end{equation*}

(2.55)

These approximations do not hold e.g. for free electrons in general! (Only e.g. for the conduction band electrons of a non degenerated semiconductor).
As an example we calculate the case \(s=3\).
First we define

\begin{equation*} v := \frac{V}{N}\;, \quad \quad \epsilon := \frac{E}{N} \; \mbox{, and} \quad \lambda := \frac{h}{p} = \frac{h}{\sqrt{2 m \epsilon}} \quad . \end{equation*}

(2.56)

\(\lambda\) is the de Broglie wave length, \(\lambda^3\) is the uncertainty of a particle in volume.
For \(v \gg \lambda^3\) we are allowed to neglect the exact quantum mechanical character of the particles and use the above approximations.
Including real numbers for e.g. Helium gas (\(N = 10^{20}\), \(T = 300^o\)K, \(m = 4\), \(m_p = 4 * 1.67* 10^{-24} g\)), we find

\begin{equation*} E = \frac{3}{2} N k T \approx \frac{1}{27} 10^{20} \mbox{eV}\; ,\quad v = \frac{24*10^3}{6*10^{23}} \mbox{cm}^3 \; \mbox{, and} \quad \lambda = \frac{h}{\sqrt{2 m \frac{1}{27}}} = 2.3*10^{-9} \mbox{cm}\; \mbox{, i.e.}\quad \frac{v}{\lambda^3} \approx \frac{1}{3} 10^7 \quad . \end{equation*}

(2.57)

So for a classical gas under normal conditions the premise is fulfilled easily.