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)

$H = \sum_{j = 1}^{s} \frac{p_{j}^{2}}{2 m}$

(2.50)

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

$Φ_{c l} (ϵ) = \int_{V_{s}} d^{s} q \int_{H \leq ϵ} d^{s} p = V_{s} K_{s} (\sqrt{2 m ϵ}) = V_{s} \frac{π^{\frac{s}{2}}}{(\frac{s}{2})!} {(2 m ϵ)}^{\frac{s}{2}} .$

(2.51)

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

$ψ_{k} (q) = a exp (i \sum_{j = 1}^{s} k_{j} q_{j}),$

(2.52)

with $k_{j} = \frac{2 π n_{j}}{l}$ , respectively $p_{j} = ℏ k_{j} = \frac{h n_{j}}{l}$ , $n_{j} = 0, \pm 1, \pm 2, . . .$
In an $s$ -dimensional momentum space the Eigenvalues have a lattice distance of $h ∕ l$ .
The number of momentum Eigenvalues with $0 \leq ϵ_{k} \leq ϵ$ equals the number of Eigenvalues within the sphere $K_{s} (\sqrt{2 m ϵ})$ . This number we can determine just by counting. As an approximation we substitute the counting by calculating the number of volume elements ${(\frac{h}{l})}^{s}$ within the sphere $K_{s} (\sqrt{2 m ϵ})$ :

$Φ_{q m} (ϵ) = \frac{V_{s}}{h^{s}} K_{s} (\sqrt{2 m ϵ}) = \frac{V_{s}}{h^{s}} \frac{π^{\frac{s}{2}}}{(\frac{s}{2})!} {(2 m ϵ)}^{\frac{s}{2}} .$

(2.53)

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

$Φ_{q m} (ϵ) = \frac{Φ_{c l} (ϵ)}{h^{s}} .$

(2.54)

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

$Φ_{q m} (ϵ, V_{s}, N) = \frac{Φ_{c l} (ϵ, V_{s}, N)}{N! h^{N s}} .$

(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

$v : = \frac{V}{N}, ϵ : = \frac{E}{N}, and λ : = \frac{h}{p} = \frac{h}{\sqrt{2 m ϵ}} .$

(2.56)

$λ$ is the de Broglie wave length, $λ^{3}$ is the uncertainty of a particle in volume.
For $v ≫ λ^{3}$ 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 ( $N = 1 0^{20}$ , $T = 30 0^{o}$ K, $m = 4$ , $m_{p} = 4 * 1.67 * 1 0^{- 24} g$ ), we find

$E = \frac{3}{2} N k T \approx \frac{1}{27} 1 0^{20} eV, v = \frac{24 * 1 0^{3}}{6 * 1 0^{23}} {cm}^{3}, and λ = \frac{h}{\sqrt{2 m \frac{1}{27}}} = 2.3 * 1 0^{- 9} cm, i.e. \frac{v}{λ^{3}} \approx \frac{1}{3} 1 0^{7} .$

(2.57)

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

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