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PHY452H1S Basic Statistical Mechanics. Lecture 7: Ideal gas and SHO phase space volume calculations. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on February 5, 2013

Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

Review. Classical phase space calculation

\begin{aligned}E_{\mathrm{ideal}} = \sum_i \frac{\mathbf{p}_i^2}{2 m}\end{aligned} \hspace{\stretch{1}}(1.2.1)

From this we calculated $\gamma(E)$, and

\begin{aligned}\frac{d\gamma(E)}{dE} = \Omega_{\mathrm{classical}}(E)\end{aligned} \hspace{\stretch{1}}(1.2.2)

Fudging with a requirement that $\Delta x \Delta p \sim h$, we corrected this as

\begin{aligned}\Omega_{\mathrm{quantum}}(E) = \frac{\Omega_{\mathrm{classical}}(E)}{N! h^{3N}}\end{aligned} \hspace{\stretch{1}}(1.2.3)

Now let’s do the quantum calculation.

Quantum calculation

Recall that for the solutions of the Quantum free particle in a box, as in (Fig 1), our solutions are

Fig 1: 1D Quantum free particle in a box

\begin{aligned}\Psi_n(x) = \sqrt{\frac{2}{L}} \sin\left( \frac{ n \pi x}{L} \right),\end{aligned} \hspace{\stretch{1}}(1.2.4)

where $n = 1, 2, \cdots$, and

\begin{aligned}\epsilon_n = \frac{n^2 h^2}{8 m L^2}\end{aligned} \hspace{\stretch{1}}(1.2.5)

.

In three dimensions, with $n_i = 1, 2, \cdots$ we have

\begin{aligned}\Psi_{n_1, n_2, n_3}(x, y, z) = \left( \frac{2}{L} \right)^{3/2} \sin\left( \frac{ n_1 \pi x}{L} \right)\sin\left( \frac{ n_2 \pi x}{L} \right)\sin\left( \frac{ n_3 \pi x}{L} \right)\end{aligned} \hspace{\stretch{1}}(1.2.6)

and

\begin{aligned}\epsilon_{n_1, n_2, n_3} = \frac{h^2}{8 m L^2} \left( n_1^2 + n_2^2 + n_3^2 \right)\end{aligned} \hspace{\stretch{1}}(1.2.7)

\begin{aligned}\gamma^{3d}_{\mathrm{classical}}(E) = \underbrace{V}_{L^3}\int d^3 p \theta \left( E - \frac{\mathbf{p}^2}{2m} \right)= V \frac{4 \pi}{3} (2 m E)^{3/2}\end{aligned} \hspace{\stretch{1}}(1.2.8)

so that

\begin{aligned}\gamma^{3d}_{\mathrm{corrected}}(E) = V \frac{4 \pi}{3} \frac{(2 m E)^{3/2}}{h^3}\end{aligned} \hspace{\stretch{1}}(1.2.9)

\begin{aligned}\gamma^{3d}_{\mathrm{quantum}}(E) = \sum_{n_1, n_2, n_3} \Theta(E - \epsilon_{n_1, n_2, n_3} ).\end{aligned} \hspace{\stretch{1}}(1.2.10)

How do the multiplicities scale by energy? We’ll have expect something like (Fig 2).

Provided the energies $E \gg 3h^2/(8 m L)$ are large enough, we can approximate the sum with

\begin{aligned}\sum_{n_1, n_2, n_3} \sim \int_0^\infty dn_1 dn_2 dn_3\end{aligned} \hspace{\stretch{1}}(1.2.11)

So

\begin{aligned}\gamma^{3d}_{\mathrm{quantum}} \left( E \gg \frac{h^2}{8 m L^2} \right) \approx\int_0^\infty dn_1 dn_2 dn_3 \Theta \left( E - \frac{h^2}{8 m L^2} \left( n_1^2 + n_2^2 + n_3^2 \right) \right)=\frac{1}{{8}}\frac{4 \pi}{3} \left( \frac{8 m L^2 E}{h^2} \right)^{3/2}=L^3\frac{4 \pi}{3} \frac{\left( 2 m E \right)^{3/2}}{h^3}\end{aligned} \hspace{\stretch{1}}(1.2.12)

Harmonic oscillator in 1D.

Our phase space is of the form (Fig 3).

Fig 3: 1D classical SHO phase space

Where the number of states in this classical picture are found with:

\begin{aligned}\gamma^{\mathrm{classical}}(E) = \int dx dp \theta\left( E - \left( \frac{1}{{2}} k x^2 + \frac{1}{{2m }} p^2 \right) \right).\end{aligned} \hspace{\stretch{1}}(1.2.13)

Rescale

\begin{aligned}\tilde{x} = x \sqrt{ \frac{k}{2}}\end{aligned} \hspace{\stretch{1}}(1.0.14a)

\begin{aligned}\tilde{p} = \frac{p}{\sqrt{2m}}\end{aligned} \hspace{\stretch{1}}(1.0.14b)

so that we have

\begin{aligned}\gamma^{\mathrm{classical}}(E) = \int d\tilde{x} d \tilde{p} \sqrt{\frac{2 \times 2 m}{k}} \theta\left( E - \tilde{x}^2 - \tilde{p}^2 \right)=2 \sqrt{\frac{m}{k}} \pi E= 2 \pi \sqrt{\frac{m}{k}} E.\end{aligned} \hspace{\stretch{1}}(1.0.15)

\begin{aligned}\gamma^{\mathrm{SHO}}_{\mathrm{corrected}}(E) = 2 \pi \sqrt{\frac{m}{k}} \frac{E}{h}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

We have for the energy

\begin{aligned}E_n^{\mathrm{SHO}} = \left( n + \frac{1}{{2}} \right) \hbar \omega\end{aligned} \hspace{\stretch{1}}(1.0.17a)

\begin{aligned}\omega = \sqrt{\frac{k}{m}}\end{aligned} \hspace{\stretch{1}}(1.0.17b)

\begin{aligned}\hbar = \frac{h}{2 \pi}\end{aligned} \hspace{\stretch{1}}(1.0.17c)

graphing the counts (Fig 4), we again have stepping as a function of energy, but no multiplicities this time

Fig 4: 1D quantum SHO states per energy level

\begin{aligned}\gamma_{\mathrm{quantum}}(E) = \sum_{n = 0}^\infty \Theta\left( E - \left( n + \frac{1}{{2}} \hbar \omega \right) \right)\end{aligned} \hspace{\stretch{1}}(1.0.18)

we make the continuous approximation for the summation again, and throwing away the zero point energy, we have

\begin{aligned}\gamma_{\mathrm{quantum}}(E \gg \hbar \omega) \approx\int_{0}^\infty dn \Theta\left( E - n \hbar \omega \right)= 2 \pi \frac{E}{h} \sqrt{\frac{m}{k}}\end{aligned} \hspace{\stretch{1}}(1.0.19)

Why $N!$?

We have a problem with out counting here. Consider some particles in a box as in (Fig 5).

Fig 5: Three particles in a box

1. particle $1$ at $\mathbf{x}_1$
2. particle $2$ at $\mathbf{x}_2$
3. particle $3$ at $\mathbf{x}_3$

or

1. particle $1$ at $\mathbf{x}_2$
2. particle $2$ at $\mathbf{x}_3$
3. particle $3$ at $\mathbf{x}_1$

This is fine in the classical picture, but in the quantum picture with an assumption of indistinguishability, no two particles (say electrons) cannot be labelled in this fashion.

\begin{aligned}\underbrace{S_{\mathrm{ideal}}^{(\mathrm{E}, \mathrm{N}, \mathrm{V})}}_{\text{Statistical entropy}}= k_{\mathrm{B}} \ln \left( \frac{\Omega_{\mathrm{classical}}}{h^{3N}} \right)\underbrace{\approx}_{N \gg 1} k_{\mathrm{B}} \left( N \ln V + \frac{3 N}{2} \ln \left( \frac{4 \pi m E }{3 N h^2} \right) + \frac{3 N}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.20)

Suppose we double the volume as in (Fig 6), then our total entropy for the bigger system would be

Fig 6: Gibbs volume doubling argument. Two identical systems allowed to mix

\begin{aligned}S_{\mathrm{total}}^{(\mathrm{E}, \mathrm{N}, \mathrm{V})}= k_{\mathrm{B}} \ln \left( \frac{\Omega_{\mathrm{classical}}}{h^{3N}} \right)\approx k_{\mathrm{B}} \left( (2 N) \ln (2 V) + \frac{3 (2 N)}{2} \ln \left( \frac{4 \pi m (2 E) }{2 ( 2 N) h^2} \right) + \frac{3 (2 N)}{2} \right).\end{aligned} \hspace{\stretch{1}}(1.0.21)

We have

\begin{aligned}S_{\mathrm{total}} = S_1 + S_2 + k_{\mathrm{B}} (2 N) \ln 2= S_1 + S_2 + k_{\mathrm{B}} \ln 2^{2 N}.\end{aligned} \hspace{\stretch{1}}(1.0.22)

This is telling us that each particle could be in either the left or the right side, but we know that this uncertainty shouldn’t be in the final answer. We must drop this $k_{\mathrm{B}}$ term.

So, if we assume that these particles are identical, and divide $\Omega$ by $N!$, then we find

\begin{aligned}S_{\mathrm{ideal}} = k_{\mathrm{B}} \left( N \ln \frac{V}{N} + \frac{3 N}{2} \ln \left( \frac{4 \pi m E }{3 N h^2} \right) + \frac{5 N}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.23)