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Classically, one can easily show the following relation using a straightforward canonical ensemble computation for a non-interacting gas:

$$\frac{PV}{\langle E\rangle}=\frac{2}{3}$$

Now, apparently the same relation is valid also for a quantum ideal gas. Trying to reproduce it I obtained the grand canonical (bosonic) partition function:

$$Z_\alpha=\prod_{i}\left(\frac{1}{1-e^{-\beta(\alpha_i\epsilon_i-\mu)}}\right)$$

where $\alpha_i$ are auxiliary parameters I introduced for convenience (normally one would set $\alpha_i=1$).

Per definition we now have:

$$PV=\left(\frac{1}{\beta}\ln Z_\alpha\right)|_{\alpha_i=1}$$

And

$$\langle E\rangle=\left(-\frac{1}{\beta}\sum_j \frac{\partial}{\partial\alpha_j}\ln Z_\alpha\right)|_{\alpha_i=1}$$

For both expressions I do not see how one should proceed with the evaluation. Any suggestions? The sums appear to be too difficult to solve (especially since it is not clear which values $\epsilon_i$ actually take on). Also, it might very well be that I did something wrong on the way there. A solution would be much appreciated.

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There's a proof for Fermions in "Introduction to statistical mechanics" by Silivo Salinas. The proof for Bosons is more or less identical and I give it below.

From the Grand Canonical Partition Function you gave, the Grand Canonical Potential, $\Phi = -pV$, is $$ \Phi = -\tfrac{1}{\beta} \ln Z = \tfrac{1}{\beta} \sum_i \ln (1 - e^{-\beta(\epsilon_i-\mu)}) $$ The energy levels $\epsilon_i$ depend upon the container, but the level spacing is very dense so we can replace the sum by an integral over the density of levels, $\rho(\epsilon)$. In three dimensions the density of levels is $\rho = A \epsilon^{1/2}$ where $A$ is some constant. Thus we obtain $$ pV = \tfrac{A}{\beta} \int_0^\infty \epsilon^{1/2} \ln (1 - e^{-\beta(\epsilon-\mu)})d\epsilon $$ Integration by parts (integrating the $\epsilon^{1/2}$ and differentiating the $\ln(\ldots)$) gives $$ pV = A\int_0^\infty \frac{2}{3}\epsilon^{3/2}f(\epsilon,\mu,\beta)d \epsilon $$ where $f(\epsilon,\mu,\beta)$ is the Bose-Einstein distribution function. The integrand is just $2/3$ times the density of levels, times the B-E distribution (which is the level occupation number), times $\epsilon$ (which is the level energy), and is therefore equal to $2/3$ the total energy: $$ pV = \frac{2}{3} \int_0^\infty \rho(\epsilon) f(\epsilon,\mu,\beta) \epsilon d \epsilon = \frac{2}{3}\langle E \rangle $$ The proof for Fermions works in exactly the same way.

Edit: why density of levels is $\propto \epsilon^{1/2}$:

This is relatively straightforward for a cubic box, and for a macroscopic box of any shape you can always imagine it of made up of small cubes where the side length of each cube is still much greater than the characteristic de-Broglie wavelength of the particles, so this is good enough.

For a cubic box of side-length $L$ with wavefunctions vanishing on the box boundary the quantum states are: $$ \epsilon = \frac{\hbar^2 |k|^2}{2m} \quad \mathrm{with}\ \vec{k}= \tfrac{2\pi}{L} (n_x,n_y,n_z) $$ where $n_x,n_y,n_z$ are integers. In $\vec{k}$-space the possible states form a cubic lattice with the 'volume' of one cube equal to $(2\pi/L)^3 = (2\pi)^3/V$. States with equal energy lie on a sphere centred at $\vec{k}=0$. Therefore the number of states with energies between $\epsilon$ and $\epsilon + \delta \epsilon$ is approximately $V/(2\pi)^3$ times the volume of the spherical shell with $\sqrt{2m\epsilon}/\hbar < |k| < \sqrt{2m(\epsilon + \delta \epsilon)}/\hbar$. Taking the limit for small $\delta \epsilon$ shows that the density of levels is given by $$ \rho(\epsilon) = \frac{(2m)^{3/2} V}{4\pi^2\hbar^3} \epsilon^{1/2} $$ I.e. $\rho \propto \epsilon^{1/2}$ as required.

Or the same argument more briefly: (number of levels with energy $< \epsilon) \propto$ (volume of sphere in $\vec{k}$-space) $\propto k^3 \propto \epsilon^{3/2}$. Density of levels = $\frac{d}{d\epsilon}$ (number of levels) $\propto \frac{d}{d\epsilon} \epsilon^{3/2} \propto \epsilon^{1/2}$.

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  • $\begingroup$ This is correct for non-relativistic particles. For relativistic particles, you have $pV = \frac{1}{3} \langle E \rangle$ $\endgroup$
    – Trimok
    Jan 23, 2014 at 11:14
  • $\begingroup$ I thought about that, but I was not sure how to get the state density. Why is it proportional to $\epsilon^{1/2}$ in three dimensions? What is the general idea here? $\endgroup$
    – Kagaratsch
    Jan 24, 2014 at 0:13
  • $\begingroup$ I mean, looking at microcanonical ensemble and $\epsilon(p,q)=\frac{p^2}{2m}$ to get the density $D\propto \sqrt{\epsilon}$ is not an option, since that is classical approach, and we want to get a quantum result, right? In using this we would just fall back to the classical case... $\endgroup$
    – Kagaratsch
    Jan 24, 2014 at 0:32
  • $\begingroup$ I've added an explanation of why the density of levels is $\propto \sqrt{\epsilon}$. Note that @Trimok is talking about the ultra-relativistic case, i.e. when $p\gg mc$ (here $p$ is typical particle momentum, not pressure). If $p \sim mc$, which is still relativistic, the answer is more complicated. $\endgroup$
    – Mark A
    Jan 24, 2014 at 1:08

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