While studying quantum gases (fermions, bosons), equation of state written were $PV = k_B T Z_{gr}$, where $Z_{gr}$ is the partition function of grand canonical ensemble. $P$ and $V$ are pressure and volume, respectively. How can we derive this equation of state? Is it same equation that we use for ideal gases ($PV=NRT$) or is it different?

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    $\begingroup$ Did you find an expression for $Z_{gr}$? $\endgroup$ – WarreG Jan 12 '19 at 10:19

When you got the expression of the grand canonical partition function, which depends on whether you are studying bosons or fermions, you can use the link with the grand potential $ \Phi = -k_BT \ln (Z_{gr})$.

From this you can use the relation $$ P = -\frac{\partial \Phi}{\partial V},$$ which will give you the pressure in function of the partition function. From here it should be little calculation to get the form $$PV = k_BTZ_{gr}.$$

  • $\begingroup$ Are you sure it's $F$? I think it's the grand canonical potential. $\endgroup$ – FGSUZ Jan 12 '19 at 11:52
  • $\begingroup$ Yes,I also found a source where that is the grand canonical potential. But it doesn't really matter because it differs by a factor which does not depend on $V$. So when taking the partial derivative to $V$ it vanishes. I could be wrong but in my class it was 'derived' like that. $\endgroup$ – WarreG Jan 12 '19 at 11:56
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    $\begingroup$ It's true that it generates the same formula, but I think that saying "it doesn't matter" is a bit bold. It can lead to confusion, and other magnitudes might differ. I'm pretty sure that $\Phi=-k_BT\ln(\mathcal{Z}_{gc})$. See Callen, for example. $\endgroup$ – FGSUZ Jan 12 '19 at 12:22
  • $\begingroup$ I've edited it. I wonder, does the same relation $P = - \frac{\partial \Phi}{\partial V}$ hold? I'm starting to doubt now. $\endgroup$ – WarreG Jan 12 '19 at 13:25
  • $\begingroup$ $\Phi=U-TS-\mu N = F - \mu N$, so yes, it does. $\endgroup$ – FGSUZ Jan 12 '19 at 13:33

When dealing with quantum gases you have to apply a correction to your result from the ideal gas!

Let us take a non-relativistic ideal gas, in which each particle has an energy of

$$ \epsilon=\frac{p^2}{2m}. $$

The ideal gas law follows from

$$ N(T,V,\mu)=\frac{1}{\beta}\frac{\partial\ln Y}{\partial\mu}, $$

so we need an expression for $\ln Y$. For bosons, we can write:

$\begin{align} Y(T,V,\mu)&=\sum_r\exp\left(-\beta\left(E_r\left(V,N_r\right)-\mu N_r\right)\right)\\ &=\prod_i\sum_{n_{p_i}}\exp(-\beta(\epsilon_{p_i}-\mu)n_{p_i})\\ &=\prod_p\frac{1}{1-\exp(-\beta(\epsilon_p-\mu))}\\ \Rightarrow\ln Y&=-\sum_p\ln(1-\exp(-\beta(\epsilon_p-\mu))) \end{align}$

A similar calculation yields $$\ln Y=2\sum_p\ln(1+\exp(-\beta(\epsilon_p-\mu)))$$ for fermions.

To find the ideal gas law and its quantum mechanical corrections we need to do a series expansion of those expressions:

$$ \ln Y = (2s+1)\sum_p\left(\exp(-\beta(\epsilon_p-\mu))\pm\frac{1}{2}\exp(-2\beta(\epsilon_p-\mu))+...\right) $$

with $s=0$ for bosons and $s=1/2$ for fermions.

The possible momentum states are very dense, therefore we can treat the sum as an integral, which yields:

$\begin{align} \ln Y &= (2s+1)\frac{V}{(2\pi\hbar)^3}\int\!\mathrm{d}^3p\,\exp(\beta\mu)\left(\exp\left(-\frac{p^2}{2mk_BT}\right)\pm\frac{1}{2}\exp(\beta\mu)\exp\left(-\frac{p^2}{mk_BT}\right)+...\right)\\ &=(2s+1)\frac{V}{(2\pi\hbar)^3}\exp(\beta\mu)\frac{1}{2}\sqrt{\pi}\sqrt{mk_BT}\left(2\sqrt{2}+\exp{\beta\mu}+...\right)\\ &:=(2s+1)\frac{V}{\lambda^3}\left(\exp(\beta\mu)\pm2^{-5/2}\exp(2\beta\mu)+...\right) \end{align}$

in the last step I just summed together some constants, to get the thermal wavelength $\lambda$, as mentioned in another answer.

Now we use $N(T,V,\mu)=\frac{1}{\beta}\frac{\partial\ln Y}{\partial\mu}$. In the zeroth-order approximation we find:

$$ N(T,V,\mu)=\frac{1}{\beta}\frac{\partial\ln Y}{\partial\mu}=\ln Y = \frac{PV}{k_BT}, $$

the ideal gas law. For higher orders we find certain corrections, such that

$$ \ln Y = N\mp\frac{2s+1}{2^{5/2}}\frac{V}{\lambda^3}\exp(2\beta\mu)+... $$


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