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?

  • 2
    $\begingroup$ Did you find an expression for $Z_{gr}$? $\endgroup$
    – WarreG
    Jan 12, 2019 at 10:19

3 Answers 3


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

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


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, 2019 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, 2019 at 11:56
  • 1
    $\begingroup$ 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, 2019 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, 2019 at 13:25
  • $\begingroup$ $\Phi=U-TS-\mu N = F - \mu N$, so yes, it does. $\endgroup$
    – FGSUZ
    Jan 12, 2019 at 13:33

All your questions are related to the choice of the independent variables. The shape of the functions might vary if you work with the canonical ensemble $(T,V,N)$ or the grand-canonical ensemble $(T,V,\mu)$. However, in the thermodynamic limit, all ensembles are equivalent and you should be able to go from one to the other through a change of variables. However, I get to a different expression for the equation of state, are you sure about its validity?

How can we derive this equation of state?

Say you have a very general system in which the canonical partition function in Boltzman's approximation (a.k.a. classical limit) reads:

$Z(T,N,V)=\frac{(V \varphi(T))^N}{N!}$

Where $\varphi$ depends on the system in question, for a three dimensional ideal gas, $\varphi(T)=(\frac{2\pi m k_B T}{h^2})$.

The grand-canonical partition function will then be:

$Z_{gr}(T,\mu,V)=\sum_N e^{\frac{\mu N}{k_B T}}Z(T,N,V)=e^{e^{\frac{\mu }{k_B T}}V \varphi(T)}$.

The associated thermodynamic potential is:

$J(T,\mu,V)=k_BT\log(Z_{gr})=k_B T e^{\frac{\mu }{k_B T}}V \varphi(T)$.

Therefore, pressure is:

$P(T,\mu,V)=(\frac{\partial J}{\partial V})_{\mu,T}= k_B T e^{\frac{\mu }{k_B T}} \varphi(T)$

Now the only thing is to get $\varphi(T)$. From the canonical partition function of one particle: $\varphi(T)=\frac{Z_1(T,V)}{V}$, from the grand-canonical partition function: $\varphi(T)=\frac{\log(Z_{gr}(T,V,\mu))}{V }e^{-\frac{\mu }{k_B T}}$ or from the grand-canonical potential: $\varphi(T)=\frac{J(T,V,\mu)}{k_B T V }e^{-\frac{\mu }{k_B T}}$. The one that approaches the most to your expression is:

$P(T,\mu,V) V=k_B T \log (Z_{gr})$.

One important thing to notice is that all possible ways of getting $\varphi$ lead to the same result (through a different functional expression).

Is it same equation that we use for ideal gases $(PV=NRT)$ or is it different?

Is the same! In order to check it, you have to go from $P(T,\mu,V)$ to $P(T,N,V)$.

$N(T,\mu,V)=(\frac{\partial J(T,\mu,V)}{\partial \mu})_{_{V,T}}=\frac{J(T,\mu,V)}{k_B T} $.

Since $P(T,\mu,V)=\frac{J(T,\mu,V)}{V}$ in one of its forms, combining the last two equations we get:

$P(T,N,V) =\frac{N k_B T}{V}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.