In condensed matter we were studying the Sommerfeld model where we had these expressions for a gas of fixed N electrons in a metal, where V is also constant but temperature T is allowed to vary.

We had: $$ E_{total} = V \int_0^\infty d\epsilon \space\epsilon g(\epsilon) \space n_F(\beta(\epsilon-\mu)) $$

where the chemical potential was treated as being defined implicitly by:

$$ N= V \int_0^\infty d\epsilon \space g(\epsilon) \space n_F(\beta(\epsilon-\mu)) $$

It was said that we can use these two formulae to numerically evaluate the variance of $E_{total}$ with $T$and thus differentiate to obtain the heat capacity.

My question is, since we have essentially enforced a 'fixed N' constraint onto what was a grand canonical ensemble, will the resulting variance $E_{total}(T)$ being completely equivalent to that if we just treated this as a canonical ensemble?

If so, is it possible to show this equivalent just in the maths, or is it too difficult to show due to the nature of these integrals?


2 Answers 2


The correct starting point to answer this question is to recall that the definition of canonical and grand-canonical ensembles is fixed.

In the grand-canonical ensemble for a simple system, the independent variables are $T$, $V$, and $\mu$ (temperature, volume, and chemical potential), or any equivalent choice (in many cases the activity $z=\exp \left(\frac{\mu}{k_B T} \right) $ is a more convenient choice). In the canonical ensemble, the independent variables are $T$, $V$, and $N$.

The two ensembles provide the same thermodynamic information only at the thermodynamic limit, in practice, for macroscopic systems. At any finite size, the results are not the same. More important for the present question, if in a grand-canonical ensemble we select a subset of the microscopic states, according to the criterion of a fixed number of particles, we are using only part of the microscopic states. The resulting ensemble cannot describe an equilibrium thermodynamic system anymore at fixed $T$, $V$, and $\mu$ but it may be interpreted as a constrained system, where, in addition to fixing the thermodynamic variables, also the number of particles in the volume $V$ is fixed. In general, this constrained system will not provide the same thermodynamics as the unconstrained, where the average number of particles is not fixed arbitrarily but is the outcome of averaging over all the microscopic states.

The second of your formulas is not fully correct. The $N$ on the left-hand side of the equation should be intended as $\langle N \rangle$, the average number of particles. This should be clear once one recalls that the Fermi distribution $n_F$ represents the average occupation of a single state. Therefore, it is not correct to say that we fix $N$. The formula provides the average number of particles in the volume $V$ at fixed $T$ and $\mu$.

The reason for introducing the expression for $\langle N \rangle$ is that the object we want to extract, the constant-volume specific heat, is usually given as a function of the density and not of the chemical potential. For that, the canonical ensemble would be more appropriate. Still, due to the constraints induced by Pauli's principle, calculations for an ideal Fermi gas are easier in the grand-canonical ensemble.

  • $\begingroup$ I mostly understand your answer, but I don't understand why you are correcting N as <N>. I agree that in the derivation of the result whilst we are working in the grand canonical ensemble we assume $\mu$ and not $N$ is fixed. However I think we can (and are) fixing $N$ now that we have this formula (as we are forcing a fixed total number of electrons in the metal -otherwise where are we assuming these new electrons are coming from?), at the price of letting $\mu$ now vary to accomodate $N$ being fixed. Do you agree? $\endgroup$
    – Alex Gower
    Jan 6, 2021 at 17:30
  • $\begingroup$ However I think your point about us using the grand-canonical ensemble because the calculations are much easier is a good point. $\endgroup$
    – Alex Gower
    Jan 6, 2021 at 17:30
  • $\begingroup$ @AlexGower We can not fix $N$. If we did, in most of the cases we would find a contradiction with the fixed value of $\mu$. What we could do is to evaluate the average $N$ and use it as if it were fixed, on the basis of the vanishing relative fluctuations of $N$ for macroscopic systems. However, conceptually it is definitely not possible to claim that both $\mu$ and $N$ are fixed. $\endgroup$ Jan 6, 2021 at 18:30
  • $\begingroup$ Exactly, that's why I'm claiming that in the usual grand canonical ensemble $\mu$ is fixed and $N$ is not fixed. But instead now we can change it so that $N$ is fixed but $\mu$ is not fixed so that if we evaluated the formula above $\mu$ would take a different value for each $T$ (perfectly to keep $N$ fixed). In this case, the only reason we have used $\mu$ at all (as an intermediate) is to make the calculations easier as you say. Would this be possible? $\endgroup$
    – Alex Gower
    Jan 6, 2021 at 18:35

In the grand canonical ensemble, the total number of particles N in the system+bath is constant. When you apply statistical mechanics to electron gases in a solid, you can define a grand canonical partition function for one particular single-particle state, which can exchange particles with all other single-particle states. You can then use this statistics to evaluate the average occupation $\langle n_i\rangle$ of the state $i$. The sum of the average occupation for all single-particle states gives $N$, and this fixes the values of the chemical potential. This explains your puzzle.

From the grand canonical partition function for a certain state you can almost trivially derive the Fermi-Dirac occupation function for this state if you assume that the occupation is limited to either 0 or 1 particle and you assume that the energy is linear in the number of particles. Electrons in a solid are in bands. If the number of electrons in a band remains fixed, what you have is a a so-called quasi-chemical-potential (called quasi-Fermi level) for each band. If the bands can exchange electrons, then all quasi-Fermi levels merge into a single Fermi level.


Your Answer

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