# Practical difference between canonical and grand canonical ensembles

I'm currently doing some calculations which require evaluating various standard thermal expectation values in the canonical ensemble (both bosons and fermions). Now, in order to make my theoretical machinations easier, I am actually using the grand canonical ensemble, where the chemical potential acts as a Lagrange multiplier enforcing the constraint $\langle \hat{N} \rangle = N$, where $N/V$ is the fixed density of the physical system. The justification for this is that the relative fluctuations in $\langle \hat{N} \rangle$ should vanish in the thermodynamic limit, in which case I expect fixing the average number to be physically equivalent to fixing the number once and for all. (Also, this approach seems to be adopted by a several presumably trustworthy references, see for example Simons & Altland Section 6.3.) This intuition seems reasonable, but I wonder if matters may be more subtle than this argument implies.

Do thermal averages in the thermodynamic limit of the grand canonical and canonical ensembles coincide?

I'm hoping for either a more rigorous justification supporting this procedure, or examples where it can go horribly wrong. Pointers to appropriate references would also be much appreciated.

• Related StackExchange questions: this, this, this and possibly this. – Mark Mitchison Oct 8 '13 at 15:57
• For a specified problem , at fixed temperature, $\langle \hat{N} \rangle$ depends on the chemical potential. For perfect fermion gas, perfect boson gas, at fixed temperature, one find that $\langle \hat{N} \rangle/V$ depends on the chemical potential (except in the case of photons, where the chemical potential is zero) – Trimok Oct 8 '13 at 16:53
• @Trimok Right, sorry, should have been clearer that I want the density to be fixed. Ultimately $N$ will always appear in the combination $N/V$ anyway. Have edited accordingly. The reason I phrased it thus is that experimentally these systems will obviously have finite $N$ and $V$. It's simply that these quantities will be large enough that I expect the thermodynamic limit to predict the correct behaviour. – Mark Mitchison Oct 8 '13 at 16:56
• @Trimok Also, I don't really understand the purpose of your comment. Are you saying what I'm doing does/doesn't make sense? Or just making an observation? – Mark Mitchison Oct 8 '13 at 17:16
• I just noted that, at fixed temperature, fixing $\langle \hat{N} \rangle/V$ means fixing the chemical potential, but it is not in contradiction with the disparition of relative fluctuations for large $N$ or large $V$. – Trimok Oct 8 '13 at 17:30

## 1 Answer

As you know, the thermodynamic limit requires the system to grow to infinite size while keeping the same density, which lets you neglect surface effects. It also requires the lack of long-range interactions so that distant parts can act independently. So, you need to neglect gravitational interactions, allow the system to be charge neutral, etc.

Even in the thermodynamic limit, the different ensembles can behave differently near phase transitions. For example, if you park the grand canonical ensemble at a liquid-gas boundary, then the system is free to be filled with liquid or gas or a mixture. So, you get a giant and non-negligible fluctuation in the system's energy and particle number. By comparison the phase transition is extended in the canonical ensemble, with a range over which you have a liquid-gas mixture. Another example is the boson condensate: once the condensate has occurred, the total particle number in the grand canonical ensemble has a geometric distribution (!) and so there are giant fluctuations.

There are probably some other issues but I can't think of them off the top of my head. Anyway, just avoid long-range interactions, extreme conditions, and critical phenomena, and your calculations should be fine.

• Thanks. I understand the issue with critical phenomena. Regarding long-range interactions I understand why the thermodynamic limit does not formally exist due to instability. However I don't see why specifically long-range interactions would affect the fundamental issue: the number fluctuations. Is it not the case that when $N$ is extremely large (but not formally infinite) expectation values calculated in the two ensembles should converge? For example, how about large finite-temperature systems that are not charge neutral, but appropriately confined so they don't just explode? – Mark Mitchison Oct 8 '13 at 16:51
• I think you're right, there are some charge imbalanced systems that wouldn't behave too badly in the thermodynamic limit. I guess the problem is really of extensivity. By doubling the canonical ensemble, you'd arrive at a different result than doubling the grand canonical ensemble. What I mean is, if I had a charge imbalanced piece of metal and attached it to a duplicate piece, but didn't allow electrons to leave, then the chemical potential would have to change. If I instead ask the chemical potential to remain the same (grand canonical ensemble), then the average charge would have to change. – Nanite Oct 9 '13 at 16:57
• Aside from the extensivity, the number fluctuations would be okay. In either case if you assume the electrons are confined to the metal (let's assume an infinitely strong metal with infinite work function), then they would just run away to the edges of the metal. You'd get a divergence of energy in the thermodynamic limit, but it wouldn't be a horrible divergence. – Nanite Oct 9 '13 at 17:07
• Thanks that's very helpful. For the problem I'm interested in, I expect on physical grounds that the relevant parameter should always be the density, not the size or number separately. The picture I have in mind, which is experimentally the most relevant, is that the volume is fixed, then one could play around with the density as a control parameter. In order to change the density I must of course modify the chemical potential appropriately. The density could always be thought of as being controlled by an external reservoir with adjustable chemical potential. – Mark Mitchison Oct 9 '13 at 17:41
• Nevertheless I think this approximation must get progressively worse as the density drops, and possibly might introduce uncontrollable errors. So I might have to do it the hard way... – Mark Mitchison Oct 9 '13 at 17:44