1. For grand partition ensemble, is it true that the introduction of chemical potential allows us to have the sum of number of the particles in each state to be the total number of particles ("On average")? and this fluctuation in the constraint of $N$ will shrink as the number of particles you got is large.

  2. Next question is about Boson gas, why would we have to consider the chemical potential to be negative or zero? Is there anything to do with the reference to the ground state energy level?

  3. What would it mean if we were to integrate the energy of those states below ground states?

NB For last question, I mean when I were to find the total energy of this boson gas system, which I know below a critical temperature Tb, the Bose Einstein condensation becomes important, and we would do the separate sum thing as usual enter image description here(see first Pic) , Now I wonder why when we do the sum of the energy we integrate the states above ground state , does it mean all the particles at ground state have zero energy?? and we only have to concern those above the ground state?

enter image description here

  • $\begingroup$ These questions seem pretty much unrelated. Please split it up into three separate posts - multiple questions per post is not recommended on this site. $\endgroup$
    – N. Virgo
    Commented Jan 5, 2014 at 6:04
  • $\begingroup$ I think they are quite relevant , they are all central to Bose Einstein Statistics $\endgroup$ Commented Jan 5, 2014 at 6:36

1 Answer 1


By passing to the grandcanonical ensemble, the particle number is allowed to fluctuate. You can imagine coupling your system to a 'bath' with which particles can be exchanged.

Nevertheless, in thermal equilibrium the relative fluctuation of the particle number behaves as $$\Delta N/N \propto 1/\sqrt{\langle N \rangle} $$ thus vanishing in the thermodynamical limit $N\rightarrow\infty$

This is analogous to passing from microcanonics to the canonical ensemble by fixing temperature and thus allowing the energy to fluctuate. Both temperature and chemical potential play the roles of Lagrange multipliers for the constraints on energy and particle number.

In the derivation of the Bose-Einstein-distribution one usually sums up a geometric series of the form $$ \sum_{n=0}^\infty (e^{-\beta (E-\mu)})^n $$ To ensure convergence, it must be $\mu\leq E_0$ ; $E_0$ the groundstate, often chosen as 0. This is sort of an 'artifact' from the idealization of perfectly non-interacting bosons. If one takes an interaction into account, this behavior is (often) regularized and $\mu$ may take values larger than 0. In situations without particle conservation, such as the photon-gas, the chemical potential is zero.

Shifting the Hamiltonian by a constant in order to make the groundstate have zero energy, doesn't affect anything since you are considering the difference $H-\mu$. Any offset in $H$ might be absorbed into $\mu$. Even in the canonical setting, this merley produces a shift in the free energy.

In a one-particle setting, shifting the Hamiltonian produces an additional global phase of the wavefunction. This also does not lead to observable effects.

But: Zero-point energies do bear physical significance. At least in settings with non-trivial boundary conditions. The most prominent example is the Casimir-Effect.

As to your last question: That's what BEC means. A macroscopic number of particles occupying the groundstate. Their constribution to the energy is $n_0\epsilon_0$ which can be takes a zero after everything said.

  • $\begingroup$ Thank you very much for your answer. That's very clear explanation to me. I would also like to ask why people often take the ground state energy as 0? For SHO the ground state energy is hbar*omega / 2 , is that an approximation? $\endgroup$ Commented Jan 5, 2014 at 3:33
  • $\begingroup$ The SHO ground state energy is $H_0 = \frac 12\hbar\omega + V_0$, where $V_0$ is the minimum of the potential well. We can choose $V_0 = -\frac 12\hbar\omega$ to make the ground state have total energy zero, or make $V_0 = 0$, or we can just leave the $V_0$ as an undetermined offset. You can make your own opinion which one is more elegant/useful. :) $\endgroup$
    – Nanite
    Commented Jan 9, 2014 at 18:13

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