I derived the averaged energy for phonon mode with frequency $\omega$ in canonical ensemble and in grand canonical ensemble.

Averaged energy derived in canonical ensemble is $E_c=\frac{1}{2}\hbar\omega+\frac{\hbar\omega}{e^{\hbar\omega\beta}-1}$.

Averaged energy derived in grand canonical ensemble is $E_g=\sum_{n=0}^{\infty}\frac{\hbar\omega(n+1/2)}{e^{\hbar\omega\beta(n+1/2)-1}}$.

I believe they should be the same, since observable quantity should be independent of the ensemble we choose. But how to show they are the same??

(In the following, I provide my derivation of the averaged energy.

The partition function in canonical ensemble is

$Z_c=\sum_{n=0}^{\infty}e^{-\hbar\omega\beta(n+1/2)}=\frac{e^{-\hbar\beta\omega/2}}{1-e^{\hbar\omega\beta}}$, so $E_c=-\frac{\partial}{\partial\beta}\ln Z_c=\frac{1}{2}\hbar\omega+\frac{\hbar\omega}{e^{\hbar\omega\beta}-1}$.

The partition function in grand canonical ensemble is

$Z_g=\Pi_{n=0}^{\infty}\sum_{a=0}^{\infty}e^{-\beta\omega\hbar(n+1/2)a}=\Pi_{n=0}^{\infty}\frac{1}{1-e^{-\hbar\omega\beta(n+1/2)}}$, so $E_g=-\frac{\partial}{\partial\beta}\ln Z_g=\sum_{n=0}^{\infty}\frac{\hbar\omega(n+1/2)}{e^{\hbar\omega\beta(n+1/2)-1}}$. Where a is the occupation number for each energy level.)

Thank you in advance.


3 Answers 3


What does the $a$ in your second formula refer to?

For a phonon mode, the formulas for the canonical and grand canonical will be the same, just the interpretation will be different. In BOTH cases the formula should be $$Z = \sum_{n=0}^\infty e^{-\beta \hbar \omega n}$$ (note that we can drop the zero-point energy $1/2 \hbar \omega$ as it's just a constant shift).

For the canonical ensemble, the interpretation is that you look at one oscillator, and this oscillator has energy levels $\epsilon_n = \hbar \omega n$. We sum over all of those energy levels.

For the grand canonical ensemble, the interpretation is that there is only one mode, and that one mode has energy $\hbar \omega$, and now we sum over all possible occupation numbers.

  • $\begingroup$ Hi, a is the occupation number in each energy level $\hbar\omega(n+1/2)$. I don't think we can interpretation the grand canonical ensemble as a single level system, since the energy level quantization is a result of quantum mechanics, which should not dependent on the ensemble we choose. $\endgroup$
    – Blue
    Jan 19, 2014 at 20:09
  • $\begingroup$ Yes, but in the grand canonical picture, you look at these equidistant energy levels and interpret them as individual particles, the phonons. Instead of saying "The oscillator is in the $n$-th excited state and thus has energy $\hbar \omega n$, you say "There are $n$ phonons in the system, each with energy $\hbar \omega$. What you're doing wrong in the grand canonical treatment is double-counting the energy levels: You're assuming that each energy level of the oscillator can be multiply occupied. $\endgroup$
    – Lagerbaer
    Jan 20, 2014 at 5:24

For both canonical and grand canonical ensemble, my formula provided in the question part are correct. I will show later that they will give the same expression for averaged energy.

For canonical ensemble, the total particle number is fixed. Let us denote it as N. Since each boson, the particle in each energy level, has no interaction, the canonical ensemble for the system should be $Z_{ctotal}=Z_c^N$. The averaged energy is $E_c=-\frac{1}{N}\frac{\partial}{\partial\beta}\ln Z_{ctotal}=-\frac{\partial}{\partial\beta}\ln Z_{c}$, which is the same as given in the question part.

For grand canonical ensemble, the particle number is not fixed, the averaged energy should be total energy over total particle number. The partition function can be written by two ways, the first is exactly the same as provided in the question section. It is also equivalent to write $Z_{gtotal}=\sum_{N=0}^{\infty}\frac{1}{N!}[\sum_{a=0}^{a=N}e^{-\beta(\hbar\omega(n+1/2)-\mu)a}]=Exp[e^{\beta\mu}z_c]$.

The meaning is that we focus on N particles at one time, and derive their contribution to the partition function. The total partition function is the sum of them. We should divide a factor $N!$ because they are N identical particles.

The particle number is $X=\beta\frac{\partial}{\partial\mu}\ln Z_{gtotal}=e^{\beta\mu}z_c=z_c|_{\mu\rightarrow0}$. The averaged energy is $E_g=\frac{-1}{X}\frac{\partial}{\partial\beta}\ln Z_{gtotal}=\frac{-1}{z_c}\frac{\partial}{\partial\beta}z_c=-\frac{\partial\ln z_c}{\partial\beta}$.

In the above, we choose chemical potential to be zero, because phonons can be create or annihilate freely.

So, canonical ensemble and grand canonical ensemble give the same answer.


The answer you wrote above doesn't add up.

\begin{equation} Z_{gc}=Exp[e^{\beta \mu}Z_c]=\sum_{N=0}^{\infty}\frac{1}{N!}e^{\beta \mu N}\big[\sum_{n=0}^{\infty}e^{-\beta \hbar\omega(n+1/2)}\big]^N \neq \sum_{N=0}^{\infty}\frac{1}{N!}\sum_{a=0}^{N}e^{-\beta [\hbar\omega(n+1/2)-\mu]a} \end{equation}

These expressions are clearly not equal- for one you are not summing over $n$ in the last expression. You can also see this by writing the sum over $a$ in closed form using the geometric series formula and comparing both expressions.

Furthermore, if you added the factor of $1/N!$ in the expression for $Z_{gc}$ due to indistinguishability, you must do the same for $Z_c$ as the $N$ phonons are indistinguishable. This is because $Z_{gc}=\sum\limits_{N=0}^{\infty}e^{\beta \mu}Z_c$ and the factor of $1/ N!$ comes from $Z_c$.

I am not even sure you can write the above equations as it is not clear what $N$ is- it would mean that the energy of a phonon is $\hbar \omega (n+1/2)$ and not just $\hbar \omega$.


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