I am trying to understand Section 3.8.3, "Independent particles", of Piers Coleman's Introduction to Many-Body Physics (self-study, mathematics background). He considers "a system of independent particles with many energy levels $E_\lambda$," with Hamiltonian
$$H – \mu N = \sum_{\lambda} (E_\lambda – \mu) \hat{n}_\lambda.$$
He then writes
the partition function is given by a product of the individual partition functions:
$$Z = \text{Tr} \left[\prod_{\lambda \, \otimes} e^{-\beta (E_\lambda – \mu) \hat{n}_\lambda} \right].$$
Since the trace of an (exterior) product of matrices, this is equal to the product of their individual traces ($\text{Tr} \prod_{\lambda \otimes} = \prod_\lambda \text{Tr}$),
$$Z = \prod_\lambda \text{Tr} [e^{-\beta (E_\lambda – \mu)\hat{n}_\lambda}] = \prod_{\lambda} Z_\lambda.$$
I am not sure how to understand this abstract presentation (for example, I do not know what should be understood by an "exterior product of matrices", or the use of the $\otimes$ symbol), and when I attempt a concrete example to gain more intuition, I am unable to reproduce the result.
For example, I can consider a two-level system containing three particles, so that the three-particle eigenstates of $H$ are $|30\rangle$, $|21\rangle$, $|12\rangle$, and $|03\rangle$. I find
\begin{align*}
Z &= \text{Tr} \, e^{-\beta (H-\mu N)} \\
&= \langle 30| e^{-\beta \sum_{\lambda=1,2} (E_\lambda – \mu) n_\lambda} | 30 \rangle +
\langle 21| e^{-\beta \sum_{\lambda=1,2} (E_\lambda – \mu) n_\lambda} | 21 \rangle + \cdots + \langle 03| e^{-\beta \sum_{\lambda=1,2} (E_\lambda – \mu) n_\lambda} | 03 \rangle \\
&= e^{-\beta [3 (E_1 – \mu)]} + e^{-\beta [2(E_1 – \mu)+(E_2-\mu)]} + e^{-\beta [(E_1 – \mu)+2(E_2-\mu)]} + e^{-\beta [3 (E_2 – \mu)].}
\end{align*}
On the other hand,
\begin{align*}
Z_1 &= \text{Tr} \, e^{-\beta (E_1-\mu)\hat{n}_1} \\
&= \langle 30| e^{-\beta (E_1-\mu)\hat{n}_1} | 30 \rangle +
\langle 21| e^{-\beta (E_1-\mu)\hat{n}_1} | 21 \rangle + \cdots + \langle 03| e^{-\beta (E_1-\mu)\hat{n}_1} | 03 \rangle \\
&= e^{-3 \beta (E_1 – \mu)} + e^{- 2 \beta (E_1 – \mu)} + e^{- \beta (E_1 – \mu)} + 1,
\end{align*}
and similarly $Z_2 = e^{-3 \beta (E_2 – \mu)} + e^{- 2 \beta (E_2 – \mu)} + e^{- \beta (E_2 – \mu)} + 1$. Evidently, $Z \neq Z_1 Z_2$, if my calculation is correct.
Where am I going wrong, and how can I understand Coleman's explanation in more concrete terms?
Best Answer
The method is not applicable in your example. The issue is that you are thinking in the canonical ensemble with fixed number of particles. Colman rather works in the grand canonical ensemble with fixed chemical potential that you then tune to match the expected number of particles. It is only in the latter that the independence can be used. You also need to be careful as you are implicitly assuming that the particles are bosons, Coleman's approach is more general and works for both.
I'll detail how the situations differ in the case of two orbitals of energies $E_1,E_2$.
The grand canonical approach (Coleman's) gives: $$ Z = Z_1Z_2 $$ with: $$ Z_i = \text{Tr}[e^{-\beta(E_i-\mu)n_i}] $$ For bosons, this becomes: $$ Z_i = \frac{1}{1-e^{-\beta(E_i-\mu)}} $$ while for fermions: $$ Z_i = 1+e^{-\beta(E_i-\mu)} $$ If you think in terms of fixed expected number of particles, you'll need to solve for the chemical potential: $$ N = \sum N_i $$ with: $$ N_i = \frac{1}{Z_i}\text{Tr}[n_ie^{-\beta(E_i-\mu)n_i}] $$ which for bosons is: $$ N_i = \frac{1}{e^{\beta(E_i-\mu)}-1} $$ and for fermions: $$ N_i = \frac{1}{e^{\beta(E_i-\mu)}+1} $$ In the case of two orbitals, you can explicitly solve this by using the quadratic formula.
The canonical approach (your approach) needs to be distinguished between bosons and fermions. For bosons: $$ \begin{align} Z &= \sum_{N_1=0}^N e^{-\beta[E_1N_1+E_2(N-N_1)]} \\ &= \frac{e^{-\beta E_2 (N+1)}-e^{-\beta E_1(N+1)}}{e^{-\beta E_2}-e^{-\beta E_1}} \end{align} $$ which does not have an obvious factorisation, and is certainly not factored into independent factors.
In the case of fermions, it's faster to just enumerate the possible partition functions: $$ \begin{align} N &= 0 & Z &= 1 \\ N &= 1 & Z &= e^{-\beta E_1}+e^{-\beta E_2} \\ N &= 2 & Z &= e^{-\beta (E_1+E_2)} \end{align} $$
The two approaches agree in the thermodynamic limit, but for three particles in the boson case will not give the same results.
Hope this helps.