The deduction of Fermi-Dirac and Bose-Einstein distrbiutions

[Lebnm]

I am studyng the deduction of Fermi-Dirac and Bose-Einstein distribution, but I'm not understanding one part. If we have a system of $N$ identical non-interaction particles, with energies levels $\varepsilon _{l}$ and occupation number $n_{l}$ (this is the number of particles with the same energy $\varepsilon_{l}$), the total energy and the partion function in the canonical emsemble are $$E(\{n_{l}\}) = \sum_{l} \varepsilon _{l} n_{l}, \ \ \ Z_{N} = \sum _{\{n_{l}\}} e^{-\beta E(\{n_{l}\})} = \sum _{\{n_{l}\}}\prod _{l}e^{-\beta n_{l}\varepsilon_{l}},$$where $\{n_{l}\}$ means tha the sum is over all possible sets of values of number occupations, such that $\sum_{l} n_{l} = N$. Now, the partion function in the grand canonical emsemble is $$Q = \sum _{N} e^{\beta \mu N} Z_{N}.$$ The book I am reading say that this is equal to $$Q = \prod _{l} \sum _{n_{l}}[e^{-\beta(\varepsilon - \mu)}]^{n_{l}}$$ but I don't understand why. Note that the sum passed from a sum over the set $\{n_{l}\}$ to a sum over the values of $n_{l}$.

 

[stevendaryl]

Okay, let me see if I can go through it in more detail.

Let's let a "state" $s$ be an assignment of occupation numbers to each energy level $l$. So $n_{sl}$ is the occupation number of energy level $l$ in state $s$. Then in terms of $s$ we can write:

$Z_N = \sum_s e^{-\beta \sum_l n_{sl} \epsilon_l} \delta_{N - \sum_l n_{sl}}$
$= \sum_s (\Pi_l e^{-\beta n_{sl} \epsilon_l}) \delta_{N - \sum_l n_{sl}}$

where the $\delta_{N - \sum_l n_{sl}}$ returns 0 unless $\sum_l n_{sl} = N$

So instead of only summing over states with $N$ particles, I'm summing over all possible states, but the ones with a number of particles different from $N$ make no contribution, because of the $\delta$.

Now, let's form the grand canonical partition function:

$Q = \sum_N e^{\beta \mu N} Z_N$
$= \sum_N e^{\beta \mu N} \sum_s (\Pi_l e^{-\beta n_{sl} \epsilon_l}) \delta_{N - \sum_l n_{sl}}$
$= \sum_N \sum_s e^{\beta \mu N} (\Pi_l e^{-\beta n_{sl} \epsilon_l}) \delta_{N - \sum_l n_{sl}}$

Now, assuming that we can switch the order of summation (which we can, unless there are convergence issues, which I'm assuming there aren't), we can write:

$Q = \sum_s \sum_N e^{\beta \mu N} (\Pi_l e^{-\beta n_{sl} \epsilon_l}) \delta_{N - \sum_l n_{sl}}$

We can do the inner sum over $N$, and because of the presence of the $\delta$, the only contributions are when $N = \sum_l n_{sl}$. So we can get rid of $N$ to get:

$Q = \sum_s e^{\beta \mu \sum_l n_{sl}} (\Pi_l e^{-\beta n_{sl} \epsilon_l})$
$Q = \sum_s \Pi_l e^{-\beta (\epsilon_l - \mu) n_{sl}}$

The final simplification is to bring the summation inside the product. Since summing over all possible states $s$ is equivalent to summing over all possible assignments to $n_1, n_2, ...$, we can rewrite this as:

$Q = \sum_{n_1} \sum_{n_2} ... e^{-\beta (\epsilon_l - \mu) n_{1}} e^{-\beta (\epsilon_l - \mu) n_{2}} ...$
$= \sum_{n_1} e^{-\beta (\epsilon_l - \mu) n_{1}} \sum_{n_2} e^{-\beta (\epsilon_l - \mu) n_2} ...$
$= (\sum_{n_1} e^{-\beta (\epsilon_l - \mu) n_{1}}) (\sum_{n_2} e^{-\beta (\epsilon_l - \mu) n_2}) ...$
$= \Pi_l (\sum_n e^{-\beta (\epsilon_l - \mu) n})$

 

[stevendaryl]

The mathematical fact that can be used is this:

$\sum_{n_1} \sum_{n_2} ... \sum_{n_M} f(1, n_1) f(2, n_2) ... f(M, n_M) = \Pi_l \sum_n f(l, n)$

 

[Lebnm]

thank you!

 

[vanhees71]

Here, it's derived using (non-relativistic) QFT:

https://th.physik.uni-frankfurt.de/~hees/publ/stat.pdf