# The deduction of Fermi-Dirac and Bose-Einstein distrbiutions

• I
• Lebnm
In summary, the partion function in the grand canonical emsemble is equal to the product of the partion function in the canonical ensemble and the function that sums over all possible state assignments.

#### 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}##.

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})##

Lebnm
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)##

thank you!