The deduction of Fermi-Dirac and Bose-Einstein distrbiutions

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  • #1
Lebnm
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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}##.
 

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  • #2
stevendaryl
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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})##
 
  • #3
stevendaryl
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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)##
 
  • #4
Lebnm
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thank you!
 

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