Quote by Troy124
Hi,
I have a question about the partition function.
It is defined as ## Z = \sum_{i} e^{\beta \epsilon_{i}} ## where ##\epsilon_i## denotes the amount of energy transferred from the large system to the small system. By using the formula for the Shannonentropy ##S =  k \sum_i P_i \log P_i## (with ##k## a random constant or ##k_B## in this case), I end up with the following: $$ S =  k \sum_i P_i \log P_i = (k \sum_i P_i \beta \epsilon_i) + (k \sum_i P_i \log Z) = \frac{U}{T} + k \log Z $$
This simplifies to ##Z = e^{\beta F}## by using the Helmholtz free energy defined as ##F = U  T S##. But Boltzmann's formula for entropy states ##S = k \log \Omega##, where ##\Omega## denotes the number of possible microstate for a given macrostate. So we will get $$ \Omega = e^{S/k} = e^{\beta (U  F)} = Z e^{\beta U} $$
So the partition function is related to the number of microstates, but multiplied by a factor ##e^{\beta U}##. And this bring me to my question: why is it multiplied by that factor? Maybe the answer is quite simple, but I can't seem to think of anything.

Boltzmann's formula ##S = k_B \ln \Omega## is applicable
only to the case of a microcanonical ensemble  a system in which every microstate is equally likely. Note that setting ##P_i = 1/\Omega## in ##S = k_B \sum_{i=1}^\Omega P_i \ln P_i## gives Boltzmann's formula.
The partition function ##Z = \sum_i \exp(\beta \epsilon_i)## corresponds to a canonical ensemble. The microstates in a canonical ensemble are
not equally likely, so Boltzmann's formula ##S = k_B \ln \Omega## does not apply. (However, the more general formula, ##S = k_B \sum_{i=1}^\Omega P_i \ln P_i##, does still apply).
You can thus not equate ##\Omega## to ##Ze^{\beta U}##, as the two formulas you used for entropy are not simultaneously true.