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I know that for the canonical ensemble:

$Z=\Sigma_n e^{-\beta E_n}$

By using the Lagrange Lagrange multiplier method, one can find for the probability of the system being in an microstate:

$Pr_n=\frac 1 Z e^{-\beta (E_n)}$

But if we consider that each microstate is $\Omega_n$ times degenerated, then we'd have:

$Z=\Sigma_{n,\Omega_n} e^{-\beta E_n}=\Sigma_n \Sigma_{k=1}^{\Omega_n}e^{-\beta E_n}$.

And consequently $Pr_n=\Omega_n \frac 1 Z e^{-\beta (E_n)}$. But how do we get $\Omega_n$? Even when I consider the lagrange multiplier, I still don't get the factor $\Omega_n$.

For the Grand Canonical Ensemble:

$$Z_G=\Sigma_n e^{-\beta(E_n - \mu N_n)}=\Sigma_n e^{-\beta\Sigma_{n_i}(\epsilon_i - \mu)}$$

$$\Sigma_{\{n_i \}}e^{-\beta\Sigma_{n_i}(\epsilon_i - \mu)}=\Pi_i \Sigma_{n_i=0}^{n_{max}}e^{-\beta{n_i}(\epsilon_i - \mu)}$$

And to simplify the expression, if we consider fermions for example, our expression for the Partition function of the grand canonical ensemble would be:

$$Z_G=\Pi_i[1+ e^{-\beta{n_i}(\epsilon_i - \mu)}]$$

Now if we consider that each energy level is g-times degenerated, we would have:

$$Z_G=\Sigma_{\{n_i \},g}e^{-\beta\Sigma_{n_i}(\epsilon_i - \mu)}$$.

$$Z_G=\Sigma_{\{n_i \},} \cdot g \cdot e^{-\beta\Sigma_{n_i}(\epsilon_i - \mu)}$$.

And this changes to:

$$Z_G=\Pi_i \Pi_{k=1}^g[1+ e^{-\beta{n_i}(\epsilon_i - \mu)}]$$.

Why do we get a product of the degeneracy, when as I showed above, it comes out as a constant?

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  • $\begingroup$ But how do we get Ωn? Somewhat strange question for me. $\Omega_n$ is the number of repeated terms with the same $E_n$. We group them together, changing the indices accordingly when summing $\endgroup$ Feb 16 at 14:01

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The degeneracy $\Omega_n$ is a property of the quantum system in question, just like its energy levels $E_n$. Neither of them are obtained from statistical analysis; both are obtained by solving the Schrodinger equation for the system. So they are 'given' quantities when we then approach the task of finding properties of the macrostate having maximum entropy (which is what statistical methods are largely about).

When presenting this, it is best to keep clear in the notation the distinction between counting quantum states and counting energy levels. So for example I would not use the same index letter for both counts. Better to write it like this: $$ Z = \sum_{n \in \text{states}} e^{-\beta E_n} = \sum_{r \in \text{levels}} \Omega_r e^{-\beta E_r} $$

Your questions about the grand partition function seem to be based on a confusion between a quantum state (which by definition is just one quantum state) and a set of states all having the same energy. When we evoke the antisymmetry requirement for fermions, leading to at most one fermion per state, we cannot then claim that the state under consideration was really a set of states. Rather, the Fermi-Dirac distribution function is what it is, and in the expression for the partition function, the state energies $\epsilon_i$ do not all have to be different.

I did not understand the final remark "it comes out as a constant" so I can't answer that part. Perhaps try rewriting that part of the question.

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  • $\begingroup$ Thank you for the explanation. I understand the difference between the state and the energy level, something like, the state is the label and you can find multiple building parts of the system having this particular label. Regarding the last part:If you expand in the penultimate equation the sum in the exponent and then also the general sum $\Sigma_{\{n_i \}}$, at the same time you can pull out $g$ as a simple constant, and there would be no reason to have $\Pi_{k=1}^g$ in the last expression. But ofc I am wrong, since the formula is correct $\endgroup$
    – imbAF
    Feb 16 at 19:35
  • $\begingroup$ One more thing. Then is it correct to say: $Pr(E_r)=\frac 1 Z \Omega_r e^{\beta E_r}$? I assume yes. $\endgroup$
    – imbAF
    Feb 16 at 19:37
  • $\begingroup$ @imbAF with minus sign in the exponent: yes (because probabilities for independent outcomes add) $\endgroup$ Feb 16 at 19:53

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