Why Does the Entropy Formula for a Fermionic Gas Include a Degeneracy Factor?

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SUMMARY

The entropy formula for a fermionic gas is defined as S = -k_B Σ_i (1-f_i)ln(1-f_i) + f_i ln(f_i), utilizing the Fermi-Dirac distribution where f_i = n_i/g_i. The discussion clarifies that the degeneracy factor g_i is a bookkeeping element, allowing for multiple states at the same energy level. It is established that in quantum mechanics, each state cannot be degenerate, thus g_i can be set to 1 when considering individual states. Proper notation, including the correct placement of k_B, is essential for accurate representation of the equations.

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Homework Statement


The question asked me to show that the entropy of a fermionic gas is
##S = -k_B \Sigma_i (1-f_i)\ln(1-f_i) +f_i\ln(f_i)##

Using the Fermi-Dirac distribution so ##f_i = \frac{n_i}{g_i}##.

Homework Equations

The Attempt at a Solution


The number of microstates ##\Omega## is
##\Omega =## Π##\frac{g_i!}{n_i!(g_i-n_i)!}## and using ##S = k_B ln(\Omega)## I've arrived at the expression:
##S = -k_B \Sigma_i g_i (1-f_i)\ln(1-f_i) +f_i\ln(f_i)##
So there's an unwanted factor of ##g_i##. I'm told it is meant to be there, and I need to explain why it can be set to 1. Something to do with i being the state index and a quantum state can't be degenerate. Could someone explain that to me? Any help is much appreciated! :)
 
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Hi Kara386

First observation here is that you need some big brackets in your equations - for example in your first equation, the k_B should be multiplying everything, not just the first term.

The business with the g_i is just bookkeeping. In their formula they're using the index i to label different states, and in your formula you're using the index i to label different energies. Your formula includes a g_i factor in it to allow for the possibility that there might be lots of states with the same energy.
 
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Oxvillian said:
Hi Kara386

First observation here is that you need some big brackets in your equations - for example in your first equation, the k_B should be multiplying everything, not just the first term.

The business with the g_i is just bookkeeping. In their formula they're using the index i to label different states, and in your formula you're using the index i to label different energies. Your formula includes a g_i factor in it to allow for the possibility that there might be lots of states with the same energy.
Thanks! I thought that might be it, because although originally the sum was over multiple states taken to be at the same energy, I think the introduction of the average ##f_i## allows the sum to be over individual states.

Thanks for your help!
 

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