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The chemical potential is defined as: $$ \mu = -T\frac{\partial{S(N,V,E)}}{\partial{N}} $$ It seems to me that this is completely independent of where I put the reference point of energy, because only the difference of entropy is relevant (and also temperature is defined as a difference in entropy).

However the Fermi-Dirac distrbution is: $$<n_r> = \frac{1}{exp(\beta(\epsilon_r-\mu))+1}$$

But if I change the value of the reference point of energy, the value of $<n_r>$ changes, which causes a contradiction. In my book about statistical mechanics, they state though that $\epsilon_r-\mu$ is independent of the reference point of energy, but I see not why, because $\mu$ seems to be independent, while $\epsilon_r$ doesn't seem independent of the reference point.

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up vote 5 down vote accepted

Good question. I think that you're incorrect to assume the chemical potential is independent of the zero point.

I think it's easiest to see that the chemical potential must change by using $\mu = (\partial U/\partial N)_{S,V}$. Suppose I first define the zero point as 0 so that the energy is $U(N)$. Then I redefine the zero-point energy (for a particle) as $\epsilon_0$, so that $U' = U + N\epsilon_0$. Now,

\begin{equation} \mu' = \left(\frac{\partial U'}{\partial N}\right)_{S',V} = \left(\frac{\partial U}{\partial N}\right)_{S,V} + \epsilon_0 = \mu + \epsilon_0 \end{equation}

Then if $\epsilon_r' = \epsilon_r + \epsilon_0$, you have that $\epsilon_r' - \mu' = \epsilon_r - \mu$, i.e. the difference is independent of the zero point.

In fact, the formula $\mu = -T(\partial S/\partial N)_{U,V}$ doesn't imply that chemical potential is independent of zero point. If you change the zero-point energy, you must adjust the form of $S$ accordingly. For instance, you need to have $S(U=0) = S'(U' = N\epsilon_0)$ and so you see that $S'$ is different that $S$.

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Thank you! That solved the question. – yarnamc May 23 '13 at 9:00

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