# Proof that Fermi level is constant throughout a system in thermal equilibrium

The explainations I saw:

1. The one I was referring to in this question. I described why I find this explaination not satisfying in the question.
2. Consider $$2$$ systems in contact. The rate of particles going from system $$1$$ to system $$2$$ is proportional to the number of vacant states in sys. $$2$$ and number of occupied states in sys. $$1$$. Similarly for the rate of particles going from system $$2$$ to system $$1$$. Now we for some unknown reason assume that the constant of proportionality is equal, rewrite the states in terms of density of states and Fermi-Dirac distribution and arrive at equality of Fermi levels. This is not rigorous and convincing. Has many assumptions.

Before I get attacked by "It is trivial, Fermi level is the potential and the gradient in potential causes charge flow". A similar reasonong to this and to what I described as explaination $$2$$ above can be found in Neamen's book. However, he states that this is just an explaination, and the proof will not be given in the book. Since the book is really comprehensive and treats every aspect with profound rigor, it is logical to assume that the actual proof is way too tedious and involved. I'm guessing it must have something to do with low-level Fermi-Dirac statistical mechanics and thermodynamics.

So how to actually rigorously mathematically derive that Fermi level is constant throughout any system in thermal equilibrium? Without any unreasonable and unnecessary assumptions.

Note that I don't want answers like "because the potential must be constant". I want the proof of this very fact.

This follows from thermodynamics. Consider a system with energy $$U$$, volume $$V$$ and number of particles (electrons) $$N$$. The first law of thermodynamics is $$dU = TdS - PdV + \mu dN$$ where $$T$$ is the temperature, $$P$$ is the pressure and $$\mu$$ is the chemical potential (the Fermi level for electrons). Rearranging, $$dS = \frac{1}{T}dU + \frac{P}{T}dV - \frac{\mu}{T}dN$$ and thus to first order, a small fluctuation in the energy, volume or the number of particles of the system will yield $$\Delta S \approx \frac{1}{T}\Delta U + \frac{P}{T}\Delta V - \frac{\mu}{T}\Delta N$$

Now consider a closed system consisting of two subsystems $$A$$ and $$B$$. $$A$$ and $$B$$ are in contact and are free to come to thermodynamic equilibrium. They might be two chunks of solids that are in contact and can exchange electrons. The change (to first order) in the entropy of the combined system in response to an exchange of energy, volume or particles between $$A$$ and $$B$$ is given by $$\Delta S= \Delta S_A + \Delta S_B \approx \left(\frac{1}{T_A} - \frac{1}{T_B}\right)\Delta U_A + \left(\frac{P_A}{T_A} - \frac{P_B}{T_B}\right)\Delta V_A - \left(\frac{\mu_A}{T_A} - \frac{\mu_B}{T_B}\right)\Delta N_A$$ where we used the fact that the total system has constant energy $$U=U_A + U_B$$, volume $$V=V_A + V_B$$ and number of particles (electrons) $$N=N_A + N_B$$.

Since $$\Delta U_A$$, $$\Delta V_A$$ and $$\Delta N_A$$ can be either positive or negative, unless their multipliers in this equation are zero, they can be chosen such that $$\Delta S$$ is positive, i.e. the system can evolve in a way that increases the entropy. According to the second law of thermodynamics, the equilibrium state of the combined system maximizes the total entropy, so if the system is to be in equilibrium, $$\Delta S$$ must always be negative or zero. This requires that $$T_A = T_B$$, $$P_A = P_B$$ and $$\mu_A = \mu_B$$.

In summary, the reason the chemical potential is constant throughout is the same as the reason two systems allowed to "exchange volume" have the same pressure at equilibrium. If the chemical potential is not uniform, particles will rearrange in such a way as to increase the total entropy.

• How did you obtain $$\Delta S= \Delta S_A + \Delta S_B \approx \left(\frac{1}{T_A} - \frac{1}{T_B}\right)\Delta U_A + \left(\frac{P_A}{T_A} - \frac{P_B}{T_B}\right)\Delta V_A - \left(\frac{\mu_A}{T_A} - \frac{\mu_B}{T_B}\right)\Delta N_A$$? Shouldn't it be $$\Delta S= \Delta S_A + \Delta S_B \approx \left(\frac{1}{T_A} + \frac{1}{T_B}\right)\Delta U_A + \left(\frac{P_A}{T_A} + \frac{P_B}{T_B}\right)\Delta V_A - \left(\frac{\mu_A}{T_A} + \frac{\mu_B}{T_B}\right)\Delta N_A$$
– Sgg8
Commented Mar 20, 2023 at 8:21
• @Sgg8 No. $U = U_A+U_B$ is constant, so $\Delta U = \Delta U_A + \Delta U_B = 0$ and $\Delta U_B = -\Delta U_A$, and so on.
– Puk
Commented Mar 20, 2023 at 8:25
• is there a necessity in approximmation up to first order? Why doesn't this ardument hold with differentials instead of deltas, that is, more generally and rigorously?
– Sgg8
Commented Mar 21, 2023 at 12:04