# Limit of Fermi-Dirac distribution as $T$ goes to zero

Hopefully this is a simple question, I just can't seem to get my mind around it.

I'm to take the limit of the Fermi-Dirac distribution for $T \rightarrow 0$.

In this limit the chemical potential is equal to the Fermi energy $\mu = \epsilon_F$, and all states of energy below the Fermi energy is occupied, while all states above are empty.

Following this argument I would say, that the Fermi-Dirac distribution tends to a step-function with argument $\epsilon_F - \epsilon$, such that

$$f \rightarrow \Theta(\epsilon_F - \epsilon) \quad \text{for} \quad T \rightarrow 0,$$ which is one for $\epsilon < \epsilon_F$ and zero for $\epsilon > \epsilon_F$.

My problem is that I have found the results stated in a textbook and a couple of other cases, where it's stated as

$$f \rightarrow \Theta(\epsilon - \epsilon_F) \quad \text{for} \quad T \rightarrow 0.$$

Can someone tell me which result is correct and maybe explain why the second result is correct if it is so.

• Which textbook did you find it in? Your expression is correct by the way. – Olaf Dec 13 '12 at 17:39
• Thank you.. I believe I found it in Quantum Theory of the Electron Liquid by Giuliani & Vignale and afterwards stated the same way at least 1 or 2 places on the internet (physics forums), but I'm not a 100 % sure. I will check up on it, when I get back after Christmas. Either way if both of you agrees with my intuition, I'll stick with that. – Rasmus Søgaard Christensen Dec 22 '12 at 22:23

$$f_{FD}(\epsilon) ~\longrightarrow ~\Theta(\epsilon_F - \epsilon) \qquad \text{for}\qquad T ~\longrightarrow ~0^{+},$$
where $\Theta$ is the Heaviside step function.