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

Question

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.

Answer 1

If we neglect the possibility of negative temperature, then OP is right:

The Fermi-Dirac distribution

$$f_{FD}(\epsilon) ~\longrightarrow ~\Theta(\epsilon_F - \epsilon) \qquad \text{for}\qquad T ~\longrightarrow ~0^{+}, $$

where $\Theta$ is the Heaviside step function.