# Fermi-Dirac distribution for $E\to 0$ but $T >0$

I have found a lot of graphs in which you can see that the limit of the fermi dirac-distribution always tends to zero when $$T \neq 0$$. But if you look at the Fermi-Dirac Distribution, you get:

$$f(E)=\frac{1}{\exp\left(\beta \cdot(E-\mu)\right)+1}.$$

Now, $$\lim\limits_{E \rightarrow 0, T \neq 0}{f(E)}=\frac{1}{\exp\left(\beta \cdot(-\mu)\right)+1} \neq 1$$

So, are all the textbooks not very precise or am I missing something?

• It seems to me that there is some confusion here. Do you want to compute $\lim_{E \to 0}$ or $\lim_{E \to \infty}$? Jul 8, 2016 at 14:25
• Anyway, your limit is correct. $f(0)$ is not $1$ in general, it is $1$ only if $T=0$. On the other hand $\lim_{E \to \infty} f(E)$ is always $0$, as you can easily verify. Jul 8, 2016 at 14:28
• thanks for your answer! I wanted to compute $\lim\limits_{E \rightarrow 0, T >0}{f(E)}$. This should correspond to probability that the ground state is occupied for temperatures larger than zero. Thanks Jul 8, 2016 at 14:39

You shouldn't expect the occupation of the ground state to be one (or zero). That would imply that, for $T>0$, the ground state is ALWAYS (never) occupied. That isn't necessarily true, as you can see from your formula. What is true is that at ZERO temperature states with $E<\mu$ are always occupied, while states with $E>\mu$ are never occupied. At $T>0$, this is blurred.
I'll also add that $E=0$ isn't even the ground state. There are plenty of systems with negative energy ground states (say, spins in a magnetic field), and plenty of systems with positive energy ground states (say, a collection of oscillators). Since $E=0$ isn't a special point, you shouldn't expect anything special to happen there. What actually matters, if you look at the function, is the difference between $E$ and $\mu$.
What IS true is that $\lim_{E\rightarrow-\infty}f(E)=1$ and $\lim_{E\rightarrow\infty}f(E)=0$. These limits have nothing to do with the value of the function at $E=0$.