# Fermi-Dirac Statistics

1. Oct 21, 2004

### tyco05

Hey kids,

The question I'm having trouble with (this time) is as follows:

Show that the Fermi-Dirac distribution function,

$$f_{FD}(E)=\frac{1}{e^{(\frac{E-E_f}{kT})}+1}$$

Has the following functional form at T= 0K
(see attachment)

Now, the first thing that screamed at me was the division by T in the exponential bit. If T=0, what is going on!?

The obvious things are:

E>Ef then f(E) = 0

and

E<Ef then f(E) = 1.

I'm just really confused at how I can show that the function has that form at T=0K

Any ideas?

Cheers

#### Attached Files:

• ###### FD-distribution.bmp
File size:
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2. Oct 21, 2004

### Galileo

I don't understand your question. Didn't you just show the function has that form?
$$\lim_{T \rightarrow 0}f_{FD}(E)=\left\{ \begin{array}{ll}1 & \mbox{if} E<E_f\\ \frac{1}{2} & \mbox{if} E=E_f \\0 & \mbox{if} E>E_f$$

3. Oct 22, 2004

### tyco05

I'm glad somebody else doesn't understand the question either.

They want me to 'show' that the distribution has the (attached pic) form at T=0.

The real problem I'm having is how do I "show" that it has that form? Via two lines of maths? That's it?

4. Oct 22, 2004

### Galileo

You HAVE just shown it. By taking the limits.
So yeah, that's it. :)

5. Oct 22, 2004

### marlon

Well, just use the definition of the Fermi-level... It is the the maximum energy-level at T = 0 K. Just fill up all the available energy-levels with all available electrons. The last electron is placed at the highest energylevel which is called the Fermi-level. Ofcourse all this is done at zero Kelvin. That is why the distribution function has the drawn form. All levels are filled up (probability one) till the Fermi-level. Above this level there are no filled levels since the Fermi-level is the highest. It is just by QM-definition of the Fermi-level

regards
marlon