Fermi Dirac distribution function

[Physicslad78]

I have a question that is puzzling me as always...The Fermi-Dirac distribution function is (at T=0):

f\epsilon=\frac{1}{e^{\beta(\epsilon-\epsilon_{F})}+1} and we know that we can subsitute f\epsilon by 1 for \epsilon< \epsilon_{F} and 0 otherwise. However what is f(-\epsilon)? The answer is easy when \epsilon_{F}=0 but what if \epsilon_{F} is not zero. what would be ff(-\epsilon)? for \epsilon< \epsilon_{F}?


Thanks

 

[Physics Monkey]

Hi Physicslad78,

I'm a bit confused by your question. You state correctly that the Fermi function at zero temperature implies that any state with energy less than the chemical potential is occupied with probability one. This statement also applies to any negative energy states that may exist below the chemical potential. Is this what you were confused about?

 

[Mute]

As T \rightarrow 0,

f(\varepsilon) = \frac{1}{\exp\lrft[\beta(\varepsilon - \varepsilon_F)\right] + 1} \rightarrow \Theta(\varepsilon - \varepsilon_F),
which is the Heaviside step function, which is zero for e < e_F and 1 for e > e_F. If you replace e with -e, the equalities just get reversed: it's 0 for e > - e_F and 1 for e < -e_F.

 

[Gokul43201]

If you replace e with -e, the equalities just get reversed: it's 0 for e > - e_F and 1 for e < -e_F.

If I understand the OP's question, this is not merely a question of a sign convention. Rather, it is asking what the function does for E<0, when E_F >0. That question is answered by Physics Monkey above.

 

[Physicslad78]

Thanks guys for all of your replies.. What I wanted to know is that when does f (-\epsilon) vanish and when it is one. I guess Mute answered the question but he reversed the cases. i tried to plot f (-\epsilon) and found that is it is 1 for \epsilon>-\epsilon_{F} and 0 otherwise. The situation I am in is I am trying to integrate the following:

\int_{0}^{\infty}(f(\epsilon)-f(-\epsilon))(\frac{1}{\omega\hbar-2\epsilon}-\frac{1}{\omega\hbar+2\epsilon}. I want to know how the Fermi Dirac functions behave. If we divide the first integral into two: \int_{0}^{E_{F}+\int_{E_F}^{\infty} so that the first term gives an f(\epsilon) of 1 while the second vanishes..Now for f (-\epsilon), I think i can perform the change of variable x= -\epsilon so that dx= -d\epsilon and the integral limits now go from -\infty to 0 but now we can write - \int_{-\infty}^{0} is \int_{0}^{+\infty} which then also gives 1 so that the total would be 2 trimes the remaining functions to be integrated which will give Log i guess...Is that right??

 

[DrDu]

Yes, probably, but you should keep in mind that your Hamiltonian is usually bounded from below, that is there is a minimal epsilon, which often is taken to be 0. For values of epsilon below, there is no FD function as there are no states to be occupied.

 

[turin]

Isn't this just a matter of density of states? I mean, the Fermi-Dirac function covers the entire domain -∞ < ε < +∞, regardless. However, the density of states can, for instance, make the occupation number vanish for ε < 0, right?