Fermi-Dirac distribution - holes and electrons

Question

The density of probability of an energy state $E$ being occupied by an electron is $$f(E,T)=\frac{1}{1+e^{\frac{E-E_F}{kT}}}$$ and the density of probability of an energy state being occupied by a hole is $$h(E,T)=1-f(E,T)=1-\frac{1}{1+e^{\frac{E-E_F}{kT}}}$$

Why are these two events incompatible? I mean, why if the probability of an electron living in $E$ is $1$, then the probability of a hole being in the same energy state is $0$?

There is another thing I don't understand. I've read that for $T=0$, the probability of an electron being in the conduction band is $0$ and it will be $1$ in the valence one. Why is this? Also, would this mean that for $T=0$, the probability of a hole living in the conduction band is $1$? (I think this last thing makes no sense... But I don't know.)

Answer 1

To answer your question we have to keep in mind, that holes are quasi particles. They are a mathematical formalism. They are introduced as empty states in the valence band. From a physical point of view it makes sense to construct these particles, as they really have the properties of real charge carriers. The hole energy is at its minimum at the top of the valence band and increasing for lower energies (from an electron point of view). Their creation operator in the valence band is defined as $$b_{k}^{+}=a_{k_\nu}$$ with the electron annihilation operator in the valence band $a_{k_\nu}$. So also mathematically speaking a hole is nothing else than a missing electron.

Let us now look at a state E in the valence band which is not completely filled. Or in other words it contains holes. The electrons are distributed according to the Fermi-Dirac statistic $$f(E,T)=\frac{1}{1+e^{\frac{E-E_F}{kT}}}$$ At each state we can either have an electron or a hole (i.e. not an electron). In other words, the holes are distributed as $h(E,T)=1-f(E,T)$.

So in summary the prensence of a hole excludes the presence of an electron as it is defined that way.

Answer 2

From the previous discussion, it seems to me that the original questioner is slightly unclear about the word 'state'. In this context, a given state has a given spatial wavefunction AND a given spin orientation. So for each spatial wavefunction there are two states (because electrons have spin 1/2). More generally, if there are $N$ states in some range of energy, then what we are saying is that there are $N/2$ mutually orthogonal spatial wavefunctions, each of which has associated with it two mutually orthogonal spin states (or, if you want to be precise, a Hilbert space of two dimensions in the spin degree of freedom). Finally, to say "there are $n$ holes" is precisely equivalent to saying "there are $N-n$ electrons" (in the given set of $N$ possible states). These are two ways of saying exactly the same thing.

Answer 3

I will stick my neck out without researching this. I think this is a matter of the Pauli exclusion principle.