Electron-hole symmetry in H and He I'm contemplating particle-hole symmetry, and as an example I am looking at either an electron moving along a hypothetical lattice of hydrogen ions, or a hole moving along a hypothetical lattice of helium atoms. 
According to some lecture notes I found, the hopping integral I get when I treat this in a tight-binding ansatz is positive for holes and negative for electrons, and now I want to see why this is so. 
For electrons, I can understand the result: The atomic wavefunctions are just the 1s hydrogen wavefunctions, and they are positive everywhere, so the matrix element with the kinetic energy operator gives something negative. 
Now, for holes I am not sure how I would have to proceed. Can I just say that the relevant Hamiltonian for holes is the negative of the Hamiltonian for electrons because, after all, a hole with energy $E$ would be an electron with energy $-E$ missing?
Of, if that's wrong, how would I come to the conclusion that the hopping integral for electrons is negative and for holes it's positive?
 A: Yes, a hole with energy $E$ is the same as an electron with a negative energy $E$ missing - that's why it's called a hole and that's how Paul Dirac first encountered it in the relativistic context (in the form of positrons).
A positively-charged positron may look more "particle-like" but one may describe it as the very same holes in the otherwise filled sea of negative-energy electron states.
In both cases - semiconductors and positrons - you may assume that the negatively-charged electrons are the only "real" particles. However, you will always derive the existence of positively-charged holes that behave "just like electrons". 
If you find states such that the energy $E$ is an increasing function of the momentum, the system will first try to fill the low-momentum, low-energy states, and you may add additional higher-energy, higher-momentum particles (electrons).
But some part of the spectrum may have the property that the energy $E$ is a decreasing function of the momentum $k$. In that case, the electron states with a higher value of momentum are filled first, and you're adding them "inwards". This is counterintuitive, so it's more logical to exchange the convention for what we mean by a "filled state" and what we mean by an "empty state". 
Once we do so, we also change the charge and energy of the particle in each state. Consequently, we will deal with positive-charge holes whose energy $E$ behaves just like $-E$ of the original electrons, and increases with $k$ just like for ordinary electron states. The only difference will be in the charge.
