Is tight-binding modified by double bonds? According to the tight-binding method, we can calculate the energy spectrum of a molecule,
considering the wave-function as a linear combination of atomic wave functions centered
at the atoms. For example, if we consider $sp^2$ benzene bonds, we would have:
$$\psi(r) = \sum_{i=1}^6 c_i \phi(r-r_i).$$
Is there a way to visualize doble bonding in this context? It seems that single and 
double bonding would have the same energy. Maybe energy levels get filled differently?
 A: I am not sure about the equation in your question - it looks too simple to me. I think there should be separate $\phi$ functions for $2s$ and $2p_{x,y,z}$ on each of the 6 carbon atoms as well as the $1s$ on the hydrogen. I am more used to LCAO (linear combination of atomic orbitals) than tight binding, but apparently these two methods are quite similar.
The two parts of a double bond are different.
In Molecular orbital theory there are sigma bonds and pi bonds - sigma bonds are directed along the bond and pi bonds are above and below the bond.... They are separate energy levels. When you have both a sigma and a pi you get a double bond - picture below.
In Valence bond theory the sigma is the $sp^2$ and the pi bond is $p$ for say the ethene (or ethylene) C2H4 molecule. 

Now the Pi bond in the image is above and below the plane of the molecule, but in the ethyne (acetylene) C2H2 molecule there are two pi bonds on the same bond, which are orthogonal (or at 90 degrees) to each other.... this gives us a triple bond made up of two pi and one sigma.

Ok - so the sigma and pi bonds have different energies and so each can take two electrons.... but the situation is really quite complicated to deal with properly.
If we simplify things to the hydrogen molecule then we have two 1s orbitals that combine to give a sigma and sigma* bonding and antibonding orbital...

... so in the case of the sigma and pi bonds in C2H4 and C2H2 we actually get sigma and sigma*, and pi and pi* orbitals - bonding and antibonding - the antibonding have the * and are higher in energy and are normally empty.
I do not have time (and this answer is long enough already) to start putting in the maths of how to combine atomic orbitals to make molecular orbitals, but to finish off I include a nice diagram of the molecular orbitals of the N2 molecule. Note that the sigma bond between the nitrogen atoms is a combination of 2s and 2pz - sp in valence bond language. The pi orbitals come from the 2px and 2py orbitals - they are orthogonal, but have the same energy - they are degenerate.
Note in the diagram below the $\sigma_g$ are bonding and the $\sigma_u$ are antibonding, but the $\pi_u$ are bonding and the $\pi_g$ antibonding - the $g$ and $u$ subscripts are for symmetrical and antisymmetrial respectively. 
 
