# Interpreting eigenvalues

1. Dec 4, 2013

### Arya_

Hi All,

My question is more from applied quantum mechanics. Suppose I have a 2D conductor(or semiconductor). I use eigenstate representation of hamiltonian in transverse direction and real space representation in longitudinal direction (direction of current flow). Now,

1. Hω=Eω , ω being eigenstates and E eigenvalues.

2. To find H we need kinetic energy + U (potential).

3. we can find n = electron density by ωω* . density matrix.

4. once n is found we can calculate U (Hartree potential) by Poissons equation.

1 and 4 are solved self consistently untill U satisfies both equations.

If I have the H matrix after the self consistent loop is over i.e. I have actual value of potential U. Then what is the physical interpretation for Eigenvalues of H, are they the allowed energy levels??

-Arya

2. Dec 4, 2013

### M Quack

The Eigenvalues of the Hamiltonian are the energies of the corresponding Eigenstates, i.e. the "allowed" energies of states that don't evolve as function of time other than a phase.

3. Dec 4, 2013

### Arya_

Does that mean if I look at Eigenvalues of H and find gaps in energy numbers those are bandgaps. In short can I find bandgap by looking at eigenvalues?

4. Dec 4, 2013

### M Quack

Yes. You have to calculate the Eigenvalues of H as function of wave vector k. This gives you the famous "spaghetti" band structures. The band gap is between the maximum of the highest occupied (E<0) band and the minimum of the lowest unoccupied (E>0) band.

5. Dec 4, 2013

### Arya_

Now, given a matrix H ,
[d,v] = eig(H) in matlab gives me d = diagonal eigenvalue matrix and v = matrix columns of which are eigenstates. This would mean each diagonal element in d is eigenvalue of corresponding column in v.
Thus I have a set of eigenstates and corresponding eigenvalues. Where is wave vector k in this and where is bandgap.

Sorry for my ignorance, I an a Quantum-sufferer into nano-electronic circuits :)

6. Dec 4, 2013

### M Quack

When you solve the Schrödinger equation for the Hydrogen atom, you find that n and L, S, J and m_J are good quantum numbers, and that the Eigenvalues/Energies of the Eigenstates depend on these quantum numbers.

In an infinite, periodic solid the good quantum numbers are n (band number) and k (crystal momentum) - leaving out spin for the moment. The eigenvalues depend on these quantum numbers.

When you write your Hamiltonian as a matrix, then the size of the matrix is the number of states. For an infinite crystal there is an infinite number of states, so the matrix should really be infinite. The number of possible energies is also infinite - the bands are continuous.

So you have to go a bit further in solid state theory and analyze the symmetry of the solid, in particular the translational invariance. The Fourier transform of the possible discrete displacements are the k-vectors, with the
largest k-vector (=Brillouin zone boundary) corresponding to the smallest displacement.....

7. Dec 4, 2013

### Arya_

Well now I am confused. Referring to my last reply, how would you interpret the eigenvalues obtained from [d,v] = eig(H).
Is that only for a particular K? However in setting up H in my original post we did not even talked about K . All that was considered is a potential U.

8. Dec 5, 2013

### M Quack

When you set up H for the hydrogen atom you don't talk about n and L either. But when you take a closer look at H (of the hydrogen atom), you see that it has a certain symmetry, and that because of that you can write the wave functions as a product of a radial and an angular part.

A general Hamiltonian is almost impossible to solve. Therefore one looks for symmetries - for each symmetry there is a conserved quantity, a "good quantum number".

In a crystal, the first symmetry to take advantage of is the translational invariance. I don't remember the exact details, but I think you essentially take the Fourier transform of H and solve H(k) independently for each value of k. After that there are additional symmetries, e.g. in cubic crystals, that reduce the number of k-values you have to solve to get a complete picture of the band structure.