Guide to HW 7.

1. Notes on energy of single electron states in an H atom potential:
The ground state is an energy eigenstate with an energy of about -13.6 eV.
1st excited states have an energy of about -13.6/4= -3.4 eV.
There are an infinite number of these bound-state energy levels. Their energies are given by \(E_n = -13.6/n^2 \: eV\).
There is only one ground state; the space of 1st excited states, belonging to the energy eigenvalue \(E_2\), requires 4 different 1st excited states to span it. 1+ 3 + 5=9 states are needed to span the space of 2nd excited states belonging to the energy eigenvalue \(E_3\). We say that the degeneracy of the energy eigenvalue \(E_3\) is 9, and the degeneracy of the energy eigenvalue \(E_2\) is 4.
     To sum up, there are a lot of linearly independent (orthonormal) bound states (a countably infinite number) each with an energy less than zero. The ones of primary interest are the lower energy ones, particularly those with n = 1, 2, 3 & 4.
     Each of these states is a single-electron state. We are talking about states of one electron. Even when we add states together to create a superposition state (mixed state), that is still a one-electron state. We have not introduced the way to make states of more than one-electron yet. When we do, you will definitely notice because it will be different from what we are doing now and more complicated.

Problem 8. Lz is introduced on page 157 of Griffiths as \(L_z = xp_y-yp_x\). However, to do this problem, I think it is very much preferable (necessary) to use spherical coordinates where
\(L_z = -i \hbar \frac{\partial}{\partial \phi}\) . This is derived on page 163. You don't need to replicate that derivation on your homework. In order to use that you would want to write \(\psi_{2,1,1} = \frac{1}{\sqrt{2}} (\psi_{2x} + i\psi_{2y})\) in spherical coordinates and hopefully combine the two terms to get something relatively simple before you take the derivative. If your derivative seems too involved, that would indicate that you have not gotten the state into a simple enough form yet.

Problem 9. For this problem most of the wavefunction is inside the wells. For your graphs, I would recommend only calculating the part of the wave-function or PD inside the well, and then just sketching a reasonable curve between the wells and outside. That would be much less work and I think you would actually learn more doing less work because it would give you an opportunity to think intuitively about what the WF or PD must look like in between the wells and outside the wells. So then I think for the whole problem, the only calculations you would need to do would be inside the well, and I believe those take just a few minutes on wolfram alpha.  The point of the problem is not to do a bunch of tedious calculations, but rather to:
1) gain familiarity with the finite square well, and
2) use the finite square well to understand electron tunneling, that is the motion of an electron from one well to another due to quantum interference.
Also, you can calculate the energy of the ground state using \(k_1\). 

(side note: There is apparently a section in the book on tunneling using a WKB method. I would advise you to completely ignore that. WKB is not useful for low-energy bound states as far as I know. I have never found it useful at all.)