# Expectation value in unpertubed basis

1. Feb 21, 2013

### aaaa202

I have a question regarding an exercise I am doing. It is an electron confined to move on a cylinder and I am asked to:
"Find the expectation value of Ly and Lz" in the unperturbed basis. I am just not sure what is meant by the expectation value in a basis? I know what the expectation value is if you give me some wave function at time t. Then <operator> is just:
<operator> = <ψloperatorlψ>

But what on earth is an "expectation value in a basis"? It is probably something pretty straightforward like the expectation value for the nth eigenfunction, but I just want to be sure.

2. Feb 21, 2013

### vela

Staff Emeritus
The problem is just asking you to calculate that expectation value while working in the basis of eigenstates of the unperturbed Hamiltonian. In other words, you might express the state as a linear combination of the eigenstates so that it's easy to calculate what the operator acting on that state gives.

3. Feb 21, 2013

### aaaa202

Well all I am given is the eigenstates of the unpertubed Hamiltonian, so should I just write e.g.
<Lz>=<ψlLzlψ>
as my answer? hmm nah probably not.
I am not given any initial wave function. All I have is the wave functions, and then I am asked to find the expectation value of Lz and Ly in the unpertubed basis. So which initial wave function do I take?

4. Feb 21, 2013

### vela

Staff Emeritus
Good question. It doesn't really make sense to ask for the expectation value without specifying a state. Perhaps your professor wants you to find the expectation value of the operators for each eigenstate. I'd ask him or her for clarification.

5. Feb 21, 2013

### aaaa202

well the actual question is show that the expectation values are zero in the unpertubed basis. So I guess a good idea would be to show it is zero for each eigenstate.

6. Feb 21, 2013

### vela

Staff Emeritus
Yeah, I was thinking it was something like that.

7. Feb 21, 2013

### aaaa202

okay great. Next question actually bothers me a bit too.

So I found the eigenfunctions of my hamiltonian to be degenerate (2 for each energy level). Now including spin of the electron in the problem I must show that there are four degenerate eigenvectors for each energy level.
Is this just done by multiplying by either a spin up or down state? Or could we multiply by any two linear independent vectors that span the space of spin½? - after all the hamiltonian only acts on the spatial part of the wave function right?

8. Feb 21, 2013

### vela

Staff Emeritus
If there's no spin-dependent part of the Hamiltonian, then yes.