# Introductory Quantum HW (Angular Momentum)

1. Apr 17, 2014

### stefan10

1. The problem statement, all variables and given/known data
Consider a three-dimensional system with wave function ψ. If ψ is in the l = 0 state, we already know that Lzψ=0. Show that Lxψ=0 and Lyψ=0 as well.

2. Relevant equations

[Lx,Ly]ψ = i*h-bar*Lzψ

3. The attempt at a solution

I'm having trouble figuring out where to start this. I think it should be clear and straight-forward, but for some reason I'm just not seeing how I can derive this. I tried using the above equation, to get

[Lx,Ly]ψ=0.

I assume from here I would prove that this is only true of Lxψ and Lyψ are 0. That is assuming, I'm on the right track.

2. Apr 17, 2014

### Simon Bridge

Have you tried that out to see what you get?
i.e. expand out the commutator.

Since you have $\psi:\text{L}_z\psi=0$ do you know what happens when you try to do $\text{L}_x\psi$?

3. Apr 17, 2014

### stefan10

The question states Lz ψ = 0, I don't have any specific function for ψ. I think the question makes this claim because Lz= m*h-bar, and m=0 if l=0.

Could I possibly conclude that L^2 = 0 since l=0, and therefore L^2 = Lz^2+Lx^2+Ly^2 implies Lx = Ly =0?

L^2 = l(l+1)h-bar

Last edited: Apr 17, 2014
4. Apr 17, 2014

### Simon Bridge

You don't need a specific function - you have to exploit the relationships: using your understanding of QM.
The question is telling you that the system is prepared in an eigenstate of the Lz operator.

... consider: can operators take values by themselves?
What does the $l$ quantum number refer to in this case?