Introductory Quantum HW (Angular Momentum)

stefan10
Messages
34
Reaction score
0

Homework Statement


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.


Homework Equations



[Lx,Ly]ψ = i*h-bar*Lzψ

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.
 
on Phys.org
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##?
 
  • Like
Likes   Reactions: 1 person
Simon Bridge said:
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##?

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:
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.

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?
... consider: can operators take values by themselves?
(you will need to be careful about this to do the problem)
What does the ##l## quantum number refer to in this case?

Part of the reason these questions get set is to force you to do a long-winded exploration before finding the simple solution. Thus, you are best advised to settle on a direction for your inquiry and see it through rather than take random jumps around the theory.

Until you settle on something, there's not much I can do to help.
Did you try any of the other suggestions?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 45 ·
2
Replies
45
Views
5K
Replies
10
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K