Introductory Quantum HW (Angular Momentum)

AI Thread Summary
The discussion focuses on proving that if a wave function ψ is in the l = 0 state, then Lxψ = 0 and Lyψ = 0, given that Lzψ = 0. Participants suggest using the commutation relation [Lx, Ly]ψ = iħLzψ to explore the implications of Lzψ being zero. There's a consensus that since L^2 = l(l+1)ħ, and l = 0, it follows that Lx and Ly must also be zero. The conversation emphasizes the importance of understanding quantum mechanics relationships rather than relying on specific wave functions. Ultimately, the problem encourages a thorough exploration of the theory to arrive at the solution.
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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.
 
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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##?
 
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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
 
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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?
 
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