Solving for <Lx^2> and <Ly^2>: How do I show that <Lx^2> = <Ly^2>?

[J man]

Homework Statement

If a particle is in an angular momentum eigenstate, psi, show that <Lx> = <Ly> = 0, where <Lx> is the angular momentum in the x-direction.

Also show that <Lx^2> = <Ly^2> = [(h bar)^2][l(l+1)-m^2]/2

Homework Equations

L^2 = (Lx^2)+(Ly^2)+(Lz^2)

The Attempt at a Solution

I began by rewriting my relevant equation as:

<Lx^2>+<Ly^2>= <L^2>-<Lz^2>

I then applied previous knowledge to rewrite the right side as

<Lx^2>+<Ly^2>= (h bar^2)l(l+1)psi-(h bar^2)m^2 psi

I'm then able to rewrite this as the second desired "show that" in the problem, I just have no idea how to apply this to the first. I understand that if I can show the right hand side is zero, <Lx^2> must equal <Ly^2> which would be zero, but I don't know why or how to show that part to prove part A. Thanks.

 

[arkajad]

What about using the commutator $$[L_x,L_z]=...$$ and calculating its expectation value in an eigenstate of $$L_z$$. Perhaps you will be able to deduce something from this?

 

[J man]

Okay...so I'm trying to work this out, and these are my steps:

I know (Lz) psi = (h bar)m psi

so expectation value of the commutator would be:

<psi|(Lx)(Lz)-(Lz)(Lx)|psi> = <psi|(Lx)(Lz)|psi>-<psi|(Lz)(Lx)|psi>

= <(Lx)psi|(Lz)psi>-<(Lz)psi|(Lx)psi> = <(Lx)psi|(h bar)m psi>-<(h bar)m psi|(Lx)psi>

from that point I'm not sure exactly where to go. Would that last expression be equal to -i(h bar)(Ly)?

 

[J man]

OR would it be equal to zero if Lx is hermitian?

 

[fzero]

Do you know how to express $$L_x$$ and $$L_y$$ in terms of the ladder operators $$L_\pm$$? That's the easiest way to compute expectation values.

Edit: The two expectation values on the RHS in post 3 are real and equal because $$L_x$$ is Hermitian. Can you relate that calculation to one of the expectation values that you wanted to compute?

 

[J man]

Well, looking back at my notes the only instances in class in which we used ladder operators were in calculating energy level differences, eigen-values and in re-writing L^2 as a function of the L+/- operators and Lz.

 

[J man]

The simplest way would seem to be writing L+/- = Lx +/- i Ly, then solving for Lx and Ly, would I think plug that into the commutator for the expectation value?

 

[vela]

No, just plug it into the expression for <L_x> directly.

 

[J man]

Okay, so referring to the non-ladder method, is the association I was supposed to make that the expectation value of the commutator of Lx, Lz should be equal to the expectation value of i (h bar) Ly which just reduces to the constants multiplied by the expectation value of Ly which would then be zero?

 

[vela]

You could conclude the expectation values are related by a constant, but you couldn't say yet that they're both 0.

 

[J man]

Okay, I was trying to apply my calculation of the expectation values of the commutator, which I found to be 0, with the expectation value of the normal commutator result, i(h bar)Ly. In which case, the i and h bar drop and I'm left with <Ly>=0. Is this not correct?

 

[vela]

Sorry, you're right. I didn't think about it too carefully.

 

[J man]

Thank you everyone. I've been having a hard time getting used to operator algebra so this is appreciated.