QM Angular Momentum Commutation Question

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Homework Statement



Consider a state [tex] | l, m \rangle[/tex], an eigenstate of both [tex] \hat{L}^{2}[/tex] and [tex] \hat{L}_{z}[/tex]. Express [tex] \hat{L}_{x}[/tex] in terms of the commutator of [tex] \hat{L}_{y}[/tex] and [tex] \hat{L}_{z}[/tex], and use the result to demonstrate that [tex] \langle \hat{L}_{x} \rangle [/tex] is zero.

Homework Equations



[tex] [ \hat{L}_{y}, \hat{L}_{z} ] = i\hbar \hat{L}_{x} [/tex]

The Attempt at a Solution



I'm sure this is pointing towards telling me that the commutator above is zero, but we know that of [tex] \hat{L}_{y}[/tex] and [tex] \hat{L}_{z}[/tex] aren't compatible operators.

I've tried,

[tex] \langle \hat{L}_{x} \rangle = - \frac{i}{\hbar} \langle [ \hat{L}_{y}, \hat{L}_{z} ] \rangle = - \frac{i}{\hbar} \langle l,m | \hat{L}_{y}\hat{L}_{z} - \hat{L}_{z}\hat{L}_{y} | l, m \rangle = - \frac{i}{\hbar} \left( \langle l,m | \hat{L}_{y}\hat{L}_{z}| l, m \rangle - \langle l,m | \hat{L}_{z}\hat{L}_{y} | l, m \rangle \right) [/tex]

Then,

[tex]\hat{L}_{z}| l, m \rangle = \hbar m | l, m \rangle[/tex]

[tex] \hbar m \hat{L}_{y} | l, m \rangle = 0[/tex] ***

So it follows that,

[tex] \langle \hat{L}_{x} \rangle = 0[/tex]

I'm not 100% convinced that the step labelled (***) is correct...

This is only for one mark out of ten, so I'm sure I'm missing something fairly obvious. Any pointers at all would be fantastic - thanks!
 
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Answers and Replies

  • #2
gabbagabbahey
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You aren't being asked to show that the commutator is zero (which is good....since it isn't zero!), you are being asked to use the fact that [itex]\hat{L}_x=\frac{1}{i\hbar}[\hat{L}_y,\hat{L}_z][/itex] to show that the expectation value of [itex]L_x[/itex] in the state [itex]|l,m\rangle[/itex] is zero...just use the definition of expectation value...
 
  • #3
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Ok, I edited my original post with a little extra working out and hit on a step that I wasn't too sure on...
 
  • #4
gabbagabbahey
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[tex] \hbar m \hat{L}_{y} | l, m \rangle = 0[/tex] ***

I'm not 100% convinced that the step labelled (***) is correct...

I'm not even 0.0001% convinced this step is correct...why would you think that it is?
 
  • #5
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Haha, I must be getting confused between a few different properties concerning the angular momentum operators. I'll have another read on the subject and come back to this I think. Bit disheartening, especially since I'm sure it's very simple!
 
  • #6
gabbagabbahey
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You might consider expressing [itex]L_y[/itex] in terms of the raising and lowering operators [itex]L_{\pm}[/itex]:wink:
 
  • #7
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Yeah, that's what I was thinking... The question looked like it was worded such that they wanted me to find a solution just directly using the commutation relation. I guess I'll have a word with my lecturer tomorrow morning and clear that up. Thanks very much for your time!
 
  • #8
gabbagabbahey
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Yes, it is worded that way, but I don't see any way of showing the intended result without using the raising and lowering operators at some point.
 
  • #9
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Ok, I've had a play around and I think I've got to the result we were intended to find...

[tex]
\langle \hat{L}_{x} \rangle = - \frac{i}{\hbar} \langle [ \hat{L}_{y}, \hat{L}_{z} ] \rangle = - \frac{i}{\hbar} \langle l,m | \hat{L}_{y}\hat{L}_{z} - \hat{L}_{z}\hat{L}_{y} | l, m \rangle = - \frac{i}{\hbar} \left( \langle l,m | \hat{L}_{y}\hat{L}_{z}| l, m \rangle - \langle l,m | \hat{L}_{z}\hat{L}_{y} | l, m \rangle \right)
[/tex]

Then, since [tex]\hat{L}_{z}[/tex] is Hermitian,

[tex]\langle l,m | \hat{L}_{z}\hat{L}_{y} | l, m \rangle = \langle \hat{L}_{z} l,m | \hat{L}_{y} | l, m \rangle = \hbar m \langle l,m |\hat{L}_{y} | l, m \rangle[/tex]

And we find that the quantity in the bracket becomes zero when the first term is calculated,

[tex]- \frac{i}{\hbar} \left( \langle l,m | \hat{L}_{y}\hat{L}_{z}| l, m \rangle - \hbar m \langle l,m |\hat{L}_{y} | l, m \rangle \right) = - \frac{i}{\hbar} \left(\hbar m \langle l,m | \hat{L}_{y}| l, m \rangle - \hbar m \langle l,m |\hat{L}_{y} | l, m \rangle \right) = 0[/tex]

As required.
 

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