QM Angular Momentum Commutation Question

[jazznaz]

Homework Statement

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

Homework Equations

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

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 \hat{L}_{y} and \hat{L}_{z} aren't compatible operators.

I've tried,

\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)

Then,

\hat{L}_{z}| l, m \rangle = \hbar m | l, m \rangle

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

So it follows that,

\langle \hat{L}_{x} \rangle = 0

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!

 

[gabbagabbahey]

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 \hat{L}_x=\frac{1}{i\hbar}[\hat{L}_y,\hat{L}_z] to show that the expectation value of L_x in the state |l,m\rangle is zero...just use the definition of expectation value...

 

[jazznaz]

Ok, I edited my original post with a little extra working out and hit on a step that I wasn't too sure on...

 

[gabbagabbahey]

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

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?

 

[jazznaz]

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!

 

[gabbagabbahey]

You might consider expressing L_y in terms of the raising and lowering operators L_{\pm}

 

[jazznaz]

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!

 

[gabbagabbahey]

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.

 

[jazznaz]

Ok, I've had a play around and I think I've got to the result we were intended to find...

<br /> \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) <br />

Then, since \hat{L}_{z} is Hermitian,

\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

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

- \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

As required.