QM Angular Momentum Commutation Question

1. Nov 11, 2009

jazznaz

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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!

Last edited: Nov 11, 2009
2. Nov 11, 2009

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...

3. Nov 11, 2009

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...

4. Nov 11, 2009

gabbagabbahey

I'm not even 0.0001% convinced this step is correct...why would you think that it is?

5. Nov 11, 2009

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!

6. Nov 11, 2009

gabbagabbahey

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

7. Nov 11, 2009

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!

8. Nov 11, 2009

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.

9. Nov 12, 2009

jazznaz

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

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

Then, 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.