# 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.