# Angular momentum and Expectation values (Another question)

1. Nov 15, 2009

### Ben4000

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

Using the fact that ,$$\left\langle \hat{L}_{x}^{2} \right\rangle = \left\langle \hat{L}_{y}^{2} \right\rangle$$ show that $$\left\langle \hat{L}_{x}^{2} \right\rangle = 1/2 \hbar^{2}(l(l+1)-m^{2}.$$

3. The attempt at a solution

$$L^{2} \left|l,m\right\rangle = \hbar^{2}l(l+1) \left|l,m\right\rangle$$

$$L_{z} \left|l,m\right\rangle = \hbar m \left|l,m\right\rangle$$

$$\left\langle \hat{L}^{2} \right\rangle = \hbar^{2}l(l+1)$$

$$\left\langle \hat{L}_{z}^{2} \right\rangle = (\hbar m)^{2}$$

$$\left\langle \hat{L}^{2} \right\rangle = \left\langle \hat{L}_{x}^{2} \right\rangle + \left\langle \hat{L}_{y}^{2} \right\rangle + \left\langle \hat{L}_{z}^{2} \right\rangle$$

$$\left\langle \hat{L}_{x}^{2} \right\rangle = 1/2 (\left\langle \hat{L}^{2} \right\rangle - \left\langle \hat{L}_{z}^{2} \right\rangle )$$

I dont think that this is really showing the solution since i have just stated that

$$\left\langle \hat{L}^{2} \right\rangle = \left\langle \hat{L}_{x}^{2} \right\rangle + \left\langle \hat{L}_{y}^{2} \right\rangle + \left\langle \hat{L}_{z}^{2} \right\rangle$$

and

$$\left\langle \hat{L}_{z}^{2} \right\rangle = (\hbar m)^{2}$$. Unless it is genrally true that $$\left\langle \hat{L}^{2} \right\rangle = \left\langle \hat{L} \right\rangle^{2$$

What do you think?

Last edited: Nov 15, 2009
2. Nov 16, 2009

### gabbagabbahey

In order to calculate any expectation value, you need to know the state of the system. I assume that you are trying to calculate $\langle \hat{L}_x^2\rangle$ in the state $|l,m\rangle$?

Well, certainly $\langle\hat{L}^2\rangle=\langle\hat{L}_x^2+\hat{L}_y^2+L_z^2\rangle$...What is the definition of expectation value?...Do the inner products involved satisfy properties that will allow you to conclude that $\langle\hat{L}_x^2+\hat{L}_y^2+L_z^2\rangle=\langle\hat{L}_x^2\rangle+\langle\hat{L}_y^2\rangle+\langle\hat{L}_z^2\rangle$?

Again, appeal to the definition of expectation value...

3. Nov 16, 2009

### Ben4000

Yes

$$\left\langle \hat{L}^{2} \right\rangle = \left\langle l,m\right| L^{2} \left|l,m\right\rangle$$

$$\left\langle l,m\right| L^{2} \left|l,m\right\rangle = \left\langle l,m\right| L_{x}^{2}+L_{y}^{2}+L_{z}^{2} \left|l,m\right\rangle$$

$$\left\langle l,m\right| L^{2} \left|l,m\right\rangle = \left\langle l,m\right| L_{x}^{2}\left|l,m\right\rangle + \left\langle l,m\right|L_{y}^{2}\left|l,m\right\rangle +\left\langle l,m\right| L_{z}^{2} \left|l,m\right\rangle$$

$$\left\langle \hat{L}^{2} \right\rangle = \left\langle \hat{L}_{x}^{2} \right\rangle + \left\langle \hat{L}_{y}^{2} \right\rangle + \left\langle \hat{L}_{z}^{2} \right\rangle$$

yes?

$$\left\langle l,m\right| L_{z} \left|l,m\right\rangle = \hbar m$$

$$L_{z} \left|l,m\right\rangle = \hbar m \left|l,m\right\rangle$$

$$L_{z}^{2} \left|l,m\right\rangle = (\hbar m)^{2} \left|l,m\right\rangle$$

$$\left\langle l,m\right| L_{z}^{2} \left|l,m\right\rangle = (\hbar m)^{2}$$

4. Nov 16, 2009

### gabbagabbahey

$$\langle L_{z}^{2} \rangle=\left\langle l,m\right| L_{z}^{2} \left|l,m\right\rangle=\langle l,m| (\hbar m)^{2} |l,m\rangle=\hbar^2m^2\langle l,m |l,m\rangle$$

5. Nov 16, 2009

### Ben4000

I am not sure how you can infer that $$\hat{L}_{z}^{2} = (\hbar m)^{2}$$ from $$L_{z} \left|l,m\right\rangle = \hbar m \left|l,m\right\rangle$$

Last edited: Nov 16, 2009
6. Nov 16, 2009

### gabbagabbahey

You can't!

However, you can infer that

[tex]L_z^2|l,m\rangle=L_zL_z|l,m\rangle=L_z(\hbar m|l,m\rangle)=\hbar m L_z|l,m\rangle=\hbar^2 m^2|l,m\rangle[/itex]

and that's all that's needed.