Angular momentum and Expectation values (Another question)

[Ben4000]

Homework Statement

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}.

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 don't 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?

 

[gabbagabbahey]

Homework Statement

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}.

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?

I don't 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

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?

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?

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

 

[Ben4000]

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?

Yes

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?

\left\langle \hat{L}^{2} \right\rangle = \left\langle l,m\right|<br /> L^{2} \left|l,m\right\rangle<br />

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

<br /> \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<br />

yes?

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

Still not so sure about this one...

<br /> \left\langle l,m\right| L_{z} \left|l,m\right\rangle = \hbar m<br />

<br /> L_{z} \left|l,m\right\rangle = \hbar m \left|l,m\right\rangle<br />

<br /> L_{z}^{2} \left|l,m\right\rangle = (\hbar m)^{2} \left|l,m\right\rangle<br />

<br /> \left\langle l,m\right| L_{z}^{2} \left|l,m\right\rangle = (\hbar m)^{2}<br />

 

[gabbagabbahey]

Still not so sure about this one...

<br /> L_{z}^{2} \left|l,m\right\rangle = (\hbar m)^{2} \left|l,m\right\rangle<br />

<br /> \left\langle l,m\right| L_{z}^{2} \left|l,m\right\rangle = (\hbar m)^{2}<br />

Why aren't you sure about this?

\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

 

[Ben4000]

Why aren't you sure about this?

\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

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

 

[gabbagabbahey]

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

You can't!

However, you can infer that

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] <br /> <br /> and that&#039;s all that&#039;s needed.