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## Homework Statement

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

## Homework Equations

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

## 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 [tex] \hat{L}_{y}[/tex] and [tex] \hat{L}_{z}[/tex] aren't compatible operators.

I've tried,

[tex] \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) [/tex]

Then,

[tex]\hat{L}_{z}| l, m \rangle = \hbar m | l, m \rangle[/tex]

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

So it follows that,

[tex] \langle \hat{L}_{x} \rangle = 0[/tex]

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!

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