- #1
Ben4000
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Homework Statement
Using the fact that ,[tex]\left\langle \hat{L}_{x}^{2} \right\rangle = \left\langle \hat{L}_{y}^{2} \right\rangle[/tex] show that [tex]\left\langle \hat{L}_{x}^{2} \right\rangle = 1/2 \hbar^{2}(l(l+1)-m^{2}.[/tex]
The Attempt at a Solution
[tex]L^{2} \left|l,m\right\rangle = \hbar^{2}l(l+1) \left|l,m\right\rangle[/tex]
[tex]L_{z} \left|l,m\right\rangle = \hbar m \left|l,m\right\rangle[/tex]
[tex]\left\langle \hat{L}^{2} \right\rangle = \hbar^{2}l(l+1)[/tex]
[tex]\left\langle \hat{L}_{z}^{2} \right\rangle = (\hbar m)^{2}[/tex][tex]\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[/tex]
[tex]\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 )[/tex]
I don't think that this is really showing the solution since i have just stated that
[tex]\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[/tex]
and
[tex]\left\langle \hat{L}_{z}^{2} \right\rangle = (\hbar m)^{2}[/tex]. Unless it is genrally true that [tex]\left\langle \hat{L}^{2} \right\rangle = \left\langle \hat{L} \right\rangle^{2[/tex]
What do you think?
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