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| '''Question:''' In the angular momentum basis, compute <math>\left \langle {l,m\left |{L_{x}^{2}} \right |l,m} \right \rangle </math> and <math>\left \langle {l,m\left |{L_{x}L_{y}} \right |l,m} \right \rangle </math>. | | '''(a)''' <math>\left \langle {l,m\left |{L_{x}^{2}} \right |l,m} \right \rangle </math> |
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| '''Solution:'''
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| * <math>\left \langle {l,m\left |{L_{x}^{2}} \right |l,m} \right \rangle </math>
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| Since <math>L_{x}=\frac{L_{+}+L_{-}}{2}</math>, we can obtain <math>L_{x}^{2}</math>; | | Since <math>L_{x}=\frac{L_{+}+L_{-}}{2}</math>, we can obtain <math>L_{x}^{2}</math>; |
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| * <math>\left \langle {l,m\left |{L_{x}L_{y}} \right |l,m} \right \rangle </math>
| | '''(b)''' <math>\left \langle {l,m\left |{L_{x}L_{y}} \right |l,m} \right \rangle </math> |
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| <math>L_{x}L_{y}=\frac{L_{+}+L_{-}}{2}\frac{L_{+}-L_{-}}{2i}</math> | | <math>L_{x}L_{y}=\frac{L_{+}+L_{-}}{2}\frac{L_{+}-L_{-}}{2i}</math> |
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| <math>=\frac{\hbar ^{2}}{4i}(-2m)=\frac{i\hbar ^{2}m}{2}</math> | | <math>=\frac{\hbar ^{2}}{4i}(-2m)=\frac{i\hbar ^{2}m}{2}</math> |
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(a)
Since
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By definition,
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(b)
Again,
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