Phy5645/angularmomcommutation/

Question: In the angular momentum basis, compute $\left\langle {l,m\left|{L_{x}^{2}}\right|l,m}\right\rangle$ and $\left\langle {l,m\left|{L_{x}L_{y}}\right|l,m}\right\rangle$.

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Solution:

• $\left\langle {l,m\left|{L_{x}^{2}}\right|l,m}\right\rangle$

Since $L_{x}={\frac {L_{+}+L_{-}}{2}}$, we can obtain $L_{x}^{2}$;

$L_{x}^{2}={\frac {1}{4}}(L_{+}+L_{-})^{2}={\frac {1}{4}}(L_{+}^{2}-L_{-}^{2}+2L_{+}L_{-}+2L_{-}L_{+})$

By definition, $L_{+}^{2}$ and $L_{-}^{2}$ won't contribute because $\left\langle {l,m\left|{L_{+}^{2}}\right|l,m}\right\rangle =\left\langle {l,m\left|{l,m+2}\right.}\right\rangle =0$ and $\left\langle {l,m\left|{L_{-}^{2}}\right|l,m}\right\rangle =\left\langle {l,m\left|{l,m-2}\right.}\right\rangle =0$

$\left\langle {l,m\left|{L_{x}^{2}}\right|l,m}\right\rangle ={\frac {1}{4}}\left\lbrace {\left\langle {l,m\left|{2L_{+}L_{-}+2L_{-}L_{+}}\right|l,m}\right\rangle }\right\rbrace$

$={\frac {1}{2}}\left\lbrace {\left\langle {l,m\left|{L_{+}L_{-}}\right|l,m}\right\rangle +\left\langle {l,m\left|{L_{-}L_{+}}\right|l,m}\right\rangle }\right\rbrace$

${\text{=}}{\frac {\hbar }{2}}{\sqrt {l(l+1)-m(m-1)}}\left\langle {l,m\left|{L_{+}}\right|l,m-1}\right\rangle +{\frac {\hbar }{2}}{\sqrt {l(l+1)-m(m+1)}}\left\langle {l,m\left|{L_{-}}\right|l,m+1}\right\rangle$

${\text{=}}{\frac {\hbar ^{2}}{2}}{\sqrt {l(l+1)-m(m-1)}}{\sqrt {l(l+1)-(m-1)m}}\left\langle {l,m}\right|\left.{}{l,m}\right\rangle +{\frac {\hbar ^{2}}{2}}{\sqrt {l(l+1)-m(m+1)}}{\sqrt {l(l+1)-(m+1)m}}\left\langle {l,m}\right|\left.{}{l,m}\right\rangle$

$={\frac {\hbar ^{2}}{2}}\left\lbrace {l(l+1)-m(m-1)+l(l+1)-m(m-1)}\right\rbrace$ $={\frac {\hbar ^{2}}{2}}\left\lbrace {2l(l+1)-2m^{2}}\right\rbrace$

$=\hbar ^{2}(l(l+1)-m^{2})$

• $\left\langle {l,m\left|{L_{x}L_{y}}\right|l,m}\right\rangle$

$L_{x}L_{y}={\frac {L_{+}+L_{-}}{2}}{\frac {L_{+}-L_{-}}{2i}}$

$={\frac {1}{4i}}(L_{+}+L_{-})(L_{+}-L_{-})={\frac {1}{4i}}\left\lbrace {L_{+}^{2}-L_{-}^{2}-L_{+}L_{-}+L_{-}L_{+}}\right\rbrace$ Again, $L_{+}^{2}$ and $L_{-}^{2}$ won't contribute

$={\frac {1}{4i}}\left\langle {l,m\left|{L_{-}L_{+}-L_{+}L_{-}}\right|l,m}\right\rangle ={\frac {1}{4i}}\left\lbrace {\left\langle {l,m\left|{L_{-}L_{+}}\right|l,m}\right\rangle -\left\langle {l,m\left|{L_{+}L_{-}}\right|l,m}\right\rangle }\right\rbrace$

$={\frac {\hbar }{4i}}\left\lbrace {{\sqrt {l(l+1)-m(m+1)}}\left\langle {l,m\left|{L_{-}}\right|l,m+1}\right\rangle -{\sqrt {l(l+1)-m(m-1)}}\left\langle {l,m\left|{L_{+}}\right|l,m-1}\right\rangle }\right\rbrace$

$={\frac {\hbar ^{2}}{4i}}\left\lbrace {{\sqrt {l(l+1)-m(m+1)}}{\sqrt {l(l+1)-(m+1)m}}\left\langle {l,m}\right|\left.{}{l,m}\right\rangle -{\sqrt {l(l+1)-m(m-1)}}{\sqrt {l(l+1)-(m-1)m}}\left\langle {l,m}\right|\left.{}{l,m}\right\rangle }\right\rbrace$

$={\frac {\hbar ^{2}}{4i}}\left({l(l+1)-m^{2}-m-l(l+1)+m^{2}-m}\right)$

$={\frac {\hbar ^{2}}{4i}}(-2m)={\frac {i\hbar ^{2}m}{2}}$

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