Phy5645/angularmomcommutation/

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

Since ${\hat {L}}_{x}={\frac {{\hat {L}}_{+}+{\hat {L}}_{-}}{2}},$

$L_{x}^{2}={\tfrac {1}{4}}({\hat {L}}_{+}+{\hat {L}}_{-})^{2}={\tfrac {1}{4}}({\hat {L}}_{+}^{2}+{\hat {L}}_{-}^{2}+{\hat {L}}_{+}{\hat {L}}_{-}+{\hat {L}}_{-}{\hat {L}}_{+}).$

By definition, ${\hat {L}}_{+}^{2}$ and ${\hat {L}}_{-}^{2}$ won't contribute because they reduce to an inner product of the form, $\left\langle {l,m\left|{l,m\pm 2}\right.}\right\rangle =0.$ Therefore, the only contribution is from the last two terms:

$\left\langle {l,m\left|{{\hat {L}}_{x}^{2}}\right|l,m}\right\rangle ={\tfrac {1}{4}}\left\lbrace {\left\langle {l,m\left|{{\hat {L}}_{+}{\hat {L}}_{-}+{\hat {L}}_{-}{\hat {L}}_{+}}\right|l,m}\right\rangle }\right\rbrace$

$={\tfrac {1}{4}}\left\lbrace {\left\langle {l,m\left|{{\hat {L}}_{+}{\hat {L}}_{-}}\right|l,m}\right\rangle +\left\langle {l,m\left|{{\hat {L}}_{-}{\hat {L}}_{+}}\right|l,m}\right\rangle }\right\rbrace$

$={\frac {\hbar }{4}}{\sqrt {l(l+1)-m(m-1)}}\left\langle {l,m\left|{{\hat {L}}_{+}}\right|l,m-1}\right\rangle +{\frac {\hbar }{4}}{\sqrt {l(l+1)-m(m+1)}}\left\langle {l,m\left|{{\hat {L}}_{-}}\right|l,m+1}\right\rangle$

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

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

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

(b) $\left\langle {l,m\left|{{\hat {L}}_{x}{\hat {L}}_{y}}\right|l,m}\right\rangle$

${\hat {L}}_{x}{\hat {L}}_{y}={\frac {{\hat {L}}_{+}+{\hat {L}}_{-}}{2}}{\frac {{\hat {L}}_{+}-{\hat {L}}_{-}}{2i}}$

$={\frac {1}{4i}}({\hat {L}}_{+}+{\hat {L}}_{-})({\hat {L}}_{+}-{\hat {L}}_{-})={\frac {1}{4i}}\left\lbrace {{\hat {L}}_{+}^{2}-{\hat {L}}_{-}^{2}-{\hat {L}}_{+}{\hat {L}}_{-}+{\hat {L}}_{-}{\hat {L}}_{+}}\right\rbrace$

Again, ${\hat {L}}_{+}^{2}$ and ${\hat {L}}_{-}^{2}$ won't contribute. We now evaluate the remaining terms:

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

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

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

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

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

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