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### Q: [1] pr 3.18

Compute the expectation values for the first and second powers of the angular momentum operators with respect to states \( \ket{lm} \).

### A:

We can write the expectation values for the \( L_z \) powers immediately

\begin{equation}\label{eqn:angularMomentumExpectation:20}

\expectation{L_z}

= m \Hbar,

\end{equation}

and

\begin{equation}\label{eqn:angularMomentumExpectation:40}

\expectation{L_z^2} = (m \Hbar)^2.

\end{equation}

For the x and y components first express the operators in terms of the ladder operators.

\begin{equation}\label{eqn:angularMomentumExpectation:60}

\begin{aligned}

L_{+} &= L_x + i L_y \\

L_{-} &= L_x – i L_y.

\end{aligned}

\end{equation}

Rearranging gives

\begin{equation}\label{eqn:angularMomentumExpectation:80}

\begin{aligned}

L_x &= \inv{2} \lr{ L_{+} + L_{-} } \\

L_y &= \inv{2i} \lr{ L_{+} – L_{-} }.

\end{aligned}

\end{equation}

The first order expectations \( \expectation{L_x}, \expectation{L_y} \) are both zero since \( \expectation{L_{+}} = \expectation{L_{-}} \). For the second order expectation values we have

\begin{equation}\label{eqn:angularMomentumExpectation:100}

\begin{aligned}

L_x^2

&= \inv{4} \lr{ L_{+} + L_{-} } \lr{ L_{+} + L_{-} } \\

&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + L_{+} L_{-} + L_{-} L_{+} } \\

&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + 2 (L_x^2 + L_y^2) } \\

&= \inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} + 2 (\BL^2 – L_z^2) },

\end{aligned}

\end{equation}

and

\begin{equation}\label{eqn:angularMomentumExpectation:120}

\begin{aligned}

L_y^2

&= -\inv{4} \lr{ L_{+} – L_{-} } \lr{ L_{+} – L_{-} } \\

&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – L_{+} L_{-} – L_{-} L_{+} } \\

&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – 2 (L_x^2 + L_y^2) } \\

&= -\inv{4} \lr{ L_{+} L_{+} + L_{-} L_{-} – 2 (\BL^2 – L_z^2) }.

\end{aligned}

\end{equation}

Any expectation value \( \bra{lm} L_{+} L_{+} \ket{lm} \) or \( \bra{lm} L_{-} L_{-} \ket{lm} \) will be zero, leaving

\begin{equation}\label{eqn:angularMomentumExpectation:140}

\begin{aligned}

\expectation{L_x^2}

&=

\expectation{L_y^2} \\

&=

\inv{4} \expectation{2 (\BL^2 – L_z^2) } \\

&=

\inv{2} \lr{ \Hbar^2 l(l+1) – (\Hbar m)^2 }.

\end{aligned}

\end{equation}

Observe that we have

\begin{equation}\label{eqn:angularMomentumExpectation:160}

\expectation{L_x^2}

+

\expectation{L_y^2}

+

\expectation{L_z^2}

=

\Hbar^2 l(l+1)

=

\expectation{\BL^2},

\end{equation}

which is the quantum mechanical analogue of the classical scalar equation \( \BL^2 = L_x^2 + L_y^2 + L_z^2 \).

# References

[1] Jun John Sakurai and Jim J Napolitano. *Modern quantum mechanics*. Pearson Higher Ed, 2014.