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### Q: [1] 3.17

Given a wave function

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

\psi(r,\theta, \phi) = f(r) \lr{ x + y + 3 z },

\end{equation}

- (a) Determine if this wave function is an eigenfunction of \( \BL^2 \), and the value of \( l \) if it is an eigenfunction.
- (b) Determine the probabilities for the particle to be found in any given \( \ket{l, m} \) state,
- (c) If it is known that \( \psi \) is an energy eigenfunction with energy \( E \) indicate how we can find \( V(r) \).

### A: (a)

Using

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

\BL^2

=

-\Hbar^2 \lr{ \inv{\sin^2\theta} \partial_{\phi\phi} + \inv{\sin\theta} \partial_\theta \lr{ \sin\theta \partial_\theta} },

\end{equation}

and

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

\begin{aligned}

x &= r \sin\theta \cos\phi \\

y &= r \sin\theta \sin\phi \\

z &= r \cos\theta

\end{aligned}

\end{equation}

it’s a quick computation to show that

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

\BL^2 \psi = 2 \Hbar^2 \psi = 1(1 + 1) \Hbar^2 \psi,

\end{equation}

so this function is an eigenket of \( \BL^2 \) with an eigenvalue of \( 2 \Hbar^2 \), which corresponds to \( l = 1 \), a p-orbital state.

### (b)

Recall that the angular representation of \( L_z \) is

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

L_z = -i \Hbar \PD{\phi},

\end{equation}

so we have

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

\begin{aligned}

L_z x &= i \Hbar y \\

L_z y &= – i \Hbar x \\

L_z z &= 0,

\end{aligned}

\end{equation}

The \( L_z \) action on \( \psi \) is

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

L_z \psi = -i \Hbar r f(r) \lr{ – y + x }.

\end{equation}

This wave function is not an eigenket of \( L_z \). Expressed in terms of the \( L_z \) basis states \( e^{i m \phi} \), this wave function is

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

\begin{aligned}

\psi

&= r f(r) \lr{ \sin\theta \lr{ \cos\phi + \sin\phi} + \cos\theta } \\

&= r f(r) \lr{ \frac{\sin\theta}{2} \lr{ e^{i \phi} \lr{ 1 + \inv{i}} + e^{-i\phi} \lr{ 1 – \inv{i} } } + \cos\theta } \\

&= r f(r) \lr{

\frac{(1-i)\sin\theta}{2} e^{1 i \phi}

+

\frac{(1+i)\sin\theta}{2} e^{- 1 i \phi}

+ \cos\theta e^{0 i \phi}

}

\end{aligned}

\end{equation}

Assuming that \( \psi \) is normalized, the probabilities for measuring \( m = 1,-1,0 \) respectively are

\begin{equation}\label{eqn:LsquaredLzProblem:180}

\begin{aligned}

P_{\pm 1}

&= 2 \pi \rho \Abs{\frac{1\mp i}{2}}^2 \int_0^\pi \sin\theta d\theta \sin^2 \theta \\

&= -2 \pi \rho \int_1^{-1} du (1-u^2) \\

&= 2 \pi \rho \evalrange{ \lr{ u – \frac{u^3}{3} } }{-1}{1} \\

&= 2 \pi \rho \lr{ 2 – \frac{2}{3}} \\

&= \frac{ 8 \pi \rho}{3},

\end{aligned}

\end{equation}

and

\begin{equation}\label{eqn:LsquaredLzProblem:200}

P_{0} = 2 \pi \rho \int_0^\pi \sin\theta \cos\theta = 0,

\end{equation}

where

\begin{equation}\label{eqn:LsquaredLzProblem:220}

\rho = \int_0^\infty r^4 \Abs{f(r)}^2 dr.

\end{equation}

Because the probabilities must sum to 1, this means the \( m = \pm 1 \) states are equiprobable with \( P_{\pm} = 1/2 \), fixing \( \rho = 3/16\pi \), even without knowing \( f(r) \).

### (c)

The operator \( r^2 \Bp^2 \) can be decomposed into a \( \BL^2 \) component and some other portions, from which we can write

\begin{equation}\label{eqn:LsquaredLzProblem:240}

\begin{aligned}

H \psi

&= \lr{ \frac{\Bp^2}{2m} + V(r) } \psi \\

&=

\lr{

– \frac{\Hbar^2}{2m} \lr{ \partial_{rr} + \frac{2}{r} \partial_r – \inv{\Hbar^2 r^2} \BL^2 } + V(r) } \psi.

\end{aligned}

\end{equation}

(See: [1] eq. 6.21)

In this case where \( \BL^2 \psi = 2 \Hbar^2 \psi \) we can rearrange for \( V(r) \)

\begin{equation}\label{eqn:LsquaredLzProblem:260}

\begin{aligned}

V(r)

&= E + \inv{\psi} \frac{\Hbar^2}{2m} \lr{ \partial_{rr} + \frac{2}{r} \partial_r – \frac{2}{r^2} } \psi \\

&= E + \inv{f(r)} \frac{\Hbar^2}{2m} \lr{ \partial_{rr} + \frac{2}{r} \partial_r – \frac{2}{r^2} } f(r).

\end{aligned}

\end{equation}

# References

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