### Q: [1] 3.17

Given a wave function

\label{eqn:LsquaredLzProblem:20}
\psi(r,\theta, \phi) = f(r) \lr{ x + y + 3 z },

• (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
\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} },

and

\label{eqn:LsquaredLzProblem:60}
\begin{aligned}
x &= r \sin\theta \cos\phi \\
y &= r \sin\theta \sin\phi \\
z &= r \cos\theta
\end{aligned}

it’s a quick computation to show that

\label{eqn:LsquaredLzProblem:80}
\BL^2 \psi = 2 \Hbar^2 \psi = 1(1 + 1) \Hbar^2 \psi,

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

\label{eqn:LsquaredLzProblem:100}
L_z = -i \Hbar \PD{\phi},

so we have

\label{eqn:LsquaredLzProblem:120}
\begin{aligned}
L_z x &= i \Hbar y \\
L_z y &= – i \Hbar x \\
L_z z &= 0,
\end{aligned}

The $$L_z$$ action on $$\psi$$ is

\label{eqn:LsquaredLzProblem:140}
L_z \psi = -i \Hbar r f(r) \lr{ – y + x }.

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

\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}

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

\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}

and

\label{eqn:LsquaredLzProblem:200}
P_{0} = 2 \pi \rho \int_0^\pi \sin\theta \cos\theta = 0,

where

\label{eqn:LsquaredLzProblem:220}
\rho = \int_0^\infty r^4 \Abs{f(r)}^2 dr.

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

\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}

(See: [1] eq. 6.21)

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

\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}

# References

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