## L_y perturbation

December 13, 2015 phy1520 , , , ,

### Q: $$L_y$$ perturbation. [1] pr. 5.17

Find the first non-zero energy shift for the perturbed Hamiltonian

\label{eqn:LyPerturbation:20}
H = A \BL^2 + B L_z + C L_y = H_0 + V.

### A:

The energy eigenvalues for state $$\ket{l, m}$$ prior to perturbation are

\label{eqn:LyPerturbation:40}
A \Hbar^2 l(l+1) + B \Hbar m.

The first order energy shift is zero

\label{eqn:LyPerturbation:60}
\begin{aligned}
\Delta^1
&=
\bra{l, m} C L_y \ket{l, m} \\
&=
\frac{C}{2 i}
\bra{l, m} \lr{ L_{+} – L_{-} } \ket{l, m} \\
&=
0,
\end{aligned}

so we need the second order shift. Assuming no degeneracy to start, the perturbed state is

\label{eqn:LyPerturbation:80}
\ket{l, m}’ = \sum’ \frac{\ket{l’, m’} \bra{l’, m’}}{E_{l,m} – E_{l’, m’}} V \ket{l, m},

and the next order energy shift is
\label{eqn:LyPerturbation:100}
\begin{aligned}
\Delta^2
&=
\bra{l m} V
\sum’ \frac{\ket{l’, m’} \bra{l’, m’}}{E_{l,m} – E_{l’, m’}} V \ket{l, m} \\
&=
\sum’ \frac{\bra{l, m} V \ket{l’, m’} \bra{l’, m’}}{E_{l,m} – E_{l’, m’}} V \ket{l, m} \\
&=
\sum’ \frac{ \Abs{ \bra{l’, m’} V \ket{l, m} }^2 }{E_{l,m} – E_{l’, m’}} \\
&=
\sum_{m’ \ne m} \frac{ \Abs{ \bra{l, m’} V \ket{l, m} }^2 }{E_{l,m} – E_{l, m’}} \\
&=
\sum_{m’ \ne m} \frac{ \Abs{ \bra{l, m’} V \ket{l, m} }^2 }{
\lr{ A \Hbar^2 l(l+1) + B \Hbar m }
-\lr{ A \Hbar^2 l(l+1) + B \Hbar m’ }
} \\
&=
\inv{B \Hbar} \sum_{m’ \ne m} \frac{ \Abs{ \bra{l, m’} V \ket{l, m} }^2 }{
m – m’
}.
\end{aligned}

The sum over $$l’$$ was eliminated because $$V$$ only changes the $$m$$ of any state $$\ket{l,m}$$, so the matrix element $$\bra{l’,m’} V \ket{l, m}$$ must includes a $$\delta_{l’, l}$$ factor.
Since we are now summing over $$m’ \ne m$$, some of the matrix elements in the numerator should now be non-zero, unlike the case when the zero first order energy shift was calculated above.

\label{eqn:LyPerturbation:120}
\begin{aligned}
\bra{l, m’} C L_y \ket{l, m}
&=
\frac{C}{2 i}
\bra{l, m’} \lr{ L_{+} – L_{-} } \ket{l, m} \\
&=
\frac{C}{2 i}
\bra{l, m’}
\lr{
L_{+}
\ket{l, m}
– L_{-}
\ket{l, m}
} \\
&=
\frac{C \Hbar}{2 i}
\bra{l, m’}
\lr{
\sqrt{(l – m)(l + m + 1)} \ket{l, m + 1}

\sqrt{(l + m)(l – m + 1)} \ket{l, m – 1}
} \\
&=
\frac{C \Hbar}{2 i}
\lr{
\sqrt{(l – m)(l + m + 1)} \delta_{m’, m + 1}

\sqrt{(l + m)(l – m + 1)} \delta_{m’, m – 1}
}.
\end{aligned}

After squaring and summing, the cross terms will be zero since they involve products of delta functions with different indices. That leaves

\label{eqn:LyPerturbation:140}
\begin{aligned}
\Delta^2
&=
\frac{C^2 \Hbar}{4 B} \sum_{m’ \ne m} \frac{
(l – m)(l + m + 1) \delta_{m’, m + 1}

(l + m)(l – m + 1) \delta_{m’, m – 1}
}{
m – m’
} \\
&=
\frac{C^2 \Hbar}{4 B}
\lr{
\frac{ (l – m)(l + m + 1) }{ m – (m+1) }

\frac{ (l + m)(l – m + 1) }{ m – (m-1)}
} \\
&=
\frac{C^2 \Hbar}{4 B}
\lr{

(l^2 – m^2 + l – m)

(l^2 – m^2 + l + m)
} \\
&=
-\frac{C^2 \Hbar}{2 B} (l^2 – m^2 + l ),
\end{aligned}

so to first order the energy shift is

\label{eqn:LyPerturbation:160}
\boxed{
A \Hbar^2 l(l+1) + B \Hbar m \rightarrow
\Hbar l(l+1)
\lr{
A \Hbar
-\frac{C^2}{2 B}
}
+ B \Hbar m
+\frac{C^2 m^2 \Hbar}{2 B} .
}

### Exact perturbation equation

If we wanted to solve the Hamiltonian exactly, we’ve have to diagonalize the $$2 m + 1$$ dimensional Hamiltonian

\label{eqn:LyPerturbation:180}
\bra{l, m’} H \ket{l, m}
=
\lr{ A \Hbar^2 l(l+1) + B \Hbar m } \delta_{m’, m}
+
\frac{C \Hbar}{2 i}
\lr{
\sqrt{(l – m)(l + m + 1)} \delta_{m’, m + 1}

\sqrt{(l + m)(l – m + 1)} \delta_{m’, m – 1}
}.

This Hamiltonian matrix has a very regular structure

\label{eqn:LyPerturbation:200}
\begin{aligned}
H &=
(A l(l+1) \Hbar^2 – B \Hbar (l+1)) I \\
&+ B \Hbar
\begin{bmatrix}
1 & & & & \\
& 2 & & & \\
& & 3 & & \\
& & & \ddots & \\
& & & & 2 l + 1
\end{bmatrix} \\
&+
\frac{C \Hbar}{i}
\begin{bmatrix}
0 & -\sqrt{(2l-1)(1)} & & & \\
\sqrt{(2l-1)(1)} & 0 & -\sqrt{(2l-2)(2)}& & \\
& \sqrt{(2l-2)(2)} & & & \\
& & \ddots & & \\
& & & 0 & – \sqrt{(1)(2l-1)} \\
& & & \sqrt{(1)(2l-1)} & 0
\end{bmatrix}
\end{aligned}

Solving for the eigenvalues of this Hamiltonian for increasing $$l$$ in Mathematica (sakuraiProblem5.17a.nb), it appears that the eigenvalues are

\label{eqn:LyPerturbation:220}
\lambda_m = A \Hbar^2 (l)(l+1) + \Hbar m B \sqrt{ 1 + \frac{4 C^2}{B^2} },

so to first order in $$C^2$$, these are

\label{eqn:LyPerturbation:221}
\lambda_m = A \Hbar^2 (l)(l+1) + \Hbar m B \lr{ 1 + \frac{2 C^2}{B^2} }.

We have a $$C^2 \Hbar/B$$ term in both the perturbative energy shift, and the first order expansion of the exact solution. Comparing this for the $$l = 5$$ case, the coefficients of $$C^2 \Hbar/B$$ in the perturbative solution are all negative $$-17.5, -17., -16.5, -16., -15.5, -15., -14.5, -14., -13.5, -13., -12.5$$, whereas the coefficient of $$C^2 \Hbar/B$$ in the first order expansion of the exact solution are $$2 m$$, ranging from $$[-10, 10]$$.

# References

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

## L_z and L^2 eigenvalues and probabilities for a given wave function

### 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.