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

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

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

A:

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

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

The first order energy shift is zero

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

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

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

and the next order energy shift is

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

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.

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

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

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

so to first order the energy shift is

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

Exact perturbation equation

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

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

This Hamiltonian matrix has a very regular structure

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

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

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

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

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

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.