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### Q: [1] pr 4.12

Solve the spin 1 Hamiltonian

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

H = A S_z^2 + B(S_x^2 – S_y^2).

\end{equation}

Is this Hamiltonian invariant under time reversal?

How do the eigenkets change under time reversal?

## Answer

In spinMatrices.nb the matrix representation of the Hamiltonian is found to be

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

H =

\Hbar^2

\begin{bmatrix}

A & 0 & B \\

0 & 0 & 0 \\

B & 0 & A

\end{bmatrix}.

\end{equation}

The eigenvalues are

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

\setlr{ 0, A – B, A + B},

\end{equation}

and the respective eigenvalues (unnormalized) are

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

\setlr{

\begin{bmatrix}

0 \\

1 \\

0

\end{bmatrix},

\begin{bmatrix}

-1 \\

0 \\

1

\end{bmatrix},

\begin{bmatrix}

1 \\

0 \\

1 \\

\end{bmatrix}

}.

\end{equation}

Under time reversal, the Hamiltonian is

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

H \rightarrow A (-S_z)^2 + B ( (-S_x)^2 – (-S_y)^2 ) = H,

\end{equation}

so we expect the eigenkets for this Hamiltonian to vary by at most a phase factor. To check this, first recall that the time reversal action on a spin one state is

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

\Theta \ket{1, m} = (-1)^m \ket{1, -m},

\end{equation}

or

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

\begin{aligned}

\Theta \ket{1,1} &= -\ket{1,-1} \\

\Theta \ket{1,0} &= \ket{1,0} \\

\Theta \ket{1,-1} &= -\ket{1,1}.

\end{aligned}

\end{equation}

Let’s write the eigenkets respectively as

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

\begin{aligned}

\ket{0} &= \ket{1,0} \\

\ket{A-B} &= -\ket{1,-1} + \ket{1,1} \\

\ket{A+B} &= \ket{1,-1} + \ket{1,1}.

\end{aligned}

\end{equation}

Under the reversal operation, we should have

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

\begin{aligned}

\Theta \ket{0} &\rightarrow \ket{1,0} \\

\Theta \ket{A-B} &= +\ket{1,-1} – \ket{1,1} \\

\Theta \ket{A+B} &= -\ket{1,-1} – \ket{1,1}.

\end{aligned}

\end{equation}

Up to a sign, the time reversed states match the unreversed states, which makes sense given the Hamiltonian invariance.

# References

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

Hana3 years agoThis is not correct, you haven’t calculated Hamiltonian H correctly, the contribution from Sy^2 is wrong, the Hamiltonian should be:

H = \hbar^2 {{A, 0, B}, {0, 0, 0}, {B, 0, A}}

Peeter Joot3 years agoThanks Hana. I’ll have to dig up my old Mathematica notebook for this and see what went wrong.

Peeter Joot3 years agoThanks Hana. Fixed in https://github.com/peeterjoot/phy1520-quantum/commit/560d519651339175ec948878d7192a1c4ada9362, and in https://github.com/peeterjoot/mathematica/commit/97c4aab86d3fa100edbdc0f64cf4e1da4f4ff6ff