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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.