Q: [1] pr 4.12

Solve the spin 1 Hamiltonian
\label{eqn:crystalSpinHamiltonianTimeReversal:20}
H = A S_z^2 + B(S_x^2 – S_y^2).

Is this Hamiltonian invariant under time reversal?

How do the eigenkets change under time reversal?

A:

In spinMatrices.nb the matrix representation of the Hamiltonian is found to be
\label{eqn:crystalSpinHamiltonianTimeReversal:40}
H =
\Hbar^2
\begin{bmatrix}
A+\frac{B}{2} & 0 & \frac{B}{2} \\
-\frac{i B}{\sqrt{2}} & B & -\frac{i B}{\sqrt{2}} \\
\frac{B}{2} & 0 & A+\frac{B}{2} \\
\end{bmatrix}.

The eigenvalues are
\label{eqn:crystalSpinHamiltonianTimeReversal:60}
\setlr{ \Hbar^2 A, \Hbar^2 B, \Hbar^2(A + B)},

and the respective eigenvalues (unnormalized) are

\label{eqn:crystalSpinHamiltonianTimeReversal:80}
\setlr{
\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix},
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
1 \\
-\frac{i \sqrt{2} B}{A} \\
1 \\
\end{bmatrix}
}.

Under time reversal, the Hamiltonian is

\label{eqn:crystalSpinHamiltonianTimeReversal:100}
H \rightarrow A (-S_z)^2 + B ( (-S_x)^2 – (-S_y)^2 ) = H,

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

\label{eqn:crystalSpinHamiltonianTimeReversal:120}
\Theta \ket{1, m} = (-1)^m \ket{1, -m},

or

\label{eqn:crystalSpinHamiltonianTimeReversal:140}
\begin{aligned}
\Theta \ket{1} &= -\ket{-1} \\
\Theta \ket{0} &= \ket{0} \\
\Theta \ket{-1} &= -\ket{1}.
\end{aligned}

Let’s write the eigenkets respectively as

\label{eqn:crystalSpinHamiltonianTimeReversal:160}
\begin{aligned}
\ket{A} &= -\ket{1} + \ket{-1} \\
\ket{B} &= \ket{0} \\
\ket{A+B} &= \ket{1} + \ket{-1} – \frac{i \sqrt{2} B}{A} \ket{0}.
\end{aligned}

Noting that the time reversal operator maps complex numbers onto their conjugates, the time reversed eigenkets are

\label{eqn:crystalSpinHamiltonianTimeReversal:180}
\begin{aligned}
\ket{A} &\rightarrow \ket{-1} – \ket{-1} = -\ket{A} \\
\ket{B} &\rightarrow \ket{0} = \ket{B} \\
\ket{A+B} &\rightarrow -\ket{1} – \ket{-1} + \frac{i \sqrt{2} B}{A} \ket{0} = -\ket{A+B}.
\end{aligned}

Up to a sign, the time reversed states match the unreversed states.

References

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