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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.