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Question: A symmetric real Hamiltonian ([1] pr. 2.9)

Find the time evolution for the state \( \ket{a’} \) for a Hamiltian of the form

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:20}
H = \delta \lr{ \ket{a’}\bra{a’} + \ket{a”}\bra{a”} }
\end{equation}

Answer

This Hamiltonian has the matrix representation

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:40}
H =
\begin{bmatrix}
0 & \delta \\
\delta & 0
\end{bmatrix},
\end{equation}

which has a characteristic equation of

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:60}
\lambda^2 -\delta^2 = 0,
\end{equation}

so the energy eigenvalues are \( \pm \delta \).

The diagonal basis states are respectively

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:80}
\ket{\pm\delta} =
\inv{\sqrt{2}}
\begin{bmatrix}
\pm 1 \\
1
\end{bmatrix}.
\end{equation}

The time evolution operator is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:100}
\begin{aligned}
U
&= e^{-i H t/\Hbar} \\
&=
e^{-i \delta t/\Hbar} \ket{+\delta}\bra{+\delta}
+ e^{i \delta t/\Hbar} \ket{-\delta}\bra{-\delta} \\
&=
\frac{e^{-i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
+ \frac{e^{i \delta t/\Hbar} }{2}
\begin{bmatrix}
-1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 \\
1
\end{bmatrix} \\
&=
\frac{e^{-i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & 1 \\
1 & 1
\end{bmatrix}
+\frac{e^{i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & -1 \\
-1 & 1
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) & -i\sin(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar) & \cos(\delta t/\Hbar) \\
\end{bmatrix}.
\end{aligned}
\end{equation}

The desired time evolution in the original basis is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:140}
\begin{aligned}
\ket{a’, t}
&=
e^{-i H t/\Hbar}
\ket{a’, 0} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) & -i\sin(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar) & \cos(\delta t/\Hbar) \\
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar)
\end{bmatrix} \\
&=
\cos(\delta t/\Hbar) \ket{a’,0} -i \sin(\delta t/\Hbar) \ket{a”,0}.
\end{aligned}
\end{equation}

This evolution has the same structure as left circularly polarized light.

The probability of finding the system in state \( \ket{a”} \) given an initial state of \( \ket{a’,0} \) is

\begin{equation}\label{eqn:symmetricHamiltonianEvolution:160}
P
=
\Abs{\braket{a”}{a’,t}}^2
=
\sin^2 \lr{ \delta t/\Hbar }.
\end{equation}

References

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