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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.