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Question: Heisenberg picture spin precession ([1] pr. 2.1)

For the spin Hamiltonian

\begin{equation}\label{eqn:heisenbergSpinPrecession:20}
H = -\frac{e B}{m c} S_z = \omega S_z,
\end{equation}

express and solve the Heisenberg equations of motion for \( S_x(t), S_y(t) \), and \( S_z(t) \).

Answer

The equations of motion are of the form

\begin{equation}\label{eqn:heisenbergSpinPrecession:40}
\begin{aligned}
\frac{dS_i^\textrm{H}}{dt}
&= \inv{i \Hbar} \antisymmetric{S_i^\textrm{H}}{H} \\
&= \inv{i \Hbar} \antisymmetric{U^\dagger S_i U}{H} \\
&= \inv{i \Hbar} \lr{U^\dagger S_i U H – H U^\dagger S_i U } \\
&= \inv{i \Hbar} U^\dagger \lr{ S_i H – H S_i } U \\
&= \frac{\omega}{i \Hbar} U^\dagger \antisymmetric{ S_i}{S_z } U.
\end{aligned}
\end{equation}

These are

\begin{equation}\label{eqn:heisenbergSpinPrecession:60}
\begin{aligned}
\frac{dS_x^\textrm{H}}{dt} &= -\omega U^\dagger S_y U \\
\frac{dS_y^\textrm{H}}{dt} &= \omega U^\dagger S_x U \\
\frac{dS_z^\textrm{H}}{dt} &= 0.
\end{aligned}
\end{equation}

To completely specify these equations, we need to expand \( U(t) \), which is

\begin{equation}\label{eqn:heisenbergSpinPrecession:80}
\begin{aligned}
U(t)
&= e^{-i H t /\Hbar} \\
&= e^{-i \omega S_z t /\Hbar} \\
&= e^{-i \omega \sigma_z t /2} \\
&= \cos\lr{ \omega t/2 } -i \sigma_z \sin\lr{ \omega t/2 } \\
&=
\begin{bmatrix}
\cos\lr{ \omega t/2 } -i \sin\lr{ \omega t/2 } & 0 \\
0 & \cos\lr{ \omega t/2 } + i \sin\lr{ \omega t/2 }
\end{bmatrix} \\
&=
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix}.
\end{aligned}
\end{equation}

The equations of motion can now be written out in full. To do so seems a bit silly since we also know that \( S_x^\textrm{H} = U^\dagger S_x U, S_y^\textrm{H} U^\dagger S_x U \). However, if that is temporarily forgotten, we can show that the Heisenberg equations of motion can be solved for these too.

\begin{equation}\label{eqn:heisenbergSpinPrecession:100}
\begin{aligned}
U^\dagger S_x U
&=
\frac{\Hbar}{2}
\begin{bmatrix}
e^{i\omega t/2} & 0 \\
0 & e^{-i\omega t/2}
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t/2} \\
e^{-i\omega t/2} & 0
\end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix},
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:heisenbergSpinPrecession:120}
\begin{aligned}
U^\dagger S_y U
&=
\frac{\Hbar}{2}
\begin{bmatrix}
e^{i\omega t/2} & 0 \\
0 & e^{-i\omega t/2}
\end{bmatrix}
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{i\Hbar}{2}
\begin{bmatrix}
0 & -e^{i\omega t/2} \\
e^{-i\omega t/2} & 0
\end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{i \Hbar}{2}
\begin{bmatrix}
0 & -e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix}.
\end{aligned}
\end{equation}

The equations of motion are now fully specified

\begin{equation}\label{eqn:heisenbergSpinPrecession:140}
\begin{aligned}
\frac{dS_x^\textrm{H}}{dt} &=
-\frac{i \Hbar \omega}{2}
\begin{bmatrix}
0 & -e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
\frac{dS_y^\textrm{H}}{dt} &=
\frac{\Hbar \omega}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
\frac{dS_z^\textrm{H}}{dt} &= 0.
\end{aligned}
\end{equation}

Integration gives

\begin{equation}\label{eqn:heisenbergSpinPrecession:160}
\begin{aligned}
S_x^\textrm{H} &=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} + C \\
S_y^\textrm{H} &=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & -i e^{i\omega t} \\
i e^{-i\omega t} & 0
\end{bmatrix} + C \\
S_z^\textrm{H} &= C.
\end{aligned}
\end{equation}

The integration constants are fixed by the boundary condition \( S_i^\textrm{H}(0) = S_i \), so

\begin{equation}\label{eqn:heisenbergSpinPrecession:180}
\begin{aligned}
S_x^\textrm{H} &=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
S_y^\textrm{H} &=
\frac{i \Hbar}{2}
\begin{bmatrix}
0 & – e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
S_z^\textrm{H} &= S_z.
\end{aligned}
\end{equation}

Observe that these integrated values \( S_x^\textrm{H}, S_y^\textrm{H} \) match \ref{eqn:heisenbergSpinPrecession:100}, and \ref{eqn:heisenbergSpinPrecession:120} as expected.

References

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