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## Question: Heisenberg picture position commutator ([1] pr. 2.5)

Evaluate

\begin{equation}\label{eqn:positionCommutator:20}

\antisymmetric{x(t)}{x(0)},

\end{equation}

for a Heisenberg picture operator \( x(t) \) for a free particle.

## Answer

The free particle Hamiltonian is

\begin{equation}\label{eqn:positionCommutator:40}

H = \frac{p^2}{2m},

\end{equation}

so the time evolution operator is

\begin{equation}\label{eqn:positionCommutator:60}

U(t) = e^{-i p^2 t/(2 m \Hbar)}.

\end{equation}

The Heisenberg picture position operator is

\begin{equation}\label{eqn:positionCommutator:80}

\begin{aligned}

x^\textrm{H}

&= U^\dagger x U \\

&= e^{i p^2 t/(2 m \Hbar)} x e^{-i p^2 t/(2 m \Hbar)} \\

&= \sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i p^2 t}{2 m \Hbar} }^k

x

e^{-i p^2 t/(2 m \Hbar)} \\

&= \sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k p^{2k} x

e^{-i p^2 t/(2 m \Hbar)} \\

&=

\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ \antisymmetric{p^{2k}}{x} + x p^{2k} }

e^{-i p^2 t/(2 m \Hbar)} \\

&= x +

\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \antisymmetric{p^{2k}}{x}

e^{-i p^2 t/(2 m \Hbar)} \\

&= x +

\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ -i \Hbar \PD{p}{p^{2k}} }

e^{-i p^2 t/(2 m \Hbar)} \\

&= x +

\sum_{k = 0}^\infty \inv{k!} \lr{ \frac{i t}{2 m \Hbar} }^k \lr{ -i \Hbar 2 k p^{2 k -1} }

e^{-i p^2 t/(2 m \Hbar)} \\

&= x +

-2 i \Hbar p \frac{i t}{2 m \Hbar} \sum_{k = 1}^\infty \inv{(k-1)!} \lr{ \frac{i t}{2 m \Hbar} }^{k-1} p^{2(k – 1)}

e^{-i p^2 t/(2 m \Hbar)} \\

&= x + t \frac{p}{m}.

\end{aligned}

\end{equation}

This has the structure of a classical free particle \( x(t) = x + v t \), but in this case \( x,p \) are operators.

The evolved position commutator is

\begin{equation}\label{eqn:positionCommutator:100}

\begin{aligned}

\antisymmetric{x(t)}{x(0)}

&=

\antisymmetric{x + t p/m}{x} \\

&=

\frac{t}{m} \antisymmetric{p}{x} \\

&=

-i \Hbar \frac{t}{m}.

\end{aligned}

\end{equation}

Compare this to the classical Poisson bracket

\begin{equation}\label{eqn:positionCommutator:120}

\antisymmetric{x(t)}{x(0)}_{\textrm{classical}}

=

\PD{x}{}\lr{x + p t/m} \PD{p}{x} – \PD{p}{}\lr{x + p t/m} \PD{x}{x}

=

– \frac{t}{m}.

\end{equation}

This has the expected relation \( \antisymmetric{x(t)}{x(0)} = i \Hbar \antisymmetric{x(t)}{x(0)}_{\textrm{classical}} \).

# References

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