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

Evaluate

\label{eqn:positionCommutator:20}
\antisymmetric{x(t)}{x(0)},

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

The free particle Hamiltonian is

\label{eqn:positionCommutator:40}
H = \frac{p^2}{2m},

so the time evolution operator is

\label{eqn:positionCommutator:60}
U(t) = e^{-i p^2 t/(2 m \Hbar)}.

The Heisenberg picture position operator is

\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}

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
\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}

Compare this to the classical Poisson bracket
\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}.

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.