[Click here for a PDF of this post with nicer formatting]

## Question: SHO translation operator expectation ([1] pr. 2.12)

Using the Heisenberg picture evaluate the expectation of the position operator \( \expectation{x} \) with respect to the initial time state

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

\ket{\alpha, 0} = e^{-i p_0 a/\Hbar} \ket{0},

\end{equation}

where \( p_0 \) is the initial time position operator, and \( a \) is a constant with dimensions of position.

## Answer

Recall that the Heisenberg picture position operator expands to

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

x^{\textrm{H}}(t)

= U^\dagger x U

= x_0 \cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t),

\end{equation}

so the expectation of the position operator is

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

\begin{aligned}

\expectation{x}

&=

\bra{0} e^{i p_0 a/\Hbar} \lr{ x_0 \cos(\omega t) + \frac{p_0}{m \omega}

\sin(\omega t) } e^{-i p_0 a/\Hbar} \ket{0} \\

&=

\bra{0} \lr{ e^{i p_0 a/\Hbar} x_0 \cos(\omega t) e^{-i p_0 a/\Hbar} \cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t) } \ket{0}.

\end{aligned}

\end{equation}

The exponential sandwich above can be expanded using the Baker-Campbell-Hausdorff [2] formula

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

\begin{aligned}

e^{i p_0 a/\Hbar} x_0 e^{-i p_0 a/\Hbar}

&=

x_0

+ \frac{i a}{\Hbar} \antisymmetric{p_0}{x_0}

+ \inv{2!} \lr{\frac{i a}{\Hbar}}^2 \antisymmetric{p_0}{\antisymmetric{p_0}{x_0}}

+ \cdots \\

&=

x_0

+ \frac{i a}{\Hbar} \lr{ -i \Hbar }

+ \inv{2!} \lr{\frac{i a}{\Hbar}}^2 \antisymmetric{p_0}{-i \Hbar}

+ \cdots \\

&=

x_0 + a.

\end{aligned}

\end{equation}

The position expectation with respect to this translated state is

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

\begin{aligned}

\expectation{x}

&= \bra{0} \lr{ (x_0 + a)\cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t)

}\ket{0} \\

&= a \cos(\omega t).

\end{aligned}

\end{equation}

The final simplification above follows from \( \bra{n} x \ket{n} = \bra{n} p \ket{n} = 0 \).

# References

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

[2] Wikipedia. Baker-campbell-hausdorff formula — wikipedia, the free encyclopedia, 2015. URL https://en.wikipedia.org/w/index.php?title=Baker\%E2\%80\%93Campbell\%E2\%80\%93Hausdorff_formula&oldid=665123858. [Online; accessed 16-August-2015].