L09E02.tex

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\begin{exercise} Prove Eq. $9.10$ by expanding each site and
comparing the results.
\end{exercise}
Eq. $9.10$ is:
\[
	[\bm{P}^2, \bm{X}] = \bm{P}[\bm{P}, \bm{X}] + [\bm{P}, \bm{X}]\bm{P}
\]
We're trying to compute the "velocity" of a free particle, meaning, the time
derivative of its position operator $\bm{X}$, which can be performed thanks
to an earlier formula derived in section $4.9$ - \textit{Connections to classical
mechanics}:
\[
	\frac{d}{dt}\avg{\bm{X}} = \frac{i}{\hbar}\avg{[\bm{H},\bm{X}]}
\]
To push the computation forward, we then need to compute $[\bm{H},\bm{X}]$,
which is, for a free particle, up to a constant, equivalent to
$[\bm{P}^2,\bm{X}]$. \\

Recall that the \textit{commutator} measures how much two operators defined
on a Hilbert space commute:
\[
	[\bm{A}, \bm{B}] := \bm{A}\bm{B} - \bm{B}\bm{A}
\]
Let's progressively expand both sides, keeping all expansions strictly equivalent:
\begin{equation*}\begin{aligned}
	~ &&
		[\bm{P}^2, \bm{X}] &=&&
		\bm{P}[\bm{P}, \bm{X}] + [\bm{P}, \bm{X}]\bm{P} \\
	\Leftrightarrow &&
		\bm{P}^2\bm{X} - \bm{X}\bm{P}^2 &=&&
		\bm{P}(\bm{P}\bm{X} - \bm{X}\bm{P}) + (\bm{P}\bm{X} - \bm{X}\bm{P})\bm{P} \\
	\Leftrightarrow &&
		\bm{P}\bm{P}\bm{X} - \bm{X}\bm{P}\bm{P} &=&&
		\bm{P}\bm{P}\bm{X} - \bm{P}\bm{X}\bm{P}
			+ \bm{P}\bm{X}\bm{P} - \bm{X}\bm{P}\bm{P} \\
	\Leftrightarrow &&
		\bm{P}\bm{P}\bm{X} - \bm{X}\bm{P}\bm{P} &=&&
		\bm{P}\bm{P}\bm{X} - \bm{X}\bm{P}\bm{P} \\
	\Leftrightarrow && 0 &=&& 0 \\
	\Leftrightarrow && \text{true} \qed
\end{aligned}\end{equation*}
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