The Hamiltonian ($H$) for a free particle of mass ($m$) is purely kinetic:
$H = \frac{p^2}{2m}$
where $p$ is the momentum operator.
First, compute the commutator $[x, H]$:
$[x, H] = [x, \frac{p^2}{2m}]$
Using the linearity property of commutators and the fundamental commutation relation $[x, p] = i\hbar$:
$[x, H] = \frac{1}{2m} [x, p^2]$
$= \frac{1}{2m} ([x, p]p + p[x, p])$
$= \frac{1}{2m} (i\hbar p + p(i\hbar))$
$= \frac{1}{2m} (2i\hbar p)$
$= \frac{i\hbar p}{m}$
Next, compute the double commutator $[x, [x, H]]$ using the result from the previous step:
$[x, [x, H]] = [x, \frac{i\hbar p}{m}]$
Factor out the constants and apply the fundamental commutation relation $[x, p] = i\hbar$ again:
$[x, [x, H]] = \frac{i\hbar}{m} [x, p]$
$= \frac{i\hbar}{m} (i\hbar)$
$= \frac{i^2 \hbar^2}{m}$
Since $i^2 = -1$:
$= -\frac{\hbar^2}{m}$
The commutator $[x, [x, H]]$ for the free particle Hamiltonian is $-\$\hbar^2$/m.
The wavefunction of a particle in one dimension is given by
$\psi(x) = \begin{cases} M, & -a < x < a \\ 0, & \text{otherwise.} \end{cases}$
Here $M$ and $a$ are positive constants. If $\phi(p)$ is the corresponding momentum space wavefunction, which one of the following plots best represents $|\phi(p)|^2$ ?