Finding the commutator of the Hamiltonian operator, H and the position operator, x and finding the mean value of the momentum operator, p | Quantum Science Philippines

## Finding the commutator of the Hamiltonian operator, H and the position operator, x and finding the mean value of the momentum operator, p

By Kim S. Ponce, MS Physics, MSU-IIT

Problem

Given that the Hamiltonian is $H = \frac{p_{x}^{2}}{2m} + V(x)$. Show that $[H,x] = -\frac{ihp_{x}}{m}$ and the mean value of the momentum ($$) is zero.

Solution

$\begin{array}{lll} [H,x]& = &Hx - xH\end{array}$

Substituting our Hamiltonian operator,

$\begin{array}{lll} [H,x]& = &[\frac{p_{x}^{2}}{2m} + V(x)]x - x[\frac{p_{x}^{2}}{2m} + V(x)]\\ \\&= &\frac{p_{x}^{2}}{2m}x + V(x)x - x\frac{p_{x}^{2}}{2m} - xV(x)\\ \\&= &\frac{p_{x}^{2}}{2m}x - x\frac{p_{x}^{2}}{2m}\end{array}$

Using the commutator of $p_{x}$ and $x$, this can be written in the form:

$\begin{array}{lll} [H,x]& = &\frac{1}{2m}[p_{x}^{2},x]\end{array}$

But $[p,x] = -i\hbar$, so

$\begin{array}{lll} [H,x]& = &-i\hbar\frac{1}{2m}(2p_{x})\end{array}$

Therefore,

$\begin{array}{lll} [H,x]& = &-\frac{ihp_{x}}{m}.\end{array}$

Now, we find the mean value of the momentum operator, $lates p_{x}$ which is given as

$\begin{array}{lll} & = &<\Psi|p_{x}|\Psi>\\ \\& = &<\Psi(t)|i\hbar\frac{d}{dx}|\Psi(t)>\end{array}$

But $i\hbar\frac{d}{dx}\Psi(t) = 0$ since we’re only considering the stationary state, thus

$\begin{array}{lll} & = &0.\end{array}$