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
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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 (<p_{x}>) 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} <p_{x}>& = &<\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} <p_{x}>& = &0.\end{array}

 

 

 

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