Another quick QM question

[iamalexalright]

For a particle in a potential V(x), calculate [H,p]

H = \frac{-\hbar^2}{2m}\frac{\delta^{2}}{\delta x^{2}} + V(x)
p = -i\hbar \frac{\delta}{\delta x}

[H,p] =
Hp - pH =
\frac{-\hbar^2}{2m}\frac{\delta^{2}}{\delta x^{2}} * -i\hbar \frac{\delta}{\delta x} + V(x)-i\hbar \frac{\delta}{\delta x} - -i\hbar \frac{\delta}{\delta x}\frac{-\hbar^2}{2m}\frac{\delta^{2}}{\delta x^{2}} - -i\hbar \frac{\delta}{\delta x}V(x) =
i\frac{\hbar^{3}}{2m}\frac{\delta^{3}}{\delta x^{3}} - V(x)ih\frac{\delta}{\delta x} - i\frac{\hbar^{3}}{2m}\frac{\delta^{3}}{\delta x^{3}} + i\hbar \frac{\delta}{\delta x}V(x) =
i\hbar \frac{\delta}{\delta x}V(x) - V(x)ih\frac{\delta}{\delta x}

Given \Delta A \Delta B \geq |<C>|/2 with |<C>| = [A,B], find
\Delta A \Delta B

<C> = \int \Psi^{*}(i\hbar \frac{\delta}{\delta x}V(x) - V(x)ih\frac{\delta}{\delta x})\Psi dx

Can I go from there? Is any of this correct?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[fzero]

[H,p] =
i\hbar \frac{\delta}{\delta x}V(x) - V(x)ih\frac{\delta}{\delta x}


It looks fine so far, but you need to remember this is an operator expression, so the derivative in the first term acts on everything to the RHS. Specifically, this means that

[H,p]\psi(x) = i\hbar \frac{\delta}{\delta x}(V(x) \psi(x)) - V(x)ih\frac{\delta \psi(x)}{\delta x}
= i\hbar \left(\frac{\delta V(x)}{\delta x}\right) \psi(x) + i\hbar V(x) \frac{\delta \psi(x) }{\delta x} - V(x)ih\frac{\delta \psi(x)}{\delta x}
= i\hbar \left(\frac{\delta V(x)}{\delta x}\right) \psi(x) .

This is important for computing \langle [H,p] \rangle in the 2nd part.

[iamalexalright]

alright, so the second part I would get:

<C> = \int \Psi^{*} \Psi i\hbar \frac{\delta}{\delta x}V(x)dx =
i\hbar \int \Psi{*} \Psi \frac{\delta}{\delta x}V(x)dx

Anyway to simplify this?

[fzero]

alright, so the second part I would get:

<C> = \int \Psi^{*} \Psi i\hbar \frac{\delta}{\delta x}V(x)dx =
i\hbar \int \Psi{*} \Psi \frac{\delta}{\delta x}V(x)dx

Anyway to simplify this?

If you don't know the potential and wavefunction, I don't see how you could do more than just call this i\hbar\langle \delta V(x)/\delta x \rangle.