QM Question: Calculating [H,p] for a Particle in a Potential V(x)

[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?

 

[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 $$i\hbar\langle \delta V(x)/\delta x \rangle$$.