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

 

[paranormal]

I can confirm that your calculations for [H,p] are correct. As for the second part, I cannot provide a definitive answer without more context or information. However, I can provide some insights on the uncertainty principle and the given equation.

The uncertainty principle states that the product of the uncertainties in two conjugate variables, such as position and momentum, must be greater than or equal to the reduced Planck's constant divided by 2. In the given equation, |<C>| represents the commutator of operators A and B, which is equal to the uncertainty in the measurement of these two operators. Therefore, in order to find the uncertainties in A and B, we can rearrange the equation to be:

\Delta A \Delta B \geq \frac{|<C>|}{2}

This shows that the product of the uncertainties in A and B is limited by the magnitude of the commutator. However, without knowing the specific operators A and B, we cannot determine the exact values of \Delta A and \Delta B.

In summary, your calculations for [H,p] are correct. As for the second part, more information is needed to determine the uncertainties in A and B.