Time deritivate of the expectation value of p

[cyberdeathreaper]

This is the problem:

Calculate:
$$\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}<br /> \frac {d \mean{p}}_{dt}$$

Here's a few more points to keep in mind...

(A) The assumption is that <p> is defined as:
[tex] \newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}<br /> \mean{p} = -i \hbar \int \left( \psi^* \frac {\partial {\psi}}_{\partial {x}} \right) dx<br /> [/itex]<br /> <br /> (B) The problem states that the solution is:<br /> $$\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}<br /> \frac {d \mean{p}}_{dt}} = \mean {\frac {- \partial {V}}_{\partial {x}}}$$<br /> <br /> (C) In working the problem, I am able to reach this point (based on previous examples), but am not sure how to approach moving forward:<br /> $$\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}<br /> \frac {d \mean{p}}_{dt}} = \left( \frac {\hbar^2}_{2m} \right) \int_{-\infty}^{+\infty} \left( \frac {\partial}_{\partial x} \right) \left( \frac {\partial }_{\partial {x}} \right) \left( \psi^* \left( \frac {\partial {\psi}}_{\partial {x}} \right) - \left( \frac {\partial {\psi^*}}_{\partial {x}} \right) \psi \right) dx$$<br /> <br /> Any ideas for what to do next? The book shows a similar example for finding the time derivative of <x>, but the difference for that approach is that the partial derivative was canceled by an x on the inside of the integral. Now I have two of them, and no ideas for how to simplify.<br /> <br /> Thanks.[/tex]

 

[marlon]

(B) The problem states that the solution is:
$$\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}<br /> \frac {d \mean{p}}_{dt}} = \mean {\frac {\partial {V}}_{\partial {x}}}$$

There should be a minus sign in the RHS. You are proving one part of the famous Ehrenfest Theorema here, which is fundamental in QM.

The trick is to start from the definition of the expectation value of p. Then take the time derivative of this integral. You will acquire derivatives of the wavefunction to both t and r, and replace those by the expression they are equal to in the Schrödinger equation.

In the end, you can assume that both the wavefunction as its derivative wtr position will vanish at infinities (which are the integration boundaries) when applying Green's law on one of the resulting integrals

This should get you started

marlon

 

[marlon]

Besides, there is a more 'natural' way of proving this, using commutators. But the you will first need to prove that the time-derivative of an expectation value is equal to the commutator of that operator and the Hamiltonian : ie i/hbar * expectation value of [H,p]

Just calculate this commutator by writing down the actual Hamiltonian expression.

Ofcourse you will also need to prove the above used expression

marlon

 

[dextercioby]

Here's the nice approach.

Consider the momentum operator in the Schrödinger picture $\hat{\mbox{p}}_{S}$.

You wish to compute this baby (assume for simplicity a pure state)

$$\frac{d}{dt} \langle \hat{p}_{S}\rangle _{|\psi (t)\rangle}^{S}$$

The theory tell us that

$$\frac{d}{dt} \langle \hat{p}_{S}\rangle _{|\psi (t)\rangle}_{S}=\frac{1}{i\hbar} \left\langle \left[\hat{p}_{H},\hat{H}_{H}\right]\right\rangle_{|\psi \rangle}_{H}$$

Okay?

Now,you use the coordinate representation.That gives you the possibility to evaluate that commutator not in the Heisenberg representation,but in the Schrödinger one.Why?Simple.The product of 4 complex exponentials (for each product of 2 operators)) is 1.


$$\hat{H}_{S}=\frac{\hat{p}^{2}}{2m}+V(x)\hat{1}$$

And u know that

$$\left[\hat{p},\hat{p}^{2}\right]_{S}^{-}=\hat{0}$$

Therefore,the initial animal,in the coordinate reperesentation becomes

$$\frac{d}{dt} \langle \hat{p}_{S}\rangle _{|\psi (t)\rangle}_{S}^{coord}=-\int_{\mathbb{R}} \psi^{*}(x)\frac{dV}{dx} \psi (x) +\int_{\mathbb{R}} \psi^{*}(x)\frac{d}{dx}\left[V(x)\psi(x)\right] =-\langle \frac{dV}{dx}\rangle_{S}$$

Q.e.d.

Daniel.

 

[cyberdeathreaper]

Thanks for the help. I'll have to finish tackling this after my other class is over tonight. But when I do, I'll let you know if I have any other problems...

In reference to using commutators, I agree it's a more simplistic approach, but the book is looking for the longer technique using just typical calculus at this point. They formally introduce that concept in a later chapter, in which they then re-visit the question using the technique (*I think*).

 

[dextercioby]

NOT SIMPLISTIC.It's more abstract.

Daniel.

 

[cyberdeathreaper]

Okay, I think I'm still getting stuck. Here's what I have:


$$\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}<br /> \mean{p} = -i \hbar \int \left( \psi^* \frac {\partial {\psi}}_{\partial {x}} \right) dx$$

$$\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}<br /> \frac {d \mean{p}}_{dt}} = -i \hbar \int_{-\infty}^{+\infty} \left( \frac {\partial}_{\partial x} \right) \left( \frac {\partial }_{\partial {t}} \right) \left( \psi^* \psi \right) dx$$

According to a theorem in the text,

$$\left( \frac {\partial }_{\partial {t}} \right) \left( \psi^* \psi \right) = \left( \frac {\partial}_{\partial x} \right) \left( \left( \frac {i \hbar}_{2m} \right) \left( \psi^* \left( \frac {\partial {\psi}}_{\partial {x}} \right) - \left( \frac {\partial {\psi^*}}_{\partial {x}} \right) \psi \right) \right) = \left( \frac {i \hbar}_{2m} \right) \left( \psi^* \left( \frac {\partial^2 {\psi}}_{\partial {x^2}} \right) - \left( \frac {\partial^2 {\psi^*}}_{\partial {x^2}} \right) \psi \right)$$

Therefore,

$$\left( \frac {\partial }_{\partial {x}} \right) \left( \frac {\partial }_{\partial {t}} \right) \left( \psi^* \psi \right) = \left( \frac {i \hbar}_{2m} \right) \left( \psi^* \left( \frac {\partial^3 {\psi}}_{\partial {x^3}} \right) - \left( \frac {\partial^3 {\psi^*}}_{\partial {x^3}} \right) \psi \right)$$

Plugging this back in gives the following:

$$\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}<br /> \frac {d \mean{p}}_{dt}} = \left( \frac {\hbar^2}_{2m} \right) \int_{-\infty}^{+\infty} \left( \psi^* \left( \frac {\partial^3 {\psi}}_{\partial {x^3}} \right) - \left( \frac {\partial^3 {\psi^*}}_{\partial {x^3}} \right) \psi \right) dx$$

Using integration-by-parts on the second term, this can be rewritten as:

$$\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}<br /> \frac {d \mean{p}}_{dt}} = \left( \frac {\hbar^2}_{2m} \right) \int \left( \psi^* \left( \frac {\partial^3 {\psi}}_{\partial {x^3}} \right) + \left( \frac {\partial^3 {\psi}}_{\partial {x^3}} \right) \psi^* \right) dx$$

which is simply:

$$\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}<br /> \frac {d \mean{p}}_{dt}} = \left( \frac {\hbar^2}_{m} \right) \int \left( \psi^* \left( \frac {\partial^3 {\psi}}_{\partial {x^3}} \right) \right) dx$$

This all seems logically grounded, but I'm still not seeing how that is equivalent to:

$$\newcommand{\mean}[1]{{<\!\!{#1}\!\!>}}<br /> \frac {d \mean{p}}_{dt}} = \mean {\frac {- \partial {V}}_{\partial {x}}}$$

 

[cyberdeathreaper]

Gah... nevermind, I finally got it. The trick that I just wasn't getting was to let the p operator operate from the start, then carry through the operations, isolating the V(x) part of the Schrödinger eqn.

Thanks for all your help though - between this resource and the web, I've now seen numerous ways of doing the problem, but unfortunately the book assumes that I only have the definition of momentum and the Schrödinger eqn available for methods of solving. All other approaches are touched upon at a later point.