(Quantum Mechanics) Prove that <p> = m (d<x>/dt)

[emol1414]

Homework Statement

Prove that &lt;p&gt; = m \frac{d&lt;x&gt;}{dt}

Homework Equations

Schrödinger Equation: i\hbar \frac{\partial \Psi} {\partial x} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V{} \Psi

Respective complex conjugate from equation above

Expectation Position: <x> = \int_{-\infty}^{+\infty} x\Psi {\Psi}^* dx

The Attempt at a Solution

Derive <x> with respect to t... with V real, we know that V = V*, and after some basic steps we get:

\frac {d&lt;x&gt;}{dt} = \frac{i \hbar}{2m} \int dx x[\Psi^*(\frac{\partial^2 \Psi}{\partial x^2}) - \Psi (\frac{\partial^2 \Psi^*}{\partial x^2})]

Then my problem is with the integration by parts... for
\int_{a}^{b} f \frac{dg}{dx} dx = fg {|}^{b}_{a} - \int_{a}^{b} g \frac{df}{dx} dx

I'm choosing f = x\Psi^* and g = \frac{\partial \Psi}{\partial x}, but I think I'm not getting right these limits considerations... any sugestions or enlightenments?_______________________________________________________
EDIT (\frac{\partial \Psi}{\partial} with respect to time, not position)
Schrödinger Equation: i\hbar \frac{\partial \Psi} {\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V{} \Psi

 

[vela]

What happened to the dx/dt term when you took the time derivative of x\Psi\Psi^*?

 

[emol1414]

What happened to the dx/dt term when you took the time derivative of x\Psi\Psi^*?

In QM x don't depend on t, right?

 

[emol1414]

*Fixed a typo in the equation

 

[vela]

In QM x don't depend on t, right?

You're right. The operator doesn't explicitly depend on time, so its derivative is 0.

 

[emol1414]

So, we have, for the product rule, that
\int_{a}^{b} f \frac{dg}{dx} dx = fg {|}^{b}_{a} - \int_{a}^{b} g \frac{df}{dx} dx

And choosing f = x\Psi^* and g = \frac{\partial \Psi}{\partial x}

\frac {d&lt;x&gt;}{dt} = \frac{i \hbar}{2m} {x \Psi^* \frac{\partial \Psi}{\partial x} |^{\infty}_{-\infty} - \int_{-\infty}^{\infty} \frac{\partial \Psi}{\partial x}(\Psi^* + x \frac{\partial \Psi^*}{\partial x})dx - x \Psi \frac{\partial \Psi^*}{\partial x} |^{\infty}_{-\infty} + \int_{-\infty}^{\infty} \frac{\partial \Psi^*}{\partial x}(\Psi + x \frac{\partial \Psi}{\partial x})dx}

I guess up to this point it's ok... now I don't know how to work with these limits, which considerations should I do?

 

[vela]

Assume the wave function and its derivative go to 0 as x goes to ±∞.

 

[emol1414]

Assume the wave function and its derivative go to 0 as x goes to ±∞.

Right... I'm not really sure why this is true (??), but doing so... we perform integration by parts 2 times and then

\frac {d&lt;x&gt;}{dt} = -\frac{i \hbar}{2m} {\int_{-\infty}^{\infty} \Psi^*(\frac{\partial \Psi}{\partial x}) - \Psi (\frac{\partial \Psi^*}{\partial x} )dx}

That's it?

 

[vela]

You're almost done. Remember that\langle p \rangle = \int dx\,\Psi^* \hat{p} \Psi = \int dx\,\Psi^*\left(-i\hbar\frac{\partial}{\partial x}\right) \Psi
You want to get the righthand side to look like that. One term already looks like that, but you still need to take care of the other one.

 

[emol1414]

Got it! Integration by parts only in one of the two terms left and then add to the other, so the factor 1/2 is gone... but... there's a m missing in the denominator, right?

Thank you! =)

One more thing... why is that \Psi goes to 0 when x \rightarrow \pm \infty? Is it a "single-case" fact, or is it always true?

 

[vela]

The wave function needs to vanish at infinity to be normalizable. You have to assume the function goes to 0 fast enough so that the boundary terms go to 0. There's probably a rigorous justification for it, but I don't recall it offhand.

 

[dextercioby]

[...]

One more thing... why is that \Psi goes to 0 when x \rightarrow \pm \infty? Is it a "single-case" fact, or is it always true?

It's not mandatory, but usually one picks up from L^2 functions only the Schwartz test functions and that for a good reason.