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

[emol1414]

Homework Statement

Prove that $<p> = m \frac{d<x>}{dt}$

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:

$\frac {d<x>}{dt}$ = $\frac{i \hbar}{2m}$ $\int$ $dx$ $x$[$\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 $g = \frac{\partial \Psi}{\partial x}$

$\frac {d<x>}{dt}$ = $\frac{i \hbar}{2m}$ {$x \Psi^* \frac{\partial \Psi}{\partial x} |^{\infty}_{-\infty} - \int_{-\infty}^{\infty} \frac{\partial \Psi}{\partial x}(\Psi^* + x \frac{\partial \Psi^*}{\partial x})dx - x \Psi \frac{\partial \Psi^*}{\partial x} |^{\infty}_{-\infty} + \int_{-\infty}^{\infty} \frac{\partial \Psi^*}{\partial x}(\Psi + x \frac{\partial \Psi}{\partial x})dx$}

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

$\frac {d<x>}{dt}$ = -$\frac{i \hbar}{2m}$ {$\int_{-\infty}^{\infty} \Psi^*(\frac{\partial \Psi}{\partial x}) - \Psi (\frac{\partial \Psi^*}{\partial x} )dx$}

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.