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

1. Aug 21, 2011

### emol1414

1. The problem statement, all variables and given/known data
Prove that $<p> = m \frac{d<x>}{dt}$

2. Relevant 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

3. 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?

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EDIT ($\frac{\partial \Psi}{\partial}$ with respect to time, not position)
Schrödinger Equation: $i\hbar$ $\frac{\partial \Psi} {\partial t}$ = -$\frac{\hbar^2}{2m}$ $\frac{\partial^2 \Psi}{\partial x^2}$ + $V{} \Psi$

Last edited: Aug 21, 2011
2. Aug 21, 2011

### vela

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

3. Aug 21, 2011

### emol1414

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

4. Aug 21, 2011

### emol1414

*Fixed a typo in the equation

5. Aug 21, 2011

### vela

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

6. Aug 21, 2011

### 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?

7. Aug 21, 2011

### vela

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

8. Aug 21, 2011

### emol1414

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?

9. Aug 21, 2011

### vela

Staff Emeritus
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.

10. Aug 21, 2011

### 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?

11. Aug 21, 2011

### vela

Staff Emeritus
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.

12. Aug 22, 2011

### dextercioby

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