# Time derivative of momentum expectation?

1. Feb 7, 2010

### theneedtoknow

Hello,

I am trying to learn about some basic quantum mechanics.

http://farside.ph.utexas.edu/teaching/qmech/lectures/node35.html this website shows that the time derivative of the momentum expectation d<p>/dt = -<dV/dx>
The part that i am not getting is how the writer goes from the first line of equation 157 on the site, to the second line of the same equation. He simply says he is integrating by parts.

I've tried integrating the 2 terms of the integrand in the first line by parts, but I am not really getting to the second line. I assume that integrating the 2 terms independently should give u1v1 - u2v2 = 0 so that we are left only with the integrals of v1du1 and v2du2(this is judging by the fact that the second line is entirely an integral).
Can someone clear up this step?

2. Feb 7, 2010

### Fredrik

Staff Emeritus
Ignore the first term for now and just focus on the second. The trick you're supposed to use is just that if D is the derivative operator and f and g are functions, then fDg=D(fg)-(Df)g. You'll end up with an extra term that doesn't appear in the second line because the integral over it is zero.

3. Feb 7, 2010

### xepma

Only the second term is integrated by parts, and the resulting total derivative is thrown away. In the first term the factor of (-ih) is swept under the complex conjugation.

EDIT: Fredrik beat me to it... ;)

4. Feb 7, 2010

### SpectraCat

AFAICS, he is just implicitly using the fact that the wavefunction must vanish at $$\pm\infty$$ in the derivation. Integration works as follows:

$$uv = \int vdu + \int udv$$ therefore, for a definite interval, we have

$$\int^{b}_{a}udv = uv|^{b}_{a} - \int^{b}_{a}vdu$$, but in this case, $$uv|^{b}_{a} \rightarrow \psi^{*}\frac{\partial\psi}{\partial t}|^{+\infty}_{-\infty} = 0$$, so only the vdu integral remains, as shown in the derivation.

P.S. Sorry for the crummy TeX formatting .. I am still learning it, and I couldn't figure out how to make the "evaluation bar" immediately to the right of the uv terms any bigger

EDIT: Yeah me too, but at least we all said the same thing

5. Feb 7, 2010

### theneedtoknow

Thank you guys!
So, in the first term then, he is just using (ih)* = -ih and dPsi*/dt = (dPsi/dt)*, to combine both terms in the brackets?

6. Feb 7, 2010

Yup!

7. Feb 7, 2010

### theneedtoknow

:) All clear now! Thanks again!

8. Feb 7, 2010

### theneedtoknow

One more question.... if i'm evaluationg dPsi/dx at inf or at -inf, it should be zero right? It makes sense that the only way for Psi to go to zero at inf/-inf is if the slope of psi (dPsi/dx)also goes to zero? (the tangent to psi approaches being parallel to the x-axis?)

9. Feb 7, 2010

### SpectraCat

Yes, psi and all of its position-space derivatives must go smoothly to zero at +/- infinity. This is required by one of the postulates of Q.M. ... which says that the square-modulus of the wavefunction in any volume element is proportional to the probability of observing the particle in that volume element (the Born interpretation of the wavefunction). This means that the wavefunction must be square integrable, i.e. that the square modulus of the wavefunction integrated over all space yields a finite number.