# Not following one step for Ehrenfest's theorem

1. Dec 3, 2011

### SamRoss

I was looking at this proof of Ehrenfest's theorem http://farside.ph.utexas.edu/teaching/qmech/lectures/node35.html

I'm confused about equation 158. It looks like the first term under the integral sign in the first expression is vanishing to obtain the second expression but I don't know why it should vanish. Any ideas?

2. Dec 3, 2011

### kof9595995

wavefunction is usually assumed to vanish at infinity.

3. Dec 3, 2011

### SamRoss

right, but in the proof the author was not evaluating the wavefunction at infinity. It was still under the integral sign. Unless I'm mistaken, you can't just let the wavefunction vanish every time you see it under the integral sign. If that were the case then normalizing would be impossible because you would always get zero instead of one.

4. Dec 3, 2011

### kof9595995

He is evaluating it at infinity, be careful it's an integration over a differentiation.

5. Dec 3, 2011

### dextercioby

It was a total derivative under the integral sign. By integrating, you get to evaluate the product of derivatives at +- infinity, where it will be zero, if psi is a Schwartz function.

6. Dec 3, 2011

### jewbinson

Well the integral of d/dx(f(x)) w.r.t. x is just f(x), by the Fundamental Theorem of Calculus. So we get the evaluation of f(x) between +infinity and -inifinity.

Here the author assumes that the wavefunction vanishes at + and - infinity, i.e. limx->+infinity(psi) = limx-> -infinity(psi) = 0. If you draw this on a graph, you should be able to see that the derivatives of psi and psi* tend to 0 as x tends to +infinite and as x tends to -infinity. These facts can be proven easily using calculus.

Since both terms tend to 0 as x tends to + infinity and as x tends to - infinity, the AOL shows immediately that the product tends to 0 as we take the limits x->+infinity and x-> - infinity.

7. Dec 3, 2011

### SamRoss

Surprised I missed that. Thanks everyone for your input.