Momentum Operator question - Proof found in Intro to Griffiths QM

[srabate]

Hi all

I am trying to go through the Griffiths Intro to QM book and I'm afraid I'm already stumped!

He determines the momentum operator by beginning with the following equation:

<x>=\int_{-\infty}^\infty {x|\psi(x)|^2}

He takes the time derivative and manipulates the integral:
(I'm skipping a few steps because I thought they made sense. If you want me to type them up I will)

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

Then, he randomly says "Performing integration by parts on the second term, we conclude: "

\frac{d<x>}{dt} = -\frac{i\hbar}{m}\int_{-\infty}^\infty {\psi^\ast(x)\frac{\partial\psi(x)}{\partial x}dx}

Why do we not perform integration by parts on the first term? I feel like I'm missing something really stupid. Is there some sort of assumption that would allow us to combine the first and second terms? (which would explain why the 2m becomes just m).

Thank you!
Abate

p.s. attached are the two pages from griffiths where this is found.

 

[nonequilibrium]

Hello,

All you have to note is that by performing integration by parts for the second term:
\int_{- \infty}^{+ \infty} \psi \frac{ \partial \psi^* } {\partial x} \mathrm d x = 0 + \int_{- \infty}^{+ \infty} \psi^* \frac{ \partial \psi } {\partial x} \mathrm d x
which is equal to the first term. This explain the disappearance of the factor 2.

(You could've also done integration by parts on the first term instead, in which case you would've gotten the same as the 2nd term.)

 

[srabate]

of course! I have no idea how I didn't see that.

Thanks!