What is the Simplified Integral for Average Momentum?

eck
Messages
10
Reaction score
0
I have the physics question along with the solution, but in the solution I don't understand how they evaluated the integral. I can't get my brower to preview TeX input, so I'm going to leave it without formatting, but you can find the problem http://web.mit.edu/8.05/probsets/ps1_v1.pdf". The problem I am looking at is number one. I've also got the problem and solution posted here (w/o formatting) but it's kind of hard to read:
Problem
--------------------
A particle's coordinate space wavefunction is square-integrable and real up to an arbitrary multiplicative phase:
psi(x) = exp(i * alpha) phi (x)
with alpha real and constant and phi(x) real. Prove that its average momentum is zero.
Solution
-------------------
Setting up the integral is easy, and you can pull out a couple constants. So you have an infinite integral with this inside:
dx exp(-i alpha) phi(x) exp(i alpha) d/dx[phi(x)]
Somehow, in the solution, they pull out 1/2 and leave the following in the integral:
dx d/dx[phi(x)^2]
When I look at it, I see the exponentials cancelling, but I don't understand where the 1/2 comes from and how the first phi(x) gets pulled into the derivative.
Can anyone shed any insight on how this integral was simplified?
 
Last edited by a moderator:
Physics news on Phys.org
I was looking at the problem some more, and all of a sudden it hit me. It's kind of embarrassing that I didn't see it before. If anyone else looks at it... nothing tricky is involved. It's integration by parts, but it's so obvious I didn't even see it.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top