# Two-particle QM Infinite Well

1. Feb 24, 2008

### Pythagorean

1. The problem statement, all variables and given/known data

Find the expectation value: <(x1-x2)^2> for two non-interacting particles in the infinite square well. If one is in state $$\psi_1( n \neq l)$$ and the other in state $$\psi_n$$ find the expectation value for a) distinguishable particles, b) bosons, c) fermions

2. Relevant equations

$$\psi_n = \sqrt{\frac{2}{a}}sin(\frac{n \pi x}{a})$$
$$\psi_1 = \sqrt{\frac{2}{a}}sin(\frac{ \pi x}{a})$$

3. The attempt at a solution

verification for a):

$$<{\Delta x}^2> = <x^2>_1 + <x^2>_n - 2<x>_1<x>_n$$

after integration I get:

$$<{\Delta x}^2> = 4a(\frac{1}{3} + \frac{1+\delta_{n,m=1}}{2\pi^2} - 4n\delta_{n,m=1})$$

My Problem

While I don't know whether the above is correct, assuming it is, when I do this for identical particles, I get the extra term (in addition to the above):

$$\mp 2|<x>_1,n|^2 = \int \frac{4}{a^2} sin(\frac{\pi x}{a})sin(\frac{n \pi x}{a})dx$$

which (using http://integrals.wolfram.com/index.jsp") comes to:
http://integrals.wolfram.com/Integrator/MSP?MSPStoreID=MSPStore188022599_0&MSPStoreType=image/gif [Broken]

evaluating this integral, the sins go to zero, but the cosines leave a 1 and a couple +/- 1's

Unfortunately, in the denominator, one of the terms has a (n-1) term. Does this mean that n cannot equal 1 even for Bosons? That is to say that it's meaningless to measure the spatial displacement between particles in the same state?

Last edited by a moderator: May 3, 2017