How Do Quantum Statistics Affect Particle Displacement in an Infinite Well?

[Pythagorean]

Homework Statement

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

Homework Equations

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

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

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?

 

[Jhero]

Hi, thank you for your interesting forum post. The expectation value for the spatial displacement between two non-interacting particles in the infinite square well can be calculated using the formula you provided, <(x1-x2)^2> = <x1^2> + <x2^2> - 2<x1><x2>. However, the calculation for identical particles requires a slight modification. For distinguishable particles, the probability of finding one particle at position x and the other at position x' is given by |psi1(x)|^2|psi2(x')|^2. However, for identical particles, we must take into account the fact that the particles are indistinguishable, so the probability is given by |psi1(x)|^2|psi2(x')|^2 + |psi1(x')|^2|psi2(x)|^2. This leads to an additional term in the expectation value calculation, which is the term you mentioned. This term does not necessarily mean that n cannot equal 1 for bosons, but it does indicate that there is a difference in the expectation value for bosons compared to distinguishable particles. In fact, for bosons, this term will be positive, while for fermions it will be negative. I hope this helps clarify the calculation for identical particles.