# Bosons and Fermions - interactions

1. Apr 30, 2014

### fengqiu

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

A particle of mass m is confined to the region |x| < a in one dimension by an infinite
square-well potential. Solve for the energies and corresponding normalized energy
eigenfunctions of the ground and first excited states.

(b) Two particles are confined in the same potential. The particles are bosons and do not
interact. What is the two-particle wavefunction, ψ(x1 , x2 ), of lowest energy? Is it an
eigenfunction of total energy? Explain.

(c) Answer part (b) with the two bosons replaced by two fermions (neglect spin).

(d) For each case [(b) and (c)] write down the probability density to find the two particles
at the same location in the potential well.

2. Relevant equations

3. The attempt at a solution

So I solved for the single particle in an infinite well and I get a sin function

For b) I think it should be 1/sqrt(2) *(2psi(x1)psi(x2))? but what confuses me here is, do I need the normalisation constant? and since psi 1 and psi 2 are already normalised, I feel my normalisation constant is not right...

and c) I take the anti symetric state 1/sqrt(2) (psi1(x1)psi2(x2)-psi2(x1)psi1(x2))

now d) is where I'm REALLY confused. so due to paulis exclusion principal the antisymetric case can't exist right (ie fermions can be at the same location in the potential well)

but for bosons... what do I integrate between? and do I do a double integral? dx1, dx2

Thanks!

2. Apr 30, 2014

### Simon Bridge

(a) ... as per usual, which you can verify by looking them up.

(b) you can check your normalization constant by applying the normalization condition.

(c) don't forget to explain your reasoning

(d) PEP does not restrict the particle "locations" - the potential does that already.

3. May 1, 2014

### fengqiu

Oh ok, that sounds right, how do I find the probability that the particles are at the same location? Do I just integrate between two arbitary numbers?

4. May 1, 2014

5. May 1, 2014

### fengqiu

Yeah, the modulus across x.
so.... would I integrate wrt d(x1-x2) with limits 0?

6. May 1, 2014

### Simon Bridge

The wavefunction corresponding to x1=x2=x would be $\psi(x,x)$ right?

7. May 1, 2014

### fengqiu

yes, right, So I plug in x1=x2, so.. the limits of the integration don't matter? or.. the limits are from -a to pos a..
RIGHT any place in the well.. as long as they're together

Thanks that makes sense