Fermions, Bosons, and nonidentical particles in a 1-d oscillator

[MaestroBach]

Homework Statement:

Consider two noninteracting particles of mass m in the harmonic oscillator potential well. For
the case with one particle in the single-particle state |n> and the other in state |k> with n not equal to k,
calculate the expectation value of the squared interparticle spacing <(x1-x2)^2>, assuming (a)
the particles are distinguishable, (b) the particles are identical spin-0 bosons, and (c) the particles
are identical spin-1/ 2 fermions in a spin triplet state. Use bra-ket notation as far as you
can, but you will have to do some integrals.

Relevant Equations:

N/A

I'm having a hard time understanding how to treat fermions, bosons, and distinguishable particles differently for this problem.

To the best of my understanding, I know that my overall state for bosons must be symmetric, and because they're spin-0, this means there's only one coupled spin state available for them, ie $\ket{S m_s} = \ket{0 0}$, and since their spin is symmetric then their space representation must also be symmetric, giving me $\psi_{nk} = \psi_n(x_1)\psi_k(x_2) + \psi_n(x_2)\psi_n(x_1)$, from which I can calculate $<(x_1-x_2)^2>$.

However, my confusion comes in when I'm dealing with fermions and distinguishable particles.

For fermions, as far as I understand, the spin could be both asymetric or symmetric, but that would change whether my space representation is asymetric or symmetric. How do I decide which to use? This is especially confusing for me, given that the spin does not play a part in my calculation of $<(x_1-x_2)^2>$, but choosing an asymetric or symmetric spin changes the space representation which to me seems like it would change
$<(x_1-x_2)^2>$, giving me two different possible answers.

As for distinguishable particles, I'm not even sure where to begin...

I appreciate any help!

 

[TSny]

To the best of my understanding, I know that my overall state for bosons must be symmetric, and because they're spin-0, this means there's only one coupled spin state available for them, ie $\ket{S m_s} = \ket{0 0}$, and since their spin is symmetric then their space representation must also be symmetric, giving me $\psi_{nk} = \psi_n(x_1)\psi_k(x_2) + \psi_n(x_2)\psi_n(x_1)$, from which I can calculate $<(x_1-x_2)^2>$.

Sounds good. You might want to include a normalization constant for $\psi_{nk}$

However, my confusion comes in when I'm dealing with fermions and distinguishable particles.

You are given that the spin state for the fermions is the triplet state.

For the distinguishable particles, you can distinguish the particle in state n from the particle in state k. Let $x_1$ denote the position variable for the particle in state n and let $x_2$ denote the position variable for the particle in state k. How would your express the wavefunction $\psi_{nk}$?

 

[MaestroBach]

You are given that the spin state for the fermions is the triplet state.

Wow. I can't believe I missed that. Thanks a ton.

For the distinguishable particles, you can distinguish the particle in state n from the particle in state k. Let $x_1$ denote the position variable for the particle in state n and let $x_2$ denote the position variable for the particle in state k. How would your express the wavefunction $\psi_{nk}$?

So for this, would the wavefunction $\psi_{nk}$ just be $\psi_{nk} = \psi_n(x_1)\psi_k(x_2)$?

For distinguishable particles I don't even care about spin right? Just that they're distinguishable?

 

[TSny]

So for this, would the wavefunction $\psi_{nk}$ just be $\psi_{nk} = \psi_n(x_1)\psi_k(x_2)$?

For distinguishable particles I don't even care about spin right? Just that they're distinguishable?

Yes, that's right.

 

[MaestroBach]

Yes, that's right.

Really appreciate the help!