How Do You Construct Antisymmetric Wavefunctions for 2 Free Electrons?

[Hydro666]

1. Consider 2 free electrons, with single-particle wavefunctions e^ip₁*r₁|+/-> and e^ip₂*r₂|+/->.
a) Construct the antisymmetric 2-electron wavefunction of net spin zero.
b) Construct the antisymmetric 2-electron wavefunction of net spin 1. Assume that both spins are up.
$$\Psi$$(r₁,r₂)=$$\psi$$_a(r₁)$$\psi$$_b(r₂)-$$\psi$$_b(r₁)$$\psi$$_a(r₂)
I am just confused about what happens when you exchange the indices. Does the momentum switch as well? because if so, then part A works out, but then it seems like the actual wavefunction is changing as well when you swap the indices, so the articles are swapping, and the wavefunctions are swapping, which doesn't make sense. Because I thought you had to preserve the physical configuration of the system, and you are just switching the labels of the particles.

 

[sgd37]

you are swapping the labels, both on the momenta and on the position so nothing changes you could think of the states as $$e^{i(\vec{p} \cdot \vec{x})_{1,2}} \left|\pm \right\rangle$$

 

[Hydro666]

If that's true, then part A works, but then when you try to do part B, I keep getting zero.

 

[sgd37]

as it should be, you can't have two fermions in the same state. Unless they have their own different states i.e. $$\left| \pm \right\rangle _1 , \left| \pm \right\rangle _2$$ then there is no state with net spin 1

 

[Hydro666]

I thought that for a particle to be antisymmetric, it either has to have a symmetric spatial part, then an antisymmetric spin state, or vice verca. A works because it has the antisymmetric spin state, but B assumes that the spin is symmetric, and Spatial part is antisymmetric. So they shouldn't occupy the same spatial state state for it to work, and that's why there is r₁ and r₂

 

[sgd37]

it doesn't matter what spatial state they occupy. half integer spin states are always going to be antisymmetric, you can't make them occupy the same spin state. Anyway since they are free they are not restricted to the same Hamiltonian (and depend on different spatial variables) so you can safely assume

$$e^{ip_1 \cdot r_1} \left| \pm \right\rangle _1 , e^{ip_2 \cdot r_2} \left| \pm \right\rangle _2$$

now you shouldn't get zero