Spin assumption for fermions in potential well.

[ranytawfik]

Hi,
Assume I’m solving a 2-particle (fermions) problem in a potential well. If I set the wavefunction as anti-symmetric, then by default I’m assuming that the two particles has the same spin and hence exchange interaction has to be accommodated for.

But what if the 2 fermions have different spins? Shouldn't in this case I set the wavefunction to be only symmetric (since no Pauli exclusion here)? And what would be the final wave function? Is it a normalized linear combination of the 4 cases (2 ups, 2 downs, and 2 mixed-spin)?

Thanks for your help on that.

 

[DrClaude]

For two identical spin-1/2 particles in different potential-well eigenstates, there are four possible spin states:
$$
| \alpha \alpha \rangle \\
\frac{1}{\sqrt{2}} \left( | \alpha \beta \rangle + | \beta \alpha \rangle \right) \\
| \beta \beta \rangle \\
\frac{1}{\sqrt{2}} \left( | \alpha \beta \rangle - | \beta \alpha \rangle \right)
$$
The first three are symmetric, and therefore the corresponding spatial wave function is anti-symmetric, while the last one is anti-symmetric, requiring a symmetric spatial wave function.

So to answer your question: you have to consider all possible linear combinations of spins that have a defined symmetry, and combine each with a spatial wave function of the propoer symmetry such that the total wave function obeys Pauli's principle.

 

[ranytawfik]

Thanks Dr Claude. Your answer make sense and consistent with my understanding. I'm assuming the same concept can be applied if we have more than 2 fermions (3 for example)? We have to list all the scenarios, assign correct symmetricity, and finally linearly combine the resulting wavefunctions.

 

[DrClaude]

I'm assuming the same concept can be applied if we have more than 2 fermions (3 for example)? We have to list all the scenarios, assign correct symmetricity, and finally linearly combine the resulting wavefunctions.

Yes. But as the number of fermions increases, it becomes very difficult to figure out by hand how to make the correct linear combinations. Fortunately, this is easily done using a Slater determinant.