# Total spin quantum number

1. Jul 21, 2013

### Roodles01

I have a pair of non interacting, identical 1/2 spin fermions in a one dimensional infinite square well with walls at x=0 and x=L.
One particle is in ground state, the other in first excited state.
This two-particle system has total spin quantum number S=0

I have normalized energy eigenfunctions for each and am trying to explain implication of this to symmetries of spin and spatial parts of the total wave function which I think is;

ψ1.2(t) = ψ(x1, x2, t) Ims1, ms2>

or for time t=0
ψ1.2 = ψ(x1, x2) Ims1, ms2>

So what exactly is the total spin quantum number please.

I can see that the spin quantum number of a spin 1/2-particle is 1/2. So could the total spin quantum number be when adding the second part of the pair?

Additionally, if this total spin quantum number becomes S=1 when they are both in the same eigenstate what does this relate to? Are both particles now either spin-up or both spin-down?

Last edited: Jul 21, 2013
2. Jul 22, 2013

### tom.stoer

You have to write down the states with quantum numbers n,s for two possibilities a,b and two particles 1,2 as

$|\psi\rangle = |n_a,s_a\rangle_1\,|n_b,s_b\rangle_2 - |n_b,s_b\rangle_1\,|n_a,s_a\rangle_2$

The total spin operator is

$S = S_1 \, \text{id}_2 + \text{id}_1 \, S_2$

Now you can apply the spin operator using the fact that an operator with index 1 (2) acts on the ket with index 1 (2) and is the identity w.r.t. the ket with index 2 (1).

The first of all four terms reads

$(S_1 \, \text{id}_2)(|n_a,s_a\rangle_1\,|n_b,s_b\rangle_2) = (S_1 |n_a,s_a\rangle_1)\,(\text{id}_2\,|n_b,s_b\rangle_2)$

Now your state with one particle in ground state and one particle in the first excited state and total spin S=0 means

$n_a=1,\,n_b=2,\,s_a=+1/2,\,s_b=-1/2$