# Do the Creation Operator and Spin Projection Operator Commute?

1. Dec 20, 2012

### Morberticus

I have bumped into a term

$a^\dagger \hat{O}_S | \psi \rangle$

I would really like to operate on the slater determinant $\psi$ directly, but I fear I cannot. Is there any easy manipulation I can perform?

2. Dec 21, 2012

### andrien

where you got that and is that O something expressible in terms of gamma matrices?

3. Dec 21, 2012

### daveyrocket

What is a spin projection operator? The things that come to mind are single body operators, and you have a many body wavefunction. If it's something like the total Z component of the spin, then as a many body operator it would be written as \sum_i (n_up - n_down), which is a combination of creation/annihilation operators ... take your operator and express it in terms of particle creation/annihilation operators and then you can work out the commutation relation.

4. Dec 21, 2012

### DrDu

Spin projection operators are used e.g. in quantum chemistry. A single Slater determinant is in general not an eigenstate of spin, but such an eigenstate can be obtained using a projection operator.
If $a^\dagger$ adds an electron in a spin orbital, the new state will in general be a combination of states with with new spin S-1/2 and S+1/2.
In general you can write the spin projected state as a sum of determinants again and then act with $a^\dagger$ on each of it.