Do the Creation Operator and Spin Projection Operator Commute?by Morberticus Tags: commute, creation, operator, projection, spin 

#1
Dec2012, 04:04 AM

P: 78

I have bumped into a term
[itex]a^\dagger \hat{O}_S  \psi \rangle[/itex] I would really like to operate on the slater determinant [itex]\psi[/itex] directly, but I fear I cannot. Is there any easy manipulation I can perform? 



#2
Dec2112, 01:05 AM

P: 977

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




#3
Dec2112, 03:16 AM

P: 185

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
Dec2112, 04:14 AM

Sci Advisor
P: 3,368

Do the Creation Operator and Spin Projection Operator Commute?
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 S1/2 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. 


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