Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Do the Creation Operator and Spin Projection Operator Commute?

  1. Dec 20, 2012 #1
    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. jcsd
  3. Dec 21, 2012 #2
    where you got that and is that O something expressible in terms of gamma matrices?
  4. Dec 21, 2012 #3
    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.
  5. Dec 21, 2012 #4


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook