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Boson and fermion operators

  1. Jan 10, 2014 #1
    For bosons we define states as eg.

    ln> = l1 0 1 ... >

    where the numbers denote how many particles belong to the j'th orbital.

    And similarly for fermions. We then define creation and anihillation operators which raise and lower the number of particles in the j'th orbital:

    c_j, c_j^(dagger)

    Now in many problems I have to commute bosons and fermion operators. I actually asked this question before, but I didn't completely understand the answer. My question is:
    Why does fermion and boson operators commute? starting from the wavefunction symmetrization and antisymmetrization requirement
     
  2. jcsd
  3. Jan 10, 2014 #2
    Is it possible that one only acts on bosons and one only acts on fermions so it doesn't matter which one you do first. (Just a guess as I know zero about this. But hey it's Friday night, why not look a bit silly!)
     
  4. Jan 10, 2014 #3

    CompuChip

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    It's been a while so I hope someone won't come along and tell me this is nonsense. But if I recall correctly, the reason is they live in different spaces. If you have a boson state ##|b\rangle## and a fermion state ##|f\rangle## then a combined state would be ##|b, f\rangle = |b\rangle \otimes |f\rangle##. A bosonic operator would look like ##\hat B \otimes 1## on this combined space, and a fermionic operator similarly like ##1 \otimes \hat F##.

    Typing math on a smartphone is a bit of a pain, so I'll leave it at this, but hopefully you can work it out from there.
     
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