Pseudoparticle Operators: Fermion or Boson?

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Gregory Gao
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Hi all,

I have a question which bugs me forever. Anyone happens to know about pseudoparticle approach?

In many-particle physics, suppose we have a many particle hamiltonian which has eigenstates in Fock space, i.e., with 0, 1, 2, 3, ... electrons, denoted by \{|S\rangle\}, here S represents both particle number and level. Pseudoparticle approach says that these states are generated from a pseudoparticle vacuum with p^{\dagger}_S, i.e., p^{\dagger}_S|PPV\rangle=|S\rangle. Also, sine the Fock space is complete, the condition that \sum_{Ni}|Ni\rangle\langle Ni|=I is just \sum_{S}p^{\dagger}_Sp_S=I in the pseudoparticle language.

I can understand this, but people are claiming "pseudoparticle operator is fermionic if it corresponds to a state with an odd number of fermions, bosonic if it corresponds to a state of an even number of fermions". And they claim this is derived from the property that electrons are fermions.

Could someone help give a detailed explanation on the fermion/boson property of pseudoparticle operator? Hard proof by formulation is preferred.

Thank you in advance!
 
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Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
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