- #1
kof9595995
- 679
- 2
I'm wondering, is there still a sharp distinction between Bosons and Fermions in a rigorous QFT, if exsits?
My question is motivated by the following, consider one of the equations of motion of QED:
[tex]\partial_\nu F^{\nu \mu} = e \bar{\psi} \gamma^\mu \psi[/tex]
In our familiar perturbative QED (Here I'm not 100% sure if I use the word "perturbative" correctly. I simply mean fields are quantized as free fields, and we introduce an interaction built from free fields operators, like an iteration method), LHS is made of Bosonic operators and RHS is made of Fermionic operators, and since the Bosonic sector and Fermionic sector are independent in the total Fock space, perturbative QED fails to satisfy this equation of motion.
I suppose if a rigourous QED exists, this equation of motion should be satisfied, but this in turn means the fermion operator and bosonic operator must act on a Hilbert space they share together, then is there still a sharp distinction between Bosons and Fermions?
My question is motivated by the following, consider one of the equations of motion of QED:
[tex]\partial_\nu F^{\nu \mu} = e \bar{\psi} \gamma^\mu \psi[/tex]
In our familiar perturbative QED (Here I'm not 100% sure if I use the word "perturbative" correctly. I simply mean fields are quantized as free fields, and we introduce an interaction built from free fields operators, like an iteration method), LHS is made of Bosonic operators and RHS is made of Fermionic operators, and since the Bosonic sector and Fermionic sector are independent in the total Fock space, perturbative QED fails to satisfy this equation of motion.
I suppose if a rigourous QED exists, this equation of motion should be satisfied, but this in turn means the fermion operator and bosonic operator must act on a Hilbert space they share together, then is there still a sharp distinction between Bosons and Fermions?