What's the role of field equation in QFT?

Question

For free field theory, it seems the solutions of a field equation are used to give a representation of Poincare group, and the field equation is still satisfied after quantization. However for an interacting theory, such as QED, the field equation:

$$\partial_\nu F^{\nu \mu} ~=~ e \bar{\psi} \gamma^\mu \psi.$$

I don't see any possibility of satisfying this equation using operators since LHS will only act nontrivially on bosonic sector of the Hilbert space and RHS only on fermionic sector, and I don't think perturbation technique can fix this since it's qualitatively not satisfied. Yet nobody seems to be bothered by this, so I'm wondering what's the role of field equation in QFT, especially for interacting ones.

Answer 1

'' LHS will only act nontrivially on bosonic sector of the Hilbert space and RHS only on fermionic sector'':

This would be the case in a free theory, but the field equation you wrote corresponds to an interacting theory. Its Hilbert space is not the Fock space one starts with, but is turned by renormalization into a different (and little understood) Hilbert space. (Actually, all separable Hilbert spaces in infinite dimensions are the same, but ''different'' conventially refers not to a structureless Hilbert space but to a Hilbert space with a good representation of the kinematic Lie algebra.)

Actually, renormalization also changes the interpretation of the field equation, as the products of distributional fields at the same point on the right hand side must be interpreted with care.

Edit: On a heuristic level, one can understand the ''mixing'' of bosonic and fermionic operators'' by considering naive (unrenormalized) perturbation theory of the operator equations. One finds that the solution of the field equations is a formal power series in the coupling constant, with the free field as the zero order part. The first and higher order part involves the interaction. Thus the free bosonic field acquires corrections involving the product of an even number of Fermionic fields, and therefore acts nontrivially on both the bosonic and the fermionic part of the state. Unforunately, this formal expansion is mathematically ill-defined, so this argument has only heuristic value.