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There are two basic ways to describe quantum field theory, we can treat everything as particles and work out Feynman diagrams. For example we would have transition amplitudes such as $\Delta(x,y,t,t')$ for an electron to go from point $x$ at time $t$ to $y$ at time $t'$.

Or we can do the Schrodinger wave functional approach and have a wave functional of fields $\Psi[\psi, A ,t)$ and a Hamiltonian. For example we would have transition amplitudes such as $\Delta[A,A',t,t']$ which would be the amplitude for a field starting of with configuration $A(x)$ at time $t$ to go to a field with configuration $A'(x)$ at time $t'$. (This can be done for Boson fields because they can be assigned values at each point. For Fermion "fields" this makes less sense as they are Grassman.)

What I'm looking for is an approach which treats the electrons as particles and so we construct Feynman graphs using electrons but which treats the electromagnetic field $A_\mu(x)$ as a field. So there would be transition amplitudes such as $\Delta(x,A,y,A',t,t')$ for an amplitude for an electron at point $x$ and electromagnetic field configuration $A(x)$ at time $t$ to go to an electron at point $y$ and electromagnetic field configuration $A'(x)$ at time $t'$.

Is there such a mixed approach in the literature? (In such an approach one would presumably calculate with Feynman graphs with electrons only.)

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  • $\begingroup$ My reaction to the title is "You describe that as 'classical electromagnetism'." But to be a little more serious the "particles" of quantum field theories are not what you think of as classical particles at all, they are excitations of field modes, so they are already in the wave-function approach. $\endgroup$ – dmckee Sep 20 '18 at 23:43
  • $\begingroup$ Yes, I get that. But in the Shrodinger functional formalism. You don't calculate things in terms of particles. You have transition amplitudes between field configurations. You don't calculate things in terms of photons. It's equivalent but a different way of writing it. It's not classical because you have a quantum amplitude for going from one field state to another. $\endgroup$ – zooby Sep 20 '18 at 23:50

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