Is QED without measurement a local hidden variable theory? An interesting thought experiment would be to create a large quantum wave-function describing both a pair of electrons with entangled spins and the equipment needed to measure the electrons' spins. In theory, this wavefunction could be evolved in time and monitored to see if any process mimicking measurement of the electrons' spins occurs.
However, as the QED Lagrangian is local I believe such a system could not violate Bell's inequality. Is that correct?
Does that mean that QED (with measurement) is not a self-consistent theory as it admits two methods of modelling the same physics which produce different results?
 A: "not a self-consistent theory as it admits two methods of modelling the same physics which produce different results" - this is pretty much the formulation of the generally recognized measurement problem of quantum theory (http://plato.stanford.edu/entries/qt-issues/#MeasProb). In standard quantum theory, it is usually stated that unitary evolution takes place between measurement, and the projection postulate is applicable to measurements. However, it seems strange that the evolution of the system depends on whether we call the relevant process "a measurement" or not.
As for the Bell theorem, I show in my paper http://link.springer.com/content/pdf/10.1140%2Fepjc%2Fs10052-013-2371-4.pdf (published in the European Physical Journal C) that some local realistic models have the same unitary evolution as quantum field theories. This suggests that one probably cannot derive violations of the Bell inequalities in quantum theory using just unitary evolution. Standard proofs of the violations use both unitary evolution and the projection postulate (or something similar), although these assumptions are, strictly speaking, mutually contradictory.
