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This is a continuation of this question.

What would be an example of linear quantization used on quantum electrodynamics? I ask this because QED is a nonlinear theory.

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what does a linear theory mean? – nervxxx May 28 '13 at 5:11
but... QED is an example of a quantum field theory, and it is not given by a solution to a differential equation. so there is no concept of it being linear or not. or are you talking about the classical equation of motion and saying it is a bilinear operator? that's very unnatural. perhaps you're mixing up with the nature of the gauge group? QED is an abelian gauge theory, and QCD is a non-abelian gauge theory, for example. but that's very different from being linear or not. – nervxxx May 28 '13 at 5:45

I guess this was prompted by one of my comments on this question. The point I was making is that quantization, even of a system which is defined by a Lagrangian encapsulating a nonlinear set of equations of motion (such as in interacting QED) proceeds by defining a Hilbert space of states and operators on this space which evolve unitarily $$ |\Psi(t)\rangle = e^{-iHt}|\Psi(0)\rangle $$ As soon as you have this behaviour, you have a Schroedinger equation (just differentiate it to see).

So the linearity and presence of the Schroedinger equation is an integral part of the quantum picture. That's all I was meaning in the comment in the linked question.

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Quantization is done for the free theory, without interactions, and this free theory is linear. For a scalar boson, for instance, each composant like $\phi(\vec k,t)$ is an harmonic oscillator (so the equation for $\phi(\vec k,t)$ is a linear equation), and you know how to quantize an harmonic oscillator.

When you calculate the transition amplitudes, you are going to use the interaction term (S-Matrix), but the propagators you are using (in Feynmann diagrams) are calculated from the free theory quantization.

So there is no "quantization of interacting theory".

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