Lastly, could one use your formula for $G_2$ to determine the 2-point function of the full interacting theory (e.g. a $\phi^4$ interaction)? I assume this would choosing $\delta_m$ to be the usual scalar field propagator and then you would have to find the appropriate measure $\rho$?

@user1504 In QFT treatments the standard approach is to first specify the Lagrangian of the theory and then things like the propagator and Kallen-Lehmann decomposition follow as a result. Here it seems like we are going in the other direction. We start with the Kallen-Lehmann decomposition and then build a theory from there. How would one define $\Delta_m$ if this is your starting point?

Sorry to comment on an older post but this is a great answer. I've been studying axiomatic field theory and I was a little confused on the definition of generalized free fields but your post helped clear things up. Is there any way to physically interpret the integral $\int_0^\infty d\rho(m) \Delta_m(x-y)$? $G_2$ is a distribution so I assume we are defining the integral on the RHS to be a distribution as well?

@AbdelmalekAbdesselam Thank you for all your helpful comments. Would you be able to elaborate on how we use (1) to define the multiplication operator $\hat{\phi}$? I don't see the role of (1) in the construction you described.

Why the downvotes? How can I improve this question?

If you could add more detail and mathematical expressions that would be very helpful.

@naturallyInconsistent Thank you for your comment. I think I may have poorly worded my question, but my intention was to ask it for both QM and QFT. What I mean is in QFT the equation is still second order in time and so it would require an initial momentum for the field, and supplying initial momentum is usually not desirable in physics. I assume this is not necessary because, as you mentioned, the field can be recast as a system of first order equations (one for the particles and the other for antiparticles)?

@ValterMoretti Thanks for all the help!

@ValterMoretti I see, so for example for a solution to the Dirac equation $\psi$ the field interpretation would keep all components of $\psi$, but the wavefunction (as in relativistic quantum mechanics) would only keep the first two components since they correspond to positive energies?

Thank you! To make sure I have understood your answer correctly, mathematically nothing changes, that is the equations remain unchanged when $\psi$ is viewed as field instead of a wave. It is only our interpretation that changes (which is why we're ok with negative energies appearing, for example).

How can one see that the commutator $[D_\mu, D_\nu]$ indeed measures curvature? If one uses this commutator to act upon a vector field then it seems very similar to how the Riemann curvature tensor is defined and similar to your point about holonomy, but it is missing the bracket term $D_{[\mu, \nu]}$ in the full Riemann curvature tensor to close the parallelogram.