# What is a particle in the context of QFT with interactions?

This is a crossposting of the same question from mathoverflow: https://mathoverflow.net/q/454768/

It seems that this question was not received well there, claiming that this question is not mathematically precise enough.

I'm a bit of a novice, so bear with me.

My understanding of the story is as follows.

### From Lagrangians to Irreducible Representations

The story of the types of possible particles begins with the Lagrangians. For example the Klein Gordon Lagrangian is what you get if you want the field equations to be $$E^2=m^2+p^2$$ when you graduate $$E$$ and $$p$$ to operators a la Schrodinger.

For each "free Lagrangian" (Lagrangian of a single type of particle) you get an associated Hamiltonian.

Then moving to the view of types of particles as irreducible unitary projective representations of the Poincare group, the relationship is that for each such representation $$U$$ we have that $$U(a(t),1)=e^{-itH}$$, where $$a(t)=(-t,0,0,0)$$, and $$H$$ is the associated Hamiltonian.

It seems to me therefore that going from the irreducible representation $$U$$ to the Hamiltonian $$H$$ is well defined, but not vice versa. So I guess going from the Lagrangian to the irreducible representation requires some choices. (Intermediate question: is this correct?)

One of the nice things is that the spin-statistic theorem tells you a classification of such representations.

To my understanding, the language of representations of Poincare is the only one that truly makes sense, because in the language of representations time evolution is relativistic: the state $$\psi$$ from the viewpoint of the spacetime point $$x$$ "evolves" into $$U(a,A)\psi$$ for the spacetime point $$(a,A)x$$, where $$(a,A)$$ is an element of the Poincare group. If we only have Hamiltonians, we are essentially restricted to moving along $$a(t)=(-t,0,0,0)$$.

### The Interaction Picture

It seems to me that in the textbooks you begin not with Lagrangians and not with irreducible representations, but with Hamiltonians.

You let $$H=H^{free}+H^{int}$$, where $$H^{free}$$ is the sum of Hamiltonians of "free" types of particles, with no interactions; and $$H^{int}$$ has interaction terms.

It seems to me like if things were good in the world, then this $$H$$ can be associated to an irreducible (or perhaps reducible) unitary projective representation of Poincare. But I don't see it in any of the textbooks.

It also seems unclear to me why such a representation would be at all related to the representations of each particular free particle whose Hamiltonian is a summand in $$H^{free}$$.

To make things more confusing, it seems that when physicists make any type of computation about time evolution in the interaction picture, everything is weird. They do some mix of the Heisenberg and Schrodinger pictures, letting $$H^{free}$$ evolve according to one and $$H^{int}$$ according to the other, and they often take time to minus and plus infinity, and then do a bunch of things about re-order operators to avoid things that are clearly nonsense.

### Question

What is the proper way to think of interactions? Can you think of it as a representation, or is it not known how to do that? If you could think of it as an irreducible unitary projective representation, then it would be beholden to the spin-statistic theorem, which would mean that it would have a spin -- that'd be weird!

Also, it seems like there is no way to relate this representation to the representations associated with the free particles in the theory, so that's weird also. What's the right way to think about it? Is it really just that the Hamiltonians kind of look the same, and there's no formal connection?

### Some thoughts

If I wanted to think of the interaction as a representation, it would make sense to me to have the Hilbert space be the tensor product of the Hilbert spaces associated with each free particle in the theory. But of course taking the tensor product of the representations associated with the free particles would be the same as the non-interaction picture.

We already know that we want $$U(a(t),1)=e^{-itH}$$ for $$a(t)=(-t,0,0,0)$$, where $$H$$ is the entire Hamiltonian including the free and interaction parts. But how would that determine the representation? Perhaps the Lorentz part should be completely determined by its action defined on the free parts.

How does any of this relate to the weird half-Schrodinger half-Heisenberg voodoo that they do with the $$S$$-matrix? I feel like the math here is missing something.