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.


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.

  • 4
    $\begingroup$ I'm downvoting this because I don't see a clear mathematical question. I suggest you read the first few chapters of Weinberg's book on QFT. There is nothing weird about the interaction picture and no, interactions cannot be thought of as a representation. $\endgroup$ Sep 17 at 23:43
  • 1
    $\begingroup$ While I have read other textbooks than Weinberg's, I don't see where "interactions cannot be thought of as a representation" would fit into the chapter structure. In fact this statement is precisely the mathematical question: how do you fit the interaction scenario into the representation framework, and if you can't then why not? And how to interpret the S-matrix formalism in the representation theoretic framework? I'm not at all sure Weinberg addresses these questions, and I think they have a lot of mathematical content... $\endgroup$ Sep 17 at 23:50
  • $\begingroup$ Crossposted to physics.stackexchange.com/q/780798/2451 $\endgroup$
    – Qmechanic
    Sep 18 at 19:32

1 Answer 1


In quantum field theory (QFT) the notion of a field is fundamental, not the notion of a particle. You can search for local excitations of the fields and identify these with particles, but this can only be done in an unambiguous way for free fields, where you can make a mode expansion and create a localised state as a superposition of modes with a definite momentum. This difficulty is discussed in connection with Haag's theorem by Jonathan Bain in Asymptotic Particle States and Interpolating Fields in Interacting QFT.

The "interaction representation" of the propagator is a convenient way to perform perturbation theory in the interaction strength, the issues here are essentially nonperturbative.


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