# Why is the free field operator the same with interactions present?

In free scalar theory, the field would have the expression $$\phi(x)=\int \frac{d^3p}{(2\pi)^3\sqrt{2E_p}}a_p e^{-ip_\mu x^\mu}+a^\dagger_p e^{ip_\mu x^\mu}$$ Suppose we have an interaction with a complex field $\psi$, such that the lagrangian has the interaction term $$\mathcal L = -g\psi^\dagger\psi\phi$$ My doubt is: do the field operators $\phi(x)$ and $\psi(x)$ have the same expression as in the free theory? It seems that this is true, but I'm not sure why.

I though of these possibilities:

1. We are in the interaction picture, which means that the evolution of operators is governed by the free hamiltonian. That is, if we go from the Schrödinger picture to the interaction picture, we only need the free Hamiltonian. Still, the operator in the Schrödinger picture should be modified by the interaction. After all, the expression for the free field was obtained by solving Klein-Gordon's equation and quantizing the solution. In the same manner, we should solve the equaitions of motion, which will be something like $$(\square+m^2)\phi=-g\phi\psi$$ and then quantize.

2. Since we are doing perturbation theory, the field doesn't change much and it's a good approximation.

3. The field expression is independent of the lagrangian.
• Your point number 1 is correct. – Prahar Jul 12 '15 at 12:53
• @Prahar Could you elaborate? Why isn't position dependece affected by the interaction? – jinawee Jul 12 '15 at 13:03
• I think what you are looking for is called the LSZ reduction formalism. However it is rather phenomenological (since it assumes assymptotical states). As far as I know, there is no rigorous approach to this problem (mathematically, interaction picture in QFT does not even exist). Probably the best thing to do here is to accept the LSZ formalizm (despite its obvious flaws) as one that describes scattering correctly (when the coupling is small) and to forget about this issue, focusing on computing time-ordered operator product expectations. – Prof. Legolasov Jul 13 '15 at 14:51

The rigorously defined form of the interacting fields in $3+1$ dimension is not known (yet), however it will most likely be different from the free field form.

There is an a priori result that at one time clarifies and messes up things: the Haag's theorem. Before trying to hint the result and some of the consequences, I would like to point out that, contrarily to what is otherwise stated in this forum (and seem to be a widespread belief among physicists), Haag's theorem indeed:

• is something true that has to be taken into account;

• complicates the naïve interaction picture for quantum fields;

• does not prevent a good mathematical theory of QFT to be formulated, and to be in agreement with the perturbative results that theoretical physics provides.

Said that, the Haag's theorem, in a very simplified form states the following:

There are infinitely many inequivalent irrepresentations of the canonical commutation relations (for the quantum fields). Among them, the representation of a free and corresponding interacting theory (that both satisfy the Wightman axioms) are inequivalent.

The Wightman axioms are mathematical axioms that are considered as a minimal requirement for a quantum theory of fields to be mathematically well-defined, and are known to be satisfied by the free theories, and some interacting theory in low dimension (e.g. $\varphi^4$ in $2+1$ dimensions). However, no interacting theory in $3+1$ dimensions is known to satisfy the axioms.

Nevertheless, Haag's theorem provides an a priori information: free and interacting quantum fields are inequivalent representations of the CCR. That means that the interacting fields are not the same as the free fields; and probably have not the same form, i.e. they are probably not in Fock representation. I say probably, because there are Fock representations that are unitarily inequivalent between each other, so Haag's theorem does not prevent the interacting fields to be Fock, only they have to be at least an "inequivalent Fock" with respect to the free ones.

In addition, we know (from the few rigorous examples we have) that the form of the interacting fields indeed depends on the theory at hand: so for some theory the interacting field may be in a Fock representation, for some other in a non-Fock one.

Let me conclude the answer with a remark on the fact that even if the free and interacting theories are not unitarily equivalent, one can develop a scattering theory that provides the results known to physicists (e.g. the LSZ reduction formulas) but is also in agreement with Haag's theorem: this theory is called Haag-Ruelle scattering theory (and can be looked up in the third volume of Reed-Simon's books).

• Good comment! Can you recommend any literature dealing with moving into a representation free phase of the theory (td. limit)? – Hamurabi Jan 25 '16 at 16:31
• @Hamurabi I'm not sure I understand exactly your request...as far as I know the only QFT results that are completely representation free are those of algebraic quantum field theory (AQFT). The reference and founding paper on this approach is of Haag and Kastler. However I don't think that it is possible to find dynamical results within AQFT. Usually to have non-perturbative dynamical results you either need to use a path integral formulation (however it is often mathematically ill-defined), or manipulations that start from the Fock representation. – yuggib Jan 25 '16 at 17:00
• @Hamurabi However if you're looking for literature on a more specific subject I may (not sure) be able to help – yuggib Jan 25 '16 at 17:01