In QFT we can write a Hamiltonian operator for a free field. So, what is a free field/ noninteracting field?
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5$\begingroup$ It's a field with at most quadratic (bilinear) Lagrangian - which appears in no other terms (i.e. interaction terms with other fields). By varying such a Lagrangian, one gets (at most) linear equations of motion which are, after some shifts, exactly linear equations of motion. And linear equations of motion have solutions that are, in general, superpositions of some basic solutions, a basis. The superposition of wave packets looks like packets that freely propagate through each other and don't care about each other or anything else - that's why they're called free. $\endgroup$– Luboš MotlCommented Feb 26, 2013 at 9:31
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1$\begingroup$ In perturbative QFT, free field theory is usually taken as the approximation to any QFT - particles move along straight lines, freely - and this approximation is then supplemented by the interactions that are viewed as perturbations of the free field theory. In this approach, the propagators of Feynman diagrams arise from the free field (bilinear) terms and the vertices are from the interaction (higher-order) terms. The free field theory part is also called "Gaussian" because the path integral contains functions like $\exp(-ax^2)$ which are called "Gaussian". $\endgroup$– Luboš MotlCommented Feb 26, 2013 at 9:33
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$\begingroup$ @ Luboš Motl: When we say free particles we understand that there is no force acting on the particles and there are no interactions among them. Are there any physical definition of free field like this? $\endgroup$– rainmanCommented Feb 26, 2013 at 9:34
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1$\begingroup$ Apologies, I don't understand how this question differs from the original one that - I believe - I have already answered. What's the problem? The adjective "physical"? Do you feel my answer was "unphysical"? Perhaps because it involved some (rudimentary) maths? You can't do physics without any maths. Accurate enough definitions in physics always involve some mathematical objects - linear equations and/or bilinear actions is really the simplest example of mathematical objects you shouldn't protest. $\endgroup$– Luboš MotlCommented Feb 26, 2013 at 9:55
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$\begingroup$ @Ome: 'force' is an emergent phenomenon in field theory, so it's not a good idea to define QFT objects in terms of it. The same goes for interactions: really the best way to define an interacting theory is to say that it is not free. $\endgroup$– VibertCommented Feb 26, 2013 at 12:33
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2 Answers
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Let us speak specifically of electron.
Definition of a free field is like a definition of a free particle: it is a particle in a rather weak or zero external field, so no essential trajectory deviation is foreseen for such a particle. Such an understanding is correct since in an external field you have a more general equation - that with the charge involved and with a more curved trajectory that can be verified experimentally. Factually, the particle equation in an external field serves as a model for a free particle. It has the same mass, charge, and spin.
answered Feb 26, 2013 at 10:22
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3$\begingroup$ Well, except that if you place an electron in an external - classical - field (electromagnetic, gravitational, or another), QFT describes it by a bilinear Lagrangian that we would still call free, wouldn't we? So the quanta of free fields aren't necessarily particles unaffected by external fields/forces. They may be very well affected. In QFT, the "freedom" means that the fields don't interact with other (or the same) quantum fields but they're allowed to be affected by the classical background. $\endgroup$ Commented Feb 26, 2013 at 10:25
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$\begingroup$ @LubošMotl: Yes, that's right. One can say they are free from "self-action" interaction. $\endgroup$ Commented Feb 26, 2013 at 10:33
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A quantum field theory is defined by the connected n-point functions of its observables. (One can recover the n-point functions from the connected ones, and from the n-point functions you can reconstruct the Hilbert space and all the operators.)
A field is free if its connected $n$-point functions vanish for $n > 2$. Otherwise, it's interacting.
Edit: These are the basic observables of a QFT are the vacuum to vacuum matrix elements $\langle vac| \phi_1(x_1) ... \phi_n(x_n|vac\rangle$ of the products of the local observables $\phi_i(x_i)$ you get by evaluating a field $\phi_i$ at $x_i$. These numbers can be assembled into distributions $G(x_1,...,x_n)$ on the product spacetime $(\mathbb{R}^{d+1})^{n}$, the n-point functions. Intuitively, the n-point function's values represent amplitudes for having some number of field quanta appear or disappear at the $x_i$.
Connected $n$-point functions are certain combinations of the ordinary $n$-point functions that you get by subtracting out the $m$-point functions for $m \leq n$. In the case where you only have one field, the simplest is $G_c(x_1,x_2) = G(x_1,x_2) - G(x_1)G(x_2)$. You can recover the ordinary $n$-point functions from these, but they're a bit more useful physically. Intuitively, they capture the situation where our field quanta appeared and disappeared at the $x_i$ by interacting with one another.
