Interpretation of $\phi(x)|0 \rangle$ in interacting theory

I'm trying to relearn quantum field theory more carefully, and the first step is seeing what facts from free field theory also hold in the interacting theory.

In free field theory, the state $\phi(x) | 0 \rangle$ contains exactly one particle, which is localized near $\mathbf{x}$ at time $t$ where $x = (\mathbf{x}, t)$. However, typical quantum field theory books are silent on the question of what $\phi(x)$ does in the interacting theory. It seems like it's implicitly assumed to still create a particle localized near $\mathbf{x}$, because correlation functions are said to be the amplitude for particle propagation, but I haven't seen any explicit justification.

• Does $\phi(x)$ still create a field excitation localized near $x$? If so, how can we see that? (I'm already aware that the localization is not perfect in the free case, but that's a separate issue.)
• Is $\phi(x) | 0 \rangle$ still a one-particle state? It's definitely not a one-particle state using the free theory's number operator, but is it in some sense a state with one 'dressed' particle? If so, how would we formalize and prove that?

More generally, how should I think of the action of $\phi(x)$ in the interacting theory? How about for a weakly interacting theory?

• You might, or might not, consider an answer to an older question joined at the hip: "almost localization"... – Cosmas Zachos Oct 14 '17 at 21:52
• @CosmasZachos I don't think all of these links are relevant to my question. They're all about formal issues about $\phi(x) |0 \rangle$ for a free field while I care about an intuitive picture of $\phi(x) |0 \rangle$ for an interacting field. – knzhou Oct 14 '17 at 21:59
• Perhaps... the asymptotic fields in scattering are free one-oscillator-style assemblies in the Fock space; x is a terrible variable to intuit the interaction in... much easier to consider in momentum space where most of QFT happens, conserving momentum. I cannot claim I understood the question but you appear to be seeking intuition on the Haag-Ruelle picture. – Cosmas Zachos Oct 14 '17 at 23:55

3 Answers

All this can be answered by the same object: the Green`s function $\langle 0|\mathcal{T}(\phi(x_1)...\phi(x_n))|0\rangle$ and the polology of it. First, we need to do a fourrier transform:

$$G(q_1,...,q_n)=\int \frac{d^dq_1}{(2\pi)^d}...\frac{d^dq_n}{(2\pi)^d}\,e^{-iq_1.x_1}...e^{-iq_n.x_n}\langle 0|\mathcal{T}(\phi(x_1)...\phi(x_n))|0\rangle$$

Poles at on-shell $q=q_1+...+q_r$, i.e. $q²=-m²$, indicates the existence of a particle of mass $m$, the existence of a one-particle state $|\vec{p},\sigma\rangle$ in the spectra. The residue of that pole measure the projection of this one-particle state $|\vec{p},\sigma\rangle\langle\vec{p},\sigma|$ with the state: $$\int \frac{d^dq_1}{(2\pi)^d}...\frac{d^dq_r}{(2\pi)^d}e^{-iq_1.x_1}...e^{-iq_r.x_r}\langle 0 |\mathcal{T}\phi(x_1)...\phi(x_r)$$ and the state $$\int \frac{d^dq_{r+1}}{(2\pi)^d}...\frac{d^dq_n}{(2\pi)^d}e^{-iq_{r+1}.x_{r+1}}...e^{-iq_n.x_n}\mathcal{T}\phi(x_{r+1})...\phi(x_n)|0\rangle$$ In particular, one can show the LSZ reduction formula by putting each $q_i$ on-shell. You can see all this in the chapter $10$ of the first volume of Weinberg QFT textbook.

Now, if you want to know how to make sense of the state $\phi(x)|0\rangle$ you just need to plug this state into an arbitrary Green's function and work out the residue of various poles. Note that just Lorentz symmetry fixes the projection of this state with one-particle state, up to a overall normalization:

$$\langle \vec{p},\sigma|\phi_{\alpha}(x)|0\rangle = \sqrt{Z}\,u_{\alpha}(\vec{p},\sigma)e^{-ipx}$$ where this $u$-function is the polarization function, responsible to match the Lorentz indice $\alpha$ with the little group indice $\sigma$. In interacting theories, $\phi_{\alpha}(0)|0\rangle$ is not parallel to $|\vec{p},\sigma\rangle$, since by the polology and LSZ reduction formula above, this would mean that the Green's function of more than two points are zero, implying that the S-matrix are trivial.

Answering both your questions:

1. Yes, the state $\phi_{\alpha}(x)|0\rangle$ correspond to a local preparation, a local field disturbance if you like, because it has the right properties under Lorentz-Poincaré transformations.
2. No, the state $\phi_{\alpha}(x)|0\rangle$ is not a one-particle state because this would imply a trivial S-matrix by the LSZ reduction formula and polology of the Green's function, contradicting the assumption of an interacting theory.

The truth is that, we actually do not know yet what $\phi|\Omega\rangle$ is except that we can solve the interacting theory exactly. Moreover, we even do not know what the vacuum $|\Omega\rangle$ is; it is not the vacuum $|0\rangle$ in field theory. For instance, let us consider the $\phi^4$ theory $$\mathcal{L}=\frac{1}{2}(\partial_\mu\phi)\partial^{\mu}\phi-\frac{1}{2}m\phi^2-\frac{1}{4!}\phi^4.$$ According to the quantization procedure, we need to solve the EoM: $$\partial^2\phi+m^2\phi+\frac{\lambda}{3!}\phi^3=0.$$ Even for this simple equation, unfortunately, we do not know the exact solutions in four dimensions. Nevertheless, let us assume we have a complete solutions $\phi_i(x)$ where $i$ may be discrete or continuous. Then we can have a bunch of operators $a_i$, $a_i^+$ via the decomposition: $$\phi(x)=\sum (a_i\phi_i(x)+a^+_i\phi^*_i(x)).$$ Then we may construct the state $a_i^+|\Omega\rangle$, or more general, the Fock space. This state, however, may be completely different from the normal particle state. In particular, it may not be an eigenstate of the momentum operator and energy operator. Then, what really it is? Well, it is just a quantum state except we are not stuck with the particle picture. What is a black hole is? Is it particles? No, it is just a solution to Einstein equations. Needless to say, the superpostion state $\phi(x)|\Omega\rangle$ may be far away from the picture of a particle located at $x$.

When $\lambda$ is big, i.e., at the strong coupling, the picture is indeed very different from the free particles. However, when $\lambda\ll 1$, we can take the $\phi^4$ as a small perturbation and expand everything around the free particles. At the zero order, we take $\phi^{(0)}=\varphi$ where $\varphi$ is the solutions to free fields. At the first order, we approximate the EoM as $$\partial^2\phi^{(1)}+m^2\phi^{(1)}+\frac{\lambda}{3!}\varphi^3=0$$ etc..

In perturbation theory, when we say a particle, it is always the state of free fields. And complication comes with the $\phi^4$ perturbation (renormalization etc..).

• If the naive particle interpretation of $\phi(x) | \Omega\rangle$ fails, then what is the right intuition for correlation functions? I've always heard that they represent some amplitude for particle propagation, but following what you say, that can't work. – knzhou Oct 15 '17 at 22:04
• @knzhou In perturbation theory, the correlation functions indeed represent some amplitude for particle propagation. If there is no interactions, then this propagation is very clean, just a line in Feynman diagram language. Due to the interactions and the perturbation expansion, there are actually infinite diagrams in this propagation. But this picture fails to be true when the couplings become very strong. That is the reason why people say we can never understand QFT fully if they can not solve it exactly. – Wein Eld Oct 16 '17 at 10:57

I guess I can give some plausible picture of interacting fields, and it is the following. Consider the Schrödinger equation for a Hydrogen atom. It contains an interaction term. Without this term we have free proton and electron with clearly defined momenta, energy, etc. Now what is an exact solution to this equation? It is a bound state, which can be represented as a free center of mass and a quasi-particle equation (with reduced mass), the latter is in a bound state. If we deal with electron coordinates, the state of the electron depends on the state of proton due to interaction, so the bound electron is in fact in a mixed state. The same is valid for proton. Thus, interacting "fields" corresponding to the individual particles will create these particles in mixed states. It is a messy description, but it is the true physics of interaction effect. Nothing special, nothing unknown.

Now, instead we may use the center of mass and the relative motion coordinates. The equations for them are decoupled from each other - they describe quasi-particles or collective motion degrees of freedom. Ideally, the normal modes of a compound (=interacting) system are decoupled from each other, at least approximately. Only external force may make them coupled again because of acting differently on the system constituents. The fields corresponding to the collective coordinates create one-excitation states of quasi-particles, which may be in pure states in absence of external forces.

Often an inclusive picture is reduced to a "one-particle" approximation that we observe in experiment. For more details and insights see "Atom as a "Dressed" Nucleus"