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?

Answer

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.