Studying QFT, one finds the term "the field propagates" and I'm not sure I understand what it means. For example, in QED, one finds that $A_0$ "doesn't propagate" because its conjugate momentum $\Pi_0=\frac{\delta\mathcal{L}}{\delta(\partial_0 A_0)}$ is zero. What should one understand from this sentence?
From what I understand, this is used to show that the components of $A_\mu$ aren't all independent because they need to fulfill the gauge condition that $F_{\mu\nu}\rightarrow F_{\mu\nu}$, but I can't see how the term "propagate" would mean this.
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asked Aug 28, 2021 at 14:48
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2 Answers
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Technically speaking, a propagating field in QFT is by definition a field with on-shell DOF, cf. e.g. this Phys.SE post. The terminology is inspired by wave propagation in the theory of waves.
answered Aug 28, 2021 at 15:48
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Gauge invariance is a red herring here that adds extra complication without helping address your question, so let me just focus on a scalar field $\phi(x)$ for simplicity. Loosely speaking, you can think of a field operator $\phi(x)$ as creating or annihilating a particle at position $x$. In this way, the time-ordered two point function $\langle 0 | T \phi(x) \phi(y) | 0 \rangle$, can be thought of as the probability amplitude for a particle to be created at $y$ and later found at $x$. This is why the time ordered product is referred to as a propagator (specifically the Feynman propagator).
In a real-space perturbative expansion of correlation functions, you see many integrals appear like \begin{equation} \int d^4 w G_F (x,w) G_F(w,y) G_F(w,z) \end{equation} If you draw this integral as a Feynman diagram, it would correspond to an edge connecting the vertex $x$ to the vertex $w$, and two lines leaving $w$ to approach $y$ and $z$. We say this diagram represents the amplitude for the particle to "propagate" from $x$ to $w$, and then for two particles to be created at $w$ and for the two to propagate to $y$ and $z$. The idea is that the interactions between particles are local, so particles "propagate" from one interaction to another, and then eventually to asymptotic infinity where they are eventually captured by a detector.
I would not recommend taking the words too seriously. A Feynman diagram is really just a term in a perturbative expansion for a given correlation function. And, furthermore, often the momentum-space representation is more useful than the real-space one. The word "propagate" here is essentially a gimmick to give some physical intuition behind terms in this expansion.
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$\begingroup$ So what does "doesn't propagate" mean? That a particle has zero probability to be created somewhere and be found somewhere else? Also, how does your reasoning apply to just one component of a particle with spin ($A_0$ in my example)? $\endgroup$ Commented Aug 28, 2021 at 16:12
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$\begingroup$ @MauroGiliberti First, the more physical way to frame things is in terms of particles. For particles with spin, you need to form more complicated statements like "a particle at position $x$ with polarization $p$ propagates to position $y$ to with polarization $q$." Fields represent particles, and the polarization index corresponds to a spacetime index on a field -- loosely we can associate the two polarization states of a photon with the spacetime index $\mu$ on $A_\mu$. However, because representations of the Lorentz group and the Poincaire group don't align, (...) $\endgroup$– AndrewCommented Aug 28, 2021 at 16:26
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$\begingroup$ (...) there is a gauge redundancy that removes unphysical degrees of freedom from $A_\mu$. This is needed since $A_\mu$ has four components but a photon only has two polarizations. If you work out the Hamiltonian of electrodynamics, you'll find that $A_0$ appears as a Lagrange multiplier enforcing a first-class constraint (Gauss's law). The other components $A_i$ have conjugate momenta, and because of this we call them propagating degrees of freedom. The first class constraint removes one component and its conjugate momentum, and we are left with two unconstrained dofs -- 2 polarizations. $\endgroup$– AndrewCommented Aug 28, 2021 at 16:28
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1$\begingroup$ But this meaning of "propagating degree of freedom" is already present in classical field theory, and is not specific to QFT. The version I mentioned in my answer (which I guess is not what you wanted to know) is why we use the word "propagate" to represent certain terms in QFT. $\endgroup$– AndrewCommented Aug 28, 2021 at 16:30
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$\begingroup$ Alright so there are two distinct meanings! One is "propagating field" = a field whose propagator isn't zero, and one is "propagating dof" = a dof which has a conjugate momentum variable. Correct? $\endgroup$ Commented Aug 28, 2021 at 16:41