What does "propagate" mean in QFT?

Question

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.

Answer 1

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.

Answer 2

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.