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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Qmechanic
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Mauro Giliberti
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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.
Andrew
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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)? – Mauro Giliberti Aug 28 '21 at 16:12
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@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, (...) – Andrew Aug 28 '21 at 16:26
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(...) 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. – Andrew Aug 28 '21 at 16:28
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1But 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. – Andrew Aug 28 '21 at 16:30
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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? – Mauro Giliberti Aug 28 '21 at 16:41
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@MauroGiliberti I think what you'll find is that a lot of physics language is imprecise and you can't always pin down a mathematical definition. The loose picture of "propagation" is that a wave gets generated at point $A$ and propagates to point $B$; there are a lot of contexts where this comes up and mathematical ways to represent this. In the context of your question, a propagating dof is one for which you need to specify initial data for it and its conjugate momentum. This implies that it exists n the Hamiltonian formulation with a conjugate momentum, so $A_0$ (a Lagrange multiplier)... – Andrew Aug 28 '21 at 16:52
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...isn't propagating. And, some combination fo the $A_i$ is not propagating, because there is one constraint (Gauss's law) and you aren'f free to specify initial data for all 3 $A_i$ independently; only 2 of them. I don't agree with the "propagator being zero" idea -- the Feynan propagator depends on the gauge choice, and there are gauges (like Lorenz gauge) where $\langle 0 | T A_0(x) A_0(y)| 0 \rangle$ is nonzero. – Andrew Aug 28 '21 at 16:54