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I often encounter the term "classical field configuration" in the scope of QFT, but I have a hard time interpreting what it really means.

If I understood it correctly, then a general field configuration is just an explicit form of a field, e.g. a free field $\phi(x)$. If this is correct, then I wonder what a classical configuration is exactly!

Qmechanic
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Welcome_Green
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A classical field configuration of a field $\phi(x)$ is just an explicit form for $\phi(x)$ that satisfies the equations of motion.

This is in contrast to quantum fields, where $\phi(x)$ is an operator-valued distribution. Instead of thinking of the field as a single function, we want to know the probability amplitude for the field to be in a given configuration. You can characterize the distribution by looking at moments, such as $\langle \phi(x)\rangle$, $\langle \phi(x) \phi(y) \rangle$, $\langle \phi(x) \phi(y) \phi(z)\rangle$, etc, where the average is taken in a given state (typically the vacuum state, especially in introductions to QFT). Much of the formalism of quantum field theory is devoted to ways to compute these correlation functions, sometimes also called Green's functions.

NOTE I removed a parenthetical statement "(note that there is a non-zero probability amplitude for the field to be in configurations that don't satisfy the classical equations of motion)" because, upon reflection, I don't know what I meant by that. In the path integral formulation, you do sum over field configurations that do not satisfy the classical equations of motion. In the operator language, general correlation functions do not evolve according to the classical equations of motion because of contact terms; see Eq 2.13.9 of https://sites.krieger.jhu.edu/jared-kaplan/files/2016/05/QFTNotes.pdf. Thanks to user @AlexGower for discussions on this point.

Andrew
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  • Okay, so: Classical field <=> Field that satisfies equations of motion (Euler-Lagrange I suppose). And this is called classical because quantum fields (operators) do not satisfy something comparable, but rather are investigated in the scope of correlators. Is that right? – Welcome_Green Mar 23 '22 at 11:53
  • Yep good summary. – Andrew Mar 23 '22 at 11:55
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    A clarification: the quantized field as an operator-valued distributions does satisfy a similar equation of motion, $\phi( A f)=0$, for all suitable functions $f$ and $A$ being the equation of motion operator (Klein-Gordon operator for example). Otherwise there would not be any information in the correlators about the underlying theory ! – Alan Garbarz Mar 23 '22 at 12:08
  • Thanks Alan, that makes sense! So, quantum fields fulfill these other equations of motion (Klein-GOrdon etc), wherears classcial fields satisfy classical equations of motion (ELG)? – Welcome_Green Mar 23 '22 at 12:27
  • @SimonFischer No, the equations satisfied by quantum fields are not directly analogous to the equations satisfied by classical fields. Sometimes the same operators appear (meaning, $\square + m^2$ will appear in an equation that a quantum field satisfies), but the solution to these equations is a distribution over possible field configurations, not a classical field configuration. There are lots of ways to formulate QFT. One way is via the Schwinger-Dyson equations, which are equations that determine the correlation functions, and these are not equivalent to, eg, the Klein-Gordon eqn. – Andrew Mar 23 '22 at 12:31
  • https://en.wikipedia.org/wiki/Schwinger%E2%80%93Dyson_equation – Andrew Mar 23 '22 at 12:32
  • Got it, thanks! Sorry to follow up so late, but in Schwartz's argumentation these "classical fields" arise as eigenvalues of eigenstates of field operators. Is there a specific reason why exactly the eigenvalues of the fields (in his case free fields) should fulifill the classical equations of motion? – Welcome_Green Mar 23 '22 at 23:45
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    In the Heisenberg picture, the fields operators obey the classical equations of motion. This is useful for free fields, since we can actually solve the classical equations exactly. Meanwhile, the states are time independent. A field eigenstate is essentially just the initial condition of the field. So you'd have an equation like $\hat{\phi}(x, t) | \phi(x, 0) \rangle = \phi(x, t) | \phi(x, 0) \rangle$ (where $\hat{\phi}$ means operator), so the classical field appears. But usually we don't compute correlation functions in field eigenstates. – Andrew Mar 23 '22 at 23:53