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Consider a quantum harmonic oscillator that is driven for a finite time by a force $J(t)$, and work entirely in Heisenberg picture. Then we may define the 'in' and 'out' vacua $$|0_{\text{in}} \rangle, \quad |0_{\text{out}} \rangle$$ to be the ground states of the Hamiltonian at early and late times. In Schrodinger picture, the 'in' vacuum corresponds to a state in the usual QHO ground state before the driving starts, while the 'out' vacuum corresponds to a state that ends up in that state when the driving ends.

In Mukhanov and Winitzki's book the retarded Green's function is defined as a matrix element between 'in' states, $$\langle 0_{\text{in}} | \hat{q}(t) | 0_{\text{in}}\rangle = \int J(t') G_{\text{ret}}(t, t') \, dt', \quad G_{\text{ret}}(t, t') = \frac{\sin \omega(t - t')}{\omega} \theta(t - t').$$ This makes perfect sense thinking semiclassically, as $\langle \hat{q}(t) \rangle$ is just the average position of the particle, given that it was at rest in the far past; that's basically the definition of what a retarded propagator is. Similarly, one can define the advanced propagator using $|0_{\text{out}}\rangle$.

Finally, Mukhanov and Winitzki define the Feynman propagator by $$\langle 0_{\text{out}} | \hat{q}(t) | 0_{\text{in}}\rangle \propto \int J(t') G_{F}(t, t') \, dt'.$$ Now, I've been searching for an intuitive understanding of the Feynman propagator for years. Typical explanations in quantum field theory speak of "negative energy solutions" and antiparticles (e.g. here) which I've always been confused about, since they don't exist in ordinary quantum mechanics (as I asked here). But above we have a Feynman propagator for an exceptionally simple non-QFT system! So if there's an intuitive explanation at all it'll be right here, but I can't quite see what the matrix element means physically.

I have two questions: first, how is this equivalent to the usual definition of the Feynman propagator, involving a particular contour choice? Second, are there intuitive words one can drape around this definition? Does it provide any additional physical insight?

knzhou
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    I don't think the question is really meaningful. Feynman propagators are useful object that are introduced in QFT because they are natural to use there (due to time ordering). If they are also useful to introduce in QM, why not use them ? – Adam Apr 03 '18 at 13:22
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    @Adam If I only cared about whether objects were useful for abstract manipulations, outside of any physical meaning, I'd have become a mathematician! – knzhou Apr 03 '18 at 13:31
  • What I mean is that I don't think your remark that Feynman propagator are surprising in QM makes much sense. They appear each time you have operators and time ordering. This happens already in QM, for instance in a path integral formalism $\langle q(t) q(t')\rangle$ is always a Feynman propagator, because the path integral formalism always time-order operators. I'm not familiar enough in the in-out formalism to see easily how they come in in that case. – Adam Apr 03 '18 at 14:20
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    @Adam I'm trying to leave correlators out of it, because I have even less intuition for those. There are no correlators and no time ordering in this question or in my other question. – knzhou Apr 03 '18 at 14:31
  • "What's the physical meaning of an expectation value between an 'in' and 'out' state?" Is there a reason that the answer for this should be any different than any other amplitude of an operator sandwiched between two states- i.e., roughly the coupling between the states due to that operator? In other words, except for the operator used, is your question materially different from ones like this: https://physics.stackexchange.com/questions/209350/off-diagonal-elements-of-hamiltonian-matrix-h-12-h-21-energy-of-trans ? – Rococo Apr 13 '18 at 23:07
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    @Rococo You can say everything in QM is just an amplitude, but the retarded Green’s function has a much better physical interpretation than that: it’s how the position of the mass reacts when you push it, produced it started at rest, just like the classical version. So I’m asking if a similarly nice interpretation exists for the Feynman Green’s function. – knzhou Apr 14 '18 at 07:33
  • @knzhou I guess I don't see much of a 'better physical interpretation.' If the retarded Green's function, which is sandwiched between two identical states, is the expectation value of a position, then it would be my naive assumption that the Feynman Green, which instead uses two different states, corresponds to a coupling in the same sense as any off-diagonal matrix element. Those explanations seem to me to be at precisely the same 'level of goodness' as far as physical interpretation. – Rococo Apr 14 '18 at 20:28
  • Maybe someone will come along with a much more satisfying answer to this question that makes this comment seem foolish, but if no one does perhaps this is part of the problem in that your sense of a satisfying physical interpretation does not necessarily comport with others'. – Rococo Apr 14 '18 at 20:29
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    @Rococo Maybe you're right. I guess I would settle for a clear explanation of how this definition is equivalent to the usual, 'close the contour in this way' definition, and maybe that result would be enough. – knzhou Apr 15 '18 at 12:49
  • Dear knzhou. I think this is a great question. I think I also know the essence of the answer to it, unfortunately I am struggling a bit with the specific example, which I am not used to. So I answered one of your related questions (https://physics.stackexchange.com/questions/384126/where-is-the-i-epsilon-in-quantum-mechanics?noredirect=1&lq=1). Hope that helps, no bounty for me unfortunately though ;-) – Wolpertinger Apr 15 '18 at 17:15

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You've motivated me to read the nice development in Mukhanov and think a little about this issue. Here is how I rationalized it to myself:

First, how is this equivalent to the usual definition of the Feynman propagator, involving a particular contour choice?

I think you should consider the fundamental definition of the Feynman propagator not as a propagator with a particular contour choice, but as an propagator that is symmetric under exchange of spacetime coordinates. In this zero-dimensional context, that means exchange of $t$ and $t'$. In QFT language this leads to a structure like:

$$ -i \langle 0 | \Theta(t-t')\psi(t)\psi(t')+ \Theta(t'-t)\psi(t')\psi(t)|0\rangle $$ While in the driven oscillator case the form is: $$\frac{i}{2\omega} \left(e^{-i\omega(t-t')}\Theta(t-t')+e^{-i\omega(t'-t)}\Theta(t'-t)\right)$$

In either case, you can Fourier transform this and get something with one pole above and one pole below the complex $\omega$ plane, from the two Heaviside functions. But since we are supposed to be looking for a physical intuition, I suggest you think of this as a consequence of demanding that the propagator be symmetric under exchange of times.

Second, are there intuitive words one can drape around this definition? Does it provide any additional physical insight?

As far as a physical interpretation, I think you will just have to accept that the interpretation isn't going to be as straightforward in the case of the Feynman propagator as it is for the retarded propagator. Your question already sets up why this is the case: one is an expectation value, which is something that appears classically and that we have a natural feel for, while the other is an off-diagonal matrix element. The good news is that by this point in our study of quantum mechanics, we've both seen enough of these to have some sense of what they mean. For example, an amplitude of the form $\langle 0_{out}|\hat{q} |0_{in}\rangle$ is well-known in atomic physics as the dipole operator. Roughly speaking, it represents the degree to which the states $|0_{out}\rangle$ and $|0_{in}\rangle$ are coupled by the operator $J \hat{q}$. Summarizing, then, the best interpretation I can offer for the Feynman propagator in this situation is that it is the object which tells you, for a given transient current, the extent to which the resulting output state is coupled to the initial state. So it is a particular, and calculationally important, way of characterizing how much the states were changed by the driving force. Since it is not itself an observable, it is not clear to me that a more intuitive description should exist or what it might entail.

Rococo
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You might try reading Chapter 3 of Robert Klauber's nice introductory book on QFT. This chapter is a free download at http://www.quantumfieldtheory.info/. His perspective is that the Feynman propagator "can be visualized as representing a virtual particle that exists fleetingly and carries energy, momentum, and in some cases, charge from one real particle to another. Thus, it is the carrier, or mediator, of force (interaction.)" The transit of such a virtual particle from x to y and the transit of a virtual antiparticle from y to x must both be taken into account in summing up all possible ways in which two physical particles can interact. This is just what the Feynman propagator does.

Pursuer
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