As far as I understand it a propagator, $D(x-y)$, gives the amplitude for a flow of positive energy-momentum from an earlier event $y$ to a later event $x$.

Addendum: Instead of talking about energy flows I should be more careful and say that the propagator, $D(x-y)$, gives the amplitude that a virtual particle created at event $y$ will be annihilated at a later event $x$ giving up a certain amount of positive energy-momentum to the absorber. One cannot imply that the virtual particle itself ever has a definite amount of energy-momentum.

But surely in order to conserve energy-momentum one then needs to apply the conjugate operator $D^\dagger(x-y)$ giving the amplitude for a balancing flow of negative energy-momentum from event $x$ backwards in time to event $y$?

Addendum: Using more careful language one should say that to conserve energy-momentum one needs to apply $D^\dagger(x-y)$ giving the amplitude that a virtual particle created at event $x$ travels backwards in time to event $y$ giving up a balancing amount of negative energy-momentum to the emitter.

Thus the overall amplitude for the complete propagation process of a particle from emitter at $y$ to absorber at $x$ is $P(x-y)$ given by:

$$P(x-y) = D(x-y) D^\dagger(x-y) = |D(x-y)|^2$$

which is in fact the probability of a particle propagating from $y$ to $x$.

Maybe the requirement of conservation of energy-momentum actually implies the Born rule?

PS Normally energy-momentum is said to be conserved at the vertices of Feynman diagrams. But if one applies conservation of energy-momentum at a vertex where a virtual particle is created then surely one is implying that the virtual particle has a definite energy-momentum which is not allowed?

Addendum 2: Summary in terms of CPT symmetry

In QED the scattering amplitude is expressed in terms of time-ordered Feynman diagrams so that is necessarily time-asymmetric.

But Maxwell's equations are symmetric with respect to time + parity + charge conjugation (CPT).

I think the solution is that when one calculates scattering probabilities using the Born rule one multiplies the QED scattering amplitude by its complex conjugate. That complex conjugate amplitude is made up of time-reversed, parity-reversed, charge-reversed Feynman diagrams. Thus the (real) product is CPT-symmetric as it should be to be consistent with Maxwell's equations.

Is this reasoning correct?

  • $\begingroup$ Those propagators are correlators for the fields themselves, e.g. $\left\langle \phi(x) \phi(y) \right\rangle $. For energy, you would have to look at correlators of specific components of the stress energy tensor, e.g. $\left\langle T_{00}(x) T_{00}(y) \right\rangle $. $\endgroup$ – lionelbrits Nov 30 '14 at 20:15
  • $\begingroup$ I shouldn't have expressed my argument in terms of energy flow. Please see my Addenda for a more careful argument that respects the fact that the virtual particles themselves do not have a definite energy-momentum. $\endgroup$ – John Eastmond Nov 30 '14 at 21:19

You're close, but wrong.

Indeed, one should always consider both possible directions of propagation - from $x$ to $y$ and from $y$ to $x$. But we do not multiply amplitudes for different processes, we add them. The total amplitude for a propagation process is given by

$$ D(x - y) + D^\dagger(x -y) = D(x - y) + D(y - x)$$

for $D(x - y) = \langle \phi(x)\phi(y) \rangle$ the not-time-ordered, ordinary propagator. This has nothing to do with energy-momentum conservation, and everything with relativity/causality - observe that, for spacelike separated $x$ and $y$, we would not even have an invariant notion of which of $x$ and $y$ comes first, so we must treat both possible processes on equal footing (and, for spacelike separation, they should turn out to cancel in oder to preserve causality).

  • $\begingroup$ A particle propagation is space-like? $\endgroup$ – Vladimir Kalitvianski Nov 30 '14 at 20:27
  • $\begingroup$ @VladimirKalitvianski: No. That's not what I wanted to say...I'll rephrase that $\endgroup$ – ACuriousMind Nov 30 '14 at 20:29
  • $\begingroup$ But in the standard QFT formalism one only considers time-ordered propagators that propagate from early events to later events. In order to take the time-reversed propagators into account one needs to multiply by the complex conjugate of the entire scattering process in just the same way as one does when one follows the Born rule to calculate the scattering probabilities. $\endgroup$ – John Eastmond Nov 30 '14 at 21:12
  • $\begingroup$ @JohnEastmond: While the time-ordered/Feynman propagator is more important in QFT, nothing prohibits us from looking at the non-ordered propagators as we do here. The Born rule tells us that we should take the modulus of $D(x-y) + D(y-x)$ to get a probability, it tells us nothing about propagators. $\endgroup$ – ACuriousMind Nov 30 '14 at 21:19

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