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Physically, propagator represents the probability amplitude of a particle to travel from one point to another. But the photon propagator $$D_{\mu\nu}(x,y) = \langle 0 | \mathcal{T}[A_\mu(x) A_\nu(y)] | 0\rangle$$ is a second rank tensor quantity, whereas probability amplitude is a scalar quantity! So

  1. What does the $(\mu,\nu)^\text{th}$ component of a photon propagator represents physically?
  2. What is the total probability that a free photon will reach at point $x$, starting its journey from point $y$ in terms of $D_{\mu\nu}(x,y)$ only? (NOTE that free photon means no other particles in the path of the photon.)
SCh
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    Sounds like a homework question, may be this tag will be appropriate – Artem Alexandrov Jan 21 '24 at 18:45
  • No, it's just a query while going through the subject. I am getting any explanation in this context neither in any of the QFT books, I've encountered nor in any article/paper. And also the first question is just a subset-question of the second one. – SCh Jan 21 '24 at 18:48
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    "propagator represents the probability amplitude of a particle to travel from one point to another" No it doesn't. – AccidentalFourierTransform Jan 23 '24 at 15:29

1 Answers1

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The two free indices on a non-scalar propagator simply label the spin- or polarization-state of the particle. So naively, $D_{\mu\nu}(x,y)$ is about a photon with polarization vector parallel to the $\mu$-direction at $x$ propagating into a photon with polarization vector parallel to the $\nu$-direction at $y$. To get a total probability ignoring polarizations, just sum over all possible polarizations for the start and end points.

However, while this interpretation of the propagator feels intuitive, it's fraught with issues in the case of photons, so you shouldn't think that you can use the naive value of this propagator to directly say anything about how photons behave:

  • Since the photon only has 2, not 4, degrees of freedom - i.e. in observable reality there are only transverse but not longitudinal or timelike polarizations of light - not all polarizations are physically meaningful, and the allowed transversal polarizations depend on the momentum of the photon. The propagator obeys corresponding Ward-Takahashi identities so that the unphysical polarizations decouple from the observable physics.

  • The exact expression for the propagator depends on the gauge-fixing scheme using during quantization of the gauge theory of the massless vector boson, e.g. the choice of $\xi$ for the Feynman gauge-fixing term $-\frac{\xi}{2}(\partial_\mu A^\mu)^2$.

  • Since the photon is massless and hence inherently relativistic, there is no non-relativistic limit in which proper position operators exist, i.e. photons are not localizable in a non-relativistic limit like massive particles are and do not have wavefunctions describing a probability for the photon to be at a point in the sense we usually use those terms, see e.g. this question and its linked question for more on photon wave functions and their localizability.

ACuriousMind
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