The Klein Gordon propagator is (Peskin p-30) $$ D_F(x-y)=\frac{i}{p^2-m^2} $$ which is actually the Green's function of the KG field. But a photon contains additionally $g_{\mu\nu}$ in the numerator. I would expect its propagator to be the same as $D_F$ since the photon is gauge boson. Why does it have $g_{\mu\nu}$? Does this somehow follow from the Ward identity?

  • 1
    $\begingroup$ You can read "Quantization of the Electromagnetic Field" in Peskin's book(p-294). $\endgroup$
    – Simon
    Commented Mar 16, 2015 at 9:05

2 Answers 2


Photon has spin 1 :) I am serious. Check it if you don't believe me. The physical meaning of the photon propagator $\Delta_{\mu \nu}$ is the following:

$$ \Delta_{\mu \nu} (x-y) = \left< A_{\mu}(x) A_{\nu}(y) \right>, $$

where $A_{\mu}$ is the electromagnetic potential form and $\left< ... \right>$ is a shorthand for the vacuum expectation value of the time-ordered operator product (in the non-interacting theory, ofcourse), or, equivalently (in the path integral picture), the following holds for any functional $\Omega[A]$:

$$ \left< \Omega[A] \right> = \int DA \cdot e^{i S[A]} \cdot \Omega[A], $$

where path integral measure $DA$ is defined up to normalizations and the $\left< 1 \right> = 1$ normalization condition is chosen.

This picture is obscured by gauge invariance, because non-gauge-invariant expectations are ill-defined. Probably the best way to deal with this is to eliminate gauge invariance by introducing the gauge-fixing term.

P.S. imagine that you have two spin-0 fields labeled by an index $\phi^a$. The propagator is then a $2\times 2$ matrix $\Delta^{a b}$. This is natural, because $\Delta$ encodes transition rates between two separate degrees of freedom ($a$ and $b$).

P.S.2. You mentioned the Klein-Gordon propagator to be the Green's function of the $\left( \Box + m^2 \right)$ operator. Well, the same holds for the photon propagator.

In the Feynman gauge, field equations are the following:

$$ \Box A_{\mu} = 0 \quad (\forall \mu), $$

which means that the field $A_{\mu}(x)$ is annihilated by the differential operator $\delta_{\nu}^{\mu} \Box$.

The Green function is therefore labeled by two space-time positions ($x$ and $y$) and two indices ($\mu$ and $\nu$) and is exactly

$$\Delta_{\mu}^{\nu}(x - y) = \delta_{\mu}^{\nu} \cdot D_F (x - y); $$

$$\Delta_{\mu \nu}(x - y) = g_{\mu \nu} \cdot D_F (x - y). $$

  • $\begingroup$ Is you D_F the same as the OP's D_F? The OP's D_F has a mass... $\endgroup$
    – hft
    Commented Mar 13, 2015 at 16:47
  • $\begingroup$ @hft My $D_F$ is the same as the OP's $D_F$ when $m=0$. I can't imagine how this could be unclear from my answer above. $\endgroup$ Commented Mar 15, 2015 at 1:43
  • $\begingroup$ It's unclear because you never mentioned that you set m=0, whereas the explicit definition has an "m". I can't imagine how you can't imagine this couldn't be unclear. $\endgroup$
    – hft
    Commented Mar 15, 2015 at 3:28
  • $\begingroup$ Thanks for the answer. You just assume vector boson dof more than one and tell metric tensor satisfies this. In fact, I see this point But I would expect how to derive it from gauge invariant lagrangian possibly. $\endgroup$
    – aQuestion
    Commented Mar 15, 2015 at 10:20
  • $\begingroup$ @hft it should be clear to anybody intelligent, because photon is mass zero ok? I mean no offence. I just think you are carping on me for no reason. $\endgroup$ Commented Mar 16, 2015 at 5:15

[...] since photon is spin-zero gauge boson. Why does it have gμν ? Does this somehow about Ward identity?

Photon is not "spin-zero gauge boson". It is "spin one gauge boson".

  • $\begingroup$ Yes it obviously a mistake but photon still a boson and I'd expect propagator as $D_F$ $\endgroup$
    – aQuestion
    Commented Mar 12, 2015 at 18:03
  • $\begingroup$ EM field obeys the KG equation what else i expect? Dirac propagator for fermions? $\endgroup$
    – aQuestion
    Commented Mar 13, 2015 at 9:40
  • $\begingroup$ Well, for one thing, photons have no mass. But the propagator you wrote does have a mass... $\endgroup$
    – hft
    Commented Mar 13, 2015 at 16:47
  • $\begingroup$ Set m=0 that is what I expect but the propagator has $g_{\mu\nu} additionally and thats I ask. $\endgroup$
    – aQuestion
    Commented Mar 13, 2015 at 17:47
  • $\begingroup$ @Major_Tom you have not one, but four independent (in Feynman gauge) Klein-Gordon fields. Propagator is therefore a unit $4\times 4$ matrix (Kronecker's delta) times the Klein-Gordon propagator. Lower one index for convenience and you will get your metric tensor. See my answer for more details. $\endgroup$ Commented Mar 15, 2015 at 1:46

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