# Why does the photon propagator contain the metric tensor?

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?

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

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

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

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

• Is you D_F the same as the OP's D_F? The OP's D_F has a mass...
– hft
Commented Mar 13, 2015 at 16:47
• @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. Commented Mar 15, 2015 at 1:43
• 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.
– hft
Commented Mar 15, 2015 at 3:28
• 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. Commented Mar 15, 2015 at 10:20
• @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. 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".

• Yes it obviously a mistake but photon still a boson and I'd expect propagator as $D_F$ Commented Mar 12, 2015 at 18:03
• EM field obeys the KG equation what else i expect? Dirac propagator for fermions? Commented Mar 13, 2015 at 9:40
• Well, for one thing, photons have no mass. But the propagator you wrote does have a mass...
– hft
Commented Mar 13, 2015 at 16:47
• Set m=0 that is what I expect but the propagator has $g_{\mu\nu} additionally and thats I ask. Commented Mar 13, 2015 at 17:47 • @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. Commented Mar 15, 2015 at 1:46