Massive vector boson propagator - Definition

[mchouza]

I'm reading Zee's QFT textbook and I'm stuck trying to understand why the $$\delta^\mu_\lambda$$ appears when he defines the propagator of a massive spin-1 boson as the inverse of a differential operator:

$$[(\partial^2 + m^2)g^{\mu\nu}-\partial^\mu\partial^\nu]D_{\nu\lambda} = \delta_\lambda^\mu \delta^{(4)}(x)$$

In particular, why only one vector index must be contracted between the propagator and the differential operator? I'm sure that the reason is very simple, but it's still eluding me.

Thanks!

 

[haushofer]

I'm reading Zee's QFT textbook and I'm stuck trying to understand why the $$\delta^\mu_\lambda$$ appears when he defines the propagator of a massive spin-1 boson as the inverse of a differential operator:

$$[(\partial^2 + m^2)g^{\mu\nu}-\partial^\mu\partial^\nu]D_{\nu\lambda} = \delta_\lambda^\mu \delta^{(4)}(x)$$

In particular, why only one vector index must be contracted between the propagator and the differential operator? I'm sure that the reason is very simple, but it's still eluding me.

Thanks!

Well, intuitively because Zee considers his analysis as a "continuation of matrix operators":

$$[D^{-1}]^{\mu\nu}(x)D_{\nu\rho}(x) = \delta^{\mu}_{\rho}\delta^4(x)$$
The 4D delta distribution is there because the propagator is a distribution; you integrate them over spacetime agains functions of spacetime.

 

[mchouza]

Thanks for your help. I was confused with the idea of "inverting a tensor", but now I realize that the differential operator works here as a bilinear form (like the metric) and not as a general tensor:

$$J_\mu M^{\mu\nu} J_\nu = J_\mu M^\mu\!_\nu J^\nu$$

So it makes sense to invert it:

$$M^\mu\!_\nu [M^{-1}]^\nu\!_\lambda = \delta^\mu_\lambda$$

$$M^{\mu\nu} [M^{-1}]_{\nu\lambda} = \delta^\mu_\lambda$$