# Deriving the photon Propagator

This has been asked before (see Deriving photon propagator), but in deriving the photon propagator, when we arrive at:

$[-g^{\mu\nu}k^2 + \frac{\alpha - 1}{\alpha}k^\mu k^\nu] \tilde{D}_{\nu\lambda}(k) = \delta^\mu _\lambda$

We are supposed to invert the operator on the left to get the propagator. I know that we are supposed to use the ansatz:

$\tilde{D}_{\nu\lambda}(k) = Ag_{\nu\lambda} + Bk_\nu k_\lambda$

to determine the coefficients A and B. But don't we need two equations to determine two variables? If so, what is the second equation?

Also I could not get very far using this condition above, all I got was:

$3A - \frac{A+B}{\alpha} = \frac{4}{k^2}$

Is this correct at all? If so, how to proceed next? A detailed solution would be very appreciated.

• When you say it has been asked before , do you mean this: physics.stackexchange.com/questions/137577/… it's a good idea to put in the link you refer to, in your post thanks – user108787 Nov 7 '16 at 10:51
• No need for apology with my sort of remarks , you need only apologise if/when you really get things wrong on a personal level :), best of luck with your question. – user108787 Nov 7 '16 at 12:56

As the original poster of that question I think I might be able to help. I will use the same notation as in the original question.

Starting from the equation $$\left(-k^2g_{\mu\nu}+(1-\frac{1}{\xi})k_\mu k_\nu\right)D^{\nu\rho}(k)=i\delta^\rho_\mu\tag{1}$$

we make the Ansatz

$$D^{\mu\rho}(k)=A g^{\mu\rho}+B k^\mu k^\rho. \tag{2}$$

Inserting this Ansatz into the Eq. $(1)$ we get

$$\left[-k^2 g_{\mu\nu}+\left(1-\frac{1}{\xi} \right)k_\mu k_\nu\right]\left[A g^{\nu\rho} +B k^\nu k^\rho\right]=i\delta^\rho_\mu.\tag{3}$$

(You are missing the factor of $i$ from the right side of $(3)$ in your question, but I'll put it here.)

What you need to do is not solve $A$ and $B$ by isolation, but actually compare the coefficients on both sides of Eq. $(3)$. Expanding the product and after a little algebra, you should arrive at

$$A=-\frac{i}{k^2},\quad B=\frac{i}{k^4}(1-\xi),$$

which will give you the coefficients for the inverse $(2)$.

• I now see I got it completely wrong. Thanks a lot for the detailed answer. – Razor Nov 7 '16 at 19:52

Since you're using tensors with Greek indices, I want to point out that convention implies that $\bar D_{\nu\lambda}(k)=Ag_{\nu\lambda}+Bk_\nu k_\lambda$ is technically 16 equations, which is more than enough to isolate for $A$ and $B$. If $g_{\nu\lambda}$ is the metric and $k_\nu$ is a non-operator-valued momentum tensor, then both terms in the r.h.s. should be symmetric, which means you have at most 10 equations (less over-determined is better).

In theory, you should be able to isolate $A$ and $B$ by entering in the different components of $g$ and $k$. Not knowing what those tensors are comprised of, I can't help there. But I can tell you this is not a case of having too few equations (in fact, I'd be upset for having too many equations).