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.

  • $\begingroup$ 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 $\endgroup$
    – user108787
    Nov 7, 2016 at 10:51
  • $\begingroup$ 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. $\endgroup$
    – user108787
    Nov 7, 2016 at 12:56

2 Answers 2


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 $(2)$.

  • $\begingroup$ I now see I got it completely wrong. Thanks a lot for the detailed answer. $\endgroup$
    – Razor
    Nov 7, 2016 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.