Hey from my notes in my PS book it seems I have solved this some time in the past, but I cannot seem to get the indices straight this time around. So in deriving the Feynman photon-propagator which includes a general parameter $\xi$ (see PS equation 9.58 page 297) PA find the solution of the following equation

$${\large[}-k^2g_{\mu\nu}+\underbrace{(1-\xi^{-1})}_{-\chi}k_\mu k_\nu{\large]}\tilde{D}_F^{\nu\rho} = \mathrm{i}\delta_\mu^{\rho}\tag{9.57b}$$

I try to rewrite this as $$ {\large[}g_{\mu\nu}+\chi \frac{k_\mu k_\nu}{k^2}{\large]}\tilde{D}_F^{\nu\rho} = \frac{-\mathrm{i}}{k^2+\mathrm{i}0}\delta_\mu^{\rho}$$

and then try to find the inverse of

$${\large[}g_{\mu\nu}+\chi \frac{k_\mu k_\nu}{k^2}{\large]}$$

by using the identity for matrices

$$\tag{$\star$}(A+B)^{-1} = A^{-1} - \frac{1}{1+g}A^{-1}BA^{-1}$$

where $g= \mathrm{tr}(BA^{-1})$ (see this MSE answer).

As suggested in this related Phys.SE post, one can simply write the most general expression (respecting the symmetries of the theory) for $\tilde{D}_F^{\mu\nu} = Ag^{\mu\nu}+B k^\mu k^\nu$, plug into equation $(9.57\mathrm{b})$ and find the functions $A$ and $B$.

But the question remains whether one can use a theorem from linear algebra such as $(\star)$ above? What trace should one use in that case, etc?

The answer is btw the familiar Feynman photon propagator

$$\tilde{D}_F^{\mu\nu}(k) = \frac{-\mathrm{i}}{k^2+\mathrm{i}0} {\large[}g^{\mu\nu}-(1-\xi) \frac{k^\mu k^\nu}{k^2}{\large]}.\tag{9.58}$$


2 Answers 2


While a correct answer has been provided, based on Ken Miller's theorem on the inversion of sums of matrices, it can also be solved by the more specific, Sherman-Morrison formula.

It asserts under certain conditions that,

$$(A_{\mu\nu} + U_\mu V_\nu)^{-1} = A^{\mu\nu} - \frac{A^{\mu\lambda}U_\lambda U_\sigma A^{\sigma \nu}}{1 + V_\mu A^{\mu\nu}U_\nu}.$$

To simplify matters, we redefine,

$$\tilde{k}^\mu := \frac{\sqrt{\chi}}{|k|}k^\mu$$

so that we must find,

$$(\eta_{\mu\nu} + \tilde k_\mu \tilde k_\nu)^{-1} = \eta^{\mu\nu} - \frac{\tilde k^\mu \tilde k^\nu}{1+\tilde k \cdot \tilde k} = \eta^{\mu\nu} - \frac{\chi}{1+\chi} \frac{k^\mu k^\nu}{k^2}$$

and noting that,

$$\frac{\chi}{1+\chi} \bigg\vert_{\chi = \frac{1}{\xi}-1} = 1 - \xi$$

we recover the desired result,

$$\eta^{\mu\nu} + (\xi-1) \frac{k^\mu k^\nu}{k^2}.$$


Clearly from your question above, we identify $$ A_{\mu \nu} = g_{\mu \nu}, B_{\mu \nu} = \chi \frac{k_\mu k_\nu}{k^2} $$

Since the Minkowski metric is its own inverse, we know that $$(A_{\mu \nu})^{-1} = A^{\mu \nu} = g^{\mu \nu}$$

We then find that $$g = \mathrm{tr}(B A^{-1}) = \mathrm{tr}(B_{\mu\nu} A^{\nu \rho} ) = B_{\mu\nu} A^{\nu \mu} = \chi \frac{k_\mu k_\nu g^{\nu \mu}}{k^2} = \chi $$

And $$ (A_{\mu \nu} + B_{\mu \nu})^{-1} = A^{\mu \nu} - \frac{1}{1+g} A^{\mu \rho} B_{\rho \sigma} A^{\sigma \nu} $$

Here I have introduced new indices $\rho$ and $\sigma$ to make sure the matrix multiplications are carried out the right way. Plugging everything in, we find $$ \left(g_{\mu \nu} + \chi \frac{k_\mu k_\nu}{k^2} \right)^{-1} = g^{\mu \nu} - \frac{\chi}{1+\chi} \frac{k^\mu k^\nu}{k^2} $$

Rewriting this into the form of (9.58) is left as an exercise for the reader ;).


Your Answer

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