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I have read an equation as follow $$[-(k^2-m^2)g^{\mu\nu}+k^{\mu}k^{\nu}]D_{\nu\lambda}(k)=\delta^{\mu}_{\lambda}$$ The solution is given as: $$D_{\nu\lambda}(k)=\large{\frac{-g_{\nu\lambda}+k_{\nu}k_{\lambda}/m^2}{k^2-m^2}}$$ I can only verify the answer but do not know how to yield the result step by step... are there any standard procedures to yield the solution?

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We can make use of the projection operators \begin{align} & P^{T}_{\mu \nu} = g_{\mu \nu} - \frac{k_{\mu}k_{\nu}}{k^{2}}, \\ & P^{L}_{\mu \nu} = \frac{k_{\mu}k_{\nu}}{k^{2}}, \end{align} where $P^{T}_{\mu \nu}$ and $P^{L}_{\mu \nu}$ are the transverse and longitudinal projectors. $P^{T}_{\mu \nu}$ and $P^{L}_{\mu \nu}$ are the projection operators onto vectors orthogonal and parallel to $k^{\mu}$, respectively. You can verify by yourself that they satisfy the properties of projectors.

We then write \begin{align} -(k^{2}-m) g^{\mu \nu} + k^{\mu}k^{\nu} = A \cdot P^{T,\mu \nu} + B \cdot P^{L,\mu \nu} \end{align} where we can find \begin{align} A = -(k^{2} - m^{2}),\ B = m^{2}. \end{align} Now, to find the inverse $D_{\mu \nu}(k)$ we simply invert the $A$ and $B$ and then lower the $\mu$ and $\nu$ in $A \cdot P^{T,\mu \nu} + B \cdot P^{L,\mu \nu}$, where we yield \begin{align} D_{\mu \nu}(k) & = \frac{1}{A} P^{T}_{\mu \nu} + \frac{1}{B} P^{L}_{\mu \nu} = \frac{1}{-(k^{2}-m^{2})}P^{T}_{\mu \nu} + \frac{1}{m^{2}} P^{L}_{\mu \nu} \\ & = \frac{1}{-(k^{2}-m^{2})}\left( g_{\mu \nu} - \frac{k_{\mu}k_{\nu}}{k^{2}} \right) + \frac{1}{m^{2}} \frac{k_{\mu}k_{\nu}}{k^{2}} \\ & = \frac{1}{k^{2}-m^{2}}\left( -g_{\mu \nu} + \frac{k_{\mu}k_{\nu}}{m^{2}} \right). \end{align} When solving this sort of equations it is always very helpful to use $P^{T}_{\mu \nu}$ and $P^{L}_{\mu \nu}$, since the inverting process is very simple. For more reference you can see "Quantum Field Theory: Lectures of Sidney Coleman". In chapter 28.6 he actually present the same calculation that I did here. He also gave the identities verifying that $P^{T}_{\mu \nu}$ and $P^{L}_{\mu \nu}$ are indeed projectors.

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  • $\begingroup$ thanks a lot... $\endgroup$
    – Li Chiyan
    Feb 27 at 4:30
  • $\begingroup$ @LiChiyan if this answer is good enough to answer your question (which I think it is), you should upvote and accept it. $\endgroup$ Feb 28 at 4:23

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