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About spin 2 polarization tensors $$\varepsilon_{\mu\nu}^{(a)}, $$ it is claimed that $$\sum_{a} \varepsilon_{\mu\nu}^{(a)}(k)\varepsilon_{\lambda\sigma}^{(a)}(k) = A(G_{\mu\lambda}G_{\nu\sigma}+G_{\mu\sigma}G_{\nu\lambda}) + B(G_{\mu\nu}G_{\lambda\sigma}),$$ with normalization by setting $$ \sum_a \varepsilon_{12}^{(a)}(k)\varepsilon_{12}^{(a)}(k) = 1$$ by putting $$\mu=\lambda=1,~ \nu=\sigma=2. $$ I have figured out $$A=1$$ But I do not know how to fix $$B=-\frac{2}{3}.$$ Thanks.

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  • $\begingroup$ What are properties of $\varepsilon_{\mu \nu}$? $\endgroup$ – nwolijin Mar 2 at 9:33
  • $\begingroup$ I have figure it out by using tracelessness of the tensor: $$g^{\mu\nu}\varepsilon_{\mu\nu}^{(a)}=0 $$, $$g^{\mu\nu}\sum_{a} \varepsilon_{\mu\nu}^{(a)}(k)\varepsilon_{\lambda\sigma}^{(a)}(k)=0$$, $$g^{\mu\nu}(g_{\mu\lambda}-\frac{k_{\mu}k_{\lambda}}{m^2})(g_{\nu\sigma}-\frac{k_{\nu}k_{\sigma}}{m^2})+g^{\mu\nu}(g_{\mu\sigma}-\frac{k_{\mu}k_{\sigma}}{m^2})(g_{\nu\lambda}-\frac{k_{\nu}k_{\lambda}}{m^2})+ B g^{\mu\nu}(g_{\mu\nu}-\frac{k_{\mu}k_{\nu}}{m^2})(g_{\lambda\sigma}-\frac{k_{\lambda}k_{\sigma}}{m^2})= ...$$ and finally get $$B=-\frac{2}{3}$$ but is too troublesome... are there easier ways? $\endgroup$ – Li Chiyan Mar 3 at 5:31

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