It is mentioned in this answer that the completeness relation for the polarization vectors of a (massive) electromagnetic field $$ \sum_\lambda \varepsilon^\mu(\lambda,k) \varepsilon^{\nu*}(\lambda,k) = - \eta^{\mu\nu} + \frac{k^\mu k^\nu}{M^2}. $$ can be understood since $P_\epsilon^{\mu\nu} := \sum_\lambda \epsilon^\mu_\lambda(k)\epsilon^{*\nu}_\lambda(k)$ is a projection operator. How can this be understood? In particular, on whom does $P_\epsilon^{\mu\nu} $ act in order to project out what?
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Applying the operator $P^{\mu\nu}$ to a generic vector $v_\nu$,
$$ -P^{\mu\nu} v_\nu = v^\mu - \frac{k^\nu v_\nu}{M^2} k_{\mu} . $$
That is, up to a sign, $P$ "removes" (projects out) from $v$ its component parallel to $k$ (assuming $k_\mu k^\mu = M^2$). Therefore, it projects any vector to the surface orthogonal to $k$.
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1$\begingroup$ perfect thanks! This explains how it works. Is there a good reason why it works? It's not entirely clear to me, why the sum over the polarization vectors should yield such a projection operator. $\endgroup$ – jak Nov 4 at 16:03
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$\begingroup$ I figured it out. Nevermind. This is really just the standard linear algebra definition of a projection operator that projects onto the subspace spanned by the vectors. $\endgroup$ – jak Nov 4 at 16:48
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$\begingroup$ Yes, the polarisation vectors are indeed a basis of the subspace orthogonal to $k$. $\endgroup$ – fqq Nov 4 at 17:12
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Hint: try to act with $P_\epsilon^{\mu\nu}$ on $k_\nu$.
