On-shell Condition for physical particles in QFT Last year Prof. Ashoke Sen proved Soft Theorems for gravitons for generic theories of gravity. The papers are exceptionally well-written and I think I understand most of it (partly due to the lecture series available online by himself) but there's a trivial thing that I'm stuck on. He says that the on-shell condition for a particle with polarisation $\epsilon_i(p_i)$ is $\epsilon^T.K = 0$. Here $K$ is the kinetic energy operator (or the 2 point coupling matrix). I understand that I must be missing something trivial I just can't figure out what. I appreciate all the help I can get.
 A: This is merely a long comment.
Let's begin with a simpler question: why does a spin-1 boson of mass $m>0$ satisfy $k^\mu\epsilon_\mu^{(a)}=0$ for polarization vectors $\epsilon_\mu^{(a)}$? By Lorentz invariance we need only check the condition in the rest frame, which simplifies the condition to $m \epsilon_0^{(a)}=0$; and since $V_0=0$ has a vector space of solutions, each $\epsilon_\mu^{(a)}$ is easily obtained in this reference frame, viz. $\epsilon_\mu^{(a)}=\delta_\mu^a$. Since we're working in the particle's rest frame, the constraints $k\cdot\epsilon^{(a)}=0$ still leave us with the spin degrees of freedom.
The basic idea above admits a number of generalisations. Sec. I.5 of Zee's Quantum Field Theory in a Nutshell discusses the above argument, and does a similar degrees-of-freedom count for a massive spin-2 boson, viz. $k^\mu\epsilon_{\mu\nu}^{(a)}=0$. In this case, the extension from the rest frame to other reference frames uses Lorentz covariance (or is it contravariance?) instead of Lorentz invariance. Sen has added a spacetime index to the kinetic factor rather than the polarization tensor, but the principle remains the same.
And in each case, the extension to massless particles follows from continuing the massive physics to the massless case, roughly speaking. There is an important discontinuity in the analysis, though it doesn't ruin any of the equations so far. Since $\epsilon_\mu\to\epsilon_\mu+\lambda(k) k_\mu$ preserves the constraint if $k^2=0$, a massless particle has one fewer degree of freedom. See also Sec. 4.4.1 here.
