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In P&S book, in Charpter 7: radiative corrections p. 212 (attached beneath, the completeness relation: (Equation 7.2) $$\textbf{1}=\left\vert \Omega\right\rangle\left\langle\Omega\right\vert+\sum\limits_{\lambda}\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_\textbf{p}} \left\vert \textbf{p}\right\rangle\left\langle\textbf{p}\right\vert.\tag{7.2}$$ This is in analogy with Equation 7.1: $$(\textbf{1})_{1-particle}=\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_\textbf{p}} \left\vert \textbf{p}\right\rangle\left\langle\textbf{p}\right\vert.\tag{7.1}$$ My question is:what exactly $\sum\limits_{\lambda}$ mean? P&S say: "the sum runs over all zero-momentum states $\left\vert \textbf{p}\right\rangle\\$. I don't really get this, does this mean we use $\lambda$ to denote different particles that are present in scattering processes? Such as: $\lambda=1$ denotes electron and $\lambda=2$ denotes positron?

The original Page 212 of P&S book:

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    $\begingroup$ @Andrew Thanks! I just fix this, sorry about the mess! $\endgroup$
    – ffz
    Commented Nov 28, 2022 at 14:24

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$|\lambda_0\rangle$ runs over all the single particle states at zero momentum. This will encode spin degrees of freedom -- so for a spin-1/2 particle, you'd sum over $m=1/2$ and $m=-1/2$ states, where $m$ is the spin quantum number (as well as antiparticle states).

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  • $\begingroup$ $\lambda_0$ also runs over states with more than two particles. $P$ denotes total momentum of the system. $\endgroup$
    – Keyflux
    Commented May 14, 2023 at 11:16

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