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I'm reading chapter 9.4 "Quantization of the electromagnetic field" of Peskin's and Schroeder's book.

When proving the unitarity of the gauge-invariant $S$-matrix, a trick is used.

$$ SS^\dagger=P_0S_\text{FP}P_0S_\text{FP}^\dagger P_0=P_0S_\text{FP}S_\text{FP}^\dagger P_0. $$

The $S$ on the LHS is a gauge-invariant $S$-matrix while the $S$ on the RHS is unitary but not gauge-invariant. $P_0$ is a projection onto the subspace of the space of asymptotic states in which all particles are either electrons, positrons, or transverse photons.

In the formula shown above, how is the $P_0$ in the middle removed?

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"Let $S_{FP}$ be the $S$-matrix between general asymptotic states, computed from the Fadeev-Popov procedure". If I understand it correctly when $S_{FP}$ acts on asymptotic states it produces also asymptotic states. Therefore, the projection in the middle is redundant.

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    $\begingroup$ I think what you mean is that apply $SS^\dagger$ to asymptotic states, as the $P_0$ on the side of $S_{FP}S_{FP}^\dagger$ already projected the space of the state into the subspace, therefore the internal $P_0$ is redundant. Did I get it right? $\endgroup$
    – David Shaw
    Commented Feb 23, 2022 at 15:32
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Equation (9.60), that was proven earlier in Section 5.5, states that you can calculate the cross section based on all polarizations because the unphysical timelike and longitudinal ones will be canceled automatically, due to the Ward-Takahashi identity. This is the information that you require to remove the projector in the middle of (9.61) : the effect of the removal will be that all polarizations will take part in the FP S-matrix calculation instead of just the physical ones, but (9.60) will guarantee that the result of the expression will remain unchanged. From then on, proving unitarity is straightforward.

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