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I am reading P&S (Peskin's and Schroeder's book on QFT), Chapter 16.3 entitled Ghosts and Unitarity. The authors employ the optical theorem to calculate the imaginary part of a $f\bar{f}\rightarrow f\bar{f}$ amplitude that also contains a gauge boson loop. Diagrams of this sort are given by taking the diagrams in Figure 16.2 and gluing them together with diagrams that correspond to the reverse processes (i.e. 2 incoming bosons $\rightarrow$ 1 fermion and 1 antifermion) in all the possible ways.

According to the authors, applying the optical theorem results in the imaginary part of this combined amplitude being reduced to a multiplication of sub-amplitudes corresponding to processes of the form $f\bar{f}\rightarrow bb$ (i.e. sum wrt all the possible ways in which the process $f\bar{f}\rightarrow bb$ can occur) with sub-amplitudes corresponding to the processes of the form $bb\rightarrow f\bar{f}$ (i.e. sum wrt all the possible ways in which the process $bb\rightarrow f\bar{f}$ can occur), integrated over the phase space corresponding to the momenta of the gauge bosons (and also divided by a symmetry factor related to the exchange of the two abovementioned momenta).

  1. My first question is related in my observation that in the optical theorem [formula (7.49)], there is a complex conjugate of the amplitude, corresponding to an interaction going from the final state to an intermediate state. On the other hand, when the authors apply the optical theorem, they have an amplitude (not it's complex conjugate) that corresponds to an intermediate state going to the final state. In other words, according to Eq. (7.49), we should have $$\mathcal{M}^*(f\bar{f}\rightarrow bb)$$ multiplied with $\mathcal{M}(f\bar{f}\rightarrow bb)$ and integrated over the momenta $k_1$ and $k_2$ (gauge boson momenta) in Eq.(16.37), but instead we have $$\mathcal{M}(bb\rightarrow f\bar{f})$$ Is there a relation between the two? How do I see this mathematically?

  2. My second question has to do with the $i\mathcal{M}_{\text{ghost}}$ (representing the $g\bar{g}\rightarrow f\bar{f}$ procedure) amplitude in Eq. (16.42). The latter is given by $$i\mathcal{M}_{\text{ghost}}= ig\bar{\upsilon}(p_+)\gamma_{\mu}t^cu(p) \frac{(-i)}{(k_1+k_2)^2}(gf^{abc}k_1^{\mu})$$ whereas the corresponding $i\mathcal{M}_{\text{ghost}}'$ (where the prime denotes the amplitude representing the procedure $f\bar{f}\rightarrow g\bar{g}$) is given by $$i\mathcal{M}_{\text{ghost}}'= ig\bar{u}(p)\gamma_{\mu}t^c\upsilon(p_+) \frac{(-i)}{(k_1+k_2)^2}(-gf^{abc}k_2^{\mu})$$ The authors claim that the relative minus sign is somehow related to the ghost loop, but I do not see that that way. I think that when we have a vertex with a gauge boson, a ghost and an anti-ghost and when the latter two "particles" are outgoing, then we have a factor $(gf^{abc}k^{\mu})$, where $k$ is the momentum of the anti-ghost. If on the other hand, the two latter particles are ingoing, then there is a minus sign in the Feynman rule corresponding to the vertex at hand and this is my explanation for the minus sign. Are the two explanations related? Or is mine wrong?

Any help will be appreciated.

P.S.#1: $f$ stands for a fermion, $\bar{f}$ stands for an anti-fermion and $b$ stands for a gauge boson, whereas $g$ stands for a ghost and $\bar{g}$ stands for an anti-ghost.

P.S.#2: I have seen the possible related post Form of the optical theorem in non-Abelian theory, but this is focused on other aspects of the calculation...

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  1. The two amplitudes are equivalent, since the complex conjugate of the incoming fermion-antifermion spinors ($u, \bar{v}$) are equivalent to the outgoing ones ($\bar{u}, v$) with $\gamma^{0}$ matrices ordered appropriately. Therefore:

\begin{equation} -i\mathcal{M}^{*} (f \bar{f} \rightarrow bb) = i\mathcal{M} (bb \rightarrow f\bar{f}) \end{equation}

  1. What you are saying is true, but this isn't what the text is referring to. The relative minus sign between $\mathcal{M}_{ghost}$ and $\mathcal{M}_{ghost}'$ is indeed due to the gauge group indices granting a minus sign when the ghosts are outgoing instead of incoming. But the minus sign they imply is the relative sign between the ghost amplitudes and those of the gauge bosons. The expression (16.42) of P&S has the same sign as the term they produce for the gauge bosons amplitude. The negative sign which you have already accounted for in your question is what comes from the ghost loop due to anti-commutation.
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  • $\begingroup$ 1) Your equation $\mathcal{M}^*(f\bar{f}\rightarrow bb)=\mathcal{M}(bb\rightarrow f\bar{f})$ is actually wrong. I set as a counter example the amplitude corresponding to the diagram that includes the three gauge boson vertex (labelled as $i\mathcal{M}_3^{\mu\nu}\epsilon^*_{1\mu}\epsilon^*_{2\nu}$ in P&S). But I assume you meant $[i\mathcal{M}(f\bar{f}\rightarrow bb)]^*=[i\mathcal{M}(bb\rightarrow f\bar{f})]$, which is true, based on your logic, for amplitudes that are purely imaginary. I wanted to ask whether or not this is always true (or even better, under what circumstances does it hold)? $\endgroup$
    – schris38
    Commented Jul 15, 2022 at 14:02
  • $\begingroup$ 2) So, if I understand correctly, you seem to suggest that $i\mathcal{M}_{\text{ghost}}$ and $i\mathcal{M}_{\text{ghost}}'$ differ by a sign because of the fact that the ghosts are incoming rather than outgoing in $i\mathcal{M}_{\text{ghost}}'$ (as opposed to $i\mathcal{M}_{\text{ghost}}$), correct? But you also state that this is not what the authors mean when they talk about the minus sign in p.517, but instead, they are talking about a minus sign of the amplitude that corresponds to a $f\bar{f}\rightarrow f\bar{f}$ diagram with a ghost loop. Is that correct? $\endgroup$
    – schris38
    Commented Jul 15, 2022 at 14:09
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    $\begingroup$ 1) You are correct. The amplitudes I wrote implied both the contraction with the polarization vectors and the $i$ in the front. But I will edit the text to make the latter more explicit. It is not always true, as you said it is necessary for the amplitudes to be purely imaginary. There's no general rule in regards to when it holds, one has to examine the particular interaction at hand. Usually vertices are either purely real or purely imaginary which greatly simplifies this, since in most cases the equivalency is upheld with at most a corrective change of sign for the second amplitude. $\endgroup$
    – rhomaios
    Commented Jul 15, 2022 at 14:40
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    $\begingroup$ 2) Correct for both of your questions in the second comment. The reason why they talk about full diagrams while they compute half diagrams is that the optical theorem is applied to the full diagram via Cutkosky's rules, and any minus sign that occurs from loops of anti-commuting particles is still going to be accounted for. They compute piece by piece, but it is still implicit that they are calculating the full diagram where the minus sign would have to exist anyway. $\endgroup$
    – rhomaios
    Commented Jul 15, 2022 at 14:43
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    $\begingroup$ okay thank you so much! $\endgroup$
    – schris38
    Commented Jul 15, 2022 at 14:58

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