Getting the wrong sign in optical theorem

Question

In Peskin&Schroeder chapter 16.3 we are showing how ghosts ensure unitarity in non-Abelian theories. My problem comes when I try to calculate a specific amplitude. In equation 16.37 we have the term $$ \frac{1}{2}(i\mathcal{M}^{\mu\nu})g_{\mu\rho}g_{\nu\sigma}(i\mathcal{M}'^{\rho\sigma}) $$ where the amplitude $i\mathcal{M}^{\mu\nu}$ corresponds to the annihillation of a fermion=anti-fermion pair to a pair of gauge bosons while the amplitude $i\mathcal{M}'^{\rho\sigma}$ corresponds to the creation of a fermion-anti-fermion pair from a gauge boson pair. My problem is in the calculation of the second interation.

Before going there let me explain that the above expression is within a phasespace integral and using the completeness relation for the metrics we get $$ i\mathcal{M}^{\mu\nu}\left(\epsilon^{-*}_\mu\epsilon^+_\rho+\epsilon^{+*}_\mu\epsilon^-_\rho-\sum_{i=1}^2\epsilon^T_{i\mu}\epsilon_{i\rho}^{*T}\right)\left(\epsilon^{-*}_\nu\epsilon_\sigma^{+}+\epsilon^{+*}_\nu\epsilon_\sigma^{-}-\sum_{i=1}^2\epsilon_{i\nu}^T\epsilon_{i\sigma}^{*T}\right)i\mathcal{M}'^{\rho\sigma} \tag{1} $$ where most of the terms are zero and the only contributions we have are $$ i\mathcal{M}^{\mu\nu}\epsilon^{-*}_\mu\epsilon_\nu^{+*}\epsilon^+_\rho\epsilon^-_\sigma i\mathcal{M}'^{\rho\sigma}+i\mathcal{M}^{\mu\nu}\epsilon_\mu^{+*}\epsilon_\nu^{-*}\epsilon_\rho^-\epsilon_\sigma^+ i\mathcal{M}'^{\rho\sigma}=\\ \left(i\mathcal{M}^{\mu\nu}\epsilon^{-*}_\mu(k_1)\epsilon^{+*}_\nu(k_2)\right)\left(i\mathcal{M}'^{\rho\sigma}\epsilon^+_\rho(k_1)\epsilon^-_\sigma(k_2)\right)+\left(i\mathcal{M}^{\mu\nu}\epsilon^{+*}_\mu(k_1)\epsilon^{-*}_\nu(k_2)\right)\left(i\mathcal{M}'^{\rho\sigma}\epsilon^-_\rho(k_1)\epsilon^+_\sigma(k_2)\right) $$ where we omit contributions of the transverse polarization vectors because they will vanish in our calculation of the Ward Identity.

Now I haven't found anywhere a clear calculation of the amplitude $i\mathcal{M}'^{\rho\sigma}$ so I tried to go through it on my own. The calculation is identical to $i\mathcal{M}^{\mu\nu}$ but there is an extra negative sign which I don't understand where it comes from.

The expression I got is $$ i\mathcal{M}'^{\rho\sigma}\epsilon_\sigma(k_2)k_{1\rho}=g^2\bar{u}(p')\gamma_\lambda t^cv(p_+')\frac{1}{k_3^2}f^{abc}\epsilon_\sigma(k_2)k_2^\sigma k_2^\lambda $$ Then we recall we only use the forward/backward polarisation vectors, and using $$ \epsilon_\rho^+(k_1)=\frac{k_{1\rho}}{\sqrt{2}|\vec{k_1}|}\Rightarrow k_{1\rho}=\sqrt{2}|\vec{k_1}|\epsilon_\rho^+(k_1)\\ \epsilon^-_\sigma(k_2)=\frac{1}{\sqrt{2}|\vec{k_2}|}(k_2^0,-\vec{k_2})\Rightarrow \epsilon^-_\sigma(k_2)k_2^\sigma=\sqrt{2}|\vec{k_2}| $$

we have $$ \begin{aligned}i\mathcal{M}'^{\rho\sigma}\epsilon^-_\sigma(k_2)k_{1\rho}&=g^2\bar{u}(p')\gamma_\lambda t^cv(p_+')\frac{1}{k_3^2}f^{abc}\epsilon^-_\sigma(k_2)k_2^\sigma k_2^\lambda\Rightarrow\\ i\mathcal{M}'^{\rho\sigma}\epsilon_\sigma^-(k_2)\sqrt{2}|\vec{k_1}|\epsilon_\rho^+(k_1)&=g^2\bar{u}(p')\gamma_\lambda t^cv(p_+')\frac{1}{k_3^2}f^{abc}\sqrt{2}|\vec{k_2}| k_2^\lambda\Rightarrow\\ i\mathcal{M}'^{\rho\sigma}\epsilon^+_\rho(k_1)\epsilon_\sigma^-(k_2)&=g^2\bar{u}(p')\gamma_\lambda t^c v(p'_+)\frac{1}{k_3^2}f^{abc}\frac{|\vec{k_2}|}{|\vec{k_1}|}k_2^\lambda \end{aligned} $$ This quantity should be negative, but I don't understand where we gain that extra minus sign.

Answer 1

This is what I have in my notes. I hope it is useful. One can work out the non-transverse part of (1) as \begin{align} &\frac{1}{2}\left( i g \bar{v} (p_2)\gamma_\mu t^c u (p_1)\frac{-i}{(k_1+k_2)^2}g f^{abc} k_2^\mu \right) \times \left( i g \bar{u} (p_1)\gamma_\rho t^d v (p_2)\frac{-i}{(k_1+k_2)^2}g f^{abd} \big(- k_1^\rho \big) \right) \nonumber\\ & + (k_1 \longleftrightarrow k_2 ) \end{align} It turns out that the first part is anti-symmetric under $k_1 \longleftrightarrow k_2$. To see this first use momentum conservation $k_1+k_2=p_1+p_2$ on \begin{align} \bar{v}(p_2)\gamma_\mu (k_1+k_2)^\mu u(p) & = \bar{v}(p_2)\gamma_\mu (p_1+p_2)^\mu u(p) = \bar{v}(p_2) (\not{\!p}_1+\not{\!p}_2) u(p) \nonumber\\ &= \bar{v}(p_2) (\not{\!p}_2+m +\not{p}_1-m) u(p) =0 \end{align} We thus have $\bar{v}(p_2)\gamma_\mu k_1^\mu u(p) = - \bar{v}(p_2)\gamma_\mu k_2^\mu u(p)$ and can thus interchange $k_1$ and $k_2$ in the first factor of the first line if we include a minus sign. We have the same result for the second factor, and so overall the first line is invariant under the change of $k_1$ and $k_2$. So we find the result \begin{align} - \left( i g \bar{v} (p_2)\gamma_\mu t^c u (p_1)\frac{-i}{(k_1+k_2)^2}g f^{abc} k_2^\mu \right) \left( i g \bar{u} (p_1)\gamma_\rho t^d v (p_2)\frac{-i}{(k_1+k_2)^2}g f^{abd} k_1^\rho \right) \end{align}