# Amplitude of quark$+$antiquark $\rightarrow$ ghost$+$antighost in QCD

Since the BRST charge operator commutes with the Hamiltonian of QCD, a physical state such as $$q+\bar q$$ should not be allowed to evolve into an unphysical one like $$\chi+\bar\chi$$, where these two are the Faddeev-Popov ghost and antighost introduced for the gauge-fixing of the theory. In other words, the amplitude of the process $$q\bar q \rightarrow \chi\bar\chi$$ should vanish. Yet, nothing seems to prevent me from drawing the (unique at tree level) Feynman graph associated with it and find the following amplitude (in Feynman gauge), which at first sight seems not to be zero: $$\left<\chi\bar\chi_{\mathrm{out}}\right|\left.q\bar q_{\mathrm{in}}\right>=-g^2(t^a_f)_{ij}f^{acd}\frac{k^\mu\bar v(q)\gamma_\mu u(p)}{(p+q)^2+i0^+} + O(g^2)$$ Then, my first idea would be to try proving that $$(t^a_f)_{ij}f^{acd}=0$$ (here the $$t^a_f$$ generate the fundamental representation of the Lie algebra, in which the quarks live, and the $$f^{acd}$$ are the structure constants of this algebra) but I am not sure that it will work. Is there something that I am missing ?

I was given a nice answer to this question that may be worth writing. When we act on the incoming physical state consisting of a quark-antiquark pair $$\left|q\bar q_{\mathrm{in}}\right>$$ with the evolution operator $$S$$, the Feynman rules allow for different outgoing states at lowest order : physical ones, $$\left|q\bar q_{\mathrm{out}}\right>$$ (quark-antiquark pair) and $$\left|ij_{\mathrm{out}}\right>$$ (gluons with physical polarisations); and unphysical ones, $$\left|\chi\bar\chi_{\mathrm{out}}\right>$$ (ghost-antighost pair) and $$\left|+-_{\mathrm{out}}\right>$$ (gluons with unphysical polarization). We can summarize it as follows: $$S\left|q\bar q_{\mathrm{in}}\right> = \left|q\bar q_{\mathrm{out}}\right> \oplus\left|ij_{\mathrm{out}}\right> \oplus\left|\chi\bar\chi_{\mathrm{out}}\right> \oplus\left|+-_{\mathrm{out}}\right>$$ where the $$\oplus$$ mean that there are some coefficients in front of the terms. After doing the calculation, it turns out that the unphysical contribution can be written as the BRST charge $$q_{\mathrm{BRST}}$$ acting on some other state: $$\left|\chi\bar\chi_{\mathrm{out}}\right> \oplus\left|+-_{\mathrm{out}}\right> = -\frac{1}{\sqrt{2|\mathbf{p}|}}q_{\mathrm{BRST}}\left|\bar\chi+_\mathrm{out}\right>$$ which means that in the end, $$S\left|q\bar q_{\mathrm{in}}\right>$$ is hopefully a physical state (that is, in the kernel of the BRST charge) thanks to the fact that $$q_{BRST}^2=0$$.