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Advanced quantum field theory for particle physics course of the University of Oxford

Past exam of the advanced quantum field theory for particle physics course of the University of Oxford

2018

The <tree-level Feynman diagrams> are the two ordered quark-exchange graphs and the $s$-channel gluon-exchange graph with a three-gluon vertex. There is no two-gluon-two-quark contact term in the ordinary <QCD> <Lagrangian density>. Let $q=k_1+k_2=p_1+p_2$, $r_t=p_1-k_1$, $r_u=p_1-k_2$, and $s=q^2$, $t=r_t^2$, $u=r_u^2$. Define $i\mathcal M$ as the product of the stated <Feynman rules>.

\Image[/past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2018-gluon-fusion-and-yukawa-loop.png]
{title=The three gluon-fusion tree diagrams and the Yukawa fermion-loop correction}
{description=The first three panels show ordered quark exchange and three-gluon exchange. The last panel shows a directed closed fermion loop between two scalar insertions. Crossings without dots are not vertices.}

Suppressing external colour indices, the <quark> graphs are
$$
\mathcal M_t^{\mu\nu}=-g^2\bar u(p_1)T^aT^b\gamma^\mu\frac{\not r_t}{t}\gamma^\nu v(p_2),
\qquad
\mathcal M_u^{\mu\nu}=-g^2\bar u(p_1)T^bT^a\gamma^\nu\frac{\not r_u}{u}\gamma^\mu v(p_2).
$$
The massless <Dirac equations> give $\bar u\not p_1=0$ and $\not p_2v=0$. Consequently
$$
\frac{\bar u\not k_1\not r_t\not k_2v}{t}=-\bar u\not k_2v=\bar u\not k_1v,
\qquad
\frac{\bar u\not k_2\not r_u\not k_1v}{u}=-\bar u\not k_1v.
$$
The last equality uses $\bar u\not qv=0$. The quark-exchange sum is therefore $-g^2\bar u\not k_1[T^a,T^b]v$.

For the <gluon> graph, use all incoming <momenta> $(k_1,k_2,-q)$ in the three-gluon tensor:
$$
V^{\mu\nu\rho}=(k_1-k_2)^\rho g^{\mu\nu}
+(k_2+q)^\mu g^{\nu\rho}-(q+k_1)^\nu g^{\rho\mu}.
$$
Its double contraction is $k_{1\mu}k_{2\nu}V^{\mu\nu\rho}=s(k_2-k_1)^\rho/2$. The vertex and <gluon> <propagator> give
$$
\mathcal M_s^{\mu\nu}=-\frac{ig^2f^{abc}}s\,
\bar u\gamma_\rho T^cv\,V^{\mu\nu\rho},
\qquad
k_{1\mu}k_{2\nu}\mathcal M_s^{\mu\nu}
=ig^2f^{abc}\bar u\not k_1T^cv.
$$
Combining the graphs yields
$$
\boxed{k_1^\mu k_2^\nu\mathcal M_{\mu\nu}
=-g^2\bar u(p_1)\not k_1
\big([T^a,T^b]-if^{abc}T^c\big)v(p_2).}
$$
Thus the requested <spinor> proportionality factor is $-g^2\bar u(p_1)\not k_1v(p_2)$ when the colour matrix is displayed separately; retaining it inside the <spinor> bilinear makes its external colour indices explicit. For a convention calling the Feynman-rule product itself $\mathcal M$, rather than $i\mathcal M$, the entire result gains one factor of $i$.

Solution

Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2018 / 1 / b / i / Solution

Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2018 / 1 / b / i / Solution

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