Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2017 / 2

Use the source term $\int d^{4}x[J\varphi +\bar{\varsigma }\psi +\bar{\psi }\varsigma ]$ and left Grassmann functional derivatives. Differentiating its exponential inserts fields according toThe sign for $\bar{\psi }$ follows because a left derivative with respect to $\varsigma$ must cross $\bar{\psi }$. Keep the order $\bar{\psi }{\gamma }_5\psi \varphi$ when replacing these fields. Expand $e^{i{S}_{\mathrm{int}}}$ in the original path integral; each field monomial can now be generated by these ordered derivatives acting on the free functional. Resumming givesThe arguments refer respectively to $\varphi ,\bar{\psi },\psi$ at $x$. The free functional arises by performing the free Gaussian field integrals, and $\mathcal{N}$ may be chosen so that $Z[0]=1$.

No additional field-integration measure remains in this differential-operator formula. Such a measure is printed outside the operator in the source even though $Z_0$ has already integrated out those fields. At $\lambda =0$, that literal expression would contain $\int \mathcal{D}\varphi \mathcal{D}\bar{\psi }\mathcal{D}\psi 1$ rather than simply $Z_0$, which is not the free generating functional. Also, when ${\mathcal{L}}_{\mathrm{int}}$ denotes a density, the spacetime integral in the exponent is required.

At first order the Dyson expansion inserts $i{S}_{\mathrm{int}}=\lambda \int d^{4}x{\bar{\psi }}_{\alpha }(x)({\gamma }_5{)}_{\alpha \beta }{\psi }_{\beta }(x)\varphi (x)$. Contract the scalar field at the vertex with the external scalar. The four remaining fermion fields have two admissible pairings: both external fermions attach to the vertex, or the two vertex fermions contract with each other while the external pair contracts freely. Wick's theorem givesThe minus sign in the second term is the fermion sign of the closed contraction: $⟨T{\bar{\psi }}_{\alpha }(x){\psi }_{\beta }(x)⟩=-i{S}_F(0{)}_{\beta \alpha }$. The first term is connected; the second is a scalar-attached fermion tadpole multiplied by a disconnected free external fermion propagator. The full time-ordered expectation includes both topologies, whereas the connected generating functional retains only the first. There is no order-${\lambda }^2$ contribution in this pure Yukawa expansion because the total number of scalar fields would be odd.

For a parity-preserving free propagator, the tadpole coefficient is $\mathrm{tr}[{\gamma }_5{S}_F(0)]$, whose integrand contains only $\mathrm{tr}{\gamma }_5$ and $\mathrm{tr}({\gamma }_5{\gamma }^{\mu })$. Both vanish. Thus the second term is a legitimate formal Wick contraction but is zero with a compatible regulator; normal ordering the interaction removes it from the outset.

The different-flavour quark annihilation into a massive pair has only an $s$-channel gluon graph at tree level when the two quark flavours differ. Its amplitude contains one incoming and one outgoing colour current, with common magnitude factor $g_s^2/s$. On-shell current conservation removes the gauge-parameter-dependent longitudinal part of the gluon propagator. To display the Dirac signs transparently, evaluate spin traces with the auxiliary mostly-minus metric $\eta =-g$ and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$; the invariants $s,t$ are then the usual positive-energy Mandelstam invariants, numerically identical to those defined in the paper.

The incoming colour average and outgoing colour sum giveAveraging over the two spins of each incident particle supplies an additional $1/4$. The physical spin sums are $\sum u\bar{u}=/p+M$ and $\sum v\bar{v}=/p-M$. ThereforeThe coefficient is equivalently $8{\pi }^2{\alpha }_s^2/(9s^2)$, with ${\alpha }_s=g_s^2/(4\pi )$.

The two top projectors cannot both contain $-M$. The printed expression has that error. It is not repaired by changing the metric convention consistently: a particle and antiparticle still require the appropriate distinct projectors. In the auxiliary convention, its outgoing tensor would obey $q_{\mu }{H}_{\mathrm{printed}}^{\mu \alpha }=8M^2{q}^{\alpha }$, rather than zero for $q=p_1+p_2$. At threshold it gives a zero contracted trace, while the correctly averaged vector-current production amplitude squared has a nonzero limit $4g_s^4/9$. The phase-space cross section still vanishes at threshold because the final velocity tends to zero. The colour and spin prefactor above is consequently the prefactor for the corrected trace; no overall constant can make the misprinted massive trace the physical answer for all kinematics.

Continue with $\eta =-g$ and dot products in that auxiliary metric. Odd gamma traces vanish. The two current tensors areContracting givesLet $u=2M^2-s-t$. The on-shell products are $p_1\cdot k_1=p_2\cdot k_2=(M^2-t)/2$, $p_1\cdot k_2=p_2\cdot k_1=(M^2-u)/2$, and $k_1\cdot k_2=s/2$. Substitution yieldsThe result is already averaged over incoming colours and spins and summed over outgoing ones. For $M\to 0$, it becomes $4g_s^4(t^2+u^2)/(9s^2)$.

An additional check uses the centre-of-mass angle and velocity ${\beta }_t=\sqrt{1-4M^2/s}$:It tends to $4g_s^4/9$ at threshold and is positive throughout the physical region. Keeping both printed minus signs would instead give $4g_s^4[(M^2-t{)}^2+(M^2-u{)}^2-2M^{2}s]/(9s^2)$, displaying the incorrect mass contribution explicitly.