Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2020 / 2

Use the path integral with source termThe sources $\eta ,\bar{\eta }$ are Grassmann variables; all derivatives below are left functional derivatives. Because $\eta$ stands to the right of $\bar{\psi }$, differentiating with respect to it introduces a minus sign. The insertion rules are thereforeFor example ${\delta }_L(\bar{\psi }\eta )/\delta \eta =-\bar{\psi }$. The order of the odd functional derivatives matters: the rightmost derivative acts first, and the displayed product inserts ${\bar{\psi }}_{\alpha }{\psi }_{\beta }$ in that order.

Expand $\exp [-ie\int \bar{\psi }{\gamma }^{\mu }\psi A_{\mu }]$ in its power series, replace each field by its source derivative, and pull the derivatives outside the free path integral. Re-exponentiating givesThis identity concerns the unnormalized generating functional. Normalized correlation functions divide by its zero-source value so that vacuum contributions cancel.

First fix a convention that is implicit in the source. For the stated free generating functionalthe time-ordered Wick contractions are $S_F=⟨T\psi \bar{\psi }{⟩}_0=-iS$ and ${\Delta }_F=⟨TAA{⟩}_0=-i\Delta$. These follow directly by the source differentiation rules in part (i). Thus the kernels called propagators in the question are distinguished here from the contractions, which already include their factors of $i$.

At first order in the QED interaction, write $a$ for its integration point. The normalized three-point correlation function isThe zeroth-order term vanishes because the free photon one-point correlation function is zero. There are two nonzero Wick contractions of the four fields:The first term pairs the two external fermions directly and closes the two fields at the interaction into a fermion loop; its minus sign is the closed fermion loop sign. The second term joins the external fermions through the interaction, giving the connected contribution. This is also obtained by differentiating the quartic term $(i\bar{\eta }S\eta {)}^2/2$ in the free generating functional; the two choices of bilinear cancel its factor $1/2$.

Insert $S_F=-iS$ and ${\Delta }_F=-i\Delta$. Since $(-ie)(-i{)}^3=e$, the result is exactlyThe trace term is a disconnected product of a photon tadpole diagram with the free external fermion propagator; the other term is the connected three-point correlation function. In this QED calculation the disconnected term vanishes, as established next. With a convention using $S_F,{\Delta }_F$ themselves as the symbols for the propagators, the equivalent formula retains the prefactor $-ie$ and the connected-minus-trace ordering above; mixing these two conventions would give an erroneous overall phase.

The massive Dirac propagator isUse the gamma matrix trace identities $\mathrm{tr}{\gamma }^{\mu }=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\rho })=4g^{\mu \rho }$. They giveThe integrand is odd under $k\mapsto -k$, and dimensional regularization preserves that symmetry. The same vanishing holds for $S=i{S}_F$. One must not assert that $S(0)$ itself is zero: its term proportional to $m1$ generally is nonzero and divergent, but its trace against a single gamma matrix is zero. This one-photon fermion loop is also the simplest instance of Furry's theorem.

The two tree-level Feynman diagrams are the $s$-channel and $u$-channel quark exchange graphs, drawn in the middle row of the figure. Along the directed quark line the incoming photon is absorbed before the outgoing photon is emitted in the first graph; those two insertions are reversed in the second. Their internal four-momenta are $q_s=p_1+p_2$ and $q_u=p_1-p_4$, respectively. There is no three-photon Feynman vertex in ordinary QED, and no two-photon-two-quark contact interaction in the Dirac field Lagrangian density.

Put $Q_u=2/3$, $s=q_s^2$, $u=q_u^2$, and suppress the common colour factor ${\delta }_{ji}$. Define $i\mathcal{M}$ by the product of Feynman rules. With massless external Dirac spinors the two graphs giveThe massless Dirac equation gives $/p_{1}u(p_1)=0$ and $\bar{u}(p_3)/p_3=0$. ThusBy four-momentum conservation, $p_2=p_3-q_u$, soThe two terms cancel:This is the Ward identity for the incoming photon; replacing its photon polarization vector by its four-momentum produces zero. The outgoing photon obeys the analogous identity. Each exchange graph separately fails this test, so the cancellation requires their sum. Consequently the physical scattering amplitude is invariant under ${\epsilon }^{\mu }(p_2)\mapsto {\epsilon }^{\mu }(p_2)+c{p}_2^{\mu }$.

For massless quark Compton scattering, the color average contributes one, since $(1/N_c){\sum }_{i,j}|{\delta }_{ji}{|}^2=1$. The spin average and incoming photon polarization average supply $1/(2\cdot 2)=1/4$. The Ward identities permit use of $-g_{\mu \nu }$ for each photon polarization completeness relation, and the massless fermion spin sum is ${\sum }_r{u}_r(p){\bar{u}}_r(p)=/p$.

LetReversing the gamma matrices order gives its Dirac-conjugate matrix ${\bar{A}}^{\mu \nu }$. The squared scattering amplitude isFor clarity evaluate the four traces before dividing by their denominators. The Mandelstam variables obey $s+t+u=0$, with $t=(p_1-p_3{)}^2$, $p_1\cdot p_2=s/2$, $p_1\cdot p_4=u(-1/2)$ and $p_1\cdot p_3=-t/2$. For the squared $s$-channel contribution, the gamma matrix trace identities implyHere $/q_s/p_1/q_s=2(q_s\cdot p_1)/q_s-q_s^2/p_1=s/p_2$ and $p_3\cdot p_2=-u/2$. For the squared $u$-channel contribution, $/q_u/p_1/q_u=-u/p_4$ givesOne interference trace reduces tosince $q_s\cdot q_u=(s+t+u)/2=0$. The other interference trace is its complex conjugate and also vanishes. ThereforeEquivalently $C=512{\pi }^2{\alpha }_{\mathrm{em}}^2/81$ in terms of the fine-structure constant. In the physical noncollinear region $s>0$, $u<0$, so the result is positive. The singular massless endpoint $u=0$ reflects a collinear divergence; it is not a negative squared scattering amplitude.