Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2020 / 3 / b

A gauge-invariant path integral counts all representatives of each gauge orbit. In a perturbative region with a regular gauge fixing slice, insert the Faddeev-Popov gauge-orbit identityThe Faddeev-Popov determinant is the functional Jacobian converting integration over the gauge transformation parameter into integration over the condition. The redundant gauge group volume then factors out of normalized correlation functions.

Choose $A_{\mu }^{\omega }=A_{\mu }+{\partial }_{\mu }\omega$. Its infinitesimal variation givesThere is no derivative of $A$ in this expression: only $A_{\mu }$ is varied in $A^2$. For $\lambda =0$, the determinant is field independent and can be absorbed into normalization. For $\lambda \ne 0$, it is a functional of the gauge field, so dropping it changes the path integral, even though the gauge group is Abelian. Constant residual gauge transformations are removed by boundary conditions or by dividing out their volume, as usual in perturbative gauge fixing.

An ordinary bosonic Gaussian functional integral produces an inverse determinant. An integral over independent Grassmann fields $c,\bar{c}$ instead produces a determinant, allowing a local action to represent the Faddeev-Popov determinant. With one convenient antighost sign convention,Multiplying the operator by the field-independent factor $-i$ changes only the normalization of its determinant. After integration by parts, an equivalent density isThus the Faddeev-Popov ghost fields have a massless kinetic term and a photon-ghost-antighost interaction proportional to $\lambda$. They are auxiliary anticommuting scalar fields, not physical external particles. Closed ghost loops carry the minus sign of Grassmann variables. Replacing $\bar{c}$ by $-\bar{c}$ reverses both displayed signs without changing any observable; both the kinetic and interaction conventions must be changed together.

For fixed $A$, the delta functional sets $\sigma (x)=F[A](x)$. Therefore its weighted functional integral isSince the action enters as $e^{iS}$, the gauge fixing contribution isThe Faddeev-Popov determinant does not depend on $\sigma$ and remains in the path integral. The limit $\xi \to 0$ enforces the original condition sharply; for nonzero $\xi$ this is a weighted average of the gauge slices.

Expand the gauge fixing square:The second term is a derivative cubic photon interaction; the third is a quartic photon contact interaction. Both arise from nonlinear Abelian gauge fixing, rather than an additional physical electromagnetic force.

For $\xi =1$, differentiating the quartic term with respect to four identical gauge fields gives eight copies for each pairing, henceWith the Fourier convention $A(x)=\int d^{4}p(2\pi {)}^{-4}{e}^{ipx}A(p)$ and all four-momenta incoming, the differentiated cubic term givesmatching the supplied Feynman rules. Reversing the Fourier convention reverses this cubic rule's sign; each exchange graph contains two such Feynman vertices, so its contribution is unchanged.

There are four tree-level Feynman diagrams: three photon exchange graphs with pairings $12|34$, $13|24$, $14|23$, and the four-photon contact interaction. They are drawn in the bottom row of the figure. With no external fermions or Faddeev-Popov ghost fields, an internal matter or ghost line would have to close into a loop; it cannot add a tree-level Feynman diagram here.

Use all-incoming null four-momenta $k_i$ with ${\sum }_i{k}_i=0$. Let ${\epsilon }_i$ be the external photon polarization vectors, including the complex conjugation for physical outgoing photons, and impose $k_i\cdot {\epsilon }_i=0$. For one pairing $ij|kl$, put $K=k_i+k_j=-(k_k+k_l)$. At the first cubic Feynman vertex, the terms proportional to $k_i\cdot {\epsilon }_i$ and $k_j\cdot {\epsilon }_j$ vanish, leavingAt the other end, the incoming internal four-momentum is $+K$, giving $+2\lambda ({\epsilon }_k\cdot {\epsilon }_l)K^{\sigma }$. The Feynman gauge photon propagator then givesThe last expression is evaluated at generic kinematics, with singular channel limits taken after cancellation. Summing the three pairings produces $+4i{\lambda }^{2}P$, whereThe quartic Feynman vertex contributes $i{\mathcal{M}}_{\mathrm{contact}}=-4i{\lambda }^{2}P$. ThusThis is the tree photon-scattering cancellation in a quadratic Abelian gauge: the individual gauge fixing interactions cancel in the physical scattering amplitude. The argument uses transverse external photon polarization vectors; it does not assert that an arbitrary off shell four-point correlation function vanishes. Physical light-by-light scattering first occurs through charged-particle loop Feynman diagrams.