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

Goldstone's theorem states that in a relativistic quantum field theory, spontaneous breaking of an exact continuous global internal symmetry produces a massless Goldstone boson for each independent broken generator. More precisely, a conserved global current whose charge acts nontrivially on the vacuum expectation value has a massless pole in an appropriate current-field correlation function. The usual counting assumes a Lorentz invariant vacuum state and the standard locality and positivity hypotheses.

A local gauge symmetry is a gauge redundancy, so it does not by itself imply massless physical Goldstone bosons. In the Higgs mechanism, the corresponding scalar directions provide the longitudinal polarizations of massive gauge bosons.

For fundamental-doublet breaking of SU(2), assume $\lambda >0$ and $v>0$. The minima of the scalar potential satisfy ${\varphi }^{†}\varphi =v^2/2$, giving a three-sphere vacuum manifold. A chosen vacuum expectation value is ${\varphi }_0=(0,v{)}^T/\sqrt{2}$. No nonzero generator $a_a{\tau }^a$ annihilates ${\varphi }_0$: its squared norm is $v^2(a_1^2+a_2^2+a_3^2)/8$. At the group level, an $SU(2)$ matrix fixing the nonzero vector ${\varphi }_0$ must be the identity. Thus all three generators of the SU(2) gauge theory are broken in the usual perturbative Higgs mechanism description.

Make the radial and angular fields explicit:A local gauge transformation removes the three angular fields ${\pi }^a$. In unitary gauge, the Pauli matrices identity $\{{\tau }^a,{\tau }^b\}={\delta }^{ab}1/2$ givesThe mixed derivative term cancels between a term and its complex conjugate. The scalar potential becomesConsequently the complete physical Lagrangian density isReading the quadratic terms with their canonical factors $1/2$ givesAll physical scalar-vector and scalar self-interactions are displayed in the second line of $\mathcal{L}$. In particular the new cubic couplings proportional to $v$ are $h{A}^a{A}^a$ and $h^3$. Their Feynman vertices, together with the quartic ones, areThe three-vector and four-vector interactions are already contained in the non-Abelian gauge field strength term. There are no further physical fields or interactions in this specified model. Before fixing unitary gauge, the quadratic angular term is $\frac{1}{2}({\partial }_{\mu }{\pi }^a-m_A{A}_{\mu }^a{)}^2$. It exhibits three zero-curvature angular directions of the scalar potential, but also their mixing with the gauge fields. They supply the third polarization of each massive gauge boson, rather than massless physical particles.

The degrees of freedom count is $3\cdot 2+4=10$ before breaking and $3\cdot 3+1=10$ afterwards. If instead the symmetry were global, as in the decoupled $g=0$ scalar theory, the three angular fields would be physical massless Goldstone bosons. This is consistent with Goldstone's theorem. The language of a broken local symmetry refers to this gauge-fixed expansion; a gauge-dependent vacuum expectation value is not itself a gauge-invariant observable.

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.