Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 50 / 3

Use the same Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ as in Question 1. For the conjugate charged field, the kinetic term means $(D_{\mu }\phi {)}^{\ast }{D}^{\mu }\phi$, with $(D_{\mu }\phi {)}^{\ast }=({\partial }_{\mu }-ie{A}_{\mu }){\phi }^{\ast }$. Thus the conjugate field has the opposite charge in the gauge covariant derivative.

For any real smooth function $\chi (x)$, make the local transformationThe electromagnetic field tensor is unchanged because the mixed derivatives of $\chi$ commute. Directly,Its conjugate transforms with the inverse phase, so $(D_{\mu }\phi {)}^{\ast }{D}^{\mu }\phi$ is invariant. The mass term depends only on $|\phi {|}^2$ and is invariant as well. All terms of the action are therefore unchanged. This proves the local U(1) gauge symmetry of scalar electrodynamics.

A gauge redundancy relates different field representatives of the same physical configuration. For example, the transformation just exhibited changes $A_{\mu }$ and the scalar phase but leaves $F_{\mu \nu }$, $|\phi {|}^2$ and all other gauge-invariant operators unchanged. Fixing a gauge removes this freedom; it does not remove a physical symmetry-related state. In canonical quantization, local transformations trivial at the boundary are generated by the Gauss law constraint in gauge theory, and physical states satisfy that constraint.

A global symmetry in field theory instead has a spacetime-independent parameter and can act nontrivially on physical states. For a continuous global symmetry in field theory, Noether theorem gives a conserved charge that generates the transformation and labels the states; discrete global symmetries do not acquire a Noether current from the same infinitesimal theorem. These are statements about physical transformations, rather than merely whether a parameter is written as a constant or a function.

Boundary conditions matter in gauge theory. gauge transformations that vanish at spatial infinity are redundancies, while transformations nontrivial at infinity may act through a physical electric charge. In particular a constant phase transformation can survive as a global charge symmetry rather than being quotiented out with the local gauge redundancy. Thus local gauge redundancy is a redundancy of description; genuine global symmetry in field theory can distinguish physical states and charges, with the boundary qualification made explicit.

Expand the gauge covariant derivative in the scalar kinetic term:The cubic interaction is thereforeFor an incoming scalar momentum $p$ and outgoing momentum $q$, derivatives give $-ip$ on the incoming field and $+iq$ on its outgoing conjugate. Multiplication by the factor $i$ from the interaction action gives the scalar electrodynamics three-point vertex $-ie(p+q{)}_{\mu }$.

The other vertex comes fromDifferentiating twice with respect to the identical photon fields gives a factor of two, so the seagull vertex is $+2i{e}^2{\eta }_{\mu \nu }$. There is no additional factor of two from the scalar fields, since the complex scalar field and its conjugate are distinct legs.

At order $e^2$, scalar particle–antiparticle scattering has a $t$-channel photon exchange and an $s$-channel annihilation and recreation diagram. They are shown below, with incoming momenta $p,q$ and outgoing momenta $p^{\prime },q^{\prime }$. The seagull vertex has only two scalar legs and does not yield a four-scalar tree diagram at this order.

For the exchange graph, let $k=p-p^{\prime }=q^{\prime }-q$. Apart from overall charge signs, the two scalar currents are $J=p+p^{\prime }$ and $K=q+q^{\prime }$. The external mass shell conditions giveFor the annihilation graph, $k=p+q=p^{\prime }+q^{\prime }$, with currents $J=p-q$ and $K=p^{\prime }-q^{\prime }$, again up to vertex signs. Their contractions are $p^2-q^2=0$ and $p^{\prime 2}-q^{\prime 2}=0$. Thus each graph couples the photon to conserved currents, the relevant scalar quantum electrodynamics Ward identity.

Write $k=(\omega ,\mathbf{k})$ and ${\kappa }^2=|\mathbf{k}{|}^2$. conserved current means $\mathbf{k}\cdot \mathbf{J}=\omega J^0$ and $\mathbf{k}\cdot \mathbf{K}=\omega K^0$. For $\kappa \ne 0$, contraction with the Coulomb-gauge photon propagator gives, displaying the rational identity before the pole boundary limit,We used $k^2={\omega }^2-{\kappa }^2$. Restoring the common Feynman i-epsilon prescription yields the Coulomb-gauge propagator between conserved currents resultHence each on-shell tree amplitude can use the Lorentz-invariant propagator. This is equality after contraction, not equality of the gauge-dependent tensors. At $\mathbf{k}=0$, the separate Coulomb-gauge terms are ill-defined; combine them first and take the conserved-current limit. The combined expression has only the physical photon pole.