Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 305 / 1

The first operator is a Lorentz scalar formed from commuting translations. Since $[P_{\mu },P_{\nu }]=0$,and the vector transformation law of $P^{\mu }$ gives

The Pauli-Lubanski pseudovector commutes with every $P_{\mu }$, so $[W^2,P_{\mu }]=0$. Its Lorentz-vector commutator similarly givesbecause the two tensor-rotation terms cancel after relabelling the contracted index. Thuscommute with every generator of the Poincare algebra. They are its two algebraically independent Casimir operators in four spacetime dimensions.

For $P^2=m^2>0$, choose the future-pointing rest momentum $p^{\mu }=(m,0,0,0)$. Its little group is the rotation group $SO(3)$, or $SU(2)$ on the double cover. Its finite-dimensional unitary irreducible representations are labelled byand have dimension $2j+1$. These are the spin states of a massive particle.

For $P^2=0$, choose $p^{\mu }=(E,0,0,E)$. Its little group is the two-dimensional Euclidean groupNontrivial unitary action of the translation subgroup produces infinite-dimensional continuous-spin representations. For the finite-helicity particles observed in ordinary relativistic field theory, that subgroup acts trivially, leaving a one-dimensional representation of rotations about $\mathbf{p}$. Such representations are labelled by helicity.

The helicity of a massless state is the component of angular momentum along its momentum,When the translation part of the massless little group is trivial, the Pauli-Lubanski pseudovector obeysA rotation through angle $\theta$ about $\mathbf{P}$ acts by the phase $e^{-i\lambda \theta }$. On the double cover, a $4\pi$ rotation is the identity, soHence $\lambda =0,\pm \frac{1}{2},\pm 1,\pm \frac{3}{2},\dots$. A parity-invariant theory pairs the $\lambda$ and $-\lambda$ representations, although one chiral massless representation need not contain both.

For a null momentum $q^{\mu }$, impose transversalityThis removes one of four vector components. The remaining equivalenceremoves a second component, leaving two transverse polarizations. For momentum along the third axis they may be chosen aswhich carry helicity $\pm 1$.

The equivalence is a gauge redundancy: vectors that differ by a multiple of $q^{\mu }$ describe the same physical state rather than distinct measurable configurations. A Lorentz transformation of a chosen transverse representative can require a compensating gauge transformation. Therefore the amplitudemust be invariant under ${ϵ}_{\mu }\mapsto {ϵ}_{\mu }+\alpha q_{\mu }$. For arbitrary $\alpha$, this is exactly the Ward identity

For emission of a soft photon of momentum $q$ from external charged particles, the Soft photon theorem giveswhere ${\eta }_i=+1$ for outgoing and $-1$ for incoming particles. Contracting with $q_{\mu }$ and applying the Ward identity givesFor a nonzero underlying amplitude, total outgoing charge equals total incoming charge. Thus photon gauge redundancy enforces conservation of electric charge.

The analogous Soft graviton theorem for helicity two requires universal coupling to energy-momentum and yields conservation of total four-momentum, the seed of the Equivalence principle. For massless helicity greater than two, the corresponding soft consistency conditions demand conserved higher-rank momentum charges that generic interacting scattering cannot satisfy. Under the assumptions of a Lorentz-invariant local S-matrix with long-range interactions, such particles therefore have no nontrivial coupling; this is the soft-theorem obstruction to interacting massless higher-spin particles.