Past exam of the theoretical physics major option course of the University of Oxford / 2016 / 6

The continuous spacetime symmetries are translations, spatial rotations and Lorentz boosts, together the connected Poincaré group. The theory also has local electromagnetic U(1) gauge symmetry, which is an internal symmetry rather than a spacetime transformation. A nonzero mass and a dimensionful interaction do not give an additional exact scale invariance.

In four dimensions the action is dimensionless and $[\partial ]=1$. The two kinetic terms giveThe axion-photon operator has mass dimension five, so this is an effective field theory. Its dimensionless interaction strength grows with energy as $|g_a|E$. Without information about an ultraviolet completion, a conservative parametric validity condition is $E\ll \Lambda$ with $\Lambda \sim |g_a{|}^{-1}$, and $m_a$ must also lie below that scale for the axion to be a degree of freedom in the effective theory. Numerical factors such as $4\pi$ depend on the ultraviolet dynamics and on the chosen perturbative criterion; the Lagrangian alone does not fix them.

The axion variation gives $\partial \mathcal{L}/\partial a=-m_a^{2}a+g_a{F}_{\mu \nu }{\tilde{F}}^{\mu \nu }/4$, soFor the photon variation, the dual electromagnetic field tensor obeys $\delta (F_{\rho \sigma }{\tilde{F}}^{\rho \sigma })=2{\tilde{F}}^{\rho \sigma }\delta F_{\rho \sigma }$. Antisymmetry then givesThere is no undifferentiated $A_{\nu }$, henceThe last equality uses the electromagnetic Bianchi identity ${\partial }_{\mu }{\tilde{F}}^{\mu \nu }=0$. These signs follow from the positive $aF\tilde{F}$ interaction in the source; reversing that interaction reverses both source terms. The induced electromagnetic current is identically conserved: its divergence contracts a symmetric second derivative of $a$ with an antisymmetric tensor, while the remaining term vanishes by the same Bianchi identity.

Evaluate the two-body Lorentz-invariant phase space in the parent's rest frame, $p=(m,0)$. The spatial Dirac delta function sets ${\mathbf{q}}_2=-{\mathbf{q}}_1$. For massless daughters $E_1=E_2=q$, soThe factor $1/2$ in the last step is the derivative of the delta-function argument. Substituting into the decay width gives, for labelled or distinct daughters,The angular domain is one sphere: $0\le \phi <2\pi$ and $-1\le \cos \theta \le 1$, equivalently $0\le \theta \le \pi$. The printed negative-to-positive polar-angle interval is not a second independent coverage of the sphere.

For identical daughters, integration over the full labelled two-body phase space counts each physical final state twice. One must therefore include an additional factor $1/2!$, or equivalently integrate over an unordered half-domain. This distinction is necessary for the two identical photons in solution.

With outgoing physical photon polarization vectors, the Feynman rule givesThe overall phase does not affect the decay width. Replacing either polarization vector by its own four-momentum gives zero by antisymmetry. Thus the Ward identity permits replacement of each physical photon polarization sum by $-{\eta }^{\mu \nu }$ inside this amplitude. The printed sum is not an equality of free-standing physical two-polarization tensors: for a photon moving along $z$, its physical $00$ component is zero whereas $-{\eta }_{00}=-1$. Gauge-dependent longitudinal terms cancel only after contraction with the transverse amplitude.

The contraction of the two Levi-Civita symbols givessince $q_1^2=q_2^2=0$ and $2q_1\cdot q_2=m_a^2$. No initial spin average is needed for the spin-zero axion. The result is independent of direction in its rest frame. Including the identical-particle factor explained in solution therefore givesOmitting $1/2!$ would overestimate the physical rate by two. The normalization agrees with the standard axion-photon convention.^[ref]