Past exam of the theoretical physics major option course of the University of Oxford / 2022 / 4

The free complex Proca field has Poincaré symmetry, with $B_{\mu }$ transforming as a spacetime four-vector, and the global internal U(1) symmetryThe free theory is also invariant under charge conjugation $B_{\mu }\leftrightarrow B_{\mu }^{\ast }$ and the usual parity and time reversal transformations of a vector field. The internal real-field description consists of two equal-mass real Proca fields, with an $O(2)$ symmetry containing the phase rotations and charge conjugation.

A gradient shift of $B_{\mu }$ leaves $G_{\mu \nu }$ unchanged, because partial derivatives commute. It changes the mass term byThis is generally nonzero: for example, $B_{\mu }=0$ and real $\lambda =at$ produce $\Delta \mathcal{L}=M^2{a}^2$. Thus a massive Proca field has no such gauge symmetry. The massless kinetic term alone would have it.

Treat $B_{\mu }$ and $B_{\mu }^{\ast }$ as independent variables in the Euler-Lagrange equation. Varying $B_{\nu }^{\ast }$ gives the Proca equationThe divergence of its first term vanishes by antisymmetry and commuting partial derivatives. Since $M\ne 0$, it follows that ${\partial }_{\nu }{B}^{\nu }=0$. Substitution into the original equation givesThe complex-conjugate variation gives ${\partial }_{\mu }{G}^{\ast \mu \nu }+M^2{B}^{\ast \nu }=0$, hence $(\square +M^2)B^{\ast \nu }=0$ and ${\partial }_{\nu }{B}^{\ast \nu }=0$.

For an on-shell plane wave $B^{\mu }={\epsilon }^{\mu }{e}^{-ipx}$, the Lorenz constraint in Proca theory is $p_{\mu }{\epsilon }^{\mu }=0$. In the rest frame it sets ${\epsilon }^0=0$, leaving three complex particle polarizations. Thus the complex Proca field has three independent complex dynamical components, or six real ones; its quantized particle and antiparticle each have three physical particle polarizations. The conjugate field is not another unrelated classical field.

Use the upper-index coordinates $B^{\mu },B^{\ast \mu }$ specified in the paper. Differentiation of the Lagrangian density gives the canonical momentaIn particular, ${\pi }_i=G^{\ast 0i}$ and ${\pi }_i^{\ast }=G^{0i}$. No time derivative of $B^0$ appears, so the zero momenta are primary constraints, not invertible equations for ${\dot{B}}^0$.

For spatial indices, $G^{0i}={\dot{B}}^i+{\partial }_i{B}_0$, so ${\dot{B}}^i={\pi }_i^{\ast }+{\partial }^i{B}_0$, with the conjugate relation for $B^{\ast i}$. AlsoThe Legendre transform for the complex Proca Hamiltonian, ${\mathcal{H}}_c={\pi }_i{\dot{B}}^i+{\pi }_i^{\ast }{\dot{B}}^{\ast i}-\mathcal{L}$ therefore givesRepeated spatial indices in the positive product ${\pi }_i^{\ast }{\pi }_i$ denote an ordinary three-component sum. A total Hamiltonian additionally includes multipliers for ${\pi }_0={\pi }_0^{\ast }=0$.

The Hamiltonian printed in the source cannot follow from the given complex Lagrangian density: its missing conjugations, mass coefficients and spatial mass sign are inconsistent. A concrete check is spatially uniform initial data $B^1=b\in \mathbb{R}$, $B^0=B^2=B^3=0$, with all canonical momenta zero. The correct Hamiltonian density is $M^2{b}^2$, while the printed expression gives $-M^2{b}^2/2$. These data satisfy the initial constraints and evolve as a spatially uniform massive oscillator. Thus the discrepancy is not a harmless surface term or a different complex-field normalization.

Preserve the primary constraints using the corrected Hamiltonian from solution. After a spatial integration by parts, the $B_0$ terms in its density are $B_0{\partial }_i{\pi }_i+B_0^{\ast }{\partial }_i{\pi }_i^{\ast }-M^2{B}_0^{\ast }{B}_0$. Hamilton's equations giveWith the definitions of $E^i,E^{\ast i}$ in the paper, the actual massive Gauss law is thereforeThis is also the $\nu =0$ component of the Proca equation. The zero right-hand sides printed in the source apply only to the massless theory or to special configurations with $B_0=0$.

For a concrete counterexample, take $k\ne 0$, ${\omega }^2=k^2+M^2$, and the longitudinal on-shell plane waveIt satisfies ${\partial }_{\mu }{B}^{\mu }=0$, but $E^3=-iM{e}^{-i\omega t+ikz}$ and ${\partial }_i{E}^i=kM{e}^{-i\omega t+ikz}\ne 0$. Imposing the printed zero-divergence condition would incorrectly remove the physical longitudinal particle polarization.

The naive equal-time canonical commutation relations would bewith all coordinate-coordinate, momentum-momentum and mixed conjugate-sector commutators zero before enforcing constraints. But the $\mu =\nu =0$ relation contradicts ${\pi }_0=0$ as an operator identity: its left side would vanish and its right side would not.

The primary zero-momentum constraints and the secondary massive Gauss laws form second-class constraints; their Poisson bracket matrix contains nonzero terms proportional to $M^2$. They must be eliminated, or treated using a Dirac bracket, before quantization. In a reduced description the three complex spatial coordinates and their momenta are canonical, whileare dependent operators. Substituting them in the integrated Hamiltonian gives the positive density, up to a surface term,Thus the problem is a constrained choice of canonical variables, not an absence of a consistent free quantum field theory.

Introduce the electromagnetic electromagnetic field tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ and the charged vector field strength ${\mathcal{G}}_{\mu \nu }=D_{\mu }{B}_{\nu }-D_{\nu }{B}_{\mu }$, with $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$. Minimal coupling givesUnder $B_{\mu }^{\prime }=e^{i\alpha (x)}{B}_{\mu }$ and $A_{\mu }^{\prime }=A_{\mu }-{\partial }_{\mu }\Lambda$,Consequently the required relation isThen ${\mathcal{G}}_{\mu \nu }^{\prime }=e^{i\alpha }{\mathcal{G}}_{\mu \nu }$, its complex conjugate transforms oppositely, the mass bilinear is invariant, and $F_{\mu \nu }^{\prime }=F_{\mu \nu }$. Each term in the total Lagrangian density is therefore invariant under the electromagnetic U(1) gauge symmetry. This phase gauge symmetry does not reinstate the broken gradient-shift gauge symmetry of the massive field itself.

Set $C_{\mu \nu }=A_{\mu }{B}_{\nu }-A_{\nu }{B}_{\mu }$, so $\mathcal{G}=G+ieC$. Expanding the minimally coupled Lagrangian density givesThus there is a derivative cubic $AB{B}^{\ast }$ interaction vertex of order $e$, and a quartic $AAB{B}^{\ast }$ interaction vertex of order $e^2$. There is no pure $B^4$ contact term in this minimal coupling theory, nor an Abelian three-photon interaction vertex.

The straight arrowed lines in the original schematic denote charged massive vector bosons; the wavy lines denote photons. For two identical positively charged vector bosons, the leading scattering amplitude has the order-$e^2$ $t$-channel photon-exchange Feynman diagram, together with its $u$-channel interchange required by identical bosons. A quartic $AAB{B}^{\ast }$ interaction vertex cannot directly attach four external massive vector bosons. For a particle-antiparticle incoming pair, the leading channels instead are $t$-channel exchange and $s$-channel photon annihilation. The shown arrows track charge flow; an outgoing charged particle corresponds to the conjugate field at a vertex.