Past exam of the theoretical physics major option course of the University of Oxford / 2017 / 5

The Proca action is invariant under spacetime translations and proper continuous Lorentz transformations, together forming the connected Poincaré group. The fixed nonzero mass excludes scale invariance. In four dimensions, a dimensionless action and the canonical kinetic term implyThese are mass dimensions. The field-strength term is invariant under $A_{\mu }\to A_{\mu }+{\partial }_{\mu }\chi$, but the mass term is not: its change contains $m_A^2{A}^{\mu }{\partial }_{\mu }\chi +\frac{1}{2}{m}_A^2(\partial \chi {)}^2$. Thus the displayed massive-vector theory has no Maxwell-type local gauge invariance. The massless limit has a different constraint structure.

Varying the vector field gives the Proca equation,Its divergence uses antisymmetry of $F^{\mu \nu }$ to cancel the first term. Since $m_A\ne 0$,For an on-shell momentum this imposes $p_{\nu }{ϵ}^{\nu }=0$ on the four polarization components, leaving three independent solutions. In the rest frame, $p=(m_A,0)$ makes ${ϵ}^0=0$, leaving precisely the three spatial polarizations of a massive spin-one particle.

There is no fourth independent oscillator hidden in $A^0$. Its canonical momentum is zero because the Lagrangian has no ${\dot{A}}^0$, and the temporal equation constrains $A^0$ in terms of the spatial data. Equivalently the primary and secondary second-class constraints remove two variables from the eight-dimensional phase space, leaving six phase-space variables, or three physical degrees of freedom.

Take $p^{\mu }=(e,0,0,k)$ with $e^2=k^2+m_A^2$. A real orthonormal basis that satisfies the source's stated bilinear normalization isEach obeys $p\cdot {ϵ}_i=0$. The two transverse norms are $-1$, while ${ϵ}_L^2=(k^2-e^2)/m_A^2=-1$, and all cross products vanish. For momentum in the opposite direction, one consistent longitudinal choice is ${ϵ}_L(-k)=(k,0,0,-e)/m_A$ for $k>0$; an overall sign of any polarization is immaterial.

Here $+$ and $-$ label the two linear transverse basis vectors. If those symbols are intended to denote circular helicities, choose ${ϵ}_{\pm }^{\mathrm{hel}}=({ϵ}_x\pm i{ϵ}_y)/\sqrt{2}$ instead. The corresponding orthonormality relation must contain complex conjugation, ${ϵ}_i^{\ast }\cdot {ϵ}_j=-{\delta }_{ij}$. Without that conjugation, a circular polarization has ${ϵ}_{+}^{\mathrm{hel}}\cdot {ϵ}_{+}^{\mathrm{hel}}=0$, a concrete counterexample to interpreting the printed bilinear condition as a helicity normalization.

For longitudinal-vector annihilation through a scalar contact interaction, first use the supplied vertex $2i{g}^2{\eta }_{\mu \nu }$. Let $s=(p_1+p_2{)}^2$ and work in the center-of-mass frame. The amplitude is $\mathcal{M}=2g^2{ϵ}_1\cdot {ϵ}_2$, up to its overall phase. For two longitudinal incoming vectors,This grows as $s^2/m_A^4$, whereas transverse amplitudes stay of order $g^2$ and mixed longitudinal-transverse amplitudes vanish in this frame. Consequently the leading high-energy contribution is longitudinal.

For an unpolarized total cross section, average over the $3\times 3$ incoming polarizations. The massive-vector polarization sum givesThe identical final scalars require a factor $1/2!$ in the two-body phase space. With ${\beta }_A=\sqrt{1-4m_A^2/s}$ and ${\beta }_h=\sqrt{1-4m_h^2/s}$, the angle-independent result isFor a prepared longitudinal pair there is no incoming polarization average, giving insteadThere is no extra initial-state factor $1/2$: a cross section uses the incoming beam flux, while the final-state factor removes double counting of indistinguishable outgoing states.

Define the energy in the stated bound to be the total center-of-mass energy, $E=\sqrt{s}$. Equating the leading expressions to $4\pi /E^2$ givesfor $m_A=100\mathrm{GeV}$ and $g=0.4$. The second value shows that a longitudinal channel reaches the supplied perturbative bound earlier. If the source's unspecified $E$ denotes the energy of one incoming particle and the bound is literally $4\pi /E^2$, substitute $s=4E^2$ instead: the corresponding values are $1.83\mathrm{TeV}$ unpolarized and $1.05\mathrm{TeV}$ longitudinal. Stating the energy and polarization conventions is necessary for a numerical answer.

Interaction-normalization qualification: the written ${\mathcal{L}}_{\mathrm{int}}=g^2{A}_{\mu }{A}^{\mu }{h}^2$ differentiates to $4i{g}^2{\eta }_{\mu \nu }$, because there are $2!$ vector assignments and $2!$ scalar assignments. For example, differentiating $g^2(A^0{)}^2{h}^2$ twice in each variable gives $4g^2$. The supplied $2i{g}^2{\eta }_{\mu \nu }$ would instead follow from ${\mathcal{L}}_{\mathrm{int}}=g^2{A}_{\mu }{A}^{\mu }{h}^2/2$. Thus all cross sections above follow the explicitly requested rule. Using the literal written interaction multiplies them by four and divides every quoted threshold energy by $\sqrt{2}$. The energy growth itself is the physical issue: a contact interaction of massive longitudinal vectors needs additional dynamics before its perturbative high-energy extrapolation becomes reliable.