Past exam of the gauge field theory course of the University of Cambridge / 2013 / Paper 1 / 1 / Solution

Use natural units and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. Varying the Proca action with respect to $A_{\nu }$ gives $\partial \mathcal{L}/\partial A_{\nu }=m^2{A}^{\nu }$ and $\partial \mathcal{L}/\partial ({\partial }_{\mu }{A}_{\nu })=-F^{\mu \nu }$. The Euler-Lagrange equation is thereforeTaking a divergence annihilates the antisymmetric field strength, since commuting derivatives are symmetric in their indices. For $m\ne 0$ this gives the Lorenz constraint in Proca theory ${\partial }_{\nu }{A}^{\nu }=0$, and hence $(\square +m^2)A^{\nu }=0$. The constraint follows from dynamics; it is not an additional gauge fixing condition for this massive theory.

For four-momentum $p^{\mu }=(E,0,0,p)$, longitudinal polarization of a massive vector boson lies in the time–$z$ plane, has $p\cdot \epsilon =0$ and is normalized by $\epsilon \cdot \epsilon =-1$. Takingsatisfies both requirements because $E^2-p^2=m^2$. It can also be obtained by boosting the rest-frame unit spatial polarization along the direction of motion. Together with two transverse polarizations it gives the three physical spin states of a massive spin one particle.

In the Higgs boson rest frame, choose $p^{\mu }=(E,0,0,p^{\ast })$ and $p^{\prime \mu }=(E,0,0,-p^{\ast })$, with $E=m_H/2$ and $p^{\ast 2}=m_H^2/4-m_W^2$. Consistent normalized longitudinal polarization vectors areTheir Lorentz contraction is $(p^{\ast 2}+E^2)/m_W^2=(2p^{\ast 2}+m_W^2)/m_W^2$. The Higgs boson coupling to W bosons consequently givesUsing the two-body Lorentz-invariant phase space, $p^{\ast }=m_H\sqrt{1-4x}/2$ and $v^{-2}=\sqrt{2}{G}_F$ yields the longitudinal partial decay widthThe charged W bosons are distinguishable, so there is no identical-particle factor of $1/2$. This expression includes the entire angular phase space but only the specified longitudinal final polarizations. The Higgs decay to two W bosons summed over all polarizations instead has the factor $1-4x+12x^2$; the additional transverse contribution is $8x^2$. At $m_H\gg m_W$ the longitudinal channel dominates and the leading width grows as $G_F{m}_H^3$.