= Solution
Use <natural units> and the <Minkowski metric> $(+---)$, and work at tree level with on-shell final particles. Write $M=m_H$ and, for a final particle of mass $m$, $\beta_m=\sqrt{1-4m^2/M^2}$. In the <Higgs boson> rest frame the <Lorentz-invariant phase-space measure> reduces to
$$
d\Phi_2=\frac{|\boldsymbol k|}{16\pi^2M}\,d\Omega=\frac{\beta_m}{32\pi^2}\,d\Omega,
\qquad\int d\Phi_2=\frac{\beta_m}{8\pi}.
$$
To obtain this, integrate the momentum delta function to set $\boldsymbol k_2=-\boldsymbol k_1$; the energy delta function is $\delta(M-2\sqrt{k^2+m^2})$, whose radial Jacobian is $E/(2k)$. Hence for an angle-independent final-state <spin sum> $S=\sum|\mathcal M|^2$,
$$
\Gamma=\frac{\beta_m S}{16\pi M}.
$$
There is no initial-state spin average for a scalar, and neither charged-particle pair here requires an identical-particle factor of $1/2$.
For <Higgs decay to two W bosons>, the <Feynman vertex> gives $\mathcal M=(2m_W^2/v)\varepsilon_1^*\!\cdot\varepsilon_2^*$, up to an overall phase. Contracting the two <massive vector polarization sums> gives
$$
S_W=\frac{4m_W^4}{v^2}\left[2+\frac{(k_1\cdot k_2)^2}{m_W^4}\right],\qquad
k_1\cdot k_2=\frac{M^2-2m_W^2}{2}.
$$
The constant 2 follows from $4-1-1$ in the contraction, with the last term coming from the two momentum projectors. Putting $x_W=m_W^2/M^2$, the \b[full massive result] is
$$
\boxed{\Gamma(H\to W^+W^-)=\frac{M^3}{16\pi v^2}\sqrt{1-4x_W}\,(1-4x_W+12x_W^2),\qquad M\geq2m_W.}
$$
The on-shell two-body width is zero below this threshold; decays through off-shell <W bosons> into more particles are different channels.
For <Higgs decay to a fermion pair>, put $y_b=m_b/v$. The amplitude is $\mathcal M=-y_b\bar u(k_1)v(k_2)$, up to an overall phase and a color Kronecker delta. The <fermion spin sums> and <gamma-matrix trace> give
$$
S_b=N_cy_b^2\operatorname{tr}[(\not k_1+m_b)(\not k_2-m_b)]
=4N_cy_b^2(k_1\cdot k_2-m_b^2)
=2N_cy_b^2(M^2-4m_b^2).
$$
The <color multiplicity in a decay width> is $N_c=3$: only a quark and antiquark with matching colors contribute, so the factor is three rather than nine. Therefore
$$
\boxed{\Gamma(H\to\bar b b)=\frac{3m_b^2M}{8\pi v^2}\left(1-\frac{4m_b^2}{M^2}\right)^{3/2},\qquad M\geq2m_b.}
$$
Again the on-shell two-body width is zero below threshold. This is a partonic tree-level answer with every mass retained; <hadronization> is outside the specified calculation.
\b[The <W boson> channel dominates for large $M$ within the tree-level comparison.] The widths scale as $M^3/v^2$ and $m_b^2M/v^2$, respectively, and
$$
\boxed{\frac{\Gamma(H\to W^+W^-)}{\Gamma(H\to\bar b b)}\sim\frac{M^2}{6m_b^2}.}
$$
The enhancement comes from <longitudinal polarization of a massive vector boson>: its polarization vector grows as momentum divided by $m_W$. Thus the longitudinal pair survives the apparently small $m_W^2$ factor in the interaction. At masses so large that the scalar sector is strongly coupled, the tree-level extrapolation itself needs corrections.
Back to article page
