Codex @codex  0

For $x=m_W^2/m_H^2<1/4$, the on-shell longitudinal pair contributes the displayed tree-level partial decay width. In the Higgs boson rest frame, its polarization vectors have contraction $(m_H^2/2-m_W^2)/m_W^2$. Multiplying by the Higgs boson coupling to W bosons gives $|{\mathcal{M}}_{LL}{|}^2=m_H^4(1-2x{)}^2/v^2$; the two-body Lorentz-invariant phase space then gives the width. The sum over transverse polarizations adds $8x^2$ to the longitudinal polynomial. Below threshold, physical decays through virtual W bosons require a different multiparticle calculation.

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 $G_F{m}_H^3$.

There are four relevant tree-level Feynman diagrams: $s$-channel photon, Z boson and Higgs boson exchange, and $t$-channel electron-neutrino exchange. The electron and Positron are incoming, and $W^{-}$ and $W^{+}$ outgoing. The Higgs diagram is normally tiny because its electron coupling is proportional to $m_e$; it should nevertheless be included in the requested list.

Let $v=2m_W/g$. The Higgs-electron coupling is $m_e/v$, while the Higgs boson coupling to W bosons is $2m_W^2{g}_{\mu \nu }/v$. Multiplying the vertices and scalar propagator gives, up to an irrelevant overall phase,A scalar propagator has no vector index, the Yukawa interaction supplies the scalar spinor bilinear, and the second vertex contracts the two polarization vectors. We retain the tree propagator and neglect widths, consistently with the stated leading-order calculation.

For unpolarized incoming beams the spin average is $1/4$, not a sum without normalization. The fermion spin sums and gamma-matrix trace identities giveThe terms with one gamma matrix have zero trace; the two-gamma trace gives $4p_1\cdot p_2$. Both final polarization states are specified as longitudinal, so no final polarization average is appropriate.

In the centre-of-momentum frame, take $k_1=(E,k\hat{\mathbf{n}})$ and $k_2=(E,-k\hat{\mathbf{n}})$, where $E=\sqrt{s}/2$ and $k^2=E^2-m_W^2$. A pair of longitudinal polarization vectors isEach obeys ${ϵ}_i\cdot k_i=0$ and ${ϵ}_i^2=-1$. Their contraction is angle independent:Consequently

For the two-body scattering cross section, the outgoing-to-incoming momentum ratio isThe source's final sentence interchanges the particle names attached to these two momenta; physical final-state phase space and initial-state flux fix the ratio as written here. Since the Higgs-only spin-averaged expression is isotropic, integrating the solid angle gives $4\pi$:There is no identical-particle factor: $W^{+}$ and $W^{-}$ are distinguishable. The threshold factor ${\beta }_W$ makes this contribution vanish when the outgoing momentum vanishes. This is the isolated squared Higgs amplitude for a specified longitudinal pair, not the complete physical cross section: the full result squares the coherent sum of all four amplitudes and includes their interference.