= Solution
Use <natural units> and the <Minkowski metric> $g_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$. Varying the <Proca action> with respect to $A_\nu$ gives $\partial\mathcal L/\partial A_\nu=m^2A^\nu$ and $\partial\mathcal L/\partial(\partial_\mu A_\nu)=-F^{\mu\nu}$. The <Euler-Lagrange equation> is therefore
$$
\boxed{\partial_\mu F^{\mu\nu}+m^2A^\nu=0.}
$$
Taking a divergence annihilates the antisymmetric field strength, since commuting derivatives are symmetric in their indices. For $m\ne0$ this gives the <Lorenz constraint in Proca theory> $\partial_\nu A^\nu=0$, and hence $(\Box+m^2)A^\nu=0$. 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 $p\cdot\varepsilon=0$ and is normalized by $\varepsilon\cdot\varepsilon=-1$. Taking
$$
\boxed{\varepsilon_L^\mu=\frac1m(p,0,0,E)}
$$
satisfies 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 $p^\mu=(E,0,0,p^*)$ and $p'^\mu=(E,0,0,-p^*)$, with $E=m_H/2$ and $p^{*2}=m_H^2/4-m_W^2$. Consistent normalized longitudinal <polarization vectors> are
$$
\varepsilon_L(p)=\frac1{m_W}(p^*,0,0,E),\qquad
\varepsilon_L(p')=\frac1{m_W}(p^*,0,0,-E).
$$
Their Lorentz contraction is $(p^{*2}+E^2)/m_W^2=(2p^{*2}+m_W^2)/m_W^2$. The <Higgs boson coupling to W bosons> consequently gives
$$
|\mathcal M_{LL}|^2=\frac{v^2g^4}{4m_W^4}(2p^{*2}+m_W^2)^2
=\frac{m_H^4}{v^2}(1-2x)^2,\qquad x=\frac{m_W^2}{m_H^2}.
$$
Using the <two-body Lorentz-invariant phase space>, $p^*=m_H\sqrt{1-4x}/2$ and $v^{-2}=\sqrt2G_F$ yields the longitudinal partial <decay width>
$$
\boxed{\Gamma(H\to W_L^+W_L^-)=\frac{G_Fm_H^3}{8\pi\sqrt2}\sqrt{1-4x}\,(1-2x)^2,\qquad m_H>2m_W.}
$$
The 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 $G_Fm_H^3$.
\Image[/past-exam-of-the-gauge-field-theory-course-of-the-university-of-cambridge/2013/paper-1-higgs-ww-polarizations.png]
{title=On-shell Higgs decay: longitudinal and transverse contributions to the W-pair width}
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