Tree-level scalar normalization and the charged-vector coupling

In the stated $U=I$ configuration, $H^{†}H=(v+h{)}^2\mathrm{/}2$. Its d'Alembertian is $\mathrm{\square }(H^{†}H)={\mathrm{\partial }}_{\mu }((v+h){\mathrm{\partial }}^{\mu }h)$. Therefore the selected operator is the product $(v+h{)}^2{\mathrm{\partial }}_{\mu }h{\mathrm{\partial }}^{\mu }h\mathrm{/}2+(v+h{)}^3\mathrm{\square }h\mathrm{/}2$ before integration by parts; equivalently, writing $s=H^{†}H$, it is $s\mathrm{\square }s$.

Integrating $s\mathrm{\square }s$ over spacetime and applying the divergence theorem to ${\mathrm{\partial }}_{\mu }(s{\mathrm{\partial }}^{\mu }s)$ gives $\int s\mathrm{\square }s=-\int {\mathrm{\partial }}_{\mu }s{\mathrm{\partial }}^{\mu }s$, since the boundary integral vanishes by [A2]. Here ${\mathrm{\partial }}_{\mu }s=(v+h){\mathrm{\partial }}_{\mu }h$, so the action contribution is $-C_{H\mathrm{\square }}\int (v+h{)}^2{\mathrm{\partial }}_{\mu }h{\mathrm{\partial }}^{\mu }h\mathrm{/}{\mathrm{\Lambda }}_{\mathrm{o}\mathrm{p}}^2$. The sign uses the stated $\mathrm{\square }$ and mostly-plus contraction.

With $U=I$, write $r=v+h$ and use $T_a={\sigma }_a\mathrm{/}2$. The upper component of the covariant derivative is $(D_{\mu }H{)}_1=igr(W_{\mu }^1-i{W}_{\mu }^2)\mathrm{/}(2\sqrt{2})=igr{W}_{\mu }^{+}\mathrm{/}2$. Since $W^1,W^2$ are real, its norm squared is $g^2{r}^2{W}_{\mu }^{-}{W}^{+\mu }\mathrm{/}4$. The Higgs kinetic term therefore contains $-g^2{r}^2{W}_{\mu }^{+}{W}^{-\mu }\mathrm{/}4$. The chosen $Q_{H\mathrm{\square }}$ has no gauge field, so this coefficient is unchanged as a function of $h$; thus $m_W^2(h)=g^2(v+h{)}^2\mathrm{/}4$ and $m_W^2(0)=g^2{v}^2\mathrm{/}4$.

The renormalizable radial kinetic density is $-\frac{1}{2}{\mathrm{\partial }}_{\mu }h{\mathrm{\partial }}^{\mu }h$. Adding the result of step 2.1 gives $-\frac{1}{2}K(h){\mathrm{\partial }}_{\mu }h{\mathrm{\partial }}^{\mu }h$ with $K(h)=1+2C_{H\mathrm{\square }}(v+h{)}^2\mathrm{/}{\mathrm{\Lambda }}_{\mathrm{o}\mathrm{p}}^2$. Since $C_{H\mathrm{\square }}$ is dimensionless and both $v+h$ and ${\mathrm{\Lambda }}_{\mathrm{o}\mathrm{p}}$ have mass dimension one, $K$ is dimensionless. Under [A3], $K>0$ near zero.

On an interval with $K>0$, ${\widehat{h}}^{\mathrm{\prime }}(h)=\sqrt{K(h)}>0$ and $\widehat{h}(0)=0$. For $h_1<h_2$, $\widehat{h}(h_2)-\widehat{h}(h_1)={\int }_{h_1}^{h_2}\sqrt{K(s)}ds>0$, so $\widehat{h}$ is strictly increasing and maps a neighborhood of zero onto an interval with a continuous inverse. At every $h_0$ in this interval, the inverse difference quotient tends to $1\mathrm{/}{\widehat{h}}^{\mathrm{\prime }}(h_0)$; hence its derivative is $1\mathrm{/}\sqrt{K({\widehat{h}}^{-1}(y))}$. This derivative is continuous, and induction in that identity makes the inverse smooth because $K$ and the square root are smooth where $K>0$. Also $\widehat{h}(h)\mathrm{/}h\to \sqrt{K(0)}$ as $h\to 0$. Hence the difference quotient in the new coordinate is ${\lim }_{h\to 0}\frac{m_W^2(h)-m_W^2(0)}{\widehat{h}(h)-\widehat{h}(0)}=\frac{(d{m}_W^2\mathrm{/}dh){\mathrm{\mid }}_0}{{\widehat{h}}^{\mathrm{\prime }}(0)}=\frac{g^{2}v\mathrm{/}2}{\sqrt{K(0)}}\mathrm{.}$ Since $2m_W^2(0)\mathrm{/}v=g^{2}v\mathrm{/}2$, the defined coupling is ${\kappa }_W=1\mathrm{/}\sqrt{K(0)}$, which yields the displayed expression. If this reference point is called a vacuum, the full retained potential must satisfy the optional local-minimum condition [F5]; at this coefficient point it is $V_4$, since [A1] sets $C_H=0$ and $Q_{H\mathrm{\square }}$ vanishes on constant fields. Otherwise the result is only a local expansion coefficient. The kinetic normalization does not prove potential stationarity, local stability, or global stability. It supplies no empirical measurement or numerical bound on discarded orders.