Where does the condition come from that in models with several electroweak breaking doublets the square root of the squared sum of the VEVs should yield $246$ GeV?


The masses of W-bosons and Z-bosons are known – and the mass of W-bosons multiplied by some coupling constants was known indirectly through the force they mediate, as the Fermi's constant.

But the mass of e.g. the W-boson comes from the gauge-invariant kinetic term of the Higgs boson, $$ \frac 12 D_\mu H \cdot D^\mu H $$ The covariant derivative includes the term proportional to $W_\mu$, a component of the $SU(2)$ gauge field, in this case. So the relevant term for the W-mass is the term $$ \frac 12 g W_\mu H \cdot g W^\mu H = \frac{H^2 g^2}{2} W_\mu W^\mu $$ where $g$ is the $SU(2)$ coupling included in the covariant derivative. If the symmetry is spontaneously broken, $\langle H \rangle = v$ and this reduces to $$\frac{v^2 g^2}{2} W_\mu W^\mu $$ which is a mass term for the W-boson in the Standard Model, therefore $m_W=gv$ (it's similar but more convoluted for the Z-mass, I need to use both $SU(2)$ and $U(1)_Y$).

If there are several Higgs bosons, we need the same derivation (terms) for each of them and we produce mass terms $$\sum_i \frac{v_i^2 g_i^2}{2} W_\mu W^\mu $$ so it's the whole sum of $g_i^2 v_i^2$ that plays the role of the $gv$ in the single-Higgs case. But if we assume that all $g_i=g$ – everything has to be doublets and triplets under $SU(2)$ just like in the Standard Model, we know that $v=246\,{\rm GeV}$ in the Standard Model and $v^2$ is replaced by $\sum v_i^2$ in the many-Higgs case, and the statement you started with follows.

I suppressed the $SU(2)$ $i=1,2,3$ index and other things to focus on the core idea.

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