# Higgs Doublet Transformations

If we consider two complex Higgs doublets with $SU(3)_{c} ⊗ SU(2)_L ⊗ U(1)_Y$ quantum numbers $\phi_{A,B}$ ∼ (1,2,1), such that

$\phi_A = \begin{pmatrix} \phi_{A}^{\dagger}\\ \dfrac{1}{\sqrt2} (Re \phi_{A}^{0} + iIm \phi_{A}^{0}) \end{pmatrix}$ & $\phi_A = \begin{pmatrix} \phi_{B}^{\dagger}\\ \dfrac{1}{\sqrt2} (Re \phi_{B}^{0} + iIm \phi_{B}^{0}) \end{pmatrix}$

If we assume that they acquire the vacuum expectation values (vevs):

$<\phi_A> = \begin{pmatrix} 0\\ \dfrac{v_A}{\sqrt2} \end{pmatrix}$ & $<\phi_B> = \begin{pmatrix} 0\\ \dfrac{v_B}{\sqrt2} \end{pmatrix}$

I know that we can change the basis from {$\phi_A,\phi_B$} to a {$\phi_1,\phi_2$} basis, bearing in mind that only one doublet gains a non zero vev as follows:

$<\phi_1> = \begin{pmatrix} 0\\ \dfrac{v}{\sqrt2} \end{pmatrix}$ & $<\phi_2> = \begin{pmatrix} 0\\ 0 \end{pmatrix}$

My issue is that which particular transformation can I use to be able to do this transformation and how do I proceed from there. Basically asking for how can I express {$\phi_A,\phi_B$} in terms of {$\phi_1,\phi_2$} and relate the $v_A, v_B$ and $v$ altogether

Please someone help me out.

Thanks!

• I'm a bit busy right now, but I will try to answer it later. I believe it is just a $SU(2)_L$ rotation, and using the fact that you can rotate $(\alpha, 0)_L^T$ into any state via an $SU(2)_L$ rotation. – InertialObserver Sep 4 '18 at 21:44
• When will you be able to help? – SSS Sep 5 '18 at 11:18

## 1 Answer

I do not have time to write out the derivation here. However, you can find it in this paper . The change of basis you seek is equation (8) and the variables used in that equation are defined in the preceding equations. He states in equation (16) that this yields the vev you seek, though of course you should work it out yourself.