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

$$\phi_A = \begin{pmatrix} \phi_{A}^{\dagger}\\ \dfrac{1}{\sqrt2} ({\rm Re}\, \phi_{A}^{0} + i{\rm Im}\, \phi_{A}^{0}) \end{pmatrix} , \phi_B = \begin{pmatrix} \phi_{B}^{\dagger}\\ \dfrac{1}{\sqrt2} ({\rm Re}\, \phi_{B}^{0} + i{\rm Im}\, \phi_{B}^{0}) \end{pmatrix}$$

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

$$\langle\phi_A\rangle = \begin{pmatrix} 0\\ \dfrac{v_A}{\sqrt2} \end{pmatrix} , \langle\phi_B\rangle = \begin{pmatrix} 0\\ \dfrac{v_B}{\sqrt2} \end{pmatrix}$$

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:

$$\langle\phi_1\rangle = \begin{pmatrix} 0\\ \dfrac{v}{\sqrt2} \end{pmatrix} , \langle\phi_2\rangle = \begin{pmatrix} 0\\ 0 \end{pmatrix}$$

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.

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

$$\phi_A = \begin{pmatrix} \phi_{A}^{\dagger}\\ \dfrac{1}{\sqrt2} ({\rm Re}\, \phi_{A}^{0} + i{\rm Im}\, \phi_{A}^{0}) \end{pmatrix} , \phi_B = \begin{pmatrix} \phi_{B}^{\dagger}\\ \dfrac{1}{\sqrt2} ({\rm Re}\, \phi_{B}^{0} + i{\rm Im}\, \phi_{B}^{0}) \end{pmatrix}.$$

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

$$\langle\phi_A\rangle = \begin{pmatrix} 0\\ \dfrac{v_A}{\sqrt2} \end{pmatrix} , \langle\phi_B\rangle = \begin{pmatrix} 0\\ \dfrac{v_B}{\sqrt2} \end{pmatrix}.$$

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:

$$\langle\phi_1\rangle = \begin{pmatrix} 0\\ \dfrac{v}{\sqrt2} \end{pmatrix} , \langle\phi_2\rangle = \begin{pmatrix} 0\\ 0 \end{pmatrix}.$$

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.

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

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

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

$\langle\phi_A\rangle = \begin{pmatrix} 0\\ \dfrac{v_A}{\sqrt2} \end{pmatrix}$ , $\langle\phi_B\rangle = \begin{pmatrix} 0\\ \dfrac{v_B}{\sqrt2} \end{pmatrix}$

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:

$\langle\phi_1\rangle = \begin{pmatrix} 0\\ \dfrac{v}{\sqrt2} \end{pmatrix}$ , $\langle\phi_2\rangle = \begin{pmatrix} 0\\ 0 \end{pmatrix}$

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!

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

$$\phi_A = \begin{pmatrix} \phi_{A}^{\dagger}\\ \dfrac{1}{\sqrt2} ({\rm Re}\, \phi_{A}^{0} + i{\rm Im}\, \phi_{A}^{0}) \end{pmatrix} , \phi_B = \begin{pmatrix} \phi_{B}^{\dagger}\\ \dfrac{1}{\sqrt2} ({\rm Re}\, \phi_{B}^{0} + i{\rm Im}\, \phi_{B}^{0}) \end{pmatrix}$$

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

$$\langle\phi_A\rangle = \begin{pmatrix} 0\\ \dfrac{v_A}{\sqrt2} \end{pmatrix} , \langle\phi_B\rangle = \begin{pmatrix} 0\\ \dfrac{v_B}{\sqrt2} \end{pmatrix}$$

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:

$$\langle\phi_1\rangle = \begin{pmatrix} 0\\ \dfrac{v}{\sqrt2} \end{pmatrix} , \langle\phi_2\rangle = \begin{pmatrix} 0\\ 0 \end{pmatrix}$$

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.

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):

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

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:

$\langle\phi_1\rangle = \begin{pmatrix} 0\\ \dfrac{v}{\sqrt2} \end{pmatrix}$ & $\langle\phi_2\rangle = \begin{pmatrix} 0\\ 0 \end{pmatrix}$

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!

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

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

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

$\langle\phi_A\rangle = \begin{pmatrix} 0\\ \dfrac{v_A}{\sqrt2} \end{pmatrix}$ , $\langle\phi_B\rangle = \begin{pmatrix} 0\\ \dfrac{v_B}{\sqrt2} \end{pmatrix}$

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:

$\langle\phi_1\rangle = \begin{pmatrix} 0\\ \dfrac{v}{\sqrt2} \end{pmatrix}$ , $\langle\phi_2\rangle = \begin{pmatrix} 0\\ 0 \end{pmatrix}$

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!