Electroweak interaction: From $W^{1}_{\mu},W^{2}_{\mu},W^{3}_{\mu},B_{\mu}$ to $W^{\pm},Z_{\mu},A_{\mu}$

EDIT: Additional question at the end

I am trying to illuminate how the "unphysical" gauge bosons $W^{1}_{\mu},W^{2}_{\mu},W^{3}_{\mu},B_{\mu}$ will be the "physical" $W^{\pm},Z_{\mu},A_{\mu}$ when diagonalizing the mass matrix. Notice that it is in Euclidean time, so we do not have to care about the Lorentz indices. Furthermore $\sigma(x)$ is the Higgs field and $v$ is the vacuum expectation value.

After the symmetry breaking $$SU(2)_L\times U(1)_Y \rightarrow U(1),$$

and inserting the vacuum expecation value, I got the following Lagrangian (just the dynamical part):

$$\frac{1}{2}D_{\mu}\phi^{\dagger}D_{\mu}\phi = \frac{1}{2}\partial_{\mu}\sigma \partial_{\mu} \sigma + \frac{(v+\sigma)^2}{8}(g^2W^{1}_{\mu}W^{1}_{\mu} + g^2W^{2}_{\mu}W^{2}_{\mu} + (gW^{3}_{\mu} - g'B_{\mu})(gW^{3}_{\mu} - g'B_{\mu})) .$$ $W^{\pm}=W^{1}_{\mu}\pm W^{2}_{\mu}$ is clear, but retrieving $Z_{\mu}$ and $A_{\mu}$ not. I tryied the following, since the last part of the Lagrangian can be written like:
$$(W^{3}_{\mu},B_{\mu}) \begin{pmatrix}g^2 & -gg'\\-gg'& g'^{2} \end{pmatrix} \begin{pmatrix}W^{3}_{\mu}\\B_{\mu} \end{pmatrix}$$ The diagonlized matrix reads $$M_D=\begin{pmatrix}0 & 0\\0& g^2 +g'^{2} \end{pmatrix}$$ and does not give the right linear combinations of $Z_{\mu}$ and $A_{\mu}$, which are given in my literature as $$A_{\mu} = \frac{g'W^{3}_{\mu} + g B_{\mu}}{\sqrt{g^2+g'^2}},\qquad Z_{\mu} = \frac{gW^{3}_{\mu} - g' B_{\mu}}{\sqrt{g^2+g'^2}}$$ My question is now, how to get these combinations, it looks like I am close, but only close. And the other question where comes the normalization conditions for the field from?

Cheers!

EDIT:

I finally found the linear combinations, mass eigenstates, like they are in the literature, by inserting not only the diagonlized mass matrix $M_D$, but by inserting $M = PM_DP^{-1}$ As I was looking at the covariant derivative to find out how the fields couple to the Higgs doublet I was wondering how I could possibly turn the following matrix into mass eigenstates of the gauge fields:

$$\frac{i}{2}\begin{pmatrix}gW^{3}_{\mu} + g'B_{\mu} & 0\\ 0& g W^{3}_{\mu} + g'B_{\mu}\end{pmatrix}$$

again, cheers!

• In the last two formulae, you meant the squared couplings under the square roots and not the couplings themselves, didn't you? When you make this fix, the eigenstate combinations clearly work out. – Luboš Motl Jul 11 '14 at 9:45
• A couple more mistakes... The Higgs is a complex scalar field, so you don't need the leading factor of $\frac12$ and $W^\pm = \frac{1}{\sqrt{2}} (W^1 \pm i W^2)$ – innisfree Jul 11 '14 at 9:46