# Unitary gauge for non-abelian case

I'm reading Chapter 19 of Mandle and Shaw's Quantum field theory. In the first section it is explained that one can go with a $SU(2)$ followed by a $U(1)$ transformation from $$\begin{bmatrix}\eta_1(x) + i\eta_2(x) \\ v + \sigma(x) +i \eta_3(x)\end{bmatrix}$$ to the state $$\begin{bmatrix} 0 \\ v + \sigma(x)\end{bmatrix}.$$ I tried to first make of this first vector a 'down isospin' by multiplying with an generic element of $SU(2)$ $$\begin{bmatrix} \alpha & \beta \\ -\beta^\star &\alpha^\star\end{bmatrix},$$ where $|a|^2+|b|^2=1$.

Is there maybe a easier representation if $SU(2)$ which leads to the correct needed transformation?

($v$ is part of $\sigma$ and may be absorbed in it.)

$$R=\exp (i\theta ~\hat{n}\cdot\vec{\sigma})=I \cos \theta +i \hat{n}\cdot\vec{\sigma} \sin \theta= \begin{bmatrix}\cos \theta +i n_3\sin \theta & i\sin \theta (n_1-in_2) \\ i\sin \theta (n_1+in_2) &\cos \theta -i n_3\sin \theta \end{bmatrix}.$$
It is easier, and, in fact, more conventional, to work backwards, $$R \begin{bmatrix} 0 \\ \rho\end{bmatrix} =\rho \begin{bmatrix} \sin\theta (i n_1+n_2) \\ \cos\theta -in_3\sin \theta\end{bmatrix} = \begin{bmatrix}\eta_1(x) + i\eta_2(x) \\ \sigma(x) +i \eta_3(x)\end{bmatrix},$$
that is $$\eta_1=n_2 \rho \sin \theta, \quad \eta_2=n_1 \rho \sin \theta, \quad \eta_3=-n_3 \rho \sin \theta, \quad \sigma= \rho \cos \theta,$$ so that $~\rho^2=\sigma^2+\vec{\eta}^2~$ and $~\vec{\eta}^2=\rho^2 \sin^2\theta=\sigma^2 \tan^2\theta~$ .
So, then, $R^{-1}$ is the SU(2) group element you were seeking, with $\theta$ and $\hat{n}$ solvable as above. Note $\rho \neq\sigma$, as you sought, as easy to confirm from the modulus/length of the two respective spinors! Your text was being slightly "metaphorical".
Also, the sign of $\eta_3$ in your expression is unconventional: most texts prefer a minus sign, and interchange the roles of $\eta_1$ and $\eta_2$, to (only) then have $$(I \sigma+i~\vec{\eta}\cdot\vec{\sigma}) \begin{bmatrix} 0 \\ 1\end{bmatrix} .$$