# Transforming to unitary gauge (for real)

In the Higgs mechanism application many textbooks chooses to expand the Higgs field around the vacuum and describes it in the following way.

$$\Psi(x) = \begin{bmatrix}\xi_1 + i\xi_2\\ 1/\sqrt2(v+H(x))-i\xi_3 \end{bmatrix}=\exp(it \xi_a T^a) \begin{bmatrix}0\\ 1/\sqrt2(v+H(x)) \end{bmatrix}$$

Then one can do $SU(2)$ transformation $\Phi(x) \rightarrow \exp(-it \xi_a T^a)\Phi(x)$ and use explicit form of doublet only involving $H$ and $v$ in say your covariant derivatives.

But if I actually do the matrix multiplication of the right hand side I am failing to obtain left hand side. Assuming $T^i=1/2(\sigma^1,\sigma^2,\sigma^3)$ where $\sigma^i$ are Pauli matrices.

$$\exp(it \xi_a T^a) = 1+it \xi_a T^a+O(\xi^2)= \begin{bmatrix}1+\frac{1}{2}it\xi_3 & \frac{1}{2}t(i\xi_1+\xi_2) \\ \frac{1}{2}t(i\xi_1+\xi_2) & 1-\frac{1}{2}it\xi_3 \end{bmatrix}+O(\xi^2)$$

Hence

$$(1+it \xi_a T^a) \begin{bmatrix}0\\ 1/\sqrt2(v+H(x)) \end{bmatrix} = \begin{bmatrix}\frac{1}{2\sqrt{2}}ti(H+v)(i\xi_1+\xi_2)\\\frac{1}{\sqrt{2}}(H+v)(1-\frac{1}{2}it\xi_3)\end{bmatrix}$$

For which no value of $t$ gives fields as in left hand of my first equation. I must be missing something here?

It seems that the three of $\xi_a$ on LHS and RHS in your first equation are not the same. For simplicity let's define $$\xi_4\equiv\frac{1}{\sqrt{2}}(\upsilon+H),$$ and rename fields on the RHS to $\xi_a'$. Then $$\begin{bmatrix} \xi_1+i\xi_2\\ \xi_4-i\xi_3 \end{bmatrix} = \exp(it\xi_a'T^a)\begin{bmatrix} 0\\ \xi_4 \end{bmatrix}~.$$ Using your result we have $$\begin{bmatrix} \xi_1+i\xi_2\\ \xi_4-i\xi_3 \end{bmatrix} = \begin{bmatrix} -\frac{t}{2}\xi_1'\xi_4+\frac{it}{2}\xi_2'\xi_4\\ \xi_4-\frac{it}{2}\xi_3'\xi_4 \end{bmatrix},$$ Which means $$\xi_1=-\frac{t}{2}\xi_1'\xi_4~,\;\;\;\xi_2=\frac{t}{2}\xi_2'\xi_4,\;\;\;\xi_3=\frac{t}{2}\xi_3'\xi_4~.$$