Every element in $SU(2)$ has the form $\begin{bmatrix} \alpha& \beta\\ -\bar{\beta} & -\bar{\alpha}\end{bmatrix}$ with $\alpha, \beta\in \mathbb{C}$

Question

I want to prove that every element in $\operatorname{SU}(2)$ has the form $$ \begin{bmatrix} \alpha& \beta\\ -\bar{\beta} & -\bar{\alpha}\end{bmatrix}, $$ with $\alpha, \beta\in \mathbb{C}$, for this I took an arbitrary element $$ \begin{bmatrix} \alpha& \beta\\ \gamma & \delta\end{bmatrix}, $$ and I arrive at the following equalities $$ \alpha\delta-\beta\gamma=1, \quad \alpha\bar{\alpha}+\beta\bar{\beta}=1, \quad \bar{\alpha}\gamma+\bar{\beta}\delta=0, \quad \gamma\bar{\gamma}+\delta\bar{\delta}=1 $$

but I don't know from here how to prove that $\gamma=-\bar{\beta}$ and $\delta=-\bar{\alpha}$, any idea? Thank you.

Answer 1

Multiply the last equation by $\bar{\alpha}$ \begin{eqnarray*} \bar{\alpha} \gamma\bar{\gamma}+ \bar{\alpha} \delta\bar{\delta}=\bar{\alpha}. \end{eqnarray*} Now use the third equation \begin{eqnarray*} -\bar{\beta} \delta\bar{\gamma}+ \bar{\alpha} \delta\bar{\delta}=\bar{\alpha} \\ \delta( -\bar{\beta} \bar{\gamma}+ \bar{\alpha} \bar{\delta})=\bar{\alpha} \\ \end{eqnarray*} Now use the first equation and we have $\delta=-\bar{\alpha}$. The other equation can be derived similarly.

Answer 2

Since

$$ \bar{\alpha}\gamma + \bar{\beta}\delta = 0, $$

it follows that

\begin{align*} 0 & = \alpha\bar{\alpha}\gamma + \alpha\bar{\beta}\delta \\ & = \left|\alpha\right|^{2}\gamma + \bar{\beta}(\alpha \delta) \\ & = \left|\alpha\right|^{2}\gamma + \bar{\beta}(1 + \beta \gamma )\\ & = \left|\alpha\right|^{2}\gamma + \bar{\beta} + \left|\beta\right|^{2}\gamma \\ & = \gamma(\left|\alpha\right|^{2} + \left|\beta\right|^{2}) + \bar{\beta} \\ & = \bar{\beta} + \gamma, \end{align*}

and so $\gamma = - \bar{\beta}$. The other question is derived similarly