# General form of elements of $SU(2)$

$$SU(2)$$ is the set of $$2\times 2$$ complex matrices $$A$$ satisfying $$AA^*=I$$ and $$\det(A)=1$$ where $$A^*$$ denotes the conjugate transpose of $$A$$ and $$I$$ is the identity matrix. I've seen everywhere that the elements of $$SU(2)$$ can be represented as $$\begin{pmatrix}\alpha&\beta\\-\bar\beta&\bar\alpha\end{pmatrix}$$ where $$\vert\alpha\vert^2+\vert\beta\vert^2=1$$, but I've been having a stupidly hard time working out the arithmetic for this claim. I was able to show that if $$\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}$$ is in $$SU(2)$$ (or even just $$U(2)$$), then $$\vert\alpha\vert=\vert\delta\vert$$ and $$\vert\beta\vert=\vert\gamma\vert$$, but I'm stuck from there. I know I must use the equation $$\alpha\delta-\beta\gamma=1$$ to obtain the desired result, but I'm failing to do so.

If $$A=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}\in SU(2)$$ then its determinant is $$1$$ and so its inverse, which must also equal $$A^H$$(the transpose of its complex conjugate) is $$A^{-1}=\begin{pmatrix}\delta&-\beta\\-\gamma&\alpha\end{pmatrix}$$. Therefore $$\delta =\overline\alpha$$ and $$\gamma=-\overline\beta$$. But the determinant is $$1=\alpha\delta-\beta\gamma=\alpha\overline\alpha+\beta\overline\beta= |\alpha|^2+|\beta|^2$$.
• Ah, that is so much easier. I was trying to derive this directly from the equations $\alpha\delta-\beta\gamma=1$, $\vert\alpha\vert^2+\vert\beta\vert^2=1$, $\vert\gamma\vert^2+\vert\delta\vert^2=1$, $\bar\alpha\gamma+\bar\beta\delta=0$, and $\alpha\bar\gamma+\beta\bar\delta=0$. Thank you so much. Feb 13, 2019 at 20:12