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$SU(2)$ 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.

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2 Answers 2

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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 $1=\alpha\delta-\beta\gamma=\alpha\overline\alpha+\beta\overline\beta= |\alpha|^2+|\beta|^2$.

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    $\begingroup$ 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. $\endgroup$
    – Anonymous
    Feb 13, 2019 at 20:12
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You need two facts only:

  • A matrix is unitary iff the columns are pairwise orthonormal.
  • A unitary matrix is special unitary if the determinant is 1.
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  • $\begingroup$ I'm aware. Those are the facts that I stated and which I have been using. $\endgroup$
    – Anonymous
    Feb 13, 2019 at 20:10

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