Form of $2\times2$ unitary matrices can be given as
$$ \left(\begin{array}{c c} a & b \\ -e^{i \theta} b^{*} & e^{i \theta} a^{*} \end{array}\right)$$
where $aa^{*}+bb^{*}=1$
Now I have a bit of trouble deriving this.
Say
$$U=\left(\begin{array}{c c} a & b \\ c & d \end{array}\right) $$
Using the fact that $UU^{*}=I$ and $U^{*}U=I$, we get six equations:
$$\begin{eqnarray}aa^{*}+bb^{*}=1\\ ca^{*}+db^{*}=0\\ cc^{*}+dd^{*}=1 \end{eqnarray}$$
and
$$\begin{eqnarray}aa^{*}+cc^{*}=1\\ ab^{*}+cd^{*}=0\\ bb^{*}+dd^{*}=1 \end{eqnarray}$$
Comparing top and bottom equations in each set of three equations, we have $$\begin{eqnarray}|c|=|b|\\ |a|=|d|\end{eqnarray} $$
At this point, I tried putting $d=ae^{i\theta_1}$ and $c=be^{i\theta_2}$. Of course, since we know what form we are expecting we might as well took $d=a^{*}e^{i\theta_1}$ and $c=b^{*}e^{i\theta_2}$ (which in the end works perfectly nice, unlike my primary chocie). But the point is, once I choose $d$ and $c$ to be in that form we end up with
$$\begin{eqnarray}a^{*}be^{i\theta_2}+ab^{*}e^{i\theta_1}=0\\ ab^{*}+a^{*}be^{i(\theta_2-\theta_1)}=0\end{eqnarray} $$
which, in the end, comes down to one equation
$$ a^{*}be^{i\theta_2}+ab^{*}e^{i\theta_1}=0 $$
because second one is satisfied as long as first one is satisfied. But this doesn't lead to anything meaningful, as far as I see.
Strange thing is, had I used the other representation $d=a^{*}e^{i\theta_1}$ and $c=b^{*}e^{i\theta_2}$ both equations are satisfied nicely, but I don't see any reason why someone would go for conjugates (unless they knew what they were expecting to see) and in principle, both of these choices should work.
So what's happening here?
