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It is well known that any $2\times 2$ unitary matrix $\mathbf{U}$ can be parametrized as $$\mathbf{U}=\begin{pmatrix} 1 & 0 \\ 0 & \mathrm{e}^{\mathrm{j}\beta_1}\end{pmatrix} \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{pmatrix}\begin{pmatrix} \mathrm{e}^{\mathrm{j}\beta_2}\ & 0 \\ 0 & \mathrm{e}^{\mathrm{j}\beta_3}\end{pmatrix}.$$

The question is, if $\mathbf{U}$ is Haar-distributed, what is the distribution of $\alpha$?

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The distribution you are seeking is: $$P(\alpha)=\sin(2\alpha),\;\;\alpha\in[0,\pi/2]$$

The general way to derive the Haar measure for any parameterization of $U$ in terms of variables $\alpha_1,\alpha_2,\ldots$, is to calculate the metric tensor $$g_{mn}=-{\rm Tr}\,U^\dagger(\partial U/\alpha_m)U^\dagger(\partial U/\partial\alpha_n)$$ Then the distribution $P(\alpha_1,\alpha_2,\ldots)$ follows from $P\propto\sqrt{{\rm Det}\,g}$.

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  • $\begingroup$ Carlo, thanks for your answer. It fits well our observations. Would you mind recommending some good references where I can learn more on this topic? $\endgroup$
    – confused guy
    Commented Nov 26, 2014 at 11:58
  • $\begingroup$ well, there is not much more than I write here; if you are interested in the Haar measure for a variety of parameterizations of different compact groups, you might take a look at Appendix B.1 of arxiv.org/abs/1405.3115 $\endgroup$
    – Carlo Beenakker
    Commented Nov 26, 2014 at 12:36

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