I want to show that every special unitary transformation in two dimensions can be written as the matrix

$$ U = \left(\begin{array}{cc} e^{i(\delta + \varphi)}\cos\theta & i~e^{i(\delta - \varphi)} \sin\theta \\ i~e^{-i(\delta - \varphi)}\sin\theta & e^{-i(\delta + \varphi)}\cos\theta\end{array}\right),$$

where $\delta,\varphi,\theta \in\mathbb{R}$. I want to understand neutrino oscillations and my professor just wrote down this matrix without deriving it. Unfortunately, I cannot find the right way to do it myself. Any help is appreciated.

  • 1
    $\begingroup$ That's a special unitary transformation, i.e. one with determinant one, not a general unitary transformation. But that is also the simpler case -- do you know the explicit expression for the inverse of $U=\begin{pmatrix}a&b\\c&d\end{pmatrix}$? $\endgroup$ – Toffomat Jun 24 '17 at 11:55
  • $\begingroup$ There is a brief proof of a similar claim in Nielsen and Chuang's quantum information text, theorem 4.1. Essentially you show the rows are orthogonal, as are the columns, and from it follows that you can uniquely determine the decomposition of your unitary into these three parameters $\endgroup$ – QtizedQ Jun 24 '17 at 11:58
  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Jun 24 '17 at 13:24

For simplicity, suppose $U$ is special unitary so its determinant is $+1$. If you want full unitary simply multiply $U$ by an overall phase $e^{i\zeta}$.

Write $$ U=\left(\begin{array}{cc} a&b\\ c&d\end{array}\right)\, ,\qquad U^{-1}=\left( \begin{array}{cc} d & -b \\ -c & a \\ \end{array} \right)\, ,\qquad U^\dagger = \left(\begin{array}{cc} a^*&c^*\\ b^*&d^*\end{array}\right) $$ where $\hbox{Det}(U)=1$ has been used. Since $U^\dagger=U^{-1}$ this immediately shows that $d=a^*$, $c=-b^*$ so $$ U=\left(\begin{array}{cc} a&b\\ -b^*&a^*\end{array}\right) $$ The condition on the determinant is now $aa^*+bb^*=1$. In your case you chose $$ a=e^{i(\delta+\varphi)}\cos\theta\, ,\qquad ie^{i(\delta-\varphi)}\sin\theta. $$ The factor of $i$ in the off-diagonal terms and the general form of your solution is typical of the parametrization used in optics.

Of course this is not the only solution, indeed this is not the "standard" solution. A special unitary $2\times 2$ matrix is often written in the factorized form $$ e^{-i\alpha\sigma_z/2}e^{-i\beta\sigma_y/2}e^{-i\gamma\sigma_z/2} =\left( \begin{array}{cc} e^{-\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right) & -e^{-\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta }{2}\right) \\ e^{\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta }{2}\right) & e^{\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right) \\ \end{array} \right)\, . $$


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