I want to characterize a unitary by given rotations on the Bloch sphere.

I know, that when I send in the State $|\Psi\rangle =\begin{pmatrix}1\\0 \end{pmatrix}$, I get the state $U|\Psi\rangle=\begin{pmatrix}\cos \theta\\ e^{i\varphi}\sin \theta \end{pmatrix}.$

When I send in the state $|\Psi'\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1 \end{pmatrix}$, I get the state $U|\Psi'\rangle=\begin{pmatrix}\cos \theta'\\ e^{i\varphi'}\sin \theta' \end{pmatrix}.$

Rotations, the way I know them on the bloch sphere are defined by an axis $\vec{n}$ and an angle $\alpha$: $$U(\vec{n},\alpha)= \cos(\theta/2)\mathrm 1-i\sin(\theta/2) \vec{n}\vec{\sigma}$$

Shouldn't that be enough information to find the unitary $U$?

So basically I have 4 free parameters here.. I don't really know how to deal with this problem. I guess I could just try angles and axis and solve this as an optimization problem numerically. But shouldn't there be a more analytical way?

It would be very nice if someone could help me with that problem.


Your 2x2 unitary is mostly determined by its action on the state vector $\left(\begin{array}{c}1 \\0\end{array}\right)$. This is because 1) in a 2-d Hilbert space, for any given vector there is a single other orthogonal vector (up to a phase factor), and 2) a unitary map preserves orthogonality. Once $|u\rangle = U\left(\begin{array}{c}1 \\0\end{array}\right)$ is known, $|v\rangle = U\left(\begin{array}{c}0 \\1\end{array}\right)$ is also known as the orthogonal of $|u\rangle$ up to a phase factor, which means $U$ is completely determined by its action on a complete basis set. $$ \\ $$ Indeed, a unitary can always be parametrized as $$ U = \left(\begin{array}{cc} a & -b^* \\ b & a^* \end{array}\right) $$ with $a, b \in \mathbb C$, $|a|^2 + |b|^2 = 1$. In your case, its action on $\left(\begin{array}{c}1 \\0\end{array}\right)$, $$ \left(\begin{array}{cc} a & -b^* \\ b & a^* \end{array}\right)\left(\begin{array}{c}1 \\0\end{array}\right) = \left(\begin{array}{c} a \\ b \end{array}\right) $$ allows the identification $$ a = \cos\theta $$ $$ b = e^{i\phi}\sin\theta $$ which implies $$ U = \left(\begin{array}{cc} \cos\theta & -e^{- i\phi}\sin\theta \\ e^{i\phi}\sin\theta & \cos\theta \end{array}\right) $$ (Note that requiring $a=\cos\theta$ imposes $a = a^*$ for this case.)

This is then easily rearranged in the $\left(\cos\frac{\alpha}{2}\right) {\hat I}- i \left(\sin\frac{\alpha}{2}\right) \left({\vec n}\cdot{\hat{\vec \sigma}}\right)$ form: $$ U = \left(\begin{array}{cc} \cos\theta & -e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta & \cos\theta \end{array}\right) = $$ $$ = \left(\cos\theta\right) {\hat I} + \left(\sin\theta\right) \left(\begin{array}{cc} 0 & -e^{-i\phi} \\ e^{i\phi} & 0 \end{array}\right) = $$ $$ = \left(\cos\theta\right) {\hat I} + i \left(\sin\theta\right) \left(\sin\phi\right) {\hat \sigma}_x - i \left(\sin\theta\right) \left(\cos\phi\right) {\hat \sigma}_y $$ which identifies $$ \alpha = 2\theta,\;\; n_x = - \sin\phi,\;\;n_y = \cos\phi $$ In this case $n_z=0$ and the unitary represents a rotation of angle $2\theta$ around a vector ${\vec n}$ in the x-y plane.

  • $\begingroup$ Thanks for your reply, udrv! I see, you left out the $n_z$ component, and together with the normalization, there are only two parameters left. This unitary would determine all actions on input states on the equatorial plane, right? Any advise on how to find that rotation matrix if I just have input and output state? $\endgroup$ – Mechanix May 8 '16 at 11:13
  • $\begingroup$ Another question: Can this really rotate any point on the bloch sphere to any other point? $\endgroup$ – Mechanix May 8 '16 at 15:43
  • $\begingroup$ Welcome. The unitary in the answer acts on any state as a rotation of angle $2\theta$ around an axis lying in the x-y plane. I appended the clarification to the answer. In general, if you have an input state $\left(\begin{array}{c}c_1 \\c_2\end{array}\right)$ taken into an output state $\left(\begin{array}{c}d_1 \\d_2\end{array}\right)$, just assume $U$ in the $a$, $b$ parametrization above, apply to $\left(\begin{array}{c}c_1 \\c_2\end{array}\right)$ and rearrange the result into a system for $a$, $b^*$. It solves immediately. $\endgroup$ – udrv May 9 '16 at 0:04
  • $\begingroup$ To identify the rotation angle and axis you need to compare this result with the most general expression for $U$, when $n_z \neq 0$. For the latter just parametrize ${\vec n}$ in spherical coordinates and work backwards from the form $U = \cos(\alpha/2){\hat I}- i \sin(\alpha/2) \left({\vec n}\cdot{\hat{\vec \sigma}}\right)$ to the 2x2 matrix expression. $\endgroup$ – udrv May 9 '16 at 0:04
  • $\begingroup$ @udrv +1 for such clear answer and for including all the steps, great! (by the way thanks a lot for your reply, this time I tried to reply back asap :) ) $\endgroup$ – user929304 May 17 '16 at 15:45

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