Find unitary for given rotations on Bloch sphere 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. 
 A: 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.  
