Eigenvectors for spin matrix along arbitrary direction In Appendix A of this paper, the authors start from (equ. A1a and A1b)
$$\sigma \cdot \hat{n} |\hat{n},+\rangle = |\hat{n},+\rangle $$
$$\sigma \cdot \hat{n} |\hat{n},-\rangle = -|\hat{n},-\rangle $$
and claim that this system of linear equations is solved by (A2a and A2b):
$$|\hat{n},+\rangle = \cos(\frac{\theta}{2})e^{-i\phi/2}|+\rangle+\sin(\frac{\theta}{2})e^{i\phi/2}|-\rangle$$
$$|\hat{n},-\rangle = -\sin(\frac{\theta}{2})e^{-i\phi/2} |+\rangle + \cos(\frac{\theta}{2})e^{i\phi/2}|-\rangle$$
I do not get how they did this. In order to solve it by myself I assumed that what they meant by $\sigma \cdot \hat{n}$ was:
$$\sigma \cdot \hat{n} = \begin{bmatrix}
\cos(\phi) & \sin(\phi)e^{-i\theta} \\
\sin(\phi)e^{i\theta} & -\cos(\phi)
\end{bmatrix}$$
as e.g. shown in the answer here. Therefore $|\hat{n},+\rangle$ and $|\hat{n},-\rangle$ should simply be the eigenvectors of this matrix. I computed the eigenvectors here, but this does not yield the same result as in the paper cited above (I tried trigonometric identities but this led me nowhere). As far as I understood $|+\rangle$ and $|-\rangle$ are supposed to be the eigenvectors of the Pauli matrix $\sigma_z$, since they are ''states of spin-up and -down along a specified direction, commonly $\hat{z}$''. [i.e. the vectors are supposed to be $|+\rangle = (1,0)$ and $|-\rangle = (0,1)$, which is the basis of the matrix I assumed above. Therefore I don't see where my error can be found.]
What did I do wrong? How can I derive $|\hat{n},+\rangle$ and $|\hat{n},-\rangle$ correctly?
 A: First, note that your $\vec{\sigma}\cdot\hat{n}$ indicates that you are using $\phi$ for the polar angle and $\theta$ for the azimuthal one, while in the solution it is defined the opposite way.
Then, some hints:

*

*Using trigonometric identities you can get

$$\cot\phi-\csc\phi=\frac{\cos\phi-1}{\sin\phi}=-\frac{2}{\sin\phi}\frac{1-\cos\phi}{2}=-\frac{2}{2\sin\left(\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right)}\sin^2\left(\frac{\phi}{2}\right)=-\frac{\sin\left(\frac{\phi}{2}\right)}{\cos\left(\frac{\phi}{2}\right)}$$
and
$$\cot\phi+\csc\phi=\frac{\cos\phi+1}{\sin\phi}=\frac{2}{\sin\phi}\frac{1+\cos\phi}{2}=\frac{2}{2\sin\left(\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right)}\cos^2\left(\frac{\phi}{2}\right)=\frac{\cos\left(\frac{\phi}{2}\right)}{\sin\left(\frac{\phi}{2}\right)}$$


*The eigenvectors must be normalized.


*You can multiply a vector by a gobal phase $e^{i\gamma}$.
I think you should be able to get the result with this, and don't forget the issue with the definition of angles!
