# 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?

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:

1. 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

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

1. The eigenvectors must be normalized.

2. 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!