# Finding the direction of a Spin-State

Given a non normalized state, find the direction of the state $\chi$?: $$\chi = (1+i)\chi_{+}^{z}-(1+i\sqrt{3})\chi_{-}^{z}$$ Where $\chi_{+}^{z}, \chi_{-}^{z}$ are eigenstates of $S_{z}$.

I know the eigenstates of $S_{z}$ are given as $$|\chi_{+}^{z}\rangle = \cos\theta/2 |+\rangle + e^{i\phi}\sin\theta/2|- \rangle$$ $$|\chi_{-}^{z}\rangle = e^{-i\phi}\sin\theta/2 |+\rangle - \cos\theta/2|- \rangle$$ After solving $S_{n}$ = $\vec{S} \cdot \hat{n}$ where $$\hat{n} = \sin\theta\cos\phi \hat{i} + \sin\theta\sin\phi \hat{j} + \cos\theta \hat{k}$$ I was thinking of setting the state equal to up and down states and see if they equal each other but I'm not really able to solve the equation without knowing the correct $\phi$ and $\theta$

Your state is of the form $$\chi=z_+\chi_++z_-\chi_-=e^{i\phi_+}\vert z_+\vert \chi_++ e^{i\phi_-}\vert z_-\vert\chi_-$$ so first normalize it to $$\bar\chi= \frac{z_+\chi_++z_-\chi_-}{\sqrt{z_+z_+^*+z_-z_-^*}}$$ and then factor out the overall phase from the first factor $$\bar\chi= \frac{e^{i\phi_+}}{\sqrt{z_+z_+^*+z_-z_-^*}} \left(\vert z_+\vert\chi_++e^{i(\phi_--\phi_+)}\vert z_-\vert \chi_-\right)\, .$$ You can then compare directly $$\frac{\vert z_+\vert}{\sqrt{z_+z_+^*+z_-z_-^*}}\chi_++ \frac{e^{i(\phi_--\phi_+)}\vert z_-\vert}{\sqrt{z_+z_+^*+z_-z_-^*}} \chi_-=\cos(\textstyle\frac{\theta}{2})\chi_++e^{i\phi}\sin(\frac{\theta}{2})\chi_-$$