# On the orthogonality of the spin operator in a $z$-basis

After, in the $$S_z$$-basis $$|S_{z,\pm}\rangle$$ denoted by $$|\pm\rangle$$ and in units $$\frac{\hbar}{2}$$, finding the spin operator in a general direction $$\vec{n} = (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$$:

$$\hat{\vec{S}}\cdot\vec{n} = \sin\theta\cos\phi\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} + \sin\theta\sin\phi\begin{bmatrix} 0 & -i\\ i & 0 \end{bmatrix} + \cos\theta\begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix} = \begin{bmatrix} \cos\theta & e^{-i\phi}\sin\theta\\ e^{i\phi} \sin\theta & -\cos\theta \end{bmatrix}$$ I find the eigenvectors $$|S_{\vec{n},+}\rangle = \begin{bmatrix} \cos\theta/2\\ e^{i\phi}\sin\theta/2 \end{bmatrix}\text{ and }|S_{\vec{n},-}\rangle = \begin{bmatrix} \sin\theta/2\\ -e^{i\phi}\cos\theta/2 \end{bmatrix}$$ Now it is normally quite easy to see that these should be orthogonal but for some reason I just don't get $$\langle S_{\vec{n},+}|S_{\vec{n},-}\rangle$$ to equal zero: $$\langle S_{\vec{n},+}|S_{\vec{n},-}\rangle = \cos\frac\theta2 \sin\frac\theta2\left(1-e^{2i\phi}\right)$$ what is going wrong? thanks in advance.

Take the complex conjugate when you go from bra to ket. So $$e^{i \phi} \to e^{-i \phi}$$ and the bracket is $$(1-1)=0$$