0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.