Diagonalize a dot product with Pauli matrices How can I diagonalize the following operator?
$$\lambda \hat{\vec{\sigma}}\cdot\vec{r}$$
where $\lambda$ is a real constant, $\hat{\vec{\sigma}}=(\hat{\sigma_{x}},\hat{\sigma_{y}},\hat{\sigma_{z}}) $ is the Pauli operator and $\vec{r}=(x,y,z)$ is the position operator.
 A: This operator commutes with $\vec{p} = (p_x,p_y,p_z)$ (an arbitrary vector operator in orbital space) so lets use as basis $|\vec{p} , \pm  \rangle $, with $\pm$ being the spin eigenvalue. The matrix in this basis takes the form:
$
\lambda \left ( \begin{matrix}
p_{z} & p_{-}\\ 
p_{+} & -p_{z}
\end{matrix} \right )
$
With $p_{+} = p_x+ip_y, p_{-}=p_{+}^{*}$. The determinant is $-\lambda^2p^2$ and the trace $0$, giving us eigenvalues of $\pm \lambda p$. Knowing this, lets translate our vector $\vec{p}$ to spherical coordinates:
$\lambda p\left ( \begin{matrix}
\cos\theta_{p} & \sin\theta_{p}\mathrm{e}^{-i\phi_{p}}\\ 
\sin\theta_{p}\mathrm{e}^{i\phi_{p}} & -\cos\theta_{p}
\end{matrix} \right )$
Diagonalization and vectors proceed as usual for a matrix in this form.
$|\vec{p} , +\lambda p  \rangle = \cos\left(\frac{\theta}{2}\right)\mathrm{e}^{-i\frac{\phi}{2}} |\vec{p} , +  \rangle + \sin\left(\frac{\theta}{2}\right)\mathrm{e}^{i\frac{\phi}{2}} |\vec{p} , -  \rangle $
$|\vec{p} , -\lambda p  \rangle = -\sin\left(\frac{\theta}{2}\right)\mathrm{e}^{-i\frac{\phi}{2}} |\vec{p} , +  \rangle + \cos\left(\frac{\theta}{2}\right)\mathrm{e}^{i\frac{\phi}{2}} |\vec{p} , -  \rangle $
