Eigenvectors of \(\hat{n}\cdot\vec{\sigma}\)
Quantum Mechanics, Scattering
The straightforward way to find the eigenvectors of \(\hat{n}\cdot\vec{\sigma}\) would be to use the usual method for finding eigenvalues and then the eigenvectors. Let us try to solve the problem using another method. We have \(\hat{n}=\sin\theta\cos\phi\,\hat{x}+\sin\theta\sin\phi\,\hat{y}+\cos\theta\,\hat{z}\). Assume we start with \(\hat{n}\) pointing along \(\hat{z}\), so the state is \(|\hat{z}_{up}\rangle=\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)\) which is an eigenvector of the \(\vec{S}\cdot\hat{n}\) operator with eigenvalue \(1\). Let us rotate the state \(|\hat{z}_{up}\rangle\) around \(\hat{y}\) by angle \(\theta\) which can be done by acting with the operator; \[\begin{equation} e^{-i \sigma_y \theta/2}=\left( \begin{array}{cc} \cos(\frac{\theta}{2}) &-\sin(\frac{\theta}{2}) \\ \sin(\frac{\theta}{2}) & \cos(\frac{\theta}{2}) \\ \end{array} \right). \label{eq:ry} \end{equation}\] You can check that above equation is correct by Taylor expanding the \(e^{-i \sigma_y \theta/2}\), or you can visualize the effect as rotating a vector around \(\hat{y}\) by angle \(\theta\) keeping in mind that this is not really a vector (spin-1 particle), but it is a spinor (spin 1/2), which is reflected by the fact that we have \(\frac{\theta}{2}\) instead of \(\theta\). Next task is to rotate again, around the \(\hat{z}\) by angle \(\phi\) which can be done by acting with the operator; \[\begin{equation} e^{-i \sigma_z \phi/2}=\left( \begin{array}{cc} e^{-i \frac{\phi}{2}} &0 \\ 0& e^{i \frac{\phi}{2}} \\ \end{array} \right).\label{eq:rz} \end{equation}\] The composite operator becomes \[\begin{eqnarray} e^{-i \sigma_z \phi/2} e^{-i \sigma_y \theta/2} &=& \left( \begin{array}{cc} e^{-i \frac{\phi}{2}} &0 \\ 0& e^{i \frac{\phi}{2}} \\ \end{array} \right) \left( \begin{array}{cc} \cos(\frac{\theta}{2}) &-\sin(\frac{\theta}{2}) \\ \sin(\frac{\theta}{2}) & \cos(\frac{\theta}{2}) \\ \end{array} \right)\nonumber \\ &=& \left( \begin{array}{cc} e^{-i \frac{\phi}{2}} \cos(\frac{\theta}{2}) & -e^{-i \frac{\phi}{2}}\sin(\frac{\theta}{2}) \\ e^{i \frac{\phi}{2}}\sin(\frac{\theta}{2}) & e^{i \frac{\phi}{2}}\cos(\frac{\theta}{2}) \\ \end{array} \right). \label{eq:expsig} \end{eqnarray}\] The eigenvectors can be recovered as \[\begin{eqnarray} |\hat{n}+\rangle &=& e^{-i \sigma_z \phi/2} e^{-i \sigma_y \theta/2}|\hat{z}_{up}\rangle= \left( \begin{array}{c} e^{-i \frac{\phi}{2}} \cos(\frac{\theta}{2}) \\ e^{i \frac{\phi}{2}}\sin(\frac{\theta}{2}) \\ \end{array} \right),\nonumber\\ |\hat{n}-\rangle &=& e^{-i \sigma_z \phi/2} e^{-i \sigma_y \theta/2}|\hat{z}_{down}\rangle= \left( \begin{array}{c} -e^{-i \frac{\phi}{2}} \sin(\frac{\theta}{2}) \\ e^{i \frac{\phi}{2}}\cos(\frac{\theta}{2}) \\ \end{array} \right).\label{eq:evec} \end{eqnarray}\]
In order to find \(\langle\hat{n}\pm|\vec{S}|\hat{n}\pm\rangle\) we can use the above method to express \(|\hat{n}\pm\rangle\) in terms of \(|\hat{z}_{u,d}\rangle\). \[\begin{equation} \langle\hat{n}\pm|\vec{S}|\hat{n}\pm\rangle=\langle\hat{z}_{u,d}|e^{i \sigma_y \theta/2}e^{i \sigma_z \phi/2} \vec{S}e^{-i \sigma_z \phi/2} e^{-i \sigma_y \theta/2}|\hat{z}_{u,d}\rangle. \label{eq:estates} \end{equation}\] To simplify the relation, we will compute the object \(e^{i \sigma_j \alpha/2}\sigma_k e^{-i \sigma_j \alpha/2}\) where we will assume \(k\neq j\) (if \(k=j\), we can move \(\sigma_k\) through the exponentials to get \(\sigma_k\)). Consider \(k\neq j\) case: \[\begin{eqnarray} e^{i \sigma_j \alpha/2}\sigma_k e^{-i \sigma_j\alpha/2} &=& \left(I \cos(\frac{\alpha}{2})+i \sigma_j \sin(\frac{\alpha}{2})\right) \sigma_k \left(I \cos(\frac{\alpha}{2})-i \sigma_k \sin(\frac{\alpha}{2})\right)\nonumber\\ &=& \cos\alpha\sigma_k -\sin\alpha \epsilon_{j k m} \sigma_m =\left( \cos\alpha\delta_{k m}+\sin\alpha \epsilon_{k j m}\right)\sigma_m\nonumber\\ &\equiv& R^{(j)}_{k m}(\alpha)\sigma_m. \label{eq:expsig2} \end{eqnarray}\] This equation is nothing but the rotation equation for the vector \(\vec{\sigma}\) around the \(j\)-axis. This tells us that \(\vec{\sigma}\) indeed transforms like a vector, this is why it has a vector arrow on top! Now the problem becomes easier, \[\begin{eqnarray} \langle\hat{n}\pm|S_k|\hat{n}\pm\rangle&=&\langle\hat{z}_{u,d}| e^{i \sigma_y \theta/2}e^{i \sigma_z \phi/2} S_k e^{-i \sigma_z \phi/2} e^{-i \sigma_y \theta/2}|\hat{z}_{u,d}\rangle \nonumber\\ &=&\langle\hat{z}_{u,d}|e^{i \sigma_y \theta/2} R^{(z)}_{k m}(\phi)S_m e^{-i \sigma_y \theta/2}|\hat{z}_{u,d}\rangle \nonumber\\ &=&R^{(z)}_{k m}(\phi)R^{(y)}_{m n}(\theta)\langle\hat{z}_{u,d}|S_n|\hat{z}_{u,d}\rangle\nonumber\\ &=&\pm\frac{1}{2}R^{(z)}_{k m}(\phi)R^{(y)}_{m 3}(\theta). \label{eq:estates2} \end{eqnarray}\] We need to keep in mind that \(R^{(j)}_{k m}(\alpha)=\delta_{k m}\) for \(j=k\). Componentwise we get \[\begin{eqnarray} \langle\hat{n}\pm|S_3|\hat{n}\pm\rangle &=& \pm\frac{1}{2}R^{(z)}_{3 m}(\phi)R^{(y)}_{m 3}(\theta)= \pm\frac{1}{2}\delta_{3 m} R^{(y)}_{m 3}(\theta) =\pm\frac{1}{2}R^{(y)}_{3 3} =\pm\frac{1}{2}\cos\theta\nonumber, \\ \langle\hat{n}\pm|S_2|\hat{n}\pm\rangle &=& \pm\frac{1}{2}R^{(z)}_{2 m}(\phi)R^{(y)}_{m 3}(\theta)= \pm\frac{1}{2}\sin\theta\sin\phi,\nonumber\\ \langle\hat{n}\pm|S_1|\hat{n}\pm\rangle &=& \pm\frac{1}{2}R^{(z)}_{1 m}(\phi)R^{(y)}_{m 3}(\theta)= \pm\frac{1}{2}\sin\theta\cos\phi. \label{eq:estates3} \end{eqnarray}\] And these results can be combined into \(\langle\hat{n}\pm|\vec{S}|\hat{n}\pm\rangle =\pm\frac{1}{2}\hat{n}\) As one can argue, this is not the fastest method to solve the problem, however it provides insights to \(\sigma\)- matrices and shows why they deserve the arrow on top. This comes from the fact that structure constants (\(\epsilon_{i j k}\)) in the fundamental representation of \(SU(2)\) group (the group of \(2\times 2\) matrices generated by \(\sigma\)-matrices), become the generators of the adjoint representation, i.e., the usual vector space.