Basis spin states of spin-1/2 particles and the generator of rotations
The rotation operator \(\hat{R}(\phi\, \mathbf{k})\) rotates a ket by an angle \(\phi\) in the anticlockwise direction about the \(z-\) axis. Mathematically, it can be written as,
\[\begin{equation} \begin{aligned} \hat{R}(\phi\, \mathbf{k}) &=e^{-\frac{i}{\hbar} \phi \hat{J}_z} \\ &=\left[1+\left(-\frac{i}{\hbar} \phi \hat{J}_z\right)+\frac{1}{2 !}\left(-\frac{i}{\hbar} \phi \hat{J}_z\right)^2+\frac{1}{3 !}\left(-\frac{i}{\hbar} \phi \hat{J}_z\right)^3+\cdots\right] \\ \end{aligned} \end{equation} \label{eq: 1}\]where \(\hat{J}{_z}\) is the operator that generates rotations.
When \(\hat{R}(\phi\, \mathbf{k})\) acts on the basis state \(|+z \rangle\) it yields a new state that is a constant times the state \(|+z \rangle\). The constant is a complex number that vanishes while calculting the probability. For this reason, for all practical purposes, the new state is same as the original basis state.
Using Eq. \eqref{eq: 1}, we can write,
\[\begin{equation} \hat{R}(\phi\, \mathbf{k})|+z\rangle =\left[1+\left(-\frac{i}{\hbar} \phi \hat{J_z}\right)+\frac{1}{2 !}\left(-\frac{i}{\hbar} \phi \hat{J}_z\right)^2+\frac{1}{3 !}\left(-\frac{i}{\hbar} \phi \hat{J}_2\right)^3+\cdots\right]|+z\rangle \end{equation} \label{eq: 2}\]We want to know, under what condition, the term inside the bracket in Eq. \eqref{eq: 2} reduces to a (complex) constant. To answer this, we need to know the action of \(\hat{J}_z\) on the basis state \(|+z \rangle\).
First, let’s assume that \(|+z \rangle\) is not an eigenstate of \(\hat{J}_z\). That is, the action of \(\hat{J}_z\) on \(|+z \rangle\) yields a state that is not a constant times the state \(|+z \rangle\). Let’s say, for the sake of discussion, it yields \(|+x \rangle\). That is,
\[\begin{equation} \hat{J}_z |+z \rangle = |+ x\rangle. \end{equation} \label{eq: 3}\]With this assumption, Eq. \eqref{eq: 3} can be written as,
\[\begin{equation} \hat{R}(\phi\, \mathbf{k})|+z\rangle = |+z\rangle +\left(-\frac{i}{\hbar} \phi\right) |+x \rangle + \frac{1}{2 !}\left(-\frac{i}{\hbar} \phi \hat{J}_z\right)^2 |+z \rangle + \frac{1}{3 !}\left(-\frac{i}{\hbar} \phi \hat{J}_z\right)^3|+z\rangle + \cdots \label{eq: 4} \end{equation}\]The RHS of Eq. \eqref{eq: 4} has a term involving \(|+x \rangle\), and there is no way it will get cancelled as the other terms are in incresing powers of \(\phi\). So, the action of \(\hat{R}(\phi\, \mathbf{k})\) on \(|+z \rangle\) is resulting in a state that is not a constant times the state \(|+z \rangle\). Since this doesn’t make physical sense, our assumption that \(|+z \rangle\) is not an an eigenstate of \(\hat{J}_{z}\) must be wrong.
Now, let’s assume that \(|+z \rangle\) is an eigenstate of \(\hat{J}_z\). If \(m\) is the corresponding eigenvalue, we can write,
\[\begin{equation} \hat{J}_z |+z \rangle = m\,|+ z\rangle. \end{equation}\]Since \(|+z \rangle\) represents only the spin state of a spin-1/2 particle, it is reasonable to assume that \(m = +\frac{\hbar}{2}\) – the magnitude of the intrinsic spin angular momentum of a spin-1/2 particle represented by the ket \(|+ z \rangle\). The above equation, can now be written as,
\[\begin{equation} \hat{J}_z |+z \rangle = +\frac{\hbar}{2}\,|+ z\rangle. \label{eq: 5} \end{equation}\]From Eq. \eqref{eq: 5}, it follows that,
\[\begin{aligned} \hat{J}_z^2|+z\rangle &=\hat{J}_z \hat{J}_z|+z\rangle \\ &=\left(+\frac{\hbar}{2}\right) \hat{J}_z |+z\rangle \\ &=\left(+\frac{\hbar}{2} \right)^2|+z\rangle. \end{aligned}\]In general,
\[\begin{equation} \hat{J}_z^n |+z \rangle = \left(+ \frac{\hbar}{2}\right)^n |+z \rangle. \label{eq: 6} \end{equation}\]Combining Eqs. \eqref{eq: 2} and \eqref{eq: 6}, we get,
\[\begin{equation} \begin{aligned} \hat{R}(\phi\, \mathbf{k})|+z\rangle & =|+z\rangle + \left(-\frac{i}{\hbar} \phi \frac{\hbar}{2}\right)|+z\rangle + \frac{1}{2 !}\left(-\frac{i}{\hbar} \phi \frac{\hbar}{2}\right)^2|+z\rangle + \frac{1}{3 !}\left(-\frac{i}{\hbar} \phi \frac{\hbar}{2}\right)^3|+z\rangle + \cdots \\ % & = {\left[1+\left(-i \frac{\phi}{2}\right) + \frac{1}{2 !}\left(-i \frac{\phi}{2}\right)^2 + \frac{1}{3 !}\left(-i \frac{\phi}{2}\right)^2 + \cdots\right]|+z\rangle } \\ % & = e^{-i \frac{\phi}{2}}|+z\rangle. \end{aligned} \end{equation}\]The action of \(\hat{R}(\phi\, \mathbf{k})\) on \(|+z \rangle\) now yields a state that is a constant times \(|+z \rangle\). Hence our assumption that \(|+z \rangle\) is an eigenstate of \(\hat{J}_z\) is correct.
Hence, the basis states used to describe the spin states of a spin-1/2 particle are eigenstates of the operator that generates rotations.
Reference: A Modern Approach to Quantum Mechanics by John S Townsend.