Active and passive transformations
Operator \(\hat{R}^{\dagger}\) rotates the state \(\mid \psi \rangle\) in clockwise direction to produce a new state \(\mid \psi^{\prime} \rangle\):
\[\begin{equation} \mid\psi^{\prime}\rangle=\hat{R}^{\dagger}\mid\psi\rangle. \label{eq1} \end{equation}\]In \(S_{z}\) basis, \(\mid \psi^{\prime} \rangle\) can be written as:
\[\begin{equation} \mid \psi^{\prime}\rangle=\mid + \mathbf{z} \rangle\langle + \mathbf{z} \mid \psi^{\prime} \rangle + \mid- \mathbf{z} \rangle \langle - \mathbf{z} \mid \psi^{\prime} \rangle. \label{eq2} \end{equation}\]We can write eq. \eqref{eq2} as a column vector:
\[\begin{equation} \left|\psi^{\prime}\right\rangle \underset{S_z \text { basis }}{\longrightarrow}\left(\begin{array}{c} \left\langle+\mathbf{z} \mid \psi^{\prime}\right\rangle \\ \left\langle-\mathbf{z} \mid \psi^{\prime}\right\rangle \end{array}\right). \label{eq3} \end{equation}\]By combining equations \eqref{eq2} and \eqref{eq3}, we can write:
\[\begin{equation} \mid\psi^{\prime}\rangle \underset{S_z \text { basis }}{\longrightarrow}\left(\begin{array}{l} \langle+\mathbf{z} \mid \psi^{\prime}\rangle \\ \langle-\mathbf{z} \mid \psi^{\prime}\rangle \end{array}\right)=\left(\begin{array}{c} \langle+\mathbf{z}\mid\hat{R}^{\dagger}\mid \psi\rangle \\ \langle-\mathbf{z}\mid\hat{R}^{\dagger}\mid \psi\rangle \end{array}\right). \label{eq4} \end{equation}\]For the sake cof concreteness, let’s replace the rotation operator \(\hat{R}\) by the operator \(\hat{R}\left(\frac{\pi}{2} \mathbf{j}\right)\). Now, equations \eqref{eq1} and \eqref{eq4} respectively become,
\[\begin{equation} \mid\psi^{\prime}\rangle=\hat{R}^{\dagger}\left(\frac{\pi}{2} \mathbf{j}\right)\mid\psi\rangle \label{eq5} \end{equation}\]and
\[\begin{equation} \mid\psi^{\prime}\rangle \underset{S_z \text { basis }}{\longrightarrow}\left(\begin{array}{l} \langle+\mathbf{z} \mid \psi^{\prime}\rangle \\ \langle-\mathbf{z} \mid \psi^{\prime}\rangle \end{array}\right)=\left(\begin{array}{c} \langle+\mathbf{z}\mid\hat{R}^{\dagger}\left(\frac{\pi}{2} \mathbf{j}\right)\mid \psi\rangle \\ \langle-\mathbf{z}\mid\hat{R}^{\dagger}\left(\frac{\pi}{2} \mathbf{j}\right)\mid \psi\rangle \end{array}\right). \label{eq6} \end{equation}\]Equations \eqref{eq5} and \eqref{eq6} tell us that the rotation operator \(\hat{R}^{\dagger}\left(\frac{\pi}{2} \mathbf{j}\right)\) acts to the right on the ket \(\mid \psi \rangle\). Specifically, it rotates the state \(\mid \psi \rangle\) about the \(y-\) axis by an angle of \(90^\circ\) in the clockwise direction. This type of transformation, in which the state itself is rotated, is called active transformation.
However, there is one more way of looking at eqn. \eqref{eq6}. Instead of considering \(\hat{R}^{\dagger}\) acting to the right on ket, we can consider it to be acting to the left on bras \(\langle + \mathbf{z} \mid\) and \(\langle - \mathbf{z} \mid\), leaving the state \(\mid \psi \rangle\) unaffected.
As seen in this post, when \(\hat{R}^{\dagger}\left(\frac{\pi}{2} \mathbf{j}\right)\) acts on the state \(\langle + \mathbf{z} \mid\), it rotates the state about the the \(y-\) axis by an angle of \(90^\circ\) in counterclockwise direction, resulting in the state \(\langle + \mathbf{z} \mid\).
\[\begin{equation} \langle +\mathbf{x} \mid = \langle + \mathbf{z}\mid \hat{R}^{\dagger}\left(\frac{\pi}{2} \mathbf{j}\right) \label{eq7} \end{equation}\]With this, the column vector on the right hand side of eqn. \eqref{eq6} can be written as:
\[\begin{equation} \left(\begin{array}{c} \langle +\mathbf{x} \mid \psi\rangle \\ \langle- \mathbf{x} \mid \psi\rangle \end{array}\right). \label{eq8} \end{equation}\]This is nothing but \(\mid \psi \rangle\) represented in \(S_{x}\) basis. Note that, the rotation operator left the state \(\mid \psi \rangle\) unchanged, but instead, changed the basis vectors from \(\mid \pm \mathbf{z} \rangle\) to \(\mid \pm \mathbf{x} \rangle\). Such transformations in which the basis states themselves are rotated are called passive transformations.
Reference:
A Modern Approach to Quantum Mechanics by John S. Townsend