The y-axis centered rotation matrix is $R_{y}(\delta)=\left[\begin{matrix} \cos \frac{\delta}{2} & -\sin \frac{\delta}{2} \\ \sin \frac{\delta}{2} & \cos \frac{\delta}{2} \end{matrix} \right]$.
I tried to use Bloch sphere to understand why this rotation matrix $R_{y}(\delta)$ would rotate the qubit state around y-axis with a counterclockwise $\delta$.
Here is what I do:
Firstly, I change the basis into eigenstates of $\sigma_{y}$, and transform the rotation matrix $R_{y}(\delta)$ into the new form $R'_{y}(\delta)$ with new basis:
$$R'_{y}(\delta)= S R_{y}(\delta) S^{-1}= \frac{1}{\sqrt{2}}\left[\begin{matrix} -i & 1 \\ 1 & -i \end{matrix} \right]\left[\begin{matrix} \cos \frac{\delta}{2} & -\sin \frac{\delta}{2} \\ \sin \frac{\delta}{2} & \cos \frac{\delta}{2} \end{matrix} \right]\frac{1}{\sqrt{2}}\left[\begin{matrix} i & 1 \\ 1 & i \end{matrix} \right]=\exp(i\frac{\delta}{2}) \left[\begin{matrix} 1 & 0 \\ 0 & \exp(-i\delta) \end{matrix}\right]$$
Similar to the $z$ bases case when we use points on the Bloch sphere to describe qubit states, an arbitrary qubit state under $y$ bases can be written as:
$$|\psi\rangle=\cos\frac{\theta}{2}|+\rangle_{y}+\sin\frac{\theta}{2}\exp(i\phi)|-\rangle_{y}$$
Then, I exert $R'_{y}(\delta)$ on $|\psi\rangle$ written this way,
$$R'_{y}(\delta)|\psi\rangle=\exp(i\frac{\delta}{2})[\cos\frac{\theta}{2}|+\rangle_{y}+\sin\frac{\theta}{2}\exp[i(\phi-\delta)]|-\rangle_{y}]$$
Here is my question: if it's counterclockwise rotation, it should be ($\phi+\delta$) above. Why it's $(\phi-\delta$)?
Is it because I wrongly use the Bloch interpretation with y basis?
Thank you for your help!
