# Why the Ry rotation matrix give counterclockwise rotation?

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?

It should be $$S^{-1}R_yS$$ instead of $$SR_yS^{-1}$$. Change from one basis into another basis, you found the right matrix $$S$$ while the wrong method. For a pedagogical method about change basis(not only in quantum mechanics), you may refer to this link.