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:


Then, I exert $R'_{y}(\delta)$ on $|\psi\rangle$ written this way,


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!

  • $\begingroup$ Remember that rotating the Bloch vector and rotating the Bloch sphere are inverse operations, so you must be careful with sources telling you they implement a rotation: are they rotating the vector or the sphere (ie, does one rotate a vector or the basis in which the vector is defined)? I tend to rotate the vector; see this answer for context quantumcomputing.stackexchange.com/a/22250/15820 $\endgroup$ Dec 29, 2021 at 20:00

1 Answer 1


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.

  • $\begingroup$ Thanks! Problem solved. $\endgroup$
    – QubitTy
    Dec 29, 2021 at 16:19

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