The classical representation of a state on the Bloch sphere is given by $|\psi\rangle = \cos \frac{\theta}{2} + e^{i \varphi} \sin \frac{\theta}{2}$.

If I want to apply a rotation around the $Y$-axis, the rotation matrix is given by

$$R_Y (\tau) = \left( \begin{matrix} \cos \frac{\tau}{2} & -\sin \frac{\tau}{2} \\ \sin \frac{\tau}{2} & \cos \frac{\tau}{2} \end{matrix} \right)$$


$$\left( \begin{matrix} \cos \frac{\tau}{2} & -\sin \frac{\tau}{2} \\ \sin \frac{\tau}{2} & \cos \frac{\tau}{2} \end{matrix} \right) \begin{pmatrix} \cos \frac{\theta}{2} \\ e^{i \varphi} \sin \frac{\theta}{2} \end{pmatrix} = \begin{pmatrix} \cos \frac{\tau}{2} \cos \frac{\theta}{2} -\sin \frac{\tau}{2} \sin \frac{\theta}{2} e^{i \varphi} \\ \sin \frac{\tau}{2} \cos \frac{\theta}{2} + \cos \frac{\tau}{2} \sin \frac{\theta}{2} e^{i \varphi}\end{pmatrix}$$

I would like to express this transformed state as the classical representation i.e., $|\psi^\prime\rangle = \cos \frac{\theta^\prime}{2} + e^{i \varphi^\prime} \sin \frac{\theta^\prime}{2}$

Is it possible? Any idea how to do it? Transform the original state in the $|+\rangle$ basis, then apply the $R_Y$ rotation which is simple (as the $R_Z$ rotation on the original state), and finally transform it back to the $|0\rangle$ basis.

Thanks for any clue.


1 Answer 1


Of course, it is possible, but it's unpleasant!

The most inportant think to do at the start is to remove a global phase from the state. In other words, you equate $$ e^{i\gamma}\cos\frac{\theta'}{2}=\cos\frac{\tau}{2}\cos\frac{\theta}{2}-\sin\frac{\tau}{2}\sin\frac{\theta}{2} e^{i\varphi}. $$ Split this up using the modulus and the argument to get the two different parts.

Once you have that, you can solve the second part $$ e^{i\gamma}e^{i\varphi'}\sin\frac{\theta'}{2}=\sin\frac{\tau}{2}\cos\frac{\theta}{2}+\cos\frac{\tau}{2}\sin\frac{\theta}{2} e^{i\varphi}. $$ The modulus should automatically be correct, so it's just the argument you're looking at to find $\varphi'$.

  • $\begingroup$ Effectively, the path may be long and unpleasant, if I have to find the value of $\theta^\prime$ for which the magnitude $\cos \frac{\theta^\prime}{2}$ must be equal to the magnitude $$\sqrt{(\cos \frac{\tau}{2} \cos \frac{\theta}{2} - \sin \frac{\tau}{2} \sin \frac{\theta}{2} \cos \varphi)^2 + (\sin \frac{\tau}{2} \sin \frac{\theta}{2} \sin \varphi)^2}$$. There is no royal road ... I will look how to simplify it in the coming days. Thanks for your help. $\endgroup$
    – JMark
    Aug 16, 2023 at 12:59
  • $\begingroup$ which probably simplifies to $\sqrt{\cos\frac{\tau-\theta}{2}-\frac12\cos\tau\sin\theta\cos\varphi}$ $\endgroup$
    – DaftWullie
    Aug 17, 2023 at 1:33

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