0
$\begingroup$

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

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

Then,

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

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $\varphi'$.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.