How do I compute the square root of the $Y$ gate?



I am trying to compute the square root of the Y gate. $$Y_\theta =\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

There are many ways for doing this. I am keen on just doing the good old diagonalization and taking the square root of the eigenvalues. I compute the determinant of $Y-\lambda I$ and I set it to zero, and I get: $(cos\theta-\lambda)^2 + sin^2\theta=0$. This cannot have solutions, as a square cannot be negative.

What am I doing wrong?

Nicholas Sathripa

Posted 2020-04-25T01:17:28.843

Reputation: 162

5A square can be negative if it's a square of a complex number. Rotation gates are not self-adjoint, so they can have complex eigenvalues – Mariia Mykhailova – 2020-04-25T01:50:11.330



There are various ways of finding the square root of a gate. As the Pauli matrix $Y$ is self adjoing, we can use the Euler formula: $$R_y(\theta) = e^{-i\frac{\theta}{2}Y} = \cos\frac{\theta}{2}I -i \sin\frac{\theta}{2}Y $$

Now is easy to see that $\sqrt{R_y(\theta)}$ is just $R_y(\theta/2)=e^{-i\frac{\theta}{4}Y} $. This gives $$\sqrt{R_y(\theta)}=\begin{pmatrix} \cos(\theta/4) & -\sin(\theta/4) \\ \sin(\theta/4) & \cos(\theta/4) \end{pmatrix}$$

It is simple to check (using various trigonometric identities) that this is the correct answer. To match the angles in the question, it suffices to take $\theta/2$ instead of $\theta$, i.e.:

$$\sqrt{R_Y(\theta)} =\begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2) \\ \sin(\theta/2) & \cos(\theta/2) \end{pmatrix}$$


Posted 2020-04-25T01:17:28.843

Reputation: 403