# Calculate the square root of Euler angles

I am trying to find a nice way to represent the square root of an arbitrary single qubit unitary to implement Lemma 6.1 from this paper

Given the Euler angles: $$R_z(a)R_y(b)R_z(c) = \left(R_z\left(a'\right)R_y\left(b'\right)R_z\left(c'\right)\right)^2$$

Is there a closed form expression that relates the angles $$a',\, b',\, c'$$ to the angles $$a,b,c$$?

I have tried some simultaneous equations after I do the Euler angle exchange to have:

$$R_z(a)R_y(b)R_z(c) = R_z(a') [R_z(A)R_y(B)R_z(C)] R_z(c')$$, where $$A,B,C$$ can be related to $$a,b,c$$ via some nasty expressions found eg. on p4 of this paper.

But from there I cannot find a closed form expression for all of $$a',b',c'$$.

This seems like the sort of problem that must have been solved before - or at least it must be known whether a nice relation exists - but I cannot find work on it. Is it possible to go via quaternions or something?

• Have you written down the 4 equations you get from writing both sides as matrices and reading off each entry? If so, include in the question. – AHusain Jun 27 at 19:05

If it were me, I'd go via the fact that you can describe any $$U$$ in the form $$U=e^{i\alpha}e^{\theta\underline{n}\cdot\underline{\sigma}}.$$ There are already questions on Stack Exchange about how to set up the relation between the Euler angles and this Bloch vector representation (this one gets you most of the way there). The point is that the square root is easy to take in this representation: $$\sqrt{U}=e^{i\alpha/2}e^{\frac{\theta}{2}\underline{n}\cdot\underline{\sigma}}$$ from which you can calculate the Euler angles. I doubt the relation is going to be particularly nice (as you can perhaps tell, because I'm not giving you the answer directly), except in exceptional circumstances such as when the rotation is about an axis of the form $$(n_x,n_y,0)$$ (in which case you'll have $$c=-a=c'=-a$$ and $$b'=b/2$$)
For a $$2\times 2$$ unitary $$U$$, you can perform an eigendecomposition: for $$\Lambda = \begin{pmatrix}\lambda_+&0\\0&\lambda_-\end{pmatrix} = \begin{pmatrix}e^{i\phi_+}&0\\0&e^{i\phi_-}\end{pmatrix}$$ and $$V = \begin{pmatrix}v_+&v_+\end{pmatrix},$$ where $$\lambda_\pm$$ are the eigenvalues and $$v_\pm$$ the corresponding eigenvectors, $$U$$ can be written as $$U = V\Lambda V^\dagger$$ and the square root of $$U$$ can be written as $$U = V\sqrt{\Lambda}V^\dagger$$ as $$V$$ is also unitary and $$\sqrt{\Lambda} = \begin{pmatrix}\sqrt{\lambda_+}&0\\0&\sqrt{\lambda_-}\end{pmatrix} = \begin{pmatrix}e^{\frac{1}{2}i\phi_+}&0\\0&e^{\frac{1}{2}i\phi_-}\end{pmatrix}.$$
Another method, adapted from this Wiki page is that, denoting $$U = \begin{pmatrix}\alpha&\beta\\-e^{i\varphi}\beta^*&e^{i\varphi}\alpha^*\end{pmatrix}$$ for an arbitrary unitary (special unitary when $$\varphi = 0, 2\pi$$) transformation, which has determinant $$e^{i\varphi}$$ and trace $$\alpha+e^{i\varphi}\alpha^*$$, a square root of $$U$$ can be written as (when $$\alpha+e^{i\varphi}\alpha^*+2e^{\frac{i}{2}\varphi}\neq0$$) $$\frac{1}{\sqrt{\alpha+e^{i\varphi}\alpha^*+2e^{\frac{i}{2}\varphi}}}\left(U+e^{\frac{i}{2}\varphi}I_2\right) = \frac{1}{\sqrt{\alpha+e^{i\varphi}\alpha^*+2e^{\frac{i}{2}\varphi}}}\begin{pmatrix}\alpha+e^{\frac{i}{2}\varphi}&\beta\\-e^{i\varphi}\beta^*&e^{i\varphi}\alpha^*+e^{\frac{i}{2}\varphi}\end{pmatrix}.$$