# How to compute the axis of a rotation expressed as a combination of Pauli matrices? [closed]

An exercise asked to compute the axis and the angle of the rotation $$THTH$$.

$$(S.1)$$ is easy to understand by using the identity $$exp(i\theta A) = \cos(\theta) \mathbb1 + i \sin(\theta) A$$ for $$A$$ s.t. $$A^2 = \mathbb1$$.

What I don't get is how one can deduce from the expression of $$U_1$$ its axis and angle (and even that it is a rotation? I guess it is deduced from the expression $$THTH$$).

The only way I see to find the axis is to compute the kernel of $$(U_1-\mathbb1)$$ in the basis $$(|0>, |1>)$$, then express it on the Bloch sphere. I have no idea how to find the angle.

Is there a result that uses the fact that $$I, X, Y, Z$$ is a base for the hermitian operators, and directly gives the axis and angle as a function of the coefficients of the matrix in this base?

## closed as off-topic by Norbert Schuch, ZeroTheHero, GiorgioP, Cosmas Zachos, PhononMay 5 at 22:13

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• Possible duplicate of Find unitary for given rotations on Bloch sphere – Will May 4 at 23:50
• @Will actually, the question you linked contains the answer to my question (in the question, not the answers). So I guess it is different? – Labo May 4 at 23:58
• haha fair point, I'll retract the flag. Does the answer below explain sufficiently for you? Going through the derivation of the first expression is a good exercise, i.e., if you understand why $$\exp\left(-i\frac{\hat{n}\cdot\vec{\sigma}}{2}\theta\right)$$ is the correct expression for a rotation on a single qubit, then all will be clear to you :) The derivation can be found in most intro quantum texts: should be in Sakurai if I remember right. – Will May 5 at 0:03
• @Will thanks :) Yes, that explains it! I had seen the formula once (in Chuang & Nielsen) with the exponential and it seems "logical" so I just ignored it as a trivial fact. But the expansion is indeed how the teacher identified the axis and angle immediately. Thanks again for your help! – Labo May 5 at 0:25

$$$$\exp{\left(-i\frac{\mathbf{n\cdot\sigma}}{2}\theta\right)} = \mathbf{I}\cos\left(\theta/2\right)- i\mathbf{n\cdot\sigma}\sin\left(\theta/2\right)$$$$
This is the rotation operator for spin 1/2 systems in the $$\{\lvert + \rangle, \lvert - \rangle\}$$ basis (i.e. $$\hat{S_z}$$ basis), where $$\mathbf{n}$$ is the axis of rotation and $$\theta$$ the rotation angle.
Comparing this identity with $$U_1$$ gives the result: $$\cos\left(\theta/2\right)=\cos^2\left(\pi/8\right)$$ which implies $$\sin\left(\theta/2\right)=\sqrt{1-\cos^4\left(\pi/8\right)}=\sqrt{(1-\cos^2\left(\pi/8\right))(1+\cos^2\left(\pi/8\right))}$$ and
$$\begin{eqnarray} \mathbf{n\cdot\sigma}\sin(\theta/2) &=& \mathbf{n\cdot\sigma} \sqrt{(1- \cos^2\left(\pi/8\right))(1+\cos^2\left(\pi/8\right))} = \sin\left(\pi/8\right)\left(\cos\left(\pi/8\right)(\sigma_x+\sigma_z)+\sin\left(\pi/8\right)\sigma_y\right) \\ \iff \mathbf{n} &=& \frac{\sin\left(\pi/8\right)}{\sqrt{(1-\cos^2\left(\pi/8\right))(1+\cos^2\left(\pi/8\right))}}\left(\cos\left(\pi/8\right),\sin\left(\pi/8\right),\cos\left(\pi/8\right)\right)\\ &=& \frac{\sqrt{1-\cos^2\left(\pi/8\right)}}{\sqrt{(1-\cos^2\left(\pi/8\right))(1+\cos^2\left(\pi/8\right))}}\left(\cos\left(\pi/8\right),\sin\left(\pi/8\right),\cos\left(\pi/8\right)\right) \\ &=& \frac{1}{\sqrt{1+\cos^2\left(\pi/8\right)}}\left(\cos\left(\pi/8\right),\sin\left(\pi/8\right),\cos\left(\pi/8\right)\right) \\ \end{eqnarray}$$