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

enter image description here

$(S.1)$ 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, Phonon May 5 at 22:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Norbert Schuch, ZeroTheHero, GiorgioP, Cosmas Zachos, Phonon
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Possible duplicate of Find unitary for given rotations on Bloch sphere $\endgroup$ – Will May 4 at 23:50
  • $\begingroup$ @Will actually, the question you linked contains the answer to my question (in the question, not the answers). So I guess it is different? $\endgroup$ – Labo May 4 at 23:58
  • 2
    $\begingroup$ 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. $\endgroup$ – Will May 5 at 0:03
  • 1
    $\begingroup$ @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! $\endgroup$ – Labo May 5 at 0:25

A simple calculation shows that (see for example J.J. Sakurais' Modern Quantum Mechanics 2nd ed. eq. 3.2.44)

\begin{equation} \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) \end{equation}

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}

  • $\begingroup$ Oh thanks, it makes a lot of sense with the formula you gave! $\endgroup$ – Labo May 4 at 23:58

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