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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?

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  • $\begingroup$ Possible duplicate of Find unitary for given rotations on Bloch sphere $\endgroup$ – Will May 4 '19 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 '19 at 23:58
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    $\begingroup$ 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 '19 at 0:03
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    $\begingroup$ @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 '19 at 0:25
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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}

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  • $\begingroup$ Oh thanks, it makes a lot of sense with the formula you gave! $\endgroup$ – Labo May 4 '19 at 23:58

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