I'm studying Nielsen&Chuang Book and need some clarification on mapping arbitrary rotation operator onto geometric tranformation of vector on Bloch sphere. I thought that $R_{\vec{n}}(\theta)$ means rotation of point on Bloch sphere defined by our state vector $\psi$ around $\vec{n}$ axis for angle $\theta$. So far so good, but then I tried to geometrically "cheat" on exercise 4.40 in "universal gates" section, which includes this equation:

For arbitrary $\alpha$, $\beta$,

$E(R_{\vec{n}}(\alpha), R_{\vec{n}}(\alpha + \beta)) = |1 - exp(i\beta/2)|$, where $E(U, V) = max_{|\psi\rangle} || (U-V)|\psi\rangle||$

Okay, let's say our operators correspond to rotations of point on circle made by intersection of Bloch sphere and plane orthogonal to our rotation axis $\vec{n}$, for $\alpha$ and $\alpha + \beta$ angles, respectively. Then we need to find maximum length of difference between two resulting vectors for all $\psi$ and $\alpha$. This length is equal to length of chord $\alpha + \beta - \alpha = \beta$ angle in our circle, which radius is of course is maximum(and equal to Bloch sphere radius 1) for $\psi$ orthogonal for $\vec{n}$ i.e. our plane contains Bloch sphere center. Then our maximum chord length(simple geometry for isosceles triangle with two sides equal to 1 and angle $\beta$ between them) will be $2sin(\beta/2)$.

But this is not we see in our exercise. Actually, correct answer with some easy trigonometry transforms to $2sin(\beta/4)$, just a half of angle we found geometrically. The second strange thing is that if we substitute $\beta$ for $2\pi$ than error value in exercise answer will be equal to $|1 - exp(i\pi)| = |1-(-1)| = 2$. But how it's possible if rotations for $\alpha$ and $\alpha+2\pi$ are essentially the same? I definitely miss something, but I can't understand what...

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    $\begingroup$ does quantumcomputing.stackexchange.com/q/16533/55 address the question? $\endgroup$
    – glS
    Commented Jan 4, 2023 at 14:59
  • $\begingroup$ Not yet(cool thread anyway, thank you). Actually, I re-read definitions of $R_{\vec{x}}(\theta)$ and other rotation operators and understood that since we use ${\theta}/2$ instead of just ${\theta}$ in formulas, than, for example, $R_{\vec{x}}(2*\pi) = - R_{\vec{x}}(0)$. I understand that this is the same state up to global phase(because $-1 = exp({\pi}i)$), but why we define our rotations this way(via ${\theta}/2$) instead of just ${\theta}$? $\endgroup$ Commented Jan 9, 2023 at 10:47
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    $\begingroup$ for that, see quantumcomputing.stackexchange.com/q/4118/55 and links therein. I'm still unclear as what is actually been asked in the post though $\endgroup$
    – glS
    Commented Jan 9, 2023 at 10:48
  • $\begingroup$ Oh, this looks exactly what I needed. Thank you! If mapping vectors on Bloch sphere includes halving the angles, than all my geometrical calculations will be correct and give the same result as expected in example. $\endgroup$ Commented Jan 9, 2023 at 10:54


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