# Geometric representation of rotation operator on Bloch sphere

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...

• does quantumcomputing.stackexchange.com/q/16533/55 address the question?
– glS
Commented Jan 4, 2023 at 14:59
• 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 ${\theta}$? Commented Jan 9, 2023 at 10:47
• 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
– glS
Commented Jan 9, 2023 at 10:48
• 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. Commented Jan 9, 2023 at 10:54