# Error bound on approximating arbitrary rotation gates

I'm working on the following problem form Nielsen and Chuang (Ex. 4.40) regarding the universality of Hadamard, phase and $$\pi/8$$ gates for single qubit gates:

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

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

Show (using the previous equality) $$\forall \epsilon >0$$ , $$\exists n$$, such that $$E(R(\alpha)_{\vec{n}}, {R(\theta)_{\vec{n}}}^n) < \frac{\epsilon}{3}$$, where $$\theta$$ is an irrational multiple of $$2\pi$$.

My attempt:

Prior to the exercise, it discusses how for any arbitrary error $$\epsilon > 0$$, we can find some multiple $$(mod$$ $$2\pi)$$ of irrational $$\theta$$ (found using the Pigeonhole principle, which I think I understand, and I'm skipping and taking as given), call it $$\theta_k \in (0, 2\pi]$$, |$$\theta_k| < \epsilon$$ such that all multiples $$(mod$$ $$2\pi)$$ of $$\theta_k$$ differ by at most $$\epsilon$$. Thus $$\forall \alpha \in [0, 2\pi)$$, $$\exists m$$, such that $$|\alpha - (m\theta_k) | =|\alpha - (mk\theta)| < \epsilon$$. Let $$n = mk$$, then

$$E(R(\alpha), {R_{\vec{n}}}(\theta)^n) = E(R_{\vec{n}}(\alpha), {R_{\vec{n}}}(n\theta)) = E(R_{\vec{n}}(\alpha), {R_{\vec{n}}}(\alpha + (n\theta -\alpha))) < E(R_{\vec{n}}(\alpha), {R_{\vec{n}}}(\alpha + \epsilon)) = |1 - exp(i\epsilon/2)| = \sqrt{((1 - cos(\epsilon/2))^2 + sin(\epsilon/2)^2} = \sqrt{2(1 - cos(\epsilon/2))} < 2$$

(which is not correct).

I feel like I'm missing something very obvious in this whole exercise. A hint would be appreciated.

• Remember to think of $\epsilon$ as small. With that, you can get a much better bound for that square root. This might put an extra upper bound on $\epsilon$ at first so you don't reach $\epsilon = 2 \pi$. But you can fix that next. Mar 23, 2020 at 20:33
• @AHusain oh wow, small angle approximation. So with small angle approximation, it becomes : $\sqrt{2(1-cos(\epsilon/2))} \approx \sqrt{2(1-(1 - (\epsilon/2)^2/2))} = \epsilon/2$ This means $\epsilon$ is approximately less than $\frac{\pi}{4}$. It doesn't look like it can be made tighter. However, I get that $\epsilon$ can technically be made smaller than $\frac{\pi}{4}$. Mar 23, 2020 at 22:20

Firstly, I think there's a reason why the bit in the textbook before the question is using $$\delta$$ instead of $$\epsilon$$ for the results. So, replacing what you've written, it should really be $$E<\sqrt{2(1-\cos(\delta/2))}$$
Now, start with a double-angle formula: $$1-\cos(\delta/2)=2\sin^2(\delta/4)$$. Thus, $$E<2\sin(\delta/4)$$ Now if you apply the small $$\epsilon$$ approximation (which should be more obvious following that manipulation), you get $$E<\delta/2.$$ Finally, if you want $$E<\delta/2$$, you simply have to select $$\delta=2\epsilon/3$$.
You might ask why you want the factor of $$1/3$$ in there at all? Why not just leave it as $$1/2$$? This point is that there's going to be a sequence of three of these rotations, and you want to bound the final error by $$\epsilon$$.