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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.

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    $\begingroup$ 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. $\endgroup$ – AHusain Mar 23 '20 at 20:33
  • $\begingroup$ @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 $\frac{\pi}{4}$. $\endgroup$ – dylan7 Mar 23 '20 at 22:20
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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 $\epsilon$.

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