In the section Performance and requirements of the phase estimation algorithm of Page 224, Quantum Computation and Quantum Information by Nielsen and Chuang

Let $b$ be the integer in the range $0$ to $2^t − 1$ such that $b/2^t = 0.b_1\cdots b_t=\frac{b_1}{2}+\frac{b_2}{2^2}\cdots\frac{b_t}{2^t}$ is the best $t$ bit approximation to the required phase $φ$, which is less than $φ$. \begin{align} b&=b_12^{t-1}+b_22^{t-2}+\cdots+b_t2^0=b_1b_2\cdots b_t\\ &\implies\frac{b}{2^t}={b_1}2^{-1}+{b_2}2^{-2}\cdots{b_t}2^{-t}=0.b_1b_2\cdots b_t\\ \phi&=0.\phi_1\phi_2\cdots\phi_t\phi_{t+1}\cdots\\ &=\phi_12^{-1}+\phi_22^{-2}+\cdots+\phi_{t-1}2^{-(t-1)}+\phi_t2^{-t}+\phi_{t+1}2^{-(t+1)}+\cdots\\ &=b_12^{-1}+b_22^{-2}+\cdots+b_{t-1}2^{-(t-1)}+\phi_t2^{-t}+\phi_{t+1}2^{-(t+1)}+\cdots\\ &\implies 0\leq\delta=\phi-\dfrac{b}{2^t}\leq 2^{-t} \end{align} That is, the difference $δ ≡ φ − b/2^t$ between $φ$ and $b/2^t$ satisfies $0 ≤ δ ≤ 2^{−t}$. We aim to show that the observation at the end of the phase estimation procedure produces a result that is close to b, and thus enables us to estimate $φ$ accurately, with high probability.

The final state of the first register in the phase estimation procedure is, $\dfrac{1}{2^{t/2}}\sum_{k=0}^{2^t-1}e^{2\pi i\phi k}|k\rangle$

The action of inverse quantum Fourier transform on the state $|j\rangle$ is that, $QFT^\dagger|j\rangle=\dfrac{1}{2^{t/2}}\sum_{k=0}^{2^t-1}e^{-2\pi ijk/2^t}|k\rangle$

Therefore, applying the inverse quantum Fourier transform to the final state of the first register in the phase estimation procedure is, \begin{align} QFT^\dagger\Big(\dfrac{1}{2^{t/2}}\sum_{k=0}^{2^t-1}e^{2\pi i\phi k}|k\rangle\Big)&=\dfrac{1}{2^{t/2}}\sum_{k=0}^{2^t-1}e^{2\pi i\phi k}QFT^\dagger|k\rangle\\ &=\dfrac{1}{2^{t/2}}\sum_{k=0}^{2^t-1}e^{2\pi i\phi k}\dfrac{1}{2^{t/2}}\sum_{l=0}^{2^t-1}e^{-2\pi ikl/2^t}|s\rangle\\ &=\dfrac{1}{2^{t}}\sum_{k,l=0}^{2^t-1}e^{2\pi i\phi k}e^{-2\pi ikl/2^t}|s\rangle \tag{5.23}\label{5.23} \end{align}

Let $\alpha_l$ be the amplitude of $|(b+l)(\mod 2^t)\rangle$ then from equation \ref{5.23}, we can see \begin{align} \alpha_l&=\dfrac{1}{2^{t}}\sum_{k}^{2^t-1}e^{2\pi i\phi k}e^{-2\pi ik(b+l)/2^t}\\ &=\dfrac{1}{2^{t}}\sum_{k=0}^{2^t-1}e^{-2\pi ik(\phi-(b+l)/2^t)}\\ &=\dfrac{1}{2^{t}}\sum_{k=0}^{2^t-1}\Big(e^{-2\pi i(\phi-(b+l)/2^t)}\Big)^k\tag{5.24} \end{align} This is a sum of a geometric progression with $a=1$ and $r=e^{-2\pi i(\phi-(b+l)/2^t)}$ for which $S_n=\dfrac{a(1-r^n)}{1-r}$ with $n=2^t$, so \begin{align} \alpha_l&=\frac{1}{2^t}\Big(\frac{1-e^{2\pi i(2^t\phi-(b+l))}}{1-e^{2\pi i(\phi-(b+l)/2^t)}}\Big)\tag{5.25}\\ &=\frac{1}{2^t}\Big(\frac{1-e^{2\pi i(2^t\delta-l)}}{1-e^{2\pi i(\delta-l/2^t)}}\Big)\tag{5.26}\\ \end{align} where $\delta=\phi-b/2^t$

Suppose the outcome of the final measurement is $m$. We aim to bound the probability of obtaining a value of $m$ such that $|m − b| > e$, where $e$ is a positive integer characterizing our desired tolerance to error. The probability of observing such an $m$ is given by $$ p(|m-b|>e)=\sum_{-2^{t-1}<l\le -(e+1)}|\alpha_l|^2+\sum_{e+1\le l\le 2^{t-1}}|\alpha_l|^2\tag{5.27}\label{5.27} $$ But for any real $θ$, $|1 − exp(iθ)| ≤ 2$, so $$ |\alpha_l|\le \frac{2}{2^t|1-e^{2\pi i(\delta-l/2^t)}|}\tag{5.28} $$

Lemma: $\sin x\ge\frac{2x}{\pi}$ when $0\le x\le \pi/2$

Let $f(x)=\sin x-\frac{2x}{\pi}$

$$ f(0)=0-0=0\quad\&\quad f(\pi/2)=1-1=0\\ f'(x)=\cos x-\frac{2}{\pi}\\ f'(0)=1-\frac{2}{\pi}> 0\quad \&\quad f'(\pi/2)=-2/\pi<0\\ $$ We have $0\le x\le \pi/2\implies 1\ge\cos x\ge0\implies 0\le\sin x\le 1$, and therefore $$ f''(x)=-\sin x\le 0 $$ This concludes that $f(x)$ is $0$ at $x=0$ and is increasing till it reaches a maximum. Then it decreases and takes the value $0$ at $x=\pi/2$. $$ \implies f(x)=\sin x-\frac{2x}{\pi}\ge 0\text{ for }0\le x\le\pi/2\\ \implies \sin x\ge\frac{2x}{\pi}\text{ for }0\le x\le\pi/2\\ $$

From the lemma, we have \begin{align} &\sin|x|\ge\frac{2|x|}{\pi} \text{ when }-\pi/2\le x\le\pi/2\\ &|\sin\theta/2|\ge \frac{|\theta|}{\pi}\text{ when }-\pi/2\le \theta\le\pi/2\\ &|\sin\theta/2|\ge \frac{|\theta|}{\pi}\text{ when }-\pi\le \theta/2\le\pi\\ \end{align} \begin{align} |1-e^{i\theta}|&=|2\sin\theta/2||e^{-i\theta/2}|\\ &=2|\sin\theta/2|\ge\frac{2\theta}{2}\ge \frac{|\theta|}{\pi}\text{ when }-\pi\le \theta/2\le\pi\\ \end{align}

$\implies |1 − exp(iθ)| ≥ 2|θ|/π$ whenever $−π ≤ θ ≤ π$.

But when $−2^t−1 < l ≤ 2^t−1$ we have $−π ≤ 2π(δ − l/2^t ) ≤ π$. Thus $$ |\alpha_l|\le \frac{1}{2^{t+1}(\delta-l/2^t)}\tag{5.29}\label{5.29} $$ Combining \ref{5.27} and \ref{5.29} gives $$ p(|m-b|>e)\le \frac{1}{4}\bigg[\sum_{l=-2^{t-1}+1}^{-(e+1)}\frac{1}{(l-2^t\delta)^2}+\sum_{l=e+1}^{2^{t-1}}\frac{1}{(l-2^t\delta)^2}\bigg]\tag{5.30} $$ Recalling that $0 ≤ 2^t δ ≤ 1$, we obtain \begin{align} p(|m-b|>e)&\le \frac{1}{4}\bigg[\sum_{l=-2^{t-1}+1}^{-(e+1)}\frac{1}{l^2}+\sum_{l=e+1}^{2^{t-1}}\frac{1}{(l-1)^2}\bigg]\tag{5.31}\\ &\le\frac{1}{2}\sum_{l=e}^{2^{t-1}-1}\frac{1}{l^2}\tag{5.32}\\ &\le \frac{1}{2}\int_{e-1}^{2^{t-1}-1}dl\frac{1}{l^2}\tag{5.33}\\ &=\frac{1}{2(e-1)}\tag{5.34} \end{align}

In order to obtain Eq. \ref{5.27} we have shifted the index $l$ by subtracting $2^{t-1}-1$, thereby changing the range $0\leq l\leq 2^t-1$ to $-2^{t-1}+1\leq l\leq 2.2^{t-1}-1-2^{t-1}+1$ or $-2^{t-1}+1\leq l\leq 2^{t-1}$ or $-2^{t-1}< l\leq 2^{t-1}$.

For deriving Eq. \ref{5.29} it seems to make use of the fact that $\sin|x|\geq 2|x|/\pi$ when $-\pi/2\leq x\leq\pi/2$ since $|1-\exp(i\theta)|=2|\sin(\theta/2)|$.


\begin{align} &-2^{t-1}<l\leq 2^{t-1} \quad\&\quad 0\leq \delta\leq 2^{-t}\\ &\implies -2^{-1}<-l/2^t\leq 2^{-1}\\ &\implies -2^{-1}<\delta-l/2^t\leq 2^{-1}+2^{-t}\\ &\implies -\pi<2\pi(\delta-l/2^t)\leq \pi+2\pi2^{-t} \end{align}

How do we obtain that $-\pi\leq 2\pi(\delta-l/2^t)\leq \pi$ ?


1 Answer 1


You've got two relevant conditions in that big block of text: \begin{align*} -2^{t-1}<&l\leq 2^{t-1} \\ 0 \leq&\delta\leq2^{-t} \end{align*} So, consider $\delta-l/2^t$. You have $$ -1/2\leq \delta-l/2^t<2^{-t}+1/2. $$ (Remember to be careful with which bounds you take. For the lower bound, you need to lower bound on $\delta$ and the upper bound on $l/2^t$.) Now just multiply through by $2\pi$.

The only problem compared to what N&C show is there's an extra $2\pi 2^{-t}$ term on the upper bound. However, note that the lower bound on $l$ was $-2^{t-1}<l$. So, since $l$ is an integer, we might equally write $-2^{t-1}+1\leq l$. This subtracts an extra $2\pi/2^t$ from the upper bound, which is exactly what you need (and now the upper bound contains the possibility of equality).

  • $\begingroup$ Could you please elaborate on the reasoning. My understanding is that $−𝑎<−𝑥≤𝑎$ and $0≤𝑦≤𝑏⟹−𝑎<−𝑥≤𝑎+𝑏$. With this argument the inequality is $-𝜋<2𝜋(𝛿−𝑙/2𝑡)≤𝜋+\pi/2^t$, right ? $\endgroup$
    – Sooraj S
    Commented Nov 24, 2021 at 8:28
  • $\begingroup$ Oh, I'm sorry. I somewhat lost track of what it was you were doing and what the book did. Yes, I agree with what you did. $\endgroup$
    – DaftWullie
    Commented Nov 24, 2021 at 9:08
  • $\begingroup$ @SoorajS Hopefully that now explains the extra step you needed. $\endgroup$
    – DaftWullie
    Commented Nov 24, 2021 at 9:16

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