Quantum Phase Estimation — Examples

Here $Q=2^3=8$ and $Q\theta =8(5\mathrm{/}8)=5$, an integer in $\{0,\dots ,7\}$. By [F2], the output probabilities in label order $0,1,\dots ,7$ are $(0,0,0,0,0,1,0,0)$.

The unique measured label is $y=5$; [F1] therefore returns the exact phase estimate $y\mathrm{/}Q=5\mathrm{/}8$.

Since $Q\theta =3\mathrm{/}2$, the grid labels $1$ and $2$ are tied nearest labels, each at circular distance $1\mathrm{/}16$. For $y=1,2$, $Q(\theta -y\mathrm{/}8)=\pm 1\mathrm{/}2$, so the geometric quotient in [F1]–[F2] has numerator modulus $2$ and denominator modulus $2\sin (\pi \mathrm{/}16)$. Thus $P_{\theta }(1)=P_{\theta }(2)=1\mathrm{/}(64{\sin }^2(\pi \mathrm{/}16))$, which is at least $4\mathrm{/}{\pi }^2$ by the nearest-grid part of [F4].

The circular distances for $y=0,\dots ,7$ are $(3\mathrm{/}16,1\mathrm{/}16,1\mathrm{/}16,3\mathrm{/}16,5\mathrm{/}16,7\mathrm{/}16,7\mathrm{/}16,5\mathrm{/}16)$, by [F3]. Hence the tail event $d_{\mathbb{T}}(\theta -y\mathrm{/}8)\ge 1\mathrm{/}4$ consists exactly of $y=4,5,6,7$. For these labels the exact outcome probabilities are $P_{\theta }(4)=P_{\theta }(7)=1\mathrm{/}(64{\sin }^2(5\pi \mathrm{/}16))$ and $P_{\theta }(5)=P_{\theta }(6)=1\mathrm{/}(64{\sin }^2(7\pi \mathrm{/}16))$, again by the finite geometric quotient.

Therefore the exact total tail is $\frac{1}{32}({\sin }^{-2}(5\pi \mathrm{/}16)+{\sin }^{-2}(7\pi \mathrm{/}16))$. Here $K=\lfloor 8(1\mathrm{/}4)\rfloor =2$, so [F4] gives the proven upper bound $1\mathrm{/}2$ for this total. Numerically the exact expression is approximately $0.07769$, well below $1\mathrm{/}2$.