= Solution
Write $\phi=2\pi n/N+\epsilon$, with $|\epsilon|\le\delta\ll2\pi/N$. Decode the measured bits as above. The <probability amplitude> of label $m$ is
$$
a_m=\frac1N\sum_{k=0}^{N-1}e^{ik\Delta_m},\qquad \Delta_m=\phi-\frac{2\pi m}{N}.
$$
Using the <finite geometric series>,
$$
a_m=\frac{e^{i(N-1)\Delta_m/2}}N\frac{\sin(N\Delta_m/2)}{\sin(\Delta_m/2)},\qquad P(m)=\frac{\sin^2(N\Delta_m/2)}{N^2\sin^2(\Delta_m/2)}.
$$
Interpret removable singularities by continuity. In particular, <near-grid quantum phase estimation> returns the nearby exact label $n\bmod N$ with <probability>
$$
\boxed{P(n)=\frac{\sin^2(N\epsilon/2)}{N^2\sin^2(\epsilon/2)}=1-\frac{N^2-1}{12}\epsilon^2+O(N^4\epsilon^4).}
$$
This is near one under the stated small-offset assumption, but not exactly one when $\epsilon\ne0$.
There is also a rigorous error bound. Taking the squared modulus of the sum gives
$$
1-P(n)=\frac1{N^2}\sum_{k,l=0}^{N-1}\bigl[1-\cos((k-l)\epsilon)\bigr]\le\frac{\epsilon^2}{2N^2}\sum_{k,l=0}^{N-1}(k-l)^2=\frac{N^2-1}{12}\epsilon^2,
$$
where $1-\cos x\le x^2/2$ and the elementary sums of $k$ and $k^2$ evaluate the last expression. Hence
$$
\boxed{P\!\left(\widehat\phi=\frac{2\pi(n\bmod N)}N\right)\ge1-\frac{N^2-1}{12}\delta^2.}
$$
On that event the circular <quantum phase> error is at most $\delta$. For other labels, the exact <probabilities> are
$$
P(m)=\frac{\sin^2(N\epsilon/2)}{N^2\sin^2\!\left(\epsilon/2+\pi(n-m)/N\right)},\qquad m\ne n\bmod N.
$$
Their total is $1-P(n)=O(N^2\delta^2)$, describing the leakage from the perfect solution. The physical output bit string remains reversed in the printed circuit; this changes its decoding, not these <probabilities>.
Back to article page
