Amplitude Estimation and Counting — Examples

Substituting $N=16,M=4,k=1$ into [F1] gives the generic expression $E(t)=\frac{\pi }{2}\sqrt{t(16-t)}+{\pi }^2$.

At $t=0$, this expression is ${\pi }^2$ but [F2] sharpens the actual error to zero. At $t=1$, it is $\frac{\pi \sqrt{15}}{2}+{\pi }^2$. At $t=8$, it is $4\pi +{\pi }^2$. At $t=16$, the expression again equals ${\pi }^2$, while [F2] gives exact output $16$.

The first-order term is zero at both endpoints and largest here at the half-count $t=8$; the second-order uniform term alone does not encode the exact endpoint behavior. All four substitutions agree with the theorem, and the endpoint cases use its explicit certainty clause.

Let the target be one-dimensional and compare $U=I$ with $U^{\mathrm{\prime }}=-I$. An uncontrolled call to $U^{\mathrm{\prime }}$ differs from an uncontrolled call to $U$ by a scalar phase on the target, so every fixed sequence of uncontrolled calls has the same physical outcome probabilities after any final measurement; global phase cancels from a vector's rank-one projector.

On a control qubit, however, $C(I)=I$ while $C(-I)=\mathrm{diag}(1,-1)=Z$. On input $\mathrm{\mid }+⟩=(\mathrm{\mid }0⟩+\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$, the first returns $\mathrm{\mid }+⟩$ and the second returns $\mathrm{\mid }-⟩=(\mathrm{\mid }0⟩-\mathrm{\mid }1⟩)\mathrm{/}\sqrt{2}$; a computational-basis-independent $X$-basis measurement distinguishes them with certainty. Thus controlled access contains information absent from the uncontrolled phase-equivalence class.

The cost theorem [F2] counts supplied controlled powers or repetitions of controlled $Q$ only under the corresponding access declaration. In repeated-unitary mode the controlled-$Q$ constituents must be charged; in power-query mode each supplied power carries its declared cost vector, including any implementation constituents. This witness proves that neither mode's controlled access is free from a bare uncontrolled oracle.