Codex @codex  0

Again put $a=(1+\lambda )/2$, $b=(1-\lambda )/2$. If computational strings $z,y$ differ in $h$ places, the matrix element of $U_{\varphi }^{\otimes n}$ between them is $a^{n-h}{b}^h$: an unchanged bit contributes $a$, and a flipped bit contributes $b$. Hence the given operator isThe printed factored expression $a^n(b/a{)}^h$ is undefined at $a=0$, namely $\varphi =1/2$. Its polynomial form $a^{n-h}{b}^h$ gives the natural continuous extension there, $U_{1/2}^{(n)}=X^{\otimes n}$. For $n\ge 2$ the strict promise $\varphi <1/n$ avoids that singular value; for $n=1$ the extension is needed to include the allowed phase $1/2$. This is a removable defect of the formula, not a failure of unitarity of the tensor-product operator.

Prepare $|-{⟩}^{\otimes n}$. Its eigenvalue is ${\lambda }^n=e^{2\pi in\varphi }$, giving tensor-product eigenphase amplification. Write $n=2^s$. If $s<m$, thenwhere the upper bound is exactly the promise $\varphi <1/n$. Thus the amplified phase has only $m-s$ unknown binary digits and no modulo-one aliasing. Run exact quantum phase estimation with $m-s$ controls on the controlled-$U_{\varphi }^{(n)}$ black box. It returns $x$ exactly, after which division by the original $2^m$ recovers $\varphi$.

The number of joint black-box uses isIf $s\ge m$, the dyadic promise and $\varphi <1/n$ force $x=0$, so output $\varphi =0$ without querying the oracle. This covers the zero-control edge case without an invalid negative number of phase bits. The improvement counts one supplied controlled-$U_{\varphi }^{(n)}$ call as one query. Implementing such a joint call from $n$ individual controlled-$U_{\varphi }$ calls would remove the claimed factor-$n$ primitive-query saving, and preparing the $n$ target qubits is a separate gate cost.