Past exam of the quantum theory and quantum computers course of the University of Oxford / 2008 / 4

Put $d=2^N$ and $\lambda =e^{2\pi i/d}$, with nonnegative integer $N$ so the register exists. The geometric sum satisfiesConsequently the matrix $F_{+}$ with entries $d^{-1/2}{\lambda }^{jk}$ has $F_{+}^{\ast }{F}_{+}=I$. This proves unitarity andFor a binary output string $j={\sum }_{r=1}^N{j}_r{2}^{N-r}$, expansion of the printed tensor product gives coefficientThus the printed product is exactly $F_{+}|k⟩$. It is a correct product decomposition of a Fourier phase state, but the printed recovery identity has inconsistent signs:by the square of the quantum Fourier transform. To recover $f(j)$ instead, use $F_{-}=F_{+}^{\ast }$, replacing every product phase by its complex conjugate. Equivalently one may reverse the sign in the definition of $\hat{f}$. The discrepancy is visible for $d\ge 4$ by taking $f$ concentrated at $1$: the printed expression returns the basis vector at $d-1$. The recovery display also uses $k$ as the summation variable on its right while its summand uses $j$; that dummy variable should be $j$.

For factorization first check $\gcd (3,91)=1$. Successive residues areThe first return is $\boxed{r=6}$. Its half power is $3^3=27$, which is neither $1$ nor $-1$ modulo $91$. Therefore

In Shor's algorithm prepare a uniform exponent register of size $Q=2^n$, conveniently $91^2\le Q<2\cdot 91^2$, and reversibly compute $3^x\mathrm{mod}91$ into a second register. Measuring that second register, or tracing it out, leaves periodic exponent correlations of period $r$. Conditional on one measured residue, the first register is a superposition over an arithmetic progression of exponents separated by $r$. The quantum Fourier transform converts this periodic structure into peaks near $k/Q=s/r$. Continued fractions recover $s/r$; when $s$ is coprime to $r$, its denominator is $r$. Verify by modular exponentiation and repeat when the reduced denominator is insufficient. For this example $Q=16384$ is suitable, and useful peaks include those near $Q/6$ and $5Q/6$. Either transform sign gives the same period information, so the source's inverse-sign error does not invalidate the order-finding mechanism.

The Part C inverse-transform display also prints $j$ as its summation index while the summand uses $k$; that sum must be over $k$, as in the corrected inverse above.