Past exam of the mathematics course of the University of Cambridge 2004 ib Paper 2 8D Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2005 ib Paper 1 15G Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2007 ib Paper 3 16B Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 67 2 a Solution Created 2026-10-03 Updated 2026-10-07
Use the bit-query Boolean quantum oracle , where denotes workspace. Initially every probability amplitude is independent of , hence is a constant polynomial. An input-independent unitary gate only forms linear combinations of probability amplitudes, so it does not increase their polynomial degree.
If is an amplitude before a query, its new value isOne query increases polynomial degree by at most one. Inductively, after queries each amplitude has polynomial degree at most . The acceptance probability is a sum of squared absolute values of the accepting amplitudes, so it is a real polynomial of degree at most . Intermediate quantum measurements and classical adaptation can be retained coherently, or handled by summing unnormalized branch probabilities; either approach yields the same degree bound.
An exact algorithm has acceptance probability precisely at every Boolean input. The multilinear reduction on the Boolean cube replaces positive powers of each by , preserving these values without increasing polynomial degree. Therefore the polynomial method for quantum query lower bounds givesHere is the degree of the unique multilinear polynomial representing the Boolean function, and is its exact quantum query complexity. Uniqueness follows, for example, by evaluating successively on the indicator vectors of subsets: the value on a subset determines its coefficient once all smaller-subset coefficients are known.
Past exam of the mathematics course of the University of Cambridge 2014 ib Paper 1 15A ii Solution Created 2026-09-24 Updated 2026-10-06
Choose the prepared wavefunction real and positive on its support. Its normalization isIn the ramp-state expansion in a symmetric infinite well, write . Orthogonality givesFor , a primitive is . Evaluating both endpoints givesAn equivalent parity-separated form isThese formulas with are the requested explicit expansion. The convergence is in the square-integrable norm; at the idealized discontinuity at the right wall, a pointwise boundary value does not change the quantum state or its probability amplitudes.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 324 1 b iii Solution Created 2026-10-03 Updated 2026-10-06
Apply the same positive-exponent quantum Fourier transform to each of the two quantum registers. The amplitude at of the coset state isThe finite geometric series is when and zero otherwise. ThereforeThere are allowed pairs, with following the uniform distribution on a finite set. The factor changes the probability amplitudes by phases depending on the output label , but has no effect on their probabilities. Using an inverse rather than forward quantum Fourier transform on both quantum registers gives the same support relation.
A modular inverse of exists precisely when . In that case the discrete logarithm isIt cannot generally be determined from every allowed pair. Put . The supported linear congruence has solutions modulo , and determines only modulo ; gives no information when . For instance and are compatible with both and . One may verify a candidate by modular exponentiation, checking . Repeated discrete-logarithm Fourier sampling finds an invertible with per-trial probability , which has the same totient lower bound from the Mertens product as before. When , the group is trivial and are determined without Fourier samples.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 324 3 b ii Solution Created 2026-10-03 Updated 2026-10-06
Apply to the data and phase quantum registers, leaving the flag quantum ancilla untouched. This inverse needs no extra oracle assumption: the promised dyadic eigenvalues give , hence . The inverse of each controlled unitary gate used in can therefore be built from repeated uses of the supplied controlled-, and the known Hadamard gates and quantum Fourier transform gates can be reversed. Uncomputation is necessary to erase the eigenvalue label coherently. The resulting quantum state isA quantum measurement in the computational basis of the flag followed by postselection on one yieldsThis requires . For a normalized input and nonnegative eigenvalues, the success probability of positive quantum spectral filtering satisfies . In the general inequality, equality holds precisely when the input is supported on the minimum-eigenvalue eigenspace. Omitting uncomputation and discarding the phase quantum register would instead leave a mixed state with diagonal weights proportional to , rather than the desired coherent pure state.
The printed universal nonzero-success request needs a nonkernel-input hypothesis. An -qubit Hermitian operator has eigenvalues, counted with multiplicity. All are distinct, and the printed dyadic grid contains exactly possible values. They therefore occupy the entire grid, including zero: this is a multiplicity-free complete dyadic spectrum, and . Taking , , and satisfies every printed spectral promise but gives . No normalized output vector exists, so no algorithm can deliver it with nonzero probability.
For every input outside the kernel, the procedure above has , giving the requested strict bound on the meaningful domain. The corrected result is therefore exact quantum spectral filtering conditional on , with the boxed probability. This does not require knowing the input probability amplitudes or making extra copies of an unknown quantum state. Arbitrarily small nonzero support outside the kernel gives arbitrarily small success probability, and a failed flag measurement disturbs the input; fresh independent trials cannot be assumed when only one unknown physical input is supplied.
Past exam of the mathematics course of the University of Cambridge 2022 ib Paper 2 15B b iii Solution Created 2026-09-24 Updated 2026-09-25
The specified initial state is normalized. The requested energy is , and its probability amplitude isThus the Born rule givesUnitary evolution changes each energy coefficient only by a phase, so this probability remains at both and . If an energy measurement returns , the state collapses to the nondegenerate eigenstate ; a subsequent energy measurement therefore returns with probability
After input-bit queries, every probability amplitude is a polynomial of degree at most , because the bit-query matrix entries are affine in the input bits. An acceptance probability is a sum of squared absolute amplitudes and has polynomial degree at most . Multilinear reduction on the Boolean cube does not increase degree. Exact acceptance therefore implies . For bounded error the acceptance polynomial only approximates the Boolean function; its exact representing degree is not the relevant bound.