Both Toffoli and CNOT need little help to do universal QC(following a paper by the same title by Yaoyun Shi)

AbstractWell known fact: {CNOT,S} is universal when S is an irrational one qubit rotation

Less well known fact:S really only needs to not square to something classical

Another less well known fact:{Toffoli, Hadamard} is universal

The AgendaBackgroundCompleteness vs. UniversalityKitaev-Solovay TheoremAnother result by KitaevCompleteness (existence) proofsCompleteness: an explicit constructionConclusion

UniversalityA (real) gate library G is universal ifit can approximate any unitary (orthogonal) operator if constant inputs from the computational basis are allowedfor example, a TOFFOLI gate can approximate a CNOT gate in this sense

CompletenessA gate library G is complete ifit can approximate any unitary operator in U(2k) for some kno extra wires or constant inputs allowedCompleteness => Universality

Why completeness?The Kitaev-Solovay Theorem:Any complete gate library can efficiently approximate any 1 qubit unitary operatorspecifically, one can get within in polylog(1/) gates

Another theorem of KitaevSuppose:M is a (real) Hilbert space of dimension > 2 is a unit vectorH SO(M ) is the stabilizer of span()v O(M ), not an eigenvector of vThen: the subgroup generated by H v-1Hv is dense in SO(M )

The AgendaBackgroundCompleteness (existence) proofsCNOTs and RotationsEigenvectors & EigenvaluesWhos DenseCompleteness: an explicit constructionConclusion

A CNOT and a rotationFix an arbitrary one qubit rotation S about an angle if / is irrational, we know from general theory that {CNOT, S} is complete

So, suppose is a rational multiple of pi

A CNOT and a rotationFinally, suppose S2 does not have both 0 and 1 as eigenvectorsa theorem of Gottesman-Knill implies that:for an S failing this condition, any {S, CNOT} circuit may be efficiently simulated by a classical computerthus, such an S is not universal for QCThen {S, CNOT} is complete.

A sketch of the proof:Let U be the operator be computed by

Apply the Kitaev lemma several timesQ.E.D.

Eigenvectors & EigenvaluesCalculating Us eigenvalues gives them as1, 1, ei, e-i is incommensurable with piLet i be the orthonormal eigenvectorsU restricted to span(1, 2) is the identityU restricted to span(3, 4):=H1 is a rotation through the angle

Whos DenseU generates a dense subgroup of H1

Call SO(span(2, 3, 4)) H2H1 H2 is the stabilizer of span(2)one CNOT, C1 fixes 1, and moves span(2)

Whos DenseThe Kitaev lemma applies: {U, C1} generates a dense subset of H2

A similar argument shows {U, C1, C2} generates a dense subset of SO(4)

So, {U, C1, C2} is complete

The AgendaBackgroundCompleteness (existence) proofsCompleteness: an explicit constructionBarenkos Reductionthe Z gateGrovers AlgorithmConclusion

An Explicit ConstructionRecall {CNOT, S} is complete when S2 doesnt have both basis states as eigenvectors It is true that {TOFFOLI, S} is completewhen S doesnt have both basis states as eigenvectors a similar proof exists

An Explicit ConstructionAdditionally, Shi explicitly {TOFFOLI, S} approximates an arbitrary one qubit gate

By Barenkos decomposition, this is sufficient to approximate an arbitrary unitary matrix

Some preliminariesDefine Ut to be rotation by the angle t

Let S be the one-qubit gate in our librarydefine by S = U

Let W be the desired one qubit operator define by W = U

Reduction of the problemIt suffices to approximatethe Z gatea gate W/2 s.t. W /20k = U/2 0 0k-1

Using these gates and the TOFFOLI, one may simulate a gate W satisfyingW ( 0k-1) = U 0k-1

The Z GateHow to use S to flip a signSuppose = pi/4One can use a well known trick:

This works because: XUpi/41=-Upi/41

The Z GateFor arbitrary , its more difficultXU1 could be anywhere relative to U1

The Z GateA similar construction exists, however

U0U1 = a(11-00) + b01 + c10swap the basis vectors 11, 00this is within sqrt(b2+c2) of a sign flipsqrt(b2+c2) < 1, so do a lot of these

The W /2 GateWant: W /20k = U/2 0 0k-1

Idea ?

Prelude to Grovers AlgorithmLet 0 = 02k Use S, CNOT, to build a T such that 0T0 is small and positivedefine = T0 Let 1 be the vector perpendicular to 0 in the plane spanned by 0 ,

Using Grovers AlgorithmThe system begins in the state 00 apply ITthe state = 0 Iteratively reflect about 1 ala Groverwant: -> cos(/2 )1 + sin(/2 )0 state = 0(cos(/2 )1 + sin(/2 )0)

Using Grovers AlgorithmApply an appropriately conjugated 2k-cnot to flip the first bit if the remaining 2k are orthogonal to 0 state = 11cos(/2 ) + 00sin(/2 )

Apply a controlled-T-1 : 11 -> 10state = (cos(/2 )1 + sin(/2 )0)0

The AgendaBackgroundCompleteness (existence) proofsCompleteness: an explicit constructionConclusion

ConclusionThe CNOT needs only a one qubit rotation whose square is nonclassical to form a complete library

The Toffoli can partner with any nonclassical gate for a complete library

In the second case, we have an explicit approximation algorithm

Questions?