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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?