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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) paper

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Abstract Well known fact: Well known fact: –{CNOT,S} is universal when S is an irrational one qubit rotation Less well known fact: Less well known fact: –S really only needs to not square to something classical Another less well known fact: Another less well known fact: –{Toffoli, Hadamard} is universal

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The Agenda Background Background –Completeness vs. Universality –Kitaev-Solovay Theorem –Another result by Kitaev Completeness (existence) proofs Completeness (existence) proofs Completeness: an explicit construction Completeness: an explicit construction Conclusion Conclusion

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Universality A (real) gate library G is universal if A (real) gate library G is universal if –it can approximate any unitary (orthogonal) operator if constant inputs from the computational basis are allowed –for example, a TOFFOLI gate can approximate a CNOT gate in this sense

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Completeness A gate library G is complete if A gate library G is complete if –it can approximate any unitary operator in U(2 k ) for some k –no extra wires or constant inputs allowed Completeness => Universality Completeness => Universality

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

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

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The Agenda Background Background Completeness (existence) proofs Completeness (existence) proofs –CNOTs and Rotations –Eigenvectors & Eigenvalues –Who’s Dense Completeness: an explicit construction Completeness: an explicit construction Conclusion Conclusion

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A CNOT and a rotation Fix an arbitrary one qubit rotation S about an angle θ Fix an arbitrary one qubit rotation S about an angle θ –if θ/π is irrational, we know from general theory that {CNOT, S} is complete So, suppose θ is a rational multiple of pi So, suppose θ is a rational multiple of pi

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A CNOT and a rotation Finally, suppose S 2 does not have both 0 and 1 as eigenvectors Finally, suppose S 2 does not have both 0 and 1 as eigenvectors –a theorem of Gottesman-Knill implies that: for an S failing this condition, any {S, CNOT} circuit may be efficiently simulated by a classical computer for an S failing this condition, any {S, CNOT} circuit may be efficiently simulated by a classical computer –thus, such an S is not universal for QC Then {S, CNOT} is complete. Then {S, CNOT} is complete.

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A sketch of the proof: Let U be the operator be computed by Let U be the operator be computed by Apply the Kitaev lemma several times Apply the Kitaev lemma several times –Q.E.D. S SS S

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Eigenvectors & Eigenvalues Calculating U’s eigenvalues gives them as Calculating U’s eigenvalues gives them as –1, 1, e i , e -i – is incommensurable with pi Let i be the orthonormal eigenvectors Let i be the orthonormal eigenvectors –span( 1 , 2 ) is the identity –U restricted to span( 1 , 2 ) is the identity –U restricted to span( 3 , 4 ):=H 1 is a rotation through the angle

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Who’s Dense U generates a dense subgroup of H 1 U generates a dense subgroup of H 1 Call SO(span( 2 , 3 , 4 )) H 2 Call SO(span( 2 , 3 , 4 )) H 2 –H 1 H 2 is the stabilizer of span( 2 ) –one CNOT, C 1 fixes 1 , and moves span( 2 )

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Who’s Dense The Kitaev lemma applies: {U, C 1 } generates a dense subset of H 2 The Kitaev lemma applies: {U, C 1 } generates a dense subset of H 2 A similar argument shows {U, C 1, C 2 } generates a dense subset of SO(4) A similar argument shows {U, C 1, C 2 } generates a dense subset of SO(4) So, {U, C 1, C 2 } is complete So, {U, C 1, C 2 } is complete

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The Agenda Background Background Completeness (existence) proofs Completeness (existence) proofs Completeness: an explicit construction Completeness: an explicit construction –Barenko’s Reduction –the Z gate –Grover’s Algorithm Conclusion Conclusion

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An Explicit Construction Recall {CNOT, S} is complete Recall {CNOT, S} is complete –when S 2 doesn’t have both basis states as eigenvectors It is true that {TOFFOLI, S} is complete It is true that {TOFFOLI, S} is complete –when S doesn’t have both basis states as eigenvectors –a similar proof exists

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An Explicit Construction Additionally, Shi explicitly {TOFFOLI, S} approximates an arbitrary one qubit gate Additionally, Shi explicitly {TOFFOLI, S} approximates an arbitrary one qubit gate By Barenko’s decomposition, this is sufficient to approximate an arbitrary unitary matrix By Barenko’s decomposition, this is sufficient to approximate an arbitrary unitary matrix

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Some preliminaries Define U t to be rotation by the angle t Define U t to be rotation by the angle t Let S be the one-qubit gate in our library Let S be the one-qubit gate in our library –define θ by S = U θ Let W be the desired one qubit operator Let W be the desired one qubit operator –define by W = U

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Reduction of the problem It suffices to approximate It suffices to approximate –the Z gate –a gate W /2 s.t. W /2 0 k = U /2 0 0 k-1 Using these gates and the TOFFOLI, one may simulate a gate W satisfying Using these gates and the TOFFOLI, one may simulate a gate W satisfying –W ( 0 k-1 ) = U 0 k-1

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The Z Gate How to use S to flip a sign How to use S to flip a sign –Suppose θ = pi/4 –One can use a well known trick: –This works because: XU pi/4 1 =-U pi/4 1 SS†S† Z = 1111 1111

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The Z Gate For arbitrary θ, it’s more difficult For arbitrary θ, it’s more difficult –XU θ 1 could be anywhere relative to U θ 1

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The Z Gate A similar construction exists, however A similar construction exists, however U θ 0 U θ 1 = a( 11 - 00 ) + b 01 + c 10 U θ 0 U θ 1 = a( 11 - 00 ) + b 01 + c 10 –swap the basis vectors 11 , 00 –this is within sqrt(b 2 +c 2 ) of a sign flip –sqrt(b 2 +c 2 ) < 1, so do a lot of these

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The W /2 Gate Want: W /2 0 k = U /2 0 0 k-1 Want: W /2 0 k = U /2 0 0 k-1 Idea ? Idea ?

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Prelude to Grover’s Algorithm Let 0 = 0 2k Let 0 = 0 2k Use S, CNOT, to build a T such that Use S, CNOT, to build a T such that – 0 T 0 is small and positive –define φ = T 0 Let 1 be the vector perpendicular to 0 in the plane spanned by 0 , φ Let 1 be the vector perpendicular to 0 in the plane spanned by 0 , φ

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Using Grover’s Algorithm The system begins in the state 0 0 The system begins in the state 0 0 –apply I T –the state = 0 φ Iteratively reflect φ about 1 ala Grover Iteratively reflect φ about 1 ala Grover –want: φ -> cos( /2 ) 1 + sin( /2 ) 0 –state = 0 (cos( /2 ) 1 + sin( /2 ) 0 )

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Using Grover’s Algorithm Using Grover’s Algorithm Apply an appropriately conjugated 2k-cnot to flip the first bit if the remaining 2k are orthogonal to 0 Apply an appropriately conjugated 2k-cnot to flip the first bit if the remaining 2k are orthogonal to 0 –state = 1 1 cos( /2 ) + 0 0 sin( /2 ) Apply a controlled-T -1 : 1 1 -> 1 0 Apply a controlled-T -1 : 1 1 -> 1 0 –state = (cos( /2 ) 1 + sin( /2 ) 0 ) 0

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The Agenda Background Background Completeness (existence) proofs Completeness (existence) proofs Completeness: an explicit construction Completeness: an explicit construction Conclusion Conclusion

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Conclusion The CNOT needs only a one qubit rotation whose square is nonclassical to form a complete library The 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 The Toffoli can partner with any nonclassical gate for a complete library In the second case, we have an explicit approximation algorithm In the second case, we have an explicit approximation algorithm

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