“Both Toffoli and CNOT need little help to do universal QC” (following a paper by the same title by Yaoyun Shi) paper
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
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
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
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
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
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 )
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
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
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.
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
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
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 )
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
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
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
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
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
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
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
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
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
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 ?
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
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 )
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
The Agenda Background Background Completeness (existence) proofs Completeness (existence) proofs Completeness: an explicit construction Completeness: an explicit construction Conclusion Conclusion
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
Questions?