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Overview Four Talks • A Rosetta Stone for Quantum Computation • Quantum Algorithms & Beyond • Distributive Quantum Computing atica Inform i Per Can • Topological quantum Computing and the Jones Polynomial • A Quantum Computing Knot Theoretic Mystery -- Can be found on my webpage. Topological Quantum Computing and the Jones Polynomial This work is in collaboration with Samuel J. Lomonaco, Jr. Dept. of Comp. Sci. & Electrical Engineering Sci. Louis H. Kauffman University of Maryland Baltimore County Baltimore, MD 21250 Email: Lomonaco@UMBC.EDU WebPage: http://www.csee.umbc.edu/~lomonaco WebPage: http://www.csee.umbc.edu/~lomonaco L-O-O-P This work is supported by: This talk is about an algorithm found in: • The Defense Advance Research Projects Dorit, Aharonov, Dorit, Aharonov, Vaughan Jones,and Zeph Landau, Agency (DARPA) & Air Force Research On the Quantum Algorithm for Approximating the Laboratory (AFRL), Air Force Materiel Command, Polynomial, Jones Polynomial, USAF Agreement Number F30602-01-2-0522. http://arxiv.org/abs/quant-ph/0511096 http://arxiv.org/abs/quant- • The National Institute for Standards and Technology (NIST) We will refer to this paper as AJL • The Mathematical Sciences Research Institute (MSRI). and also about the paper • L-O-O-P The L-O-O-P Fund. Lomonaco, and Kauffman, Topological Quantum • The Institute of Scientific Interchange Polynomial, Computing and the Jones Polynomial, http://arxiv.org/abs/quant- http://arxiv.org/abs/quant- ph/0605004 1 Jones m World Polynomial Quantu ??? Quantum States Superposition atica Unitary Inform Per Can i Measurement Evolution Entanglement Quantum Algorithms Quantum Algorithms Zoo Other Algs. Algs. Fundamental Questions ??? Deutsch- Deutsch-Jozsa Grover ’s Grover’ What are the limits of Quantum Jones Simon’ Simon’s Poly. Alg . Computation ??? Amplitude Amplif. Amplif. Shor’ Shor’s Trace Algs. Algs. Estimation Q. Hidden Subgroup Algs. Algs. Quadratic New Will future quantum computers be general Speedup purpose or special purpose devices ??? Exponential Q. Random Speedup Quantum Walks Adiabatic Simulation Non- Non-Classical Algs of Q. Related Systems Behavior ??? ??? Complexity Questions ??? Why are we interested in the Jones But the Jones polynomial is in the complexity polynomial ??? class #P !!!. Why do we want to find a fast, i.e. polytime, polytime, quantum algorithm for the Jones polynomial ??? So what !!! DoD Why should the MoD be interested in the Jones polynomial ??? So what ! Who cares ? 2 The Class #P Questions ??? SAT) NP- The satisfiability problem (SAT ) is NP-complete So a #P problem is computationally harder than NP- an NP - complete problem. Boolean expr in CNF form Yes ? Decision polytime, Why do we want to find a fast, i.e. polytime, e ( x1 ,K , xn ) Satisfiable quantum algorithm for the Jones polynomial ??? ??? Problem No ? Answer: BQP. Answer: This would mean that P#P is in BQP. But since NP is in P#P, this implies that NP is in The corresponding #P class is Counting BQP. BQP. Hence, a polytime quantum algorithm for the Jones polynomial implies there exists a polytime Problem NP- quantum algorithm for NP - complete problems. Find # of r u r u e( x1 ,K , x n ) Satisfying Values { } n } # ξ∈{ 0,1 : e(ξ) =True Wow !!! Boolean expr But this is probably not the case!!! in CNF form Questions ??? On the other hand … polytime, Why do we want to find a fast, i.e. polytime, quantum algorithm for the Jones polynomial ??? held, belief: Widely held, but unproven belief : There is NP- no quantum algorithm that computes an NP- Answer: Answer: This would mean that P # P ⊆ BQP complete problem in polynomial time. But NP ⊆ P # P So can there really be a polytime quantum Ergo, NP ⊆ BQP algorithm for computing the Jones Therefore, there exists polytime quantum polynomial ??? Probably not !!! NP- algorithms for NP-Complete problems ! Wow !!! But creating a quantum algorithm for the Jones polynomial might well give us some But this is probably not the case!!! insight why NP is not in BQP ??? Questions ??? We will now discuss the polytime algorithm: • How can we compute the Jones polynomial on a quantum computer ? Knot K= Jones polynomial • How fast is the resulting quantum Integer k AFJKL Algorithm ( ) V K e 2π i / k ± ε algorithm ? with exponentially Precision ε small probability of error 3 We will now discuss the following theorem: Theorem. Theorem . (AJL) Quick For a given braid β in Bn with m crossings and for a given integer k , there is a quantum Overview algorithm which is polynomial in n, m , k , which with all but exponentially (in n, m , k) small to Knot probability, outputs a result in the closed interval ( ) ( ) Re V tr ( e2 π i / k ) − εd n −1 , Re V t r ( e 2π i / k ) + ε d n −1 β β Theory Skip to braids for ε which is inverse polynomial in n, k, m . The same is true for the imaginary part. Placement Problem: Knot Theory ¡3 Knot Diagram • Ambient space = ¡ 3 Orientation Preserving • Group G = AutoHomeo( ¡ ) 3 Labeled Vertex S1 men t#1 Place Labeled Vertex Placement#2 Labeled Vertex ~K Def. Def . • Planar four valent graph with K 1 if g ∈ G s.t . gK 1 = K 2 • Labeled vertices 2 Problem. Problem. When are two placements the same ? ? K K ~ 1 2 Reidemeister Moves Comment: R1 is not a Physical Move R0 R1 R1 R2 R3 A Discontinuous Move ! These are local moves ! 4 Differential Geometry: The Frenet Frame Differential Geometry: The Frenet Frame 3- Each point of a curve in 3- space is naturally 3- Each point of a curve in 3- space is naturally 3- frame. associated with a 3- frame, called the Frenet frame. 3- frame. associated with a 3- frame, called the Frenet frame. B B B = binormal = TxN B T T T T T = tangent B B T T N N N N T N=normal N= normal N Differential Geometry: The Frenet Frame Comment: R1 is not a Physical Move In other words, given a smooth curve R1 x ( t ) = ( x1 ( t ), x 2 ( t), x 3 ( t )) 3- in 3 - space, there is a naturally associated curve A( t ) in SO(3) A Discontinuous Move ! Example of Application of Reidemeister Moves When do two Knot diagrams represent the same or different knots ? R3 R2, R2 Reidemeister). Theorem (Reidemeister). Two knots (or links) diagrams represent the same knot (or link) iff one can be transformed R1 Sa S me me into the other by a finite sequence of Kn ot tT Reidemester moves. Ty p pe R1 5 What is a knot invariant ? Def. Def . A knot invariant I is a map I : Knots → MathematicalDomain that takes each knot K to a mathematical What Is the Braid object I(K) such that ⇒ I ( )= ( ) Group Bn ??? K I K ~ 1 Consequently, I( ) ≠ I( )⇒ ~K 2 K K 1 The Jones polynomial is a knot invariant. K K 2 K 1 1 2 2 A Braid Why is the braid group important for Q Comp ? • The representations of the Symmetric Sn are Hat Box the basic building blocks for the representation of the unitary group U used in quantum mechanics, • The braid group B n “ sits above ” the symmetric Bn group Sn , i.e., there is a natural epimorphism • Thus, new representations of the braid group Sn B n will give us new representations of the unitary group U , i.e., quantum gates • Claim: These quantum gates can be implemented in Claim: quantum systems that are resistant to decoherence 3 Strand braid β obstructions, because of topological obstructions, e.g., in terms of the fractional quantum Hall effect, anyonic systems Two Equal Braids Two Unequal Braids = ≠ β1 = β 2 β1 ≠ β 2 6 Shorthand Notation Product of Braids Hat Box Shorthand Notation Times = = 3 Strand braid β β β1 • β 2 = β 3 Inverse of of a Braid Generators of the Braid Group Bn = Bn is generated by = The braid group Times L L L L β • β −1 = 1 b1 b2 L bn−1 To construct the inverse of a braid, take the mirror image of each crossing, and then reverse the order of the crossings. Relations Among the Generators of Bn ster 3 M ov e bi bi +1bi = bi +1bbi +1 , 1 ≤ i < n i Reidemei bi bi +1bi = bi +1bbi +1 , 1 ≤ i < n i = bi bj = bjbi , for i − j ≥ 2 Skip to presentation 7 A Presentation of the Braid Group Bn bi bj = bjbi , for i − j ≥ 2 bibi +1 bi = bi+1bi bi+1 , 1 ≤ i < n −1 b1 , b2 ,K, bn−1 : = bibj = bjbi , i − j > 1,1 ≤ i , j < n − 1 Complete Generators set of Relations Almost” A Braid Is “Almost” a Permutation Braids as Words Every braid β in Bn can be written as a bbi +1bi = bi +1bi bi +1 , 1 ≤ i < n −1 product of braid generators b1 , b 2 ,K , bn − 1 i Bn = b1 , b2 ,K, bn−1 : and their inverses b1−1 , b2 −1 ,K , bn −1−1 bb j = bj bi , i − j > 1,1 ≤ i , j< n− 1 i L β= ∏b j( i ) ε (i ) = bj 1) ε (1) bj ( 2 )ε (2) L bj ( L )ε ( L) Natural i =1 Epimorphism bi bi +1 bi = bi + 1bi bi +1 , 1 ≤ i < n − 1 B n = b1, b2 ,K , bn −1 : bi 2 = 1, 1 ≤ i < n − 1 Sn bi b j = b j bi , i − j > 1, 1 ≤ i , j< n− 1 where ε (i ) = ±1 Braid word w Another Perspective Anyons: A Very Brief Overview Anyons: A Braid Represents the Movement of n Holes in a Disc Anyons are quantum systems that are confined to two dimensions. They were This braiding can be used to first proposed by Nobel Laureate F. represent Anyon exchanges Wilczek. See for example, Wilczek. A B Wilczek, F., Fractional statistics and Wilczek, superconductivity, anyon superconductivity, World Scientific Press, (1990). Anyonic braiding corresponds to Anyons can used to explain the fractional a Unitary transformation quantum Hall effect Recall: Q.M.= Qroup Rep. Theory 8 Anyons Can Also Fuse or Split Anyons: Anyons: A Very Brief Overview (Cont.) Quantum Topology gives us the tools needed to find new unitary representations based AC B on fusing and braiding These new unitary transformations are created with an object called a unitary topological modular functor which we call model. simply an anyon model. Recall: Q.M.= Qroup Rep. Theory From Braids to Knots Knots The Markov Trace from Closure of a Braid Braids Close Braid The Markov trace closure Braid β Closed Braid β tr Can Every Knot Can Be Constructed from the When Does the Closure of Two Braids Produce the Closure of Some braid? Same Link ? Theorem (Markov). Two braids β 1 and β 2 produce the same link under Markov trace braid closure iff there exists a finite sequence of Markov moves that Alexander). Theorem (Alexander). Every transforms one braid into the other. knot is the closure of a braid. Definition. Definition . Let β be a braid in B n . Then the Markov moves are defined as: M1 β a β ' = bi ±1 gβ gbim1, 1 ≤ i < n Skip Markov Moves M2 β a β g bn+1±1 ∈ Bn+1 9 The Markov 1 Move The Markov 2 Move M1 β a β ' = bi ±1g β gbi m1, 1 ≤ i < n M2 β a β g bn+1±1 ∈ Bn+1 … … bi - 1 Apply R1 … β β … … … … bn+1 bi … … From Braids to Knots The Plat Closure of a Braid The Aharonov- Freedman- Jones- Kitaev- Landau (AFJKL) Aharonov- Freedman- Jones- Kitaev- Algorithm Close Braid Braid β Closed Braid β pl Strategy for Computing Jones Polynomial • We use the Jones representation What is the ρ A : Bn → TLn( d ) Temperley-Lieb algebra Temperley- Temperley- from the braid group Bn to the Temperley- Lieb algebra TLn (d) , where d is an (d) (d) TLn(d) indeterminate complex number, and where A is a complex number defined by d= -A2-A-2. d=- ??? 10 (d) Diagrammatic Representation of TLn (d) (d) Two Equal Elements of TLn (d) Rectangle = An element X of (d) TLn (d) X1 = X2 Two Unequal Elements of TLn (d) (d) Shorthand Notation Rectangle ≠ Shorthand Notation X1 ≠ X2 X X Product Product Times = = Times = = =d X1 • X 2 = X 3 11 Temperley- (d) Generators of the Temperley-Lieb Algebra TLn (d) (d) Relations Among the Generators of TLn (d) The Temperley- Lieb algebra Temperley- (d) TLn (d) is generated by Ei E j = E j Ei , for i − j ≥ 2 L L LL L Ei Ei ±1Ei = Ei , 1 ≤ i < n 1 E1 E2 L E n−1 E i 2 = dE i , 1 ≤ i < n (d) The Markov Trace Trn : TLn (d) à C Properties of the Markov Trace Original number of strands The Markov trace is the unique trace satisfying the following conditions: • Trn ( 1 ) = 1 Trn→ = = d 2 −n • Trn ( XY ) = Trn ( YX ) • X ∈ TLn −1 ( d ) ⇒ Trn −1 ( XEn ) = 1 Tr ( X ) d n Number of Loops Strategy for Computing Jones Polynomial • We use the Jones representation How to Compute the ρ A : Bn → TLn ( d ) Temperley- from the braid group Bn to the Temperley- Jones Polynomial (d) Lieb algebra TLn (d) , where d is an indeterminate complex number, and where A d=- is a complex number defined by d= -A2-A-2. 12 Temperley- (d) Temperley-Lieb algebra TLn(d) Temperley- (d) Temperley-Lieb algebra TLn(d) Let n be an integer, and let d be a complex number Let n be an integer, and let d be a complex number Temperley- The Temperley- Lieb algebra (d) TLn (d) is the algebra Temperley- The Temperley- Lieb algebra (d) TLn (d) is the algebra generated by generated by { I , E1, E2 , K , En −1 } { I , E1, E2 , K , En −1 } with relations with relations EiE j = EjEi , i − j ≥ 2 EiE j = EjEi , i − j ≥ 2 E i E i± 1 E i = E i E i E i± 1 E i = E i E i 2 = dE i E i 2 = dE i and with an involution * defined by the conjugate and with an involution * defined by the conjugate linear extension of (E ) linear extension of (E ) ∗ ∗ i 1 E i 2 L E ir = E ir E ir − 1 L E i1 i1 E i 2 L E ir = E ir E ir − 1 L E i1 The Jones Representation Strategy for Computing Jones Polynomial • Let d be an indeterminate complex number, and let A be a complex number such that d = - A 2 - A - 2 . • We use Jone’s definition of the Jones poly Jone’ • Let B n denote the braid group, and let TLn (d ) group, (d) • Let β be an element of B t=A n -4 denote the Temperley- Lieb algebra. Temperley- algebra. • Let β be the closure of β tr • Let Tr : TL (d) à C be the standard Markov n (d) n The Jones representation trace into the complex numbers C ρ A : Bn → TLn ( d ) • Let w(β ) be the writhe of β ,i.e., w(β w(β w(β ) = sum of exponents in braid word for β is the group representation defined by bi a AI + A−1Ei = A + A−1 Then the Jones polynomial is given by bi −1 a A−1 I + AEi Vβ tr ( A−4 ) = − A2 w ( β ) d n −1Trn ( ρ A ( β ) ) = A−1 + A Strategy (Cont.) The Graph Gk 1 2 3 K-1 We now wish to compute a good approximation 0 1 0 L 0 0 of the value of the Jones poly at e2πι /k for k 1 0 1 Adjacency a positive integer. 1 Matrix Mk = M O • Let 0 1 Gk be the graph 0 L 1 0 ( k− 1 )×( k −1) … 1 2 3 K-1 Eigenvector v u λ = ( λ l ) = ( sin ( π l / k ) ) K - 1 vertices & k- 2 edges l∈{ 1,2,K ,k− 1 Corresponding } Eigenvalue 2cos(π l/ k ) 13 The Graph Gk 1 2 3 K-1 Strategy (Cont.) We will use the graph Gk to construct a • Let d be equal to the eigenvalue 2cos(π/k) Temperley- unitary representation of the Temperley- 2cos(π (d) Lieb algebra TLn (d) . • Since d = -A -A 2 , there are four -2 • Let Pn,k = set of paths in Gk of length n possible chooses for A , namely which start at the vertex 1 . A = ± e± iπ / 2 k • Let H n,k be the Hilbert space with orthonormal basis We choose A = e − iπ / 2k {p : p ∈ Pn, k} (d) Constructing a Unitary Rep of TLn(d) (d) Constructing a Unitary Rep of TLn(d) (Cont.) • We begin by identifying each path p in Pn,k with a binary string of length n , where we interpret a bit • We now construct a unitary representation left” 0 to be “to the left” (a “Zig”) , and bit 1 to be “ to Zig” right” the right” (a “Zag” ) Zag” Temperley- (d) of the Temperley-Lieb algebra TLn (d) • For each path p , let p i -1] be the subpath corresponding to the first i - 1 bits of p . Let p [i+2 Φ : TLn ( d ) → £U ( Hn ,k ) be the subpath corresponding bits i+2 to the last bit of p. Finally, let p[i..i+1] be the subpath corresponding to bits i and i+1 of p. • Let ei (p ) be the endpoint of the subpath pi - 1] . So e i ( p ) ∈ {1,2,K , k − 1} (d) Constructing a Unitary Rep of TLn(d) (Cont.) (d) Constructing a Unitary Rep of TLn(d) (Cont.) • We define each Φi as • To define the rep Φ : TL (d ) → £U ( H ) n n ,k 0 if p [ iL i +1] = 00 all that we need do is to specify the images λ ei ( p) −1 λ e ( p ) − 1λ e ( p) +1 λe ( p ) p + i i p i − 1] 10 p[ i + 2 if p[ i L i+ 1 ] = 01 Φ i = Φ ( Ei ) on each of the generators Ei i λe i ( p ) of the Temperley-Lieb algebra TLn (d) , Temperley- (d) Φi p = λe i ( p ) − 1λ ei ( p) +1 i − 1] λ p 0 1 p[ i + 2 + e i ( p ) + 1 p if p [i L i +1 ] = 1 0 and then show that the Φi ' s are compatible λe i ( p ) λe ( p ) i Temperley- with the relations defining the Temperley- if p [ i L i + 1 ] = 1 1 Lieb algebra 0 14 (d) Constructing a Unitary Rep of TLn(d) (Cont.) (d) Constructing a Unitary Rep of TLn(d) (Cont.) • We define each Φi as • We then verify that Φ is actually a representation by checking that it is [ iL i + 1] = 00 0 Zig- Zig-Zag Zag- Zag-Zig if p Temperley- campatible with the following Temperley-Lieb p + ( Blah) p 10 p algebra identities: ( Blah− ) i −1] [ i +2 if p [ iLi +1] = 01 Ei E j = E j E i , i − j ≥ 2 Φi p = ( Blah) i −1] p 01 p [i +2 + Blah+ p if p [iLi +1] = 10 E i E i± 1 E i = E i Zig-Zag Zig- Zag- Zag-Zig E i 2 = dE i 0 if p [i Li +1 ] = 11 Φi Φj = ΦjΦi , i − j ≥ 2 i.e., the following hold ΦΦi±Φ = Φ i i 1 i Φi2 = dΦi Strategy (Cont.) (d) The Unitary Rep Φ of TLn (d) • Recall that we define each Φ i as • We now seek to construct a trace ~ 0 if p [ iL i +1] = 00 Tr: Im(Φ) à Tr: Im(Φ)àC which is compatible with the ~ (d) Markov trace on TLn (d) , i.e., a trace Tr ( Blah− ) p + ( Blah) pi −1]10 p[i +2 if p[iLi +1] = 01 such that the following diagram is commutative: Φi p = ( Blah) pi −1] 01 pi +2 + Blah+ p [ if p[iLi +1] = 10 TLn ( d ) → Φ Im ( Φ ) ⊂ £U ( H n ,k ) if p [ i Li +1] = 11 ± 0 Tr ] ↓ Tr Lemma. The representation Φ : TLn ( d ) → £U ( H n, k ) Lemma. £ maps each ket p to a linear combination of Markov Compatible Trace Trace kets each labeled by a path of the same length and the same enpoint as p . Constructing a Compatible Trace (d) The Unitary Rep Φ of TLn (d) Lemma. Lemma. The representation Φ : TLn ( d ) → £U ( H n, k ) • Let Pn,k,m = set of paths in Gk of length n maps each ket p to a linear combination of kets each labeled by a path of the same which start at the vertex 1 and end at vertex m . length and the same enpoint as p . Theorem. Theorem . • Let H be the Hilbert space with n,k,m k −1 k− 1 Φ = ⊕ Φ( m) : TLn ( d ) → ⊕ £U ( Hn, k ,m ) ⊂ £U ( Hn, k ) orthonormal basis m= 1 m= 1 {p : p ∈ Pn ,k ,m } where Φ ( m) :TLn ( d ) →£U ( Hn, k, m ) k −1 Hence, Im ( Φ) ⊆ ⊕ £U ( Hn, k, m ) m =1 15 Constructing a Compatible Trace (Cont.) Constructing a Compatible Trace (Cont.) k−1 ⊕ k−1 Recall TL ( d ) → (m) Φ= Φ ⊕ £U ( H k −1 ) ⊂ £U ( Hn, k ) Φ=⊕ Φ (m) k−1 TLn ( d ) → ⊕ £U ( H ) ⊂ £U ( Hn, k ) m=1 n n ,k ,m m=1 m =1 n ,k ,m m =1 Tr ] ± Tr ] Tr ? ± Tr ? £ £ k −1 So Φ ( X ) = ⊕ Φ (X) Claim : The following linear combination is Claim: (m) m =1 compatible with the Markov trace. This give us a lot of latitude in choosing a k −1 Tr n ( Φ ( X ) ) = ± ∑λ Tr ( Φ( m ) ( X ) ) trace. We can choose from among all 1 possible linear combinations of the standard m N m =1 traces { Tr ( Φ (m) ( X )) : 1≤ m < k } where N= ∑ k −1 m=1 λ m Dim ( H n, k ,m ) Recall … Constructing a Compatible Trace (Cont.) 1 2 3 K-1 The Graph Gk k−1 ⊕1 Φ (m) Φ= k−1 0 1 0 L 0 0 TLn ( d ) → m= ⊕ £U ( H m =1 n ,k ,m ) ⊂ £U ( Hn, k ) 1 0 1 Tr ] Adjacency ± Tr ? 1 Matrix Mk = M O £ 0 1 0 L 1 0 ( k− 1 )×( k −1) Claim : The following linear combination is Claim: Eigenvector compatible with the Markove trace. v u λ = ( λ l ) = ( sin ( π l / k ) ) k −1 Tr n ( Φ ( X ) ) = ± ∑λ Tr ( Φ( m ) ( X ) ) 1 } l∈{ 1,2,K ,k− 1 m Corresponding N m =1 Eigenvalue where ∑ λ Dim ( H n, k ,m ) k −1 N= m=1 m 2cos(π l/ k ) Proving Our Trace is Capatible Where are we ??? Compatible Trace k −1 ⊕ Φ= Φ (m ) We prove each of the following: We have Bn → TL n ( d ) → ρ A m= 1 U ( H n ,k ) • ± Tr o Φ ( 1 ) = 1 ] ± • Jones Rep ↓ Tr n ± ± Tr o Φ ( XY ) = Tr o Φ ( YX ) T rn £ • Markov Trace ± 1± X ∈ TLn− 1 ( d) ⇒ Tr o Φ ( XE n −1 ) = Tr o Φ ( X ) d d = 2cos ( π / k ) Since the Markov trace is the unique trace satisfying the above three conditions, we have that the following where d = -A2-A-2 and A = e − i π / 2k diagram is commutative. t = A−4 = e 2π i / k k −1 Φ = ⊕ Φ( m ) k− 1 T Ln ( d ) m → =1 ⊕ £U (H m =1 n , k ,m ) ⊂ £U (H )n ,k Moreover, we have the Jones polynomial ( ) Vβ tr ( t ) = − A d n− 1Trn ( ρ A ( β ) ) Tr ] ± 2 w β tr ↓ Tr Markov Compatible Trace £ Trace 16 Where are we ??? Where are we ??? We need to compute Trn ( ρ A ( β ) ) All we need to do is to compute the traces of But ± Trn ( ρ A ( β )) = Trn ( Φ o ρ A ) ( β ) ( ) U ( m ) = Φ ( m ) o ρ A ( β ) for 1 ≤ m < k ± k −1 Let L = Trn ⊕ Φ (m ) o ρ A ( β ) β = ∏ b j( l )ε ( l) m =1 l =1 k −1 Tr ( Φ o ρ A ) ( β ) 1 = N ∑λ m =1 m (m ) Let U j( m ) = ( Φ ( m ) o ρ A ) b j ( ) for 1 ≤ m < k 1≤ j <n Thus, all we need to do is to compute the traces of the unitary transformations Then U ( m ) = ( Φ( m ) o ρ A ) ( β ) = ∏ U ( m ) j ( l ) ( ) L ε (l ) U ( m ) = ( Φ( m ) o ρ A ) ( β ) for 1 ≤ m < k l =1 Where are we ??? Overview Thus, all reduces to the problem of computing The algorithm consists of two phases: the traces of ∏ (U ) L ε (l ) U (m) = (m) j ( l) Phase 1: Compilation Phase – We compile l= 1 Mathematics into hardware. Program) (Braid Word = Computer Program ) for 1 ≤ m < k , where Phase 2: Execution Phase – We run our compiled program U j( m ) = ( Φ( m ) o ρ A ) bj ( ) Compilation Phase for Estimating V β Tr (e 2π i / k ) Execution Phase for Estimating V β Tr (e 2π i / k ) Select Select Algorithm AJK (n , k ) k Input k for & Trace = 0 Input e2π i / k loop m = 1 K k − 1 Software Compilation Tracem = 0 and λm = sin ( π m / k ) loop p ∈ P ,k ,m n β = ∏ j= 1 bj( i )ε ( i ) L Re = 0 and Im = 0 n Hardware Compilation loop Iteration = 1 K NumberIterations ReBit = QRe ( p ) and ImBit = QIm ( p ) {U j ( m) } :1 ≤ m < k , 1 ≤ j < n Re = Re + ( −1 ) ReBit and Im = Im − ( − 1) ImBit Hardware Compilation end Iteration-loop Trace m = Tracem + ( Re + i *Im) / NumberIterations {U (m ) : 1 ≤ m < k} end p -loop Trace = Trace + λm * Tracem Hardware Compilation end m -loop output Trace Stop {Q R em, Q I mm :1 ≤ m < k } endAlgorithm 17 Execution Phase for Estimating V β Tr (e 2π i / k ) Execution Phase for Estimating V β Tr (e 2π i / k ) Algorithm AJK (n , k ) Algorithm AJK (n , k ) Trace = 0 Trace = 0 loop m = 1 K k − 1 loop m = 1 K k − 1 Tracem = 0 and λm = sin ( π m / k ) Tracem = 0 and λm = sin ( π m / k ) loop p ∈ P ,k ,m n loop p ∈ P ,k ,m n Re = 0 and Im = 0 Re = 0 and Im = 0 loop Iteration = 1 K NumberIterations loop Iteration = 1 K NumberIterations ReBit = QRe ( p ) and ImBit = QIm ( p ) ReBit = QRe ( p ) and ImBit = QIm ( p ) Re = Re + ( −1 ) and Im = Im − ( − 1) Re = Re + ( −1 ) and Im = Im − ( − 1) ReBit ImBit ReBit ImBit end Iteration-loop end Iteration-loop Trace m = Tracem + ( Re + i *Im) / NumberIterations Trace m = Tracem + ( Re + i *Im) / NumberIterations end p -loop end p -loop Trace = Trace + λm * Tracem Trace = Trace + λm * Tracem end m -loop end m -loop output Trace output Trace endAlgorithm endAlgorithm Execution Phase for Estimating V β Tr (e 2π i / k ) The Two Quantum Subroutines QRe & QIm Algorithm AJK (n , k ) Trace = 0 • QRe – Used to compute the real part loop m = 1 K k − 1 Tracem = 0 and λm = sin ( π m / k ) of trace, i.e., Re ( Tr ( U ) ) loop p ∈ P ,k ,m n Re = 0 and Im = 0 loop Iteration = 1 K NumberIterations ReBit = QRe ( p ) and ImBit = QIm ( p ) Re = Re + ( −1 ) and Im = Im − ( − 1) ReBit ImBit end Iteration-loop Trace m = Tracem + ( Re + i *Im) / NumberIterations • QIm – Used to compute the imaginary part end p -loop of the trace, i.e., Im ( Tr ( U ) ) Trace = Trace + λm * Tracem end m -loop output Trace endAlgorithm The Quantum Subroutines QRe In Out Trace Estimation U QRe Re Tr ( U ) 18 Unitary Wiring Diagram for Unitary Transformation Transformation Controlled -U U :H → H Contr - U : H 2 ⊗ H → H 2 ⊗ H n-Dimensional Hilbert Space 2n-Dimensiomal 2n- Control-Qubit Control- HilbertSpace 0 ⊗ 0 ⊗ψ ψ U Uψ ψ U 0 ⊗ψ Do Zilch if control = 0 Output = 1 ⊗U ψ Apply U if control = 1 Unitary Unitary Transformation Controlled -U Transformation Controlled -U Contr - U : H 2 ⊗ H → H 2 ⊗ H Contr - U : H 2 ⊗ H → H 2 ⊗ H 2n- 2n-Dimensiomal 2n-Dimensiomal 2n- Control- Control-Qubit Control-Qubit Control- HilbertSpace HilbertSpace 1 a 0 +b1 ⊗ 1 ⊗U ψ ⊗ a 0 ⊗ψ +b 1 ⊗ ψ U ψ U ψ U 0 ⊗ψ Do Zilch if control = 0 0 ⊗ψ Do Zilch if control = 0 Output = Output = 1 ⊗U ψ Apply U if control = 1 1 ⊗U ψ Apply U if control = 1 Trace(U) Estimating Trace(U ) Engineering a Probability Distribution Hadamard H= 1 1 1 0 H H Output = 0 or 1 Transform 2 1 − 1 ⊗ U Measure p H H Basis elt ( Prob (0 ) − Prob ( 1) = Re U pp ) Real part of p-th diag elt If we repeat many times, then #1's − #0's U #Bits ≈ Re ( p | U | p ) = R e U pp ( ) Thus, we have constructed a probability distribution We have used Quantum Mechanics to for computing the real part of engineer a probablity distribution !!! Trace (U ) = ∑ U pp = ∑ p | U | p p p 19 The Quantum Subroutines QRe The Quantum Subroutines QIm In Out U QIm Im Tr ( U ) In Out U QRe Re Tr ( U ) 1 0 H 0 i H U Summary Question We have constructed a polytime quantum Can we transform the AFJKL quantum algorithm: algorithm using polynomial interpolation into a polytime quantum algorithm that computes the Jones polynomial V K ( t ) with exponentially Knot K= Jones polynomial small probability of error ? Integer k AFJKL ( ) V K e 2π i / k ± ε No ! Algorithm with exponentially Answer: Answer: Precision ε small probability of error Jones Polynomial ??? atica Inform i Per Can 20

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Jones polynomial, the Jones, link diagram, Alexander polynomial, skein relation, knot theory, bracket polynomial, knots and links, Kauﬀman bracket, Knot Invariants

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posted: | 5/27/2011 |

language: | English |

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