Unitary Solutions to the Yang-Baxter. Equation in Dimension Four

Transcription

1 arxiv:quant-ph/02050 v3 Aug 2003 Unitary Solutions to the Yang-Baxter Equation in Dimension Four H. A. Dye Department of Mathematics, Statistics, and Computer Science University of Illinois at Chicago 85 South Morgan St Chicago, IL August, 2003 Abstract In this paper, we determine all unitary solutions to the Yang- Baxter equation in dimension four. Quantum computation motivates this study. This set of solutions will assist in clarifying the relationship between quantum entanglement and topological entanglement. We present a variety of facts about the Yang-Baxter equation for the reader unfamiliar with the equation.

2 Acknowledgement. This effort was sponsored by the Defense Advanced Research Projects Agency (DARPA) and Air Force Research Laboratory, Air Force Materiel Command, USAF, under agreement F The U.S. Government is authorized to reproduce and distribute reprints for Government purposes notwithstanding any copyright annotations thereon. The views and conclusions contained herein are those of the author and should not be interpreted as necessarily representing the official policies or endorsements, either expressed or implied, of the Defense Advanced Research Projects Agency, the Air Force Research Laboratory, or the U.S. Government. (Copyright 2002.) Introduction In this paper we classify all unitary solutions to the Yang-Baxter equation in dimension four. These solutions represent both braid operators and quantum operators which describe topological entanglement and quantum entanglement respectively. The motivating question for this work in found in [] in which Kauffman and Lomonaco compare the concepts of entanglement in the areas of topology and quantum computing. They suggest that quantum entanglement may be described in relation to a set of entangled links. This classification describes all 4 4 unitary matrices that are both braid operators and quantum operators. Determining all unitary, 4 4 matrix solutions to the Yang-Baxter equation will assist us in studying the relationship 2

3 between these two forms of entanglement. We introduce quantum operators and braid operators in Section. We also discuss the connection between the Yang-Baxter equation and braids in this section. In Section 2, we examine the algebraic properties of the Yang-Baxter equation. We will apply these properties to determine unitary solutions to the Yang-Baxter equation. In Section 3, we describe these properties in terms of matrices. We recall facts about matrices that are conjugate to unitary matrices. We state the equation used to determine the unitary matrices in Proposition 3.4 and prove a variety of facts chosen to simplify the proof of Theorem 4.. In Section 4, we present a complete list of unitary 4 4 matrix solutions to the algebraic Yang-Baxter equation that consists of five families of unitary matrices. These families are conjugacy classes determined by matrices that preserve unitarity and the property that the conjugate is a solution to the Yang-Baxter equation. However, the four families are conjugate under a larger class of matrices that do not necessarily preserve these properties. We obtain the unitary families by analyzing each solution given in [3] and applying facts from the earlier sections. In Section 5, we determine which unitary solutions are also solutions to the bracket skein equation given in [5]. 3

4 Braid Operators and Quantum Gates Topological entanglement and quantum entanglement are non-local structural features that occur in topological and quantum systems, respectively. Kauffman introduces this fact in [] and asks the following question. What is the relationship between these two concepts? We describe these concepts as an introduction to the Yang-Baxter equation. Topological entanglement is described in terms of link diagrams and via the Artin braid group. The Artin braid group on n-strands is denoted by B n and is generated by {σ i i n }. The group B n consists of all words of the form σ ± j σ ± j 2...σ ± j n modulo the following relationships. σ i σ i+ σ i = σ i+ σ i σ i+ for all i n σ i σ j = σ j σ i such that i j > This group has the following diagrammatic representation. A n-strand braid is the immersion of n arcs with over and under crossing information at each singularity. Furthermore, any transverse arc intersects the braid at n points, except at the levels at which singularities occur. Each generator σ i of the Artin braid group is associated to a diagram as shown in Figure. Multiplication of diagrams is performed by concatenation of the diagrams from upper to lower. We show an example in Figure 2. The diagrammatic form of the relationships in the Artin braid group are shown in Figure 3. 4

5 2... n id σ i 2 i i+ n Figure : Generators of the Artin Braid Group σ σ Figure 2: Multiplicaton To obtain the braided Yang-Baxter equation, we associate a vector space V to each endpoint in the diagram. Each n-strand braid represents a linear map from V n to V n. The generator σ i is associated to a map R : V V V V in the following manner: σ i I I R I I with the R in the ith position and all the I s representing the identity map. This representation will respect the relations of the Artin braid group. In the three strand case, this corresponds to the braided Yang-Baxter equation. (R I)(I R)(R I) = (I R)(R I)(I R) () Notice that R is a linear map and may be expressed as a matrix for some basis of V. For the remainder of this paper we will refer to the Yang-Baxter equation as the ybe. 5

6 σ σ σ σ σ σ i i+ i i+ i i σ i σ j σ σ j i Figure 3: Relations We algebraically describe quantum entanglement as the action of of a unitary, linear map with certain properties. In a quantum system, quantum states are elements of V n where V is two dimensional vector space over C. Using Dirac notation, we express the basis of V as { 0 >, >}. The term qubit refers to vectors of the form α 0 > +β > such that α 2 + β 2 =. If α and β are both non-zero then the qubit is said to be in a state of superposition. This state collapses to 0 > with probability α 2 and the state > with probability β 2. The n-qubit is an element of V n. If the n-qubit is in superposition, it may simultaneously represent 2 n states of classical information with associated probabilities. The n-qubit will collapse to a single outcome when measured, but until then will carry information about all 2 n states. A quantum gate acts on a qubit so that the sum of the probabilities of all states is preserved. In quantum computing, we assemble a series of quantum 6

7 gates to perform a computation. Each quantum gate may be represented as a unitary matrix. Consider the case of a two-qubit. This is a quantum state of the form: α 0 > 0 > +β 0 > > +γ > 0 > +δ > >. This state is said to be entangled if it can not be written in the form: (x 0 > +y >)(ˆx 0 > +ŷ >). Note that this implies that αδ βγ 0. We may transform an unentangled state into a entangled state by application of a quantum gate. In general, a n-qubit state ψ is said to be entangled if ψ can not be written in the form P(ψ ψ 2 ) where P : V n V n is a permutation of the tensor factors of V n and ψ V l, ψ 2 V k such that l + k = n. The relationship between topological and quantum entanglement is not fully understood. We examine the properties of the Yang-Baxter equation in the next section. 2 The Yang-Baxter Equation Let V be a vector space over a field F. Let R be a linear map: R : V V V V Let I be the identity map on V then R is said to be a solution the braided Yang-Baxter equation if the following holds: (R I)(I R)(R I) = (I R)(R I)(I R) 7

8 Now assume that V is finite dimensional and that {v 0, v, v n } is a basis for V over F. We may denote the basis of V V as {v i v j i, j {0,, 2...n }}. Using this basis we may describe R by its action on the generators of V V : R(v i v j ) = k,l R kl ij v k v l Applying the original definition of the Yang-Baxter equation, we may describe R as a series of equations that are dependent on the choice of i, j, k and x, y, z. a,b,c R ab ij Rcz bk Rxy ac = m,n,p R np jk Rxm in Ryz mp We prove the following propositions about solutions to the braided Yang- Baxter Equations. These facts will be used to obtain unitary solutions to the braided Yang-Baxter equation. These propositions also indicate that a single solution produces a class of solutions by conjugation and scalar multiplication [3], [4]. Proposition 2.. If R is a solution to the braided Yang-Baxter equation then the following hold. i) If α F, then αr is a solution to the braided Yang-Baxter equation. ii) If Q is an invertible map, Q : V V, then (Q Q)R(Q Q) is a solution to the braided Yang-Baxter equation. iii) If R is invertible, then R is a solution to the braided Yang-Baxter equation. Proof: See [4] page 68. 8

9 We now define the algebraic Yang-Baxter equation. Let τ : V V V V be the map such that τ(u v) = (v u). Let R 2, R 3, R 23 : V V V V be linear maps. Let R 2 = (R I), R 3 = (I τ)(r I)(I τ), and R 23 = (I R). Then R is a solution to the algebraic Yang-Baxter equation if: R 2 R 3 R 23 = R 23 R 3 R 2 (2) In this paper, we will refer to the equation as the algebraic Yang-Baxter equation and do not follow the notation of [3]. In [3], the equation is refered to as the quantum Yang-Baxter equation. We obtain a complete list of solutions to the algebraic Yang-Baxter equation from [3]. The algebraic Yang-Baxter equation is related to the braided ybe by the linear transformation τ of V V. Proposition 2.2. If R is a solution to the braided Yang Baxter equation, and τ : V V V V such that τ(u v) = v v then R τ is a solution to the algebraic Yang-Baxter equation. Similarly, if ρ is a solution to the algebraic Yang-Baxter equation, then ρ τ is a solution to the braided Yang-Baxter equation. Proof: See [4]. Note that if we obtain a solution to the algebraic Yang-Baxter equation, we may apply τ to obtain a solution to the braided Yang-Baxter equation. 9

10 Using the basis for V V, we rewrite equation 2 as a sum of basis elements. R 2 R 3 R 23 (v i v j v k ) = R 2 R 3 (I R)(v i v j v k ) = R 2 R 3 R b,c j,k (v i v b v c ) b,c = R 2 (I τ)(r I) R b,c j,k (v i v c v b ) b,c = R 2 (I τ) R b,c i,c (v x v a v b ) = R 2 = x,a,b,c R b,c j,k Rx,a x,a,b,c R b,c k,j Rx,a i,c Ry,z y,z,x,a,b,c j,k Rx,a i,c (v x v b v a ) Consider the right hand side of the equation and obtain: R 23 R 3 R 2 (v i v j v k ) = R 23 R 3 m,n R m,n b,a (v x v y v z ) i,j (v m v n v k ) = R 23 (I τ)(r I) m,n = R m,n i,j R p,x m,k Ry,z x,y,z,m,n,p R m,n i,j (v m v n v k ) p,n (v x v y v z ) Therefore, a solution to the algebraic Yang-Baxter equation satisfies the following equations: R b,c k,j Rx,a i,c Ry,z a,b,c b,a (v x v y v z ) = m,n,p R m,n i,j R p,x m,k Ry,z p,n (v x v y v z ). 0

11 3 Matrix Representation We may represent solutions to the braided Yang-Baxter equation and the algebraic Yang-Baxter equation as matrices. We study matrix notation for the case where the dimension of V is two. In this section, we present facts about matrix solutions to the Yang-Baxter equation and facts about unitary matrices. We present a method that determines families of unitary solutions to the Yang-Baxter equation. Recall that we may conjugate R by Q Q or multiply by a scalar. The representation of Q Q has a specific matrix form that will assist in determining the unitary solutions. Let V be finite dimensional. If R is a solution to the quantum Yang- Baxter equation, we may rewrite R as a matrix R. that: Let R be a solution to the algebraic Yang-Baxter equation, and recall R(v i v j ) = a,b R may be written in matrix form R ab ij R ab ij v a v b where ij represents the column and ab the row of the matrix. In particular, if the dimension of V is two then V has a basis {v 0, v }. Hence V V has a basis of the form {v 00, v 0, v 0, v }. R = R00 00 R0 00 R0 00 R 00 R00 0 R0 0 R0 0 R 0 R00 0 R0 0 R0 0 R 0 R00 R0 R0 R

12 The matrix R acts on a vector of the form a b c c that represents a(v 0 v 0 ) + b(v 0 v ) + c(v v 0 ) + d(v v ). Suppose Q : V V is an invertible linear map. We may rewrite Q as the matrix Q: Q0 0 Q 0 such that Q 0 0 Q Q0 Q 0 0 Q 0 Q Let A denote the linear map Q Q : V V V V. This map may be represented as a matrix A. In the case where the dimension of V is two, A is a 4 4 matrix of the form: Q 0 0 Q0 0 Q 0 0 Q0 Q 0 Q0 0 Q 0 Q0 Q 0 0 Q 0 Q 0 0 Q Q 0 Q 0 Q 0 Q Q 0 Q0 0 Q 0 Q0 Q Q0 0 Q Q0 Q 0 Q 0 Q 0 Q Q Q 0 Q Q For each matrix solution in [3], there is a family of solutions to the algebraic Yang-Baxter equations. To obtain another element of this family, we conjugate R by A. We wish to obtain all unitary families of solutions. Note that it is not sufficient to determine which representatives are unitary. Although R may not be unitary, it is possible that ARA is unitary for some A. 2

13 Linear Algebra Facts We prove facts about matrix solutions to the algebraic Yang-Baxter equation. These facts determine a preliminary approach to evaluating whether or not a family of solutions will produce any unitary solutions. Let M be a finite dimensional matrix. M T denotes the transpose of the matrix M. Let M denote the conjugate of M. Let M denote the conjugate transpose of the matrix M. Recall that M is a unitary matrix if M = M. Proposition 3.. Recall the following matrix facts. i)let A be an invertible matrix then (A ) = (A ). ii)if A is a unitary matrix and let α C, then αa is a unitary matrix if and only if α =. iii) If A is a unitary matrix and λ is an eigenvalue of A, then λ =. Proposition 3.2. If ARA is unitary, then i) AR A = (A ) R A implying that A AR = R A A, and ii) R and R have the same set of eigenvalues. Proof: Suppose ARA is unitary. By the definition of unitary: (ARA ) = (ARA ) AR A = (A ) R A AR A = (A ) R A A AR = R A A 3

14 R and R are conjugate by A A. Therefore, they have the same set of eigenvalues. We make the following observations about an invertible matrix Q of the form Proposition 3.3. If Q = a b c d c = a b d then Q Q is diagonal. If A = Q Q then A A is also diagonal. Proof: Recall that the matrix form of Q Q is: aa ab ba bb ac ad bc bd A = ca cb da db cc cd dc dd Let H = A A. Let x = (aā + c c), y = (b b + d d), and z = (a b + c d). x 2 x z zx z 2 xz xy zz zy H = zx z z yx y z z 2 zy yz y 2 If c = a b d 4

15 then z = 0 and H is a diagonal matrix. The relationship between a matrix Q of this form and a 4 4 matrix B may be simplified using the following facts. Proposition 3.4. Let H = A A and let D = HB B H. Then D ij = k H ik B kj m B imh mj If H is a diagonal matrix, then D ij = H ii B ij B ij H jj We will apply these facts to determine unitary solutions to the algebraic Yang Baxter equation. Proposition 3.5. If H is constructed as before then H ii 0 for all i. Proof: If H i,i = 0 for some i, then either x = 0 or y = 0. This contradicts the assumption that Q is an invertible matrix. Proposition 3.6. Suppose H is constructed as before. If H ij = 0 for some i j then H ij = 0 for all i j Proof: We observe that H ij = k A ik A kj. If H ij = 0 for some i j then we observe that x = 0,y = 0, or z = 0. If x = 0 then a 2 + c 2 = 0 implying that a = c = 0 and Q is not invertible. This contradicts our definition of Q. We have a similar contradiction if y = 0. 5

16 Note that z is a factor of each off diagonal entry. We have eliminated the the other possiblities. Therefore x 0, y 0 and z = 0 which forces H to be a diagonal matrix. We apply these facts to the families of solutions in [3]. If the matrix R produces unitary solutions under conjugation, then R is of rank 4 and R and R have equivalent sets of eigenvalues. Further, if ARA is a solution to the Yang-Baxter equation, then A AR = R A A. If R is a unitary solution to the quantum Yang-Baxter, then αr is a unitary solution if and only if α =. 4 Solutions to the Yang-Baxter Equation Theorem 4.. There are five families of 4 4 unitary matrix solutions to the braided Yang-Baxter equation. Each solution has the form: kara T 6

17 where k is a scalar of norm one, Q is an invertible matrix such that: Q = a b c d a 2 ab ba b 2 ac ad bc bd A = Q Q = ca cb da db c 2 cd cd d 2 and T = Additional information about the matrices R and A are specified for each family. Family : 7

18 The matrix R has the form p 0 0 R = 0 0 q r and (3) = p p = q q = r r. The variable c in the matrix Q has the restriction that c = a b d. Family 2: The matrix R has the form: p R = q and (4) 8

19 when c a b d and pq =, p = (b b + d d)(āb + cd) (aā + c c)(a b + c d), q = (aā + c c)(a b + c d) (b b + d d)(āb + cd). Family 3: The third family consists of matrices with the form: p R = and (5) q pq =, p p = The matrix Q has the restriction that (d d) 2 (aā) 2, q q = (aā)2 (d d) 2. c = a b d. 9

20 Family 4: The matrix R has the form 0 0 (2) 0 0 (2) (2) R = 0 0 (2) (2) 0 0 (2) The matrix Q has the following restrictions: c = a b d a = d. (2) (2). (6) Family 5: The matrix R has the form R =. (7) There are no additional restrictions on the matrix Q. Remark 4.. We may transform the elements of families 2 and 3 into elements from family by conjugation by unitary matrices. However, these matrices do not have the form Q Q. 20

21 Remark 4.2. The matrices of families 2 and 3 perform quantum entanglement. The matrices from families 2 and 3 do not detect knotting but do detect linking. Family 4 performs quantum entanglement and detects knotting and linking [2]. Proof: If φ is a solution to the algebraic Yang-Baxter equation and τ : V V V V is a linear transformation such that τ(v i v j ) = v j v i, then φ τ is a solution to the braided Yang Baxter equation. We may express τ as the unitary matrix T since V V has a basis {v 0 v 0, v 0 v, v v 0, v v }. Note that R is a unitary solution to the algebraic Yang-Baxter equation if and only if RT is a unitary solution to the braided Yang-Baxter equation. To find all unitary solutions to the braided Yang-Baxter equation, it suffices to determine all unitary solutions to the algebraic Yang-Baxter and multiply by T. The families of solutions to the algebraic Yang-Baxter equation in [3] are indicated by a single element. This representative element is is conjugated by the matrix Q Q and multipled by a scalar k to produce the entire family. If a family of solutions produces any unitary solutions, the representative of the family must be invertible From [3] we obtain the following list of invertible 2

22 representatives: R0 = R02 = R03 =

23 p p q q p 2 q 2 0 p 2 + q 2 p 2 q 2 0 R = 0 p 2 q 2 p 2 + q 2 0 p 2 q p 2 2 p q q 2 p 0 0 k 0 p p q 0 R2 = 0 0 q q k 2 k p k p p q 0 k 2 0 k q R3 = 0 0 k 2 k q k p 0 0 k 0 R4 = 0 k 0 0 q

24 k k p k 2 p q 0 R2 = 0 0 k q k 2 k k p k 2 p q 0 R22 = 0 0 k q p q k p q s 0 k 0 q R23 = 0 0 k p k k p 0 0 R3 = 0 0 q s If R produces unitary solutions, then R and R are conjugate. We eliminate from this list all matrices that do not satisfy the condition that R and R have the same set of eigenvalues. From this computation, we determine only the representatives: R0, R02, R03, R2, R3, R4, R2, R22, R23, and R3 could produce unitary solutions to the Yang-Baxter equation. The eigenvalues in each of these matrices are non-zero, using this fact we have 24

25 removed one degree of freedom from each representative. Multiplying by a non-zero eigenvalue determines the new representatives used in the following cases. To determine if a representative produces unitary solutions, we refer to Proposition 3.4. If R and Q produce a unitary matrix, then D ij = 0 for all i, j. Case: R3 We consider the case of the diagonal matrix R3. Note that if R3 and R3 have the same set of eigenvalues then R3 is unitary and each eigenvalue has norm one. Further, the eigenvalues are the diagonal elements p 0 0 R3 =. 0 0 q k The inverse and conjugate transpose of the matrix are: R3 0 p 0 0 = R p =. 0 0 q q k k To determine the family of solutions given by R3 we examine the equation from Proposition 3.4. D ij = kh ik R3 k,j m R3 im H mj 25

26 To obtain a unitary R matrix under conjugation by Q, each Each D ij = 0. R3 is a diagonal matrix and by Proposition 3.3 D ij = H ij R3 jj R3 ii H ij 0 = (R3 jj R3 ii )H ij From this equation, we obtain two cases. In case one, R3 jj R3 ii = 0 for all i, j. This implies that R3 is the identity matrix, and we obtain solutions that are scalar multiples of the identity matrix. In the second case, R3 is not a multiple of the identity. If R3 is not a multiple of the identity, then (R3 jj R3 ii ) 0 for some i j. Therefore, H ij = 0 for i j. By Proposition 3.6, H is a diagonal matrix and c = a b d This produces the first family of solutions. Case: R2 The matrix R2 is not unitary. We observe that R2 and R2 have the same set of eigenvalues if p = q =. 26

27 p pq 0 R2 = 0 0 q The inverse and conjugate transpose of the matrix R2 are: R2 0 p 0 0 = R2 0 = + 0 p pq 0 pq q q To determine when R2 produces unitary solutions under conjugation, we examine: D ij = kh ik R2 kj m R2 im H mj If H and R2 produce a unitary solution, then D is the zero matrix. Examine the entry D 2 = k H k R2 k2 m R2 mh m2 0 = H 2 p H 2 0 = ( p )H 2 From this equation, we obtain two cases: either H is a diagonal matrix by 27

28 Proposition 3.6 or p =. If p = D 23 = k H 2k R2 k3 m R2 2mH m3 0 = H 22 ( q ) + H 23 q H 23 q 0 = H 22 ( q ) We conclude that q = since if H 22 = 0 then Q is not invertible. Therefore if p = then q = which is an element of family. If H is a diagonal matrix then implying that c = a b d D ij = H ii R2 ij R2 ij H jj D 23 = H 22 R2 23 R2 23H 33 D 23 = H 22 ( pq + ) D 23 = H 22 ( pq + ) Hence + = 0 or p =, which is a subcase of family. All unitary pq q solutions produced by R2 are in family. Case: R22 The matrix R22 is of the following form after scaling. Note that p = q =. 28

29 p pq 0 R22 =. 0 0 q pq The inverse and conjugate transpose of R22 have the form R22 0 p pq = + 0 and R22 0 p 0 0 = q 0 pq q pq pq To determine if R22 produces unitary solutions under conjugation, we examine: D ij = k D 2 = k H ik R22 kj m H k R22 k2 m R22 im H mj R22 mh m2 0 = p H 2 H 2 0 = ( p + )H 2 This implies H is a diagonal matrix, by Proposition 3.6, or p =. Let p =, 29

30 and examine D 23 = k H 2k R22 k3 m R22 2m H m3 0 = H 22 ( q ) + H 23 q H 23 q 0 = H 22 ( q ) If H 22 = 0 then Q is not invertible. The solution which consists of p = q = is a subcase of family. If H is a diagonal matrix, then c = a b d and D ij = H ii R22 ij R22 ij H jj D 23 = H 22 ( pq + ) 0 = H 22 ( pq + ) Hence p = q which is a subcase of family. 30

31 The matrix R23 We examine the matrix R23. We may multiply by a scalar and assume that p q s 0 0 q R23 = 0 0 p In this case: p q s + 2pq R q = 0 0 p and R23 p 0 q = q 0 p s q p We obtain D 2 = k H k R23 k2 m R23 m H m2 0 = H ( p) + H 2 H 2 0 = ph, Note that p = 0. If H = 0, the matrix Q is not invertible. We consider 3

32 the following two entries. D 3 = k H k R23 k3 m R23 m H m3 0 = H q + H 3 H 3, 0 = qh. D 4 = k H k R23 k4 m R23 m H m4 0 = H ( 2 + 2pq) H 2 p + H 3 q + H 4 H 4, 0 = ( s + 2pq)H H 2 p + H 3 q. Hence p = q = s = 0. R23 produces trivial solutions which are contained in family. Case: R2 After scaling, the matrix R2 has the form: 0 0 k 0 q 0 R2 := 0 0 q q 32

33 The inverse and the conjugate transpose of the matrix R2 are: 0 0 k q R2 0 0 q = and R = q 0 q q k 0 0 q q Observe that D 23 = k H 2k R2 k3 m R2 2m H m3 0 = H 22 ( q ) + H 23 q H 23 0 = ( q )(H 22 H 23 ). Therefore, either q = or H 22 = H 23. If q = then D 4 = k H k R2 k4 m R2 m H m4 0 = kh + H 4 H 4 0 = kh, and k = 0 since H 0. This solution is part of family. Suppose that H 22 = H 23 then det(q) = 0 contradicting the assumption that Q was invertible. 33

34 Case: R3 The matrix R3 has the form p p pq 0 0 q R3 :=. 0 0 q The inverse and conjugate transpose of R3 are p p pq R3 0 0 q =, R3 p 0 0 =. 0 0 q p pq q q We consider the equation: D 2 = k H k R3 k2 m R3,m H m2 0 = ph + H 2 H 2 0 = ph 34

35 H 0 by Proposition 3.5 If p = 0 then D 24 = k H 2k R3 k4 m R3 2m H m4 0 = qh 22 + qh 23 + H 24 H 24 0 = qh 22 + qh 23 If H 22 = H 23 then det(q) = 0. Hence, q = 0 and this solution is a subcase of family. Case: R4 The matrix has the following form after scaling: p R4 = q We assume that pq = and that the inverse and conjugate transpose are q q R = and R = p p 35

36 Consider the individual entry D 43 = k H 4k R4 k3 m R4 4m H m3 0 = H 42 ph 3 0 = (b b + d d)(a b + c d) p(aā + c c)(āb + cd). As a result or p = (b b + d d)(āb + cd) (aā + c c)(a b + c d) c = a b d. Suppose that c a b d and p = (aā+c c)(āb+ cd) then (b b+d d)(a b+c d) D 2 = k H k R4 k2 m R4 mh m2 0 = H 3 qh 42 0 = (aā + c c)(āb + cd) q(b b + d d)(a b + c d) Hence q = (aā + c c)(a b + c d) (b b + d d)(āb + cd) 36

37 Each entry is now zero in the matrix D. If p = (b b + d d)(āb + cd) (aā + c c)(a b + c d) q = (aā + c c)(a b + c d) (b b + d d)(āb + cd) we obtain the second family of solutions. Now suppose that c = a b d. All entries in D except the following two entries are zero. D 4 = k H k R4 k4 m R4 m H m4 0 = H q qh 44 0 = (aā + c c) 2 q q(b b + d d) 2 This implies that: and a 2 = q d 2. (d d + b b) 2 (ā 2 a 2 q qd 2 d2 ) qd 2 d 2 = 0 D 4 = k H 4k R4 k m R4 4m H m 0 = H 44 p ph 0 = (b b + d d) 2 p(aā + c c)2 p 37

38 This implies that: (d d + b b) 2 (p pā 2 a 2 d 2 d2 ) pd 2 d 2 = 0 and p a 2 = d 2. This produces the third family of solutions. Case: R0 Consider the matrix R R0 := The inverse and conjugate transpose of this matrix are R = R := We observe D 4 = k H 4k R0 k m R0 4,m H m, 0 = H 4 H H 4 0 = H. 38

39 If H = 0 then Q is not an invertible matrix. This matrix produces no unitary solutions. Case: R02 Consider the matrix R R02 := The inverse and conjugate transpose of this matrix are R = R := Note that R02 and R02 is not a unitary matrix. However, in [3], the solutions are families determined by constants and Q Q. We multiply the matrix R02 by the constant 2 2 and consider the modified matrix, R R :=

40 The inverse and conjugate transpose of this matrix are R = R := The matrix R02 is unitary, and we determine the family produced by R02. 2 D = k H k R02 k m R02 m H m 0 = H R02 + H 4 R02 4 R02 H R02 4H 4 0 = 2 (H + H 4 H + H 4 ) = 2 (H 4 + H 4 ). Hence, c = a b d Using this result, we determine that: D 4 = k H k R02 k4 m R02 mh m4 0 = H R H 4R02 44 R02 H 4 R02 4 H 44 0 = 2 (H H 44. This reduces to: a = d. Hence, this forms the fourth family of solutions to the Yang-Baxter equation. 40

41 Case: R03 Consider the matrix R03. This is a unitary matrix and R03 = R03 = R R03 = We examine the equation D ij = k 0 = k H ik R03 kj m R03 imh mj H ik R03 kj R03 ik H kj H 22 ( pq + ) Computations will demonstrate that D is the zero matrix. This case gives rise to the solutions of type 5. 5 Solutions to the Bracket Equation The Kauffman bracket skein relation determines solutions to the braided Yang-Baxter equation, [5]. We determine which 4 4 matrix solutions of the bracket skein relation are unitary. These solutions are a subcase of the family 3 indicated in Theorem 4. If ˆR is a 4 4 matrix solution to the bracket skein relation then it satisfies the following equation. ˆR = αi 4 + α U 4

42 where U = N K and N, K are 2 2 dimensional matrices. The operator is defined for N and K N = a b c d K = g h k l ag ah ak al bg bh bk bl N K = cg ch ck cl dg dh dk dl Refering to [5] the matrix U has the property that U 2 = (α 2 + α 2 )U. We will denote (α 2 + α 2 ) as δ. We will assume that α has norm one and that α = e iθ. We obtain the following facts from [5]. ˆR = α I 4 + αu K = N δ = a,b N ab N ab 42

43 If ˆR is unitary, we obtain the additional restriction that ˆR = ˆR. ˆR = ˆR ᾱi 4 + ᾱ U = α I 4 + αu Recall that α has norm one implying that α = ᾱ. From this fact, we obtain U = U. We determine that N = N from U = U N N = N N. We use the fact that N = N to determine the value of α. Note that this argument did not refer to the dimensionality of N. If α =, then ˆR is a 4 4 matrix. We demonstrate this for an n n matrix N by calculating δ and the trace of N N. Note that N N = I n. δ = i,j N ij Nij δ = k N kk Nkk + i j N ij Nij We compute that value of the trace of N N. trace(n N) = n = j,k N kj Njk n = k N kk Nkk + i j N ij Nji 43

44 Combining these two calculations, we determine that δ n = i j N ij Nij i j N ij Nji δ n = i j N ij N ji 2 Hence, if N is a n n matrix then δ n. This inequality contradicts the fact that δ = (α 2 + α 2 ) = (e 2iθ + e 2iθ ), unless n 2. We consider the specific case when N is a 2 2 matrix. N N = NN N N = aā + b c ba + b d āc + cd bc + d d N N = 0 0 We examine the case when N is a 2 2 matrix. Recall that δ = N a,b N a,b. a,b Note that N a,b N a,b = a,b Na,b, since a,b a,bn N = N. From these facts we obtain δ = aā + b b + c c + d d δ = b c + b b + c c + bc δ = 2 + (b c)( b c) so that δ 2. Recall that α = e iθ and δ = (α 2 + α 2 ), we compute that δ = 2cos(θ). 44

45 Implying that δ 2. These inequalities indicate that δ = 2. Using this result, we determine that α = i and b = c. With the restrictions that b = c and N = N then re ig r2 e i p 2 N = r2 e i p 2 re i(p g) We determine the family from Theorem 4 of this solution. We determine a matrix Q such that ˆR M = (Q Q) ˆR(Q Q), (8) where ˆR M is a representative of a family from Theorem 4. constructed from a matrix M. The family of the ˆR is also dependent on the value of Q. Notice that ˆ R M = (Q Q) ˆR(Q Q) αid 4 + α M M = (Q Q)αI 4 + α N N(Q Q) αid 4 + α M M = αi 4 + α (Q Q)N N(Q Q) Note that (Q Q)N N(Q Q) = (QNQ t ) (Q N(Q ) t ). The conditions of the equation 8 are equivalent to determining a matrix Q such that QNQ t = M. If Q has the following form: Q = x 0 z r /2 x 45

46 and then z = i r 2 e i( p 2 g) r M = x2 re ig 0. 0 x 2 e i(p g) We determine the family of ˆR M by considering a generic 2 2 diagonal matrix. Then M = x 0 0 y 0 0 x y M M = y 0 0 x α x y Rˆ M = αi 4 + α M M 0 α 0 0 = 0 0 α 0 α y x The matrix ˆ R M is a solution to the braided Yang-Baxter equation. To obtain a solution to the algebraic Yang-Baxter equation, we multiply the matrix ˆ R M 46

47 by the matrix T given in Theorem α x y Rˆ 0 0 α 0 M T = R M = 0 α 0 0 α y x The matrix ˆR M is an element of family 2 or family 3, dependent on the value of Q. Finally, we observe that the R matrix and Q produced satisfy the conditions of family 3 from Theorem 4. References [] Louis Kauffman and Samuel J. Lomonaco, Quantum Entanglement and Topological Entanglement New Journal of Physics, Vol. 4 (2002) no. 73 URL: stacks.iop.org/ /4/73 (electronic) quant-ph/ [2] Louis Kauffman and Samuel J. Lomonaco, Questions about Quantum and Topological Entanglement In Preperation [3] Jarmo Hietarinta, All solutions to the constant quantum Yang-Baxter equation in two dimensions Physics Letters A, Vol. 65, p North Holland, 992 [4] Christian Kassel, Quantum Groups Graduate Texts in Mathematics, Springer-Verlag, New York 995 [5] Louis Kauffman, Knots and Physics, 3rd Ed Series on Knots and Everything, World Scientific Publishing Co., Inc