Universality in perfect state transfer
Erin Connelly, Nathaniel Grammel, Michael Kraut, Luis Serazo, Christino Tamon
aa r X i v : . [ qu a n t - ph ] J a n Universality in perfect state transfer
Erin Connelly ∗ Nathaniel Grammel † Michael Kraut ‡ Luis Serazo § Christino Tamon ¶ March 14, 2018
Abstract
A continuous-time quantum walk on a graph is a matrix-valued function exp( − i At )over the reals, where A is the adjacency matrix of the graph. Such a quantum walkhas universal perfect state transfer if for all vertices u, v , there is a time where the( v, u ) entry of the matrix exponential has unit magnitude. We prove new characteriza-tions of graphs with universal perfect state transfer. This extends results of Cameron et al . (Linear Algebra and Its Applications, 455:115-142, 2014). Also, we constructnon-circulant families of graphs with universal perfect state transfer. All prior knownconstructions were circulants. Moreover, we prove that if a circulant, whose order isprime, prime squared, or a power of two, has universal perfect state transfer then its un-derlying graph must be complete. This is nearly tight since there are universal perfectstate transfer circulants with non-prime-power order where some edges are missing. Keywords : Quantum walk, perfect state transfer, circulant, cyclotomic fields. A continuous-time quantum walk on a graph is given by the one-parameter matrix-valuedmap U ( t ) = exp( − i At ), where A is the Hermitian adjacency matrix of the graph. This notionwas introduced by Farhi and Guttman [6] to study quantum algorithms for search problems.Using this formulation, Bose [2] studied problems related to information transmission inquantum spin chains. In such a quantum walk, we say that there is perfect state transfer from vertex u to vertex v at time τ if the ( v, u ) entry of U ( τ ) has unit magnitude.Kay [13] showed that if a graph, whose adjacency matrix is real symmetric, has perfectstate transfer from u to v and also from u to w , then v must be equal to w . Cameron ∗ Department of Mathematics and Statistics, Haverford College. † Department of Computer Science, NYU Polytechnic. ‡ Mathematics Department, University of California at Santa Cruz. § Mathematics and Statistics Department, Vassar College. ¶ Department of Computer Science, Clarkson University. Contact: [email protected] t al . [3] showed that it is possible to violate this “monogamy” property in graphs withcomplex Hermitian adjacency matrices. They studied graphs with a universal propertywhere perfect state transfer occurs between every pair of vertices. The smallest nontrivialexample is the circulant Circ (0 , − i , i ). Some work related to universality in state transferfor quantum computing applications may be found in [14].We may view certain graphs with Hermitian adjacency matrices as gain graphs. Theseare graphs whose “directed” edges are labeled with elements from a group Γ. If the edge( u, v ) is labeled with group element g ∈ Γ, then the reversed edge ( v, u ) is labeled with g − .If Γ is the circle group, we get the complex unit gain graphs (see Reff [16]). In our case, wesimply require that edges in opposite directions have weights which are complex conjugatesto each other.Our main goal is to characterize graphs with universal perfect state transfer. Cameron et al . [3] proved strong necessary conditions for graphs with the universal perfect state trans-fer property. They showed that such graphs must have distinct eigenvalues, their unitarydiagonalizing matrices must be type-II (see Chan and Godsil [4]), and their switching auto-morphism group must be cyclic. A spectral characterization for circulants with the universalproperty was also proved in [3].In this work, we extend some of the observations from Cameron et al . [3]. More specif-ically, we prove new characterizations of graphs with universal perfect state transfer. Thefirst characterization exploits the fact that the unitary diagonalizing matrix of the graphadmits a canonical form. This allows us to show a tight connection between the spectra ofthe graph with the perfect state transfer times. Our second characterization is on circulantswith the universal property. It involves the set of minimum times when perfect state transferoccur between pairs of vertices. We prove that these minimum times are equally spaced onthe periodic time interval when perfect state transfer returns to the start vertex if and onlyif the graph is circulant. This complements the observation in [3] which characterizes theswitching automorphism group of circulants with the universal property.Most of the examples studied in [3] were circulants whose nonzero weights are ± i . Here,we provide a construction of non-circulant graphs of composite order with the universal prop-erty. To the best of our knowledge, this is the first known example of such family of graphs.We show that these families are non-circulant by appealing to our second characterizationabove (based on spacings of the minimum perfect state transfer times).Finally, we provide a nearly tight characterization of circulants with the universal prop-erty in terms of the number of nonzero coefficients. We show that if a circulant has universalperfect state tranfer and its order is prime, square of a prime, or a power of two, then allof its off-diagonal coefficients must be nonzero. As a partial converse, we show an infinitefamily of circulants with universal perfect state transfer whose order is not a prime powerand where some off-diagonal coefficients are zero.We conclude by studying universal perfect state transfer in complex unit gain graphs.The only known examples of complex unit gain graphs with the universal property are thecirculants K and Circ (0 , − i , i ). We conjecture that this set is unique.For a recent survey and a comprehensive treatment of quantum walk on graphs, we refer2he interested reader to Godsil [8, 7]. A weighted graph G = ( V, E, w ) is defined by a vertex set V , an edge set E and a weightfunction w : E → C . If G has n vertices, we will often identify the vertex set with Z /n Z .The adjacency matrix A ( G ) of graph G is a n × n matrix defined as A ( G ) k,j = w ( j, k ) if( j, k ) ∈ E , and A ( G ) k,j = 0 otherwise. A graph is Hermitian if its adjacency matrix is (see[11]).For complex numbers a , . . . , a n − ∈ C , we let C = Circ ( a , . . . , a n − ) denote the circu-lant matrix of order n where C jk = a k − j , for j, k ∈ Z /n Z . If C is Hermitian, note that a must be real and a n − j = a j , for j = 1 , . . . , n −
1. It is known that any circulant is diago-nalized by the Fourier matrix F n defined by h j | F n | k i = ζ jkn / √ n . Here, ζ n = e π i /n denotes aprimitive n th root of unity.We call a matrix flat if all of its entries have the same magnitude. A matrix is type-IIif it is flat and unitary (see Chan and Godsil [4]). Note that the Fourier matrix is type-II.A monomial matrix is a product of a permutation matrix and an invertible diagonal matrix(see Davis [5]). Two matrices A and B are switching equivalent if M A = BM for somemonomial matrix. The switching automorphism group of a graph G , denoted SwAut ( G ), isthe group of all monomial matrices which commute with A ( G ). This generalizes the notionof an automorphism group of a graph.For more background on algebraic graph theory, see Godsil and Royle [12]. The following result shows strong necessary conditions for graphs with universal perfect statetransfer.
Theorem 1. (Cameron et al. [3])Let G be a Hermitian graph with universal perfect state transfer. Then, the following hold:1. All eigenvalues of G are distinct.2. The adjacency matrix of G is unitarily diagonalized by a flat matrix.3. The switching automorphism group of G is cyclic whose order divides the size of G . We show some additional properties of graphs with universal perfect state transfer.
Definition 1.
Let G be a graph with universal perfect state transfer. For every pair ofvertices v and w of G , we let T v,w denote the set of times where perfect state transfer occursfrom v to w . That is, T v,w := { t ∈ R + : |h w | e − i A ( G ) t | v i| = 1 } . (1)3 act 1. Let G be a graph with universal perfect state transfer. For each pair of vertices v and w , T v,w is a discrete additive subgroup of R .Proof. See Godsil [7] or Cameron et al . [3].Since T v,w is a discrete additive subgroup of the reals, it has a smallest element. We willdenote the minimum element of the above set as t v,w := min T v,w . Lemma 1.
Let G be a graph with universal perfect state transfer and let u be a vertex of G .Then, for all vertices v = u , we have t u,v < t u,u .Proof. If t u,v > t u,u , let q be the largest integer for which qt u,u < t u,v . Then, ˆ t = t u,v − qt u,u is an element of T u,v which is smaller than t u,v . Lemma 2.
Let G be a graph and let u be an arbitrary vertex of G . Then G has universalperfect state transfer if and only if perfect state transfer occurs from u to all vertices of G .Proof. It suffices to prove only one direction since the other direction is immediate. Supposethat perfect state transfer occurs from u to all vertices. By Lemma 1, the quantum walkstarting at u visits all the other vertices before returning to u . If t u,v < t u,w then perfectstate transfer occurs from v to w . But, there is also perfect state transfer from w to v sincethe quantum walk has perfect state transfer from w back to u (at time t u,u − t u,w ) and thenfrom u to v (at time t u,v ). This proves that there is perfect state transfer between every pairof vertices. A flat unitary matrix is also called a type-II matrix (see [4]). We say that a type-II matrixis in canonical form if both its first row and its first column are the all-one vector.
Lemma 3.
Let G be a Hermitian graph on n vertices with universal perfect state tranfer.Then G is unitarily diagonalized by a type-II matrix X in canonical form: X = 1 √ n . . . e i α , e i α , . . . e i α ,n − e i α , e i α , . . . e i α ,n − ... ... ... ... e i α n − , e i α n − , . . . e i α n − ,n − (2) where α j,k ∈ [0 , π ) .Proof. Suppose A is the Hermitian adjacency matrix of G that is unitarily diagonalized by Z . Then, AZ = Z Λ where Λ is a diagonal matrix of the eigenvalues of A . The columns of Z are the eigenvectors of A which we will denote as | z i , . . . , | z n − i . Let D be a diagonalmatrix defined as D jj = 1 / h | z j i . Then, ˜ Z = ZD is also a flat unitary which diagonalizes A h | ˜ z k i = 1, for each k = 0 , . . . , n −
1. Next, consider a diagonal switching matrix S defined as S jj = 1 / h j | ˜ z i . Then, we have˜ A = ( S ˜ Z )Λ( S ˜ Z ) − . (3)Note ˜ A = SAS − is switching equivalent to A . Since X = S ˜ Z is a flat unitary matrix of theclaimed form, we are done. Corollary 1.
Let M be a flat unitary matrix in canonical form. Then, except for the firstrow and the first column, the row sums and the column sums of M are zero.Proof. Since the columns are orthonormal and the first column is the all-one vector, it is clearthat the column sums must be zero. The row sums are zero since M T is unitary whenever M is.Using Lemma 3, we show a spectral characterization of graphs with universal perfectstate transfer. Theorem 2.
Let G be a n -vertex Hermitian graph with eigenvalues λ , . . . , λ n − . Suppose G is diagonalized by a canonical type-II matrix X , where X j,k = e i α j,k / √ n with α j,k ∈ [0 , π ) and α j,k = 0 if either j or k is zero. Then, G has universal perfect state transfer if and onlyif for each ℓ = 0 , . . . , n − , there is t ℓ ∈ R so that for all k = 1 , . . . , n − , we have ( λ k − λ ) t ℓ = α ℓ,k . (4) Proof.
Let A be the Hermitian adjacency matrix of G . We denote the k th column of X as | λ k i which is the eigenvector of A corresponding to eigenvalue λ k . Thus, e − i At = n − X k =0 e − i λ k t | λ k ih λ k | . (5)( ⇒ ) Assume G has universal perfect state transfer. Suppose that perfect state transfer fromvertex 0 to vertex ℓ occurs at time t ℓ ∈ R with phase e i θ ℓ . Then, h ℓ | e − i At ℓ | i = n − X k =0 e − i λ k t ℓ h ℓ | λ k ih λ k | i = 1 n n − X k =0 e − i λ k t ℓ e i α ℓ,k . (6)Moreover, we have e i θ ℓ = 1 n " e − i λ t ℓ + n − X k =1 e − i ( λ k t ℓ − α ℓ,k ) . (7)So, for each k = 1 , . . . , n −
1, we have λ k t ℓ − α ℓ,k = λ t ℓ , which shows these conditions arenecessary for universal perfect state transfer. 5 ⇐ ) Suppose that for each ℓ , there is a time t ℓ so that for each k = 0,( λ k − λ ) t ℓ = α ℓ,k . (8)Then, h ℓ | e − i At ℓ | i = 1 n n − X k =0 e − i λ k t ℓ e i α ℓ,k = e − i λ t ℓ . (9)This shows that there is perfect state transfer from 0 to ℓ . Therefore, there is perfect statetransfer from 0 to all vertices. By Lemma 2, this shows there is perfect state transfer betweenevery pair of vertices. Cameron et al . [3] showed the following result on the switching automorphism group ofcirculants with universal perfect state transfer.
Theorem 3. (Cameron et al. [3])Let G be a graph with universal perfect state transfer. Then, G is switching isomorphic to acirculant if and only if SwAut ( G ) is cyclic of order n . In what follows, we provide new characterizations of circulants with universal perfect statetransfer. The first one is based on the set of times when perfect state transfer occur. Thesecond one is based on the explicit form of allowable weights.Recall that T u,v is the set of times (positive real numbers) when perfect state transferoccur from vertex u to vertex v . Also, we denote t u,v as the smallest element of T u,v . Theorem 4.
Let G be a n -vertex graph with universal perfect state transfer. Assume that t ,k < t ,k +1 for all k = 1 , . . . , n − and that t , = min { t k,k +1 : k ∈ Z /n Z } . Then, G isswitching isomorphic to a circulant if and only if t k,k +1 = t , , (10) for all k ∈ Z /n Z .Proof. Let A be the adjacency matrix of G .( ⇒ ) Suppose that G is switching isomorphic to a circulant. Consider the set of timeswhen perfect state transfer occur in G : T = { t ∈ R + : ∃ j, k ∈ Z /n Z , |h k | e − i At | j i| = 1 } . (11)Since T is a discrete additive subgroup of R , it has a minimum. Without loss of generality,assume that t , = min T . Thus, h | e − i At , | i = γ, (12)6or some γ ∈ T . By Theorem 3, SwAut ( G ) is cyclic of order n . We may assume that ˜ P = P φ D generates SwAut ( G ) where φ = (0 1 . . . n −
1) and D is a diagonal switching matrix. Since˜ P is a switching automorphism, ˜ P − A ˜ P = A , which implies e − i At = ˜ P − e − i At ˜ P . Therefore, h | e − i At , | i = ˜ γ h | ˜ P − e − i At , ˜ P | i = ˜ γ h | e − i At , | i , (13)for some ˜ γ ∈ T . So, perfect state transfer occurs from 1 to 2 at time t , . By repeatedlyusing the same argument, we see that perfect state transfer occurs from k to k + 1 at time t , , for k = 0 , , . . . , n − ⇐ ) Suppose t k,k +1 = t , , for k ∈ Z /n Z . This implies that at time t , perfect statetransfer occurs from k to k + 1 (simultaneously) from k to k + 1 for each k ∈ Z /n Z . So,assume h k + 1 | e − i At , | k i = γ k , (14)for some complex unit weight γ k ∈ T . Thus, e − i At , = . . . γ n − γ . . . γ . . . . . . ... ...0 0 . . . γ n − = . . . . . . . . . . . . ... ...0 0 . . . γ . . . γ . . .
00 0 . . . . . . ...0 0 . . . γ n − , (15)which shows that e − i At , is a monomial matrix. Since it commutes with A , it is a switchingautomorphism of G . Moreover, it generates SwAut ( G ). Thus, by Theorem 3, G is switchingisomorphic to a circulant.Next we find explicit forms for the weights on circulants which have universal perfectstate transfer. But, first we state a spectral characterization of circulants with universalperfect state transfer proved by Cameron et al . [3]. Theorem 5. (Cameron et al. [3])Let G be a graph that is switching equivalent to a circulant. Then G has universal perfectstate transfer if and only if for some integer q coprime with n , for real numbers α, β with β > , the eigenvalues of G are given by λ k = α + β ( qk + c k n ) , k = 0 , , . . . , n − , (16) where c k are integers. Let A be the adjacency matrix of G . In Theorem 5, we may assume α = 0 by allowinga diagonal shift A + α I , which does not affect the quantum walk. Furthermore, we mayassume β = 1 by allowing the time scaling β A , which does not affect perfect state transfer.Finally, we may multiply the adjacency matrix with the multiplicative inverse of q modulo n (to cancel the factor q in qk ). In summary, we have the following.7 orollary 2. Let G be a graph that is switching equivalent to a circulant. Then G hasuniversal perfect state transfer if and only if G is switching equivalent to a graph with eigen-values λ k = k + c k n, k = 0 , , . . . , n − where c k are integers. Next, we show a general form for the coefficients of a circulant which has universal perfectstate transfer.
Theorem 6.
Let
Circ ( a , . . . , a n − ) be a circulant with universal perfect state transfer.Then, we have a j = 1 ζ − jn − n − X k =0 c k ζ − jkn , j = 1 , . . . , n − for integers c k , where k = 0 , . . . , n − .Proof. Let G = Circ ( a , . . . , a n − ). By Corollary 2, the eigenvalues of G are of the form λ k = k + c k n , for some integers c k , where k ∈ Z /n Z . Since circulants are diagonalized bythe Fourier matrix (see Biggs [1]), the coefficients of G are given by a j = 1 n n − X k =0 λ k ζ − jkn , j = 1 , . . . , n − . (19)Using the assumed form of λ k , we get a j = 1 n n − X k =0 ( k + c k n ) ζ − jkn = 1 n n − X k =0 kζ − jkn + n − X k =0 c k ζ − jkn . (20)Let U = P n − k =0 kζ − jkn and L = P n − k =1 P kℓ =1 ζ − jℓn . Note that L + U = ( n − n − X k =1 ζ − jkn = 1 − n. (21)Now, we have L = n − X k =1 k X ℓ =1 ζ − jℓn = n − X k =1 ζ − j ( k +1) n ζ − jn − − ! (22)= 1 ζ − jn − n − X k =2 ζ − jkn − ( n − ζ − jn ! (23)= 1 − n − nζ − jn − . (24)Thus, U = n/ ( ζ − jn − a j = 1 ζ − jn − n − X k =0 c k ζ − jkn . (25)8 Non-circulants with universal state transfer
In this section, we show a construction of a family of non-circulant graphs with universalperfect state transfer. This provides the first known examples of non-circulant families withuniversal perfect state transfer.
Theorem 7.
Let n = ab be an integer where a ≥ b ≥ are integers. Fix an integer β ≥ and for a positive integer d , let ϑ d be a function which maps elements of Z /n Z to the positiveintegers defined as ϑ d ( x ) := β ⌊ x/d ⌋ d + x mod d. (26) Let X be a n × n matrix whose ( j, k ) -entry is given by X j,k = 1 √ n ζ ϑ a ( j ) ϑ b ( k ) βn . (27) Then, X is type-II. Moreover, if G is the graph with eigenvalues { ϑ b ( k ) : k = 0 , , . . . , n − } whose adjacency matrix is unitarily diagonalized by X , then G has universal perfect statetransfer.Proof. First, we show that X is type-II. It is clear that X is flat from its definition. So,it suffices to show that the columns of X form an orthonormal set. In what follows, for k, ℓ ∈ Z /n Z , let Q b ( k, ℓ ) = ⌊ k/b ⌋ − ⌊ ℓ/b ⌋ and R b ( k, ℓ ) = k mod b − ℓ mod b . If X k and X ℓ arethe k th and ℓ th columns of X , then h X ℓ | X k i = 1 n n − X j =0 ζ M b ( k,ℓ ) ϑ a ( j ) βn , where M b ( k, ℓ ) = βQ b ( k, ℓ ) b + R b ( k, ℓ ) (28)= 1 n a − X r =0 ζ M b ( k,ℓ ) rβn b − X q =0 (cid:0) ζ aM b ( k,ℓ ) n (cid:1) q . (29)But, note that for any integer M = 0, provided ζ Mb = 1, we have b − X q =0 ( ζ aMn ) q = ζ abMn − ζ aMn − , (30)since ζ abn = 1. So, if R b ( k, ℓ ) = 0, then M b ( k, ℓ ) = 0 (mod b ), which implies h X ℓ | X k i = 0.On the other hand, if R b ( k, ℓ ) = 0, then Q b ( k, ℓ ) = 0, for otherwise k = ℓ . Here, we have h X ℓ | X k i = 1 n b − X q =0 ζ qβQ b ( k,ℓ )1 a − X r =0 (cid:0) ζ Q b ( k,ℓ ) a (cid:1) r (31)Thus, h X ℓ | X k i = 0 also holds. 9ext, we show that G has universal perfect state transfer. If we let α j,k = 2 πβn ϑ a ( j ) ϑ b ( k ) , (32)for j, k ∈ Z /n Z , then X j,k = e i α j,k . For each j ∈ Z /n Z , let t j = (2 π/βn ) ϑ a ( j ). Then, λ k t j = α j,k (33)holds for all k ∈ Z /n Z , since λ k = ϑ b ( k ) and λ = 0. By Theorem 2, this shows that G hasuniversal perfect state transfer.Finally, we show that G is not switching equivalent to a circulant. By Theorem 4, itsuffices to show that t j +1 − t j are not all equal. From the definition of ϑ a ( j ), we have t − t = πβn whereas t a − t a − = πβn (( β − a + 1). Example 1.
For even k ≥
2, let G k be a graph of order 4 with eigenvalues { , , k, k + 1 } .Let X be the following unitary matrix (whose columns are the eigenvectors of G k ): X = e i π/k e i π e i π ( k +1) /k e i π e i πk e i π e i π ( k +1) /k e i π ( k +1) e i π/k (34)Let A = X Λ X − , where Λ = diag(0 , , k, k +1). be the adjacency matrix of G k . By Theorem7, G k has universal perfect state transfer.Note G is a non-circulant graph with universal perfect state transfer. The eigenvaluesof G are { , , , } and its adjacency matrix is A = (1 + e − i π/ )
12 32 (1 − e − i π/ ) (1 + e i π/ ) 0 (1 − e i π/ ) (1 − e − i π/ ) 0 (1 + e − i π/ ) (1 − e i π/ )
12 32 (1 + e i π/ ) 0 . (35) In this section, we show that if a circulant has universal perfect state transfer, then all of itscoefficients must be nonzero under certain conditions on the order.
Definition 2.
A circulant
Circ ( a , a , . . . , a n − ) is called dense if a j = 0 for j = 1 , . . . , n − . Our main result in this section is the following.
Theorem 8.
Let G = Circ ( a , . . . , a n − ) be a circulant with universal perfect state transfer.If n is a prime, square of a prime, or a power of two, then G is dense.
10e will divide the proof of the theorem into several lemmas. First, we consider the casewhen the order of the circulant is prime.
Lemma 4.
For a prime p , let Circ ( a , . . . , a p − ) be a circulant with universal perfect statetransfer. Then, a j = 0 for j = 1 , . . . , p − .Proof. We may assume that λ = 0 (by a diagonal shift). Since a j = p P p − k =1 λ k ζ − jkp , wehave that a j ∈ Q ( ζ p ). Note that { ζ jp : j = 1 , . . . , p − } is a basis for the cyclotomic fieldextension Q ( ζ p ) / Q . Thus, if a j = 0 for some j = 0, then λ k = 0 for all k = 0. But, this is acontradiction to the assumed form of λ k in (17).Second, we consider universal perfect state transfer in circulants whose order is the squareof a prime. Lemma 5.
Given a prime p , let n = p . For a circulant Circ ( a , . . . , a n − ) , suppose a j ∈ Q for some j = 0 . Then, a d = a j for d = gcd( j, n ) .Proof. Let j ∈ { , . . . , n − } and d = gcd( j, n ). Then, there is ℓ ∈ ( Z /n Z ) ⋆ so that jℓ ≡ d (mod n ). To see this, note jℓ ≡ d (mod n ) is solvable since gcd( j, n ) divides d . Moreover,gcd( ℓ, n ) = 1 since gcd( ℓ, n/d ) = 1 and gcd( ℓ, d ) = 1. Here, we used the fact that n = p .Since Gal ( Q ( ζ n ) / Q ) = ( Z /n Z ) ⋆ , there is a field automorphism φ ℓ of Q ( ζ n ) which fixes Q for which φ ℓ ( ζ n ) = ζ ℓn . Since a j = (1 /n ) P n − k =0 λ k ζ − jkn , we have φ ℓ ( a j ) = 1 n n − X k =0 λ k ζ − ℓjkn = 1 n n − X k =0 λ k ζ − dkn = a d . (36)But, φ ℓ ( a j ) = a j since a j ∈ Q . This shows a j = a d . Lemma 6.
Let G = Circ ( a , . . . , a n − ) be a circulant of order n with universal perfect statetransfer. If d is an odd divisor of n and n/d is prime, then a d = 0 .Proof. Let p = n/d be prime. Then, a d = 1 n n − X k =0 λ k ζ − dkn = 1 p p − X k =0 Λ k ζ − kp , (37)where Λ k = d P d − ℓ =0 λ k + ℓp . By Corollary 2, we haveΛ k = 1 d d − X ℓ =0 ( k + ℓp + n Z ) = k + p ( d − n Z . (38)Using this in (37) combined with the fact that P p − k =0 ζ − kp = 0, we get a d = 1 p p − X k =0 (cid:18) k + p ( d − n Z (cid:19) ζ − kp (39)= 1 p p − X k =1 kζ − kp . (40)11ince ζ kp , for k = 1 , . . . , p −
1, are linearly independent, we have a d = 0.For our next lemma, we will need the following fact about connectivity in circulants. Fact 2. (Meijer [15], Theorem 4.2)A circulant
Circ ( a , . . . , a n − ) is connected if and only if gcd( { j : a j = 0 } ∪ { n } ) = 1 . Lemma 7.
For a prime p , suppose n = p . Let G = Circ ( a , . . . , a n − ) be a circulant withuniversal perfect state transfer. Then, a j = 0 for all j = 0 .Proof. Suppose a j = 0 for some j = 0. By Lemma 5, we have a d = 0 for d = gcd( j, n ). Since n = p , we have two cases to consider: d = 1 or d = p . If a = 0, then G is not connectedby Fact 2. If a p = 0, then this contradicts Lemma 6.Finally, we consider universal perfect state transfer in circulants whose order is a powerof two. Here, we use the following result of Good in a crucial manner. Fact 3. (Good [10], Theorem 1)If m is a power of two, then { e i πr/m : r = 0 , . . . , m − } is linearly independent over Q . Lemma 8.
For a positive integer d , suppose n = 2 d . Let G = Circ ( a , . . . , a n − ) be acirculant with universal perfect state transfer. Then, a j = 0 for all j = 1 , . . . , n − .Proof. Given d ∈ Z + , let n = 2 d and m = 2 d − . Recall that ζ n = e π i /n . We have a j = 1 n n − X k =0 λ k ζ − jkn = 12 m m − X ℓ =0 Λ ℓ ( j )( e i π/m ) − ℓ , (41)where Λ ℓ ( j ) = X k : jk ≡ ℓ λ k − X k : jk ≡ m + ℓ λ k (42)By Fact 3, if a j = 0 then Λ ℓ ( j ) = 0 for all ℓ = 0 , . . . , m −
1. We show that Λ = 0.We consider the case when ℓ = 0. If j is odd, then the map f j ( k ) ≡ jk (mod n ) is abijection. Thus, Λ ( j ) = λ − λ m ≡ m (mod n ). Next, suppose j is even with j = 2 e s where e ≥ s is odd. Then, the values k for which jk ≡ n ) are given by r d − e for r = 0 , , . . . , e −
1. Also, the values k for which jk ≡ m (mod n ) are given by (2 r + 1)2 d − e − for r = 0 , , . . . , e −
1. Therefore,Λ = e − X r =0 ( r d − e − (2 r + 1)2 d − e − ) ≡ m (mod n ) . (43)Thus, in both cases we have Λ n ), which implies Λ = 0. Proof. (of Theorem 8)Follows immediately from Lemmas 4, 7, and 8.12 .1 Non-dense circulants with universal state transfer
We show a partial converse of Theorem 8 by constructing non-dense circulants with universalperfect state transfer whose orders are not prime powers. This observation uses the followingfact about cyclotomic units.
Fact 4. (Washington [17], Proposition 2.8)Suppose n is a positive integer which has at least two distinct prime factors. Then − ζ n isa unit of Z [ ζ n ] . Moreover, Y For two distinct primes p and q , let n = pq . Let c , . . . , c n − be integers so n − X k =0 c k ζ − kn = 11 − ζ − n . (45) Let a = 0 and, for j = 1 , . . . , n − , let a j = 1 / ( ζ − jn − 1) + P n − k =0 c k ζ − jkn . Then, G = Circ ( a , . . . , a n − ) is a non-dense circulant with universal perfect state transfer.Proof. By the choice of the integers c k in (45), a = a n − = 0. This shows that G is notdense. Also, note that λ = n − X j =0 a j = n − X j =1 ζ − jn − − n − X k =0 c k . (46)The other eigenvalues are given by λ ℓ = P n − j =1 a j ζ jℓn , for ℓ = 0. By definition of a j , λ ℓ = n − X j =1 ζ − jn − n − X k =0 c k ζ − jkn ! ζ jℓn = n − X j =1 ζ jℓn ζ − jn − c ℓ n − n − X k =0 c k . (47)Using (46), we get λ ℓ = n − X j =1 ζ jℓn − ζ − jn − c ℓ n + λ = ( n − − n − X j =1 − ζ j ( ℓ +1) n − ζ jn + c ℓ n + λ . (48)But, note that n − X j =1 − ζ j ( ℓ +1) n − ζ jn = n − X j =1 ℓ X s =0 ζ sjn = n − − ℓ. (49)This shows that λ ℓ = ℓ + c ℓ n + λ , for ℓ = 1 , . . . , n − 1. By Theorem 5, this shows G hasuniversal perfect state transfer. 13ere, we confirm that the underlying graph of G is connected. Since 1 − ζ − p is not a unitof Z [ ζ p ], for any prime p , we have a q = 1 ζ − qn − n X k =0 c k ζ − qkn = 1 ζ − p − n − X k =0 c k ζ − kp = 0 . (50)since c k are all integers. Similarly, a p = 0. Since gcd( p, q ) = 1, Fact 2 implies G is connected. Example 2. We show a circulant G of order n = 6 with universal perfect state transferwhich is not dense. Here, we have ζ = + i √ , ζ = − + i √ , ζ = − ζ = − − i √ , ζ = − i √ . Note that a = 1 ζ − − X k =0 c k ζ − k = 1 ζ − − ζ ) = 0 . (51)So, in Theorem 6, we choose c = 1, c = c = c = c = 0 and c = − 1. The othercoefficients can be computed using a j = 1 / ( ζ − − 1) + P k =0 c k ζ − jk . By a straightforwardcomputation, a = , a = 1 − i √ , a = , a = 1 + i √ and, of course, a = 0.Hence, the adjacency matrix of G is given by − i √ i √ − i √ i √ i √ − i √ i √ − i √ − i √ i √ 00 1 − i √ i √ . (52)The eigenvalues of G are given by λ k = P j =0 a j ζ jk . We confirm the eigenvalue form in (17)that λ ℓ = ℓ +6 c ℓ , for ℓ = 0 , , . . . , 5. It can be verified that λ = 6 = 0+6 c , λ = 1 = 1+6 c , λ = 2 = 2 + 6 c , λ = 3 = 3 + 6 c , λ = 4 = 4 + 6 c , and λ = − c . T In this concluding section, we consider universal perfect state transfer in complex unit gaingraphs. First, we observe the following fact. Fact 5. Circ (0 , − i , i ) is the only graph on vertices with universal perfect state transfer,up to switching equivalence. roof. Cameron et al . [3] showed that Circ (0 , − i , i ) has universal perfect state transfer.Hence, it suffices to show that any graph on 3 vertices with universal perfect state transfermust be a circulant.Let G be a graph with Hermitian adjacency matrix A . Suppose M is a type-II matrix ofthe form M = e i α e i β e i β e i α (53)which diagonalizes A . By Corollary 1, we have that 1 + e i α + e i β = 0. This shows cos( α ) +cos( β ) = − α ) + sin( β ) = 0. This implies α = 2 π/ β = 2 α . Thus, M/ √ k , there is only a finite number of (unweighted)graphs with maximum degree k with perfect state transfer. This shows that perfect statetransfer is a rare phenomenon (in the absence of weights). This motivates our next conjecture.We say a graph has property T if all of the nonzero coefficients in its adjacency matrixare complex numbers with unit magnitude. Conjecture 1. Circ (0 , − i , i ) is the only circulant with property T which has universalperfect state transfer. Acknowledgments Research supported by NSF grant DMS-1262737. Part of this work was started while C.T.was visiting Institut Henri Poincar´e (Centre ´Emile Borel) and University of Waterloo. Thisauthor would like to thank IHP and Chris Godsil for hospitality and support. References [1] N. Biggs. Algebraic Graph Theory. Cambridge University Press , 2nd edition, 1994.[2] S. Bose. Quantum communication through an unmodulated spin chain. Physical ReviewLetters (20):207901, 2003.[3] S. Cameron, S. Fehrenbach, L. Granger, O. Hennigh, S. Shrestha, C. Tamon. UniversalState Transfer on Graphs. Linear Algebra and Its Applications :115-142, 2014.[4] A. Chan, C. Godsil. Type-II matrices and combinatorial structures. Combinatorica (1):1-24, 2010.[5] P.J. Davis. Circulant Matrices. Wiley, 1979.[6] E. Farhi, S. Gutmann. Quantum computation and decision trees. Physical Review A :915-928, 1998. 157] C. Godsil. Graph Spectra and Quantum Walks. Manuscript, 2015.[8] C. Godsil. State transfer on graphs. 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