Average mixing of continuous quantum walks

Transcription

1 Average mixing of continuous quantum walks Juergen Kritschgau Department of Mathematics Iowa State University Ames, IA April 7, 2017

2 Juergen Kritschgau (ISU) 2 of 17 This is a presentation of: Average mixing of continuous quantum walks, Journal of Combinatorial Theory, Series A June 2013, by Chris Godsil

3 Juergen Kritschgau (ISU) 3 of 17 Outline 1) Background and Definitions, particularly, the average mixing matrix (AMM) 2) How to work with the AMM 3) The AMM of paths 4) The AMM and cospectral vertices

4 Juergen Kritschgau (ISU) 4 of 17 Continuous Quantum Walk Let X be a graph on V (X ) = {1,..., n} vertices with adjacency matrix A. We will be considering the operator H(t) = exp(ıta) = k=0 1 k! (ıta)k where t R. The idea is that H(t) determines a continuous quantum walk by associating the standard basis vector e i C n with the vertex i. Doing so, yields that e T u H(t)e v 2 is the probability that the quantum walk starting at vertex v is at vertex u at time t.

5 Juergen Kritschgau (ISU) 5 of 17 Average Mixing Matrix of X The Schur Product of two n m matrices A, B, denoted by A B, is the entry-wise product of A and B.Then define the following: M X (t) := H(t) H( t) 1 T ˆM X := lim M X (t)dt T T 0 The matrix ˆM X is the average mixing matrix of X. The goal is to relate properties of ˆMX (t) to X.

6 Juergen Kritschgau (ISU) 6 of 17 Properties of the Matrix Exponential There are a few properties of the matrix exponential that are useful for us. Let B, C be complex square matrices. 1) If B is skew hermitian, then exp(b) is unitary. 2) If BC = CB, then exp(b + C) = exp(b)exp(c). 3) If B is idempotent, then expb = k=0 1 k! Bk = I + ( k=1 1 k!) B = I + (e 1)P Property 1) gives us that H(t) = exp(ıta) is unitary in virtue of (ıta) = ıta

7 Juergen Kritschgau (ISU) 7 of 17 Working with ˆM X (t) Recall that the spectral decomposition of A is given by A = r θ r E r where E r is idempotent. Then we have: H(t) = r exp(ıtθ r )E r M X (t) = r E r E r + 2 r s cos((θ r θ s )t)e r E s Lemma If A = r θ r E r, then ˆM X = r E r E r.

8 Juergen Kritschgau (ISU) 8 of 17 Connectivity Lemma If X is connected, then all entries of ˆM X are positive. Proof. Notice that E r E r is nonnegative. Suppose ( ˆM X ) u,v = 0. This implies (E r E r ) u,v = 0 for all r. Furthermore, this implies that (E r ) u,v = 0 for all r. Recall that A is a linear combination of E r and E i E j = 0 for i j. Therefore, (A k ) u,v = 0 for all k, and X is disconnected.

9 Average Mixing on Paths Lemma The idempotents of E 1,..., E n in the spectral decomposition of P n are given by (E r ) j,k = 2 ( ) ( ) jrπ krπ n + 1 sin sin. n + 1 n + 1 Proof. Recall that the adjacency eigenvalues of P n are 2 cos(β) with sin(β) sin(2β) eigenvectors z(β) = where β ranges over. sin(nβ) β = πr, r [n]. To be continued... n + 1 Juergen Kritschgau (ISU) 9 of 17

10 Juergen Kritschgau (ISU) 10 of 17 Average Mixing on Paths [FUN FACT!] If λ is a simple eigenvalue with vector x, then 1 x T x xx T is the projection onto the eigenspace of λ. proof continued. 1 Using the fun fact, we get that z(β) T z(β) = 2 n+1 and ( ) ( ) jrπ krπ (z(β)z(β) T ) j,k = sin(jβ) sin(kβ) = sin sin. n + 1 n + 1

11 Juergen Kritschgau (ISU) 11 of 17 Average Mixing on Paths Lemma If E 1,..., E n are idempotents for P n, then ˆM Pn = r E r E r = 1 (2J + I + J). 2n + 2 Proof. By the previous Lemma we have (E r E r ) j,k = 4 (n + 1) 2 sin2 Using some trigonometry we get... ( ) ( ) jrπ krπ sin 2. n + 1 n + 1

12 Juergen Kritschgau (ISU) 12 of 17 Average Mixing on Paths proof continued. (n + 1) 2 (E r ) E r ) j,k = 1 cos ( 2jrπ n+1 cos ( 2krπ n+1 ) ( ) cos 2(j+k)rπ n+1 ). ( cos 2(j k)rπ n+1 ) = 1, Summing across r and using the fact that ( n r=1 cos 2lrπ n+1 the right hand side (RHS) evaluates to 3(n + 1)/2 j = k; 3(n + 1)/2 j + k = n + 1; RHS = 2n + 2 j = k, j + k = n + 1; n + 1 otherwise.

13 Juergen Kritschgau (ISU) 13 of 17 An Example: P 3 Recall that the eigenvalues of P 3 are 0, 2, 2 with corresponding vectors [1, 0, 1] T, [1, 2, 1] T, [1, 2, 1] T. These are all simple eigenvalues, so the corresponding projections are E 1 = E 2 = E 3 = Then 2 1 3/8 2/8 3/8 ˆM P3 = E 1 /2 + E 2 /4 + E 3 /4 = 2/8 4/8 2/8 3/8 2/8 3/8

14 Juergen Kritschgau (ISU) 14 of 17 Cospectral Vertices Theorem Let X be a graph with vertices u and v, and let E 1,..., E m be the idempotents in the spectral decomposition of A(X ). TFAE: 1) (A k ) u,u = (A k ) v,v for all k 0 2) E r e u 2 = E r e v 2 3) The graphs X u and X v. If any of the conditions above hold for vertices u and v, we say that u and b are cospectral. Furthermore, if E r e u = ±E r e v for all r, then we say that u and v are strongly cospectral. Lemma If u and V are vertices in X and the eigenvalues of X are all simple, then U and v are strongly cospecral if and only if they are cospectral.

15 Juergen Kritschgau (ISU) 15 of 17 Cospectral Vertices Theorem Let ˆM X be the average mixing matrix of the graph X. Then vertices u andv are strongly cospectral if and only if ˆM x (e v e u ) = 0. Proof. Some algebra shows that if N is PSD and N(e u e v ) = 0, then N[v, u](e u e v ) = 0. ˆM X (e u e v ) = 0 implies 0 = (e u e v ) T ˆM X (e u e v ) = r (e u e v ) T (E r E r )(e u e v ). Notice that each summand E r E r is PSD, so we have (E r E r )(e u e v ) = 0 for all r. This implies ((E r ) u,u ) 2 = ((E r ) u,v ) 2 = ((E r ) v,v ) 2. Applying Cauchy-Schwarz gives that E r e u = ±E r e v.

16 Juergen Kritschgau (ISU) 16 of 17 Returning to our Example Recall that that the average mixing matrix of P 3 is 3/8 2/8 3/8 ˆM P3 = 2/8 4/8 2/8. 3/8 2/8 3/8 The first and last row are equal, and P 3 has all simple eigenvalues. Therefore, P 3 v 1 and P 3 v 3 are cospectral.

17 Juergen Kritschgau (ISU) 17 of 17 The End Thank you!