Perfect State Transfer on Distance-Regular Graphs
Gabriel Coutinho (joint work with C. Godsil, K. Guo and F. Vanhove)
2013
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Distance regular graphs
Definition 1. A graph X is distance-regular of diameter d if, in the following partition,
the numbers ci , ai and bi don’t depend on u and v.
Example 1. If X is the cycle C8 , we have (b0 , b1 , b2 , b3 ) = (2, 1, 1, 1) and (c1 , c2 , c3 , c4 ) = (1, 1, 1, 2).
Proposition 1. Let X be distance-regular. Some facts about ai , bi and ci .
1. X is regular, its valency is b0 .
2. Then b0 = ci + ai + bi for any i.
3. b0 > b1 ≥ b2 ≥ ... ≥ bd−1 > 0.
4. 1 = c1 ≤ c2 ≤ ... ≤ cd ≤ b0 .
Definition 2. Define the graphs Xi as:
• V (Xi ) = V (X)
• u ∼ v in Xi iff d(u, v) = i in X.
Example 2. If X is the cycle C8 , then X2 is the disjoint union of two copies of C4 , X3 is a copy of C8 ,
and X4 is the disjoint union of four copies of K2 .
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Proposition 2. Given X, define the matrix Ai = A(Xi ). If X is a distance regular graph, then:
AAi = bi−1 Ai−1 + ai Ai + ci+1 Ai+1
(1)
By induction using the equation above, we can prove that:
Ai Aj =
d
X
pkij Ak
(2)
k=0
for certain constants pkij .
Definition 3. Given a distance-regular graph X, define the matrix algebra A as:
A = span{A0 , ..., Ad }
Proposition 3. By equation (2), the matrices {A0 , ..., Ad } actually form a basis for A. Moreover, these
are symmetric matrices, so by (2) again, Ai Aj = Aj Ai . Hence A is a commutative algebra of symmetric
matrices.
Proposition 4. Commutative symmetric matrices can be simultaneously orthogonally diagonalized,
hence there are orthogonal projectors {E0 , ..., Em } such that:
Ai =
m
X
Pir Er
r=0
Moreover, each Er is a polynomial in Ai , so they belong to the algebra, and hence they also form a
basis (and so d = m).
Definition 4. The Schur product of matrices M and N is defined as:
(M ◦ N )ab = Mab · Nab
Theorem 1. Let A = span{A0 , ..., Ad } = span{E0 , ..., Ed }. Hence:

 (1) Ai ◦ Ai = Ai
(2) A
1. {A0 , ..., Ad } is a basis such that
Pi ◦ Aj = 0

(3)
Ai = J

 (1) Ei · Ei = Ei
(2) E
2. {E0 , ..., Ed } is a basis such that
Pi · Ej = 0

(3)
Ei = I
In particular, A is closed and commutative with respect to both products ◦ and ·. Moreover, all matrices
in A have constant row and column sums.
Proof. Only the last statement deserves a proof here. From 1.(3), J ∈ A. If M ∈ A, then M J = JM ,
as A is commutative, and hence M has constant row and column sums.
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2
Perfect state transfer
Definition 5. Given a graph X with adjacency matrix A = A(X), define:
U (t) = exp(ıtA)
Proposition 5. Note that U (t)∗ = exp(−ıtAT ) = exp(ıtA)−1 = U (t)−1 . So U (t) is unitary.
Definition 6. We say that X admits perfect state transfer between u and v at time t if:
U (t)eu = λev
for some λ ∈ C.
Proposition 6. Note that:
U (t)∗ eu = λev ⇒ λeu = U (t)ev
Proposition 7. Using the fact that
A=
d
X
θr Er
r=0
we can write:
U (t) =
d
X
eıtθr Er
r=0
making the expression more approachable.
Proposition 8. Hence uv-pst happens if and only if:
d
X
eıtθr Er eu = λ
r=0
d
X
Er ev
r=0
and multiplying both sides by Er , we have uv-pst if and only if, for all r,
eıtθr Er eu = λEr ev
Proposition 9. Because eiθr t and λ are complex numbers of norm 1, and because Er eu and Er ev are
real vectors, uv-pst implies:
Er eu = ±Er ev ,
3
(3)
Definition 7. Vertices u and v satisfying Er eu = ±Er ev for all r are called strongly co-spectral.
Proposition 10. Let X be a graph with e-values θ0 > ... > θd . Suppose vertices u and v are strongly
co-spectral. When does uv-pst occurs?
Let S+ be the set of indices r such that 0 6= Er eu = Er ev , and S− be the set of indices r such that
0 6= Er eu = −Er ev .
(Strong) interlacing implies that 0 ∈ S+ .
Pst occurs at time t if and only if:
eiθ0 t = eiθr t
for all r ∈ S+
and
eiθ0 t = −eiθs t
for all s ∈ S− .
These conditions can be restated as:
t(θ0 − θr ) = 2kπ
for all r ∈ S+
and
t(θ0 − θs ) = (2k + 1)π
3
for all s ∈ S− .
PST in DRGs
Theorem 2 (Godsil). Let X be a distance-regular graph with diameter d and adjacency matrix A =
A(X). If X admits perfect state transfer at time τ , then there is a permutation matrix T with no fixed
points and of order two such that UA (τ ) = λT for some λ ∈ C. Moreover, T = Ad .
Proof. From
U (t) =
d
X
eıtθr Er ,
r=0
we see that U (t) belongs to the matrix algebra A. If, for some τ ,
U (τ )eu = λev
with
|λ| = 1,
then consider the 0-1 matrix T which is the unique element of the basis {A0 , ..., Ad } with 1 in the (u, v)
position. We have that
d
X
U (τ ) =
α i Ai
i=0
with αi ∈ C. So the coefficient of T is λ. Because U (τ ) has exactly one non-zero entry in the uth row
and because all the matrices in {A0 , ..., Ad } have at least one non-zero entry in every row, the coefficients
of Ai 6= T are all equal to 0. Then U (τ ) = λT .
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Observe that the uth row of T is the basis vector ev . The row and column sum of T are constant
and must then be equal to 1. Therefore T is a permutation matrix. Since T 6= A0 = I, the permutation
represented by T has no fixed points. Since U (τ ) is symmetric, we have that T is a permutation matrix
of order 2.
Now let i be the index for which Ai = T . Suppose 0 < i < d, and so d > 1. Hence v is the only
vertex at distance i from u, then bi−1 = 1. By Proposition 1, this implies that bj = 1 for all j ≥ i.
In particular, there will be a unique vertex at distance d from u, and this vertex will have degree 1,
hence the graph is K2 and d = 1, a contradiction. Therefore Ad = T , and being at distance 0 or d is an
equivalence relation with classes of size 2.
Lemma 1. Let X be a distance-regular graph. If perfect state transfer occurs on X, then all its
eigenvalues are integers.
Proof. By Theorem 2, U (2τ ) = λ2 I. Hence the eigenvalues of U (2τ ) are all equal to λ2 . These
eigenvalues are:
eı2τ θr for r ∈ {0, ..., d}.
Suppose θ is not an integer. It is an algebraic integer though, and so its algebraic conjugate θe is also
an eigenvalue. Now θ0 is the largest eigenvalue and is an integer. So we have that:
2τ (θ0 − θ)
and
e
2τ (θ − θ)
are integer multiples of π, clearly a contradiction.
Proposition 11. Consider a drg X of diameter d, and projectors {E0 , ..., Ed }. If Ad is a permutation
matrix of order two, note that the eigenvalues of Ad are either +1 or −1, and so Ad Ej = ±Ej .
Definition 8. We define a partition (S+ , S− ) of {0, ..., d} as j ∈ S+ if Ad Ej = Ej , and j ∈ S− otherwise.
Definition 9. If x ∈ Z, denote by ord2 (x) the exponent of 2 in the factorization of x (here we convention
ord2 (0) = +∞).
Theorem 3. Let X be a distance-regular graph with distinct eigenvalues θ0 > ... > θd . Let
α := gcd({θ0 − θk : k ∈ {0, ..., d}}).
Perfect state transfer occurs in X at time t between vertices u and v if and only if all of the following
hold:
1. Ad is a permutation matrix of order 2 with no fixed points;
2. the eigenvalues of X are integers;
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3. ord2 (θ0 − θj ) > ord2 (α) for all j ∈ S+ and ord2 (θ0 − θ` ) = ord2 (α) for all ` ∈ S− . Moreover, t is
an odd multiple of απ .
Proof. By Theorem 2 and Lemma 1, conditions (i) and (ii) are necessary for X to have perfect state
transfer at time t. Suppose conditions (i) and (ii) hold.
We may write
X
X
eıθ` t E`
eıθj t Ej +
U (t) =
j∈S+
and, for any λ ∈ C,
λAd =
X
`∈S−
λEj +
j∈S+
X
−λE` .
`∈S−
Since U (t) and Ad are both elements A and the idempotents Ej form a basis for this algebra, we see
that U (t) = λAd for some t and some λ if and only if the corresponding coefficients of every Ej and E`
are equal. We saw that 0 ∈ S+ , so λ = eıθ0 t . Thus, for all j ∈ S+ , eıθj t = λ if and only if t(θ0 − θj ) is a
multiple of 2π; and, for all ` ∈ S− , eıθ` t = −λ if and only if t(θ0 − θ` ) is an odd multiple of π. A time
t satisfying these conditions can be chosen if and only if condition (iii) holds. Moreover, the minimum
time t satisfying these conditions is τ = απ . Perfect state transfer happens at another time τ1 if and only
if τ1 is an odd multiple of τ .
Using the theory of orthogonal polynomials, we can actually determine the sets S+ and S− a priori.
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Orthogonal polynomials
Definition 10. Given a distance-regular graph X, parameters {a1 , ..., ad }, {b0 , ..., bd−1 } and {c0 , ..., cd },
define the following family of matrices Mi , for i = 1 to i = d.


0 b0 0 . . .
...
0
 c1 a1 b1 0
...
0 


 0 c2 a2 b2

0
.
.
.
0



.. 
.. ..
..
..

.
.
.
.
. 
Mi+1 =  0 0
(4)
.
 ..

.. . .
 .
.
.
0 


 0 0 . . . 0 ci−1 ai−1 bi−1 
0 0 ...
0
ci
ai
Proposition 12. Observe that the matrix Mi+1 is similar to the matrix:
√

0
b0 .c1 √ 0
...
...
0
√
 b0 .c1
a
b
.c
0
.
.
.
0
1
1
2

√
√

0
b
.c
a
b
.c
0
.
.
.
0
1 2
2
2 3


..
.
.
.
.
..
..
..
..
0
0
0
.
Mi+1
=


..
..
.
..

.
.

p
p 0

bi−2 .ci−1 pai−1
bi−1 .ci
0
0
...
0
0
0
...
0
bi−1 .ci
ai
6






,





which is symmetric. Hence the Perron-Frobenius theorem states that the eigenvalues Mi+1 interlace the
eigenvalues of Mi+2 for all i.
Definition 11. Now we define a sequence of polynomials.
• Let ω−1 (x) = 0.
• Let ω0 (x) = 1.
• For i = 1 until i = d (convention bd = 1), we make:
bi ωi+1 (x) = xωi (x) − ai ωi (x) − ci ωi−1 (x).
Proposition 13.
1. The polynomials ωi (x) and ωi+1 (x) have no roots in common, otherwise a reverse induction would
conclude that ω0 (x) has degree at least 1.
2. Polynomial ωi (x) is the characteristic polynomial Mi . (easy proof by induction).
3. Hence, by interlacing, there is exactly one root of ωi (x) between each pair of consecutive roots of
ωi+1 (x).
4. And the roots of ωi (x) are simple for all i. In particular Md+1 has d + 1 distinct eigenvalues.
Proposition 14. If X is a distance-regular graph, and its distinct eigenvalues are θ0 > ... > θd , note
from interlacing that the sign of ωd (θi ) is (−1)i .
Proposition 15. The eigenvalues of Md+1 are the eigenvalues of A1 = A(X).
Proof. Let Q be the matrix whose columns are indexed by numbers {0, ..., d} and rows are indexed by
vertices of X (in the same order as in A(X)). Fix a vertex u. Put a 1 in entry (i, v) if d(u, v) = i, 0
otherwise. Note that:
AQ = QMd+1
If θ is an eigenvalue for Md+1 with eigenvector v, then θ is an eigenvalue for A with eigenvector Qv.
Because both Md+1 and A have d + 1 distinct eigenvalues the result follows.



Proposition 16. If θ is an eigenvalue for Md+1 , its corresponding eigenvector is 

ω0 (θ)
ω1 (θ)
..
.
ωd (θ)
7



.

Proof. Follows from the definition of the polynomials:
θωi (θ) = ci ωi−1 (θ) + +ai ωi (θ) + bi ωi+1 (θ)



Proposition 17. If θ is an eigenvalue for X, one of its corresponding eigenvectors is Q 

ω0 (θ)
ω1 (θ)
..
.



.

ωd (θ)
Proposition 18. Any eigenvector for A is an eigenvector for Ad , because Ad can be written as a
polynomial in A. This is a consequence of Proposition 2.



Proposition 19. Therefore v(θ) = Q 

ω0 (θ)
ω1 (θ)
..
.



 is an eigenvector for Ad .

ωd (θ)
If Ad is a permutation matrix of order two and no fixed points, Ad v(θ) = ±v(θ). The first entry of
v(θ) is 1, and the last entry is ωd (θ), which then has to be ±1. The sign of the eigenvalue is determined
by the sign of ωd (θ).
Proposition 20. If X is a distance regular graph, θ0 > ... > θd its eigenvalues, and Ad is a permutation
matrix of order 2 and no fixed points, we have that:
Ad Ei = (−1)i Ei
Proof. It is a combination of Propositions 19 and 14.
Theorem 4. Let X be a distance-regular graph with distinct eigenvalues θ0 > ... > θd . Let
α := gcd({θ0 − θk : k ∈ {0, ..., d}}).
Perfect state transfer occurs in X at time t between vertices u and v if and only if all of the following
hold:
1. Ad is a permutation matrix of order 2 with no fixed points;
2. the eigenvalues of X are integers;
3. ord2 (θ0 − θj ) > ord2 (α) for all j even and ord2 (θ0 − θ` ) = ord2 (α) for all ` odd. Moreover, t is an
odd multiple of απ .
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