New Examples of Perfect State Transfer - IME-USP

Transcription

New Examples of
Perfect State Transfer
Gabriel Coutinho
University of Waterloo
A conference to celebrate the work of Chris Godsil.
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Perfect state transfer:
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Motivation from quantum computing.
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Example, definition, some known results.
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Our methods to find new examples.
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Model: To each vertex of a graph, assign a 2-dim vector space.
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A 2-dim space is a basic unit of quantum information, or a qubit.
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Graph is associated to a vector space of dimension 2|V | .
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Graph is associated to a vector space of dimension 2|V | .
State:
corresponding to
1
0
0
0
⊗
⊗
⊗
0
1
1
1
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Next step: define a Hamiltonian to govern the time-evolution of
this quantum system.
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That is, a Hermitian matrix H of dimension 24 .
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That is, a Hermitian matrix H of dimension 24 .
Choice: a time independent Hamiltonian that encodes precisely the
adjacency of the graph.
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We are interested in a perfect transfer of state, ie, a time t such that
evolves to (up to a factor λ ∈ C)
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We are interested in a perfect transfer of state, ie, a time t such that
evolves to (up to a factor λ ∈ C)
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This is related to the
problem of transferring
information in certain models
of quantum computing.
The action of our Hamiltonian is equivalent to the action of the
adjacency matrix A = A(X ).
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By Schrödinger’s Equation, at time t ∈ R+ , this means that the
evolution will be determined by
exp(itA) =
X (itA)k
k≥0
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k!
.
By Schrödinger’s Equation, at time t ∈ R+ , this means that the
evolution will be determined by
exp(itA) =
X (itA)k
k≥0
k!
.
So we are looking for a time t such that
   
1
0
0  0 
  
exp(itA) 
0 = λ
0
0
where |λ| = 1.
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Formalization:
We say that X admits perfect state transfer from vertex u to
vertex v if there is a time t ∈ R+ and a phase λ ∈ C such that
exp(itA)eu = λev
or equivalently
| exp(itA)u,v | = 1.
We say that u is periodic if u = v above.
The existence of periodic vertices depends only on the eigenvalues,
whereas state transfer depends also on the structure of the graph.
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Suppose X = K2 . In this case,
0 1
0 1
(odd)
A=
A
=
1 0
1 0
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(even)
A
1 0
=
0 1
0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
So it is easy to see that
exp(itA) =
X (itA)k
k≥0
k!
= cos(t)
Hence
exp(itA)
1 0
0 1
+ i sin(t)
0 1
1 0
1
cos(t)
=
0
i sin(t)
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0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
exp(itA) =
X (itA)k
k≥0
Hence
k!
1 0
0 1
= cos(t)
+ i sin(t)
0 1
1 0
1
cos(t)
exp(itA)
=
0
i sin(t)
t=0
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0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
exp(itA) =
X (itA)k
k≥0
Hence
k!
1 0
0 1
= cos(t)
+ i sin(t)
0 1
1 0
1
cos(t)
exp(itA)
=
0
i sin(t)
t=
9
π
10
0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
exp(itA) =
X (itA)k
k≥0
Hence
k!
1 0
0 1
= cos(t)
+ i sin(t)
0 1
1 0
1
cos(t)
exp(itA)
=
0
i sin(t)
t=
9
2π
10
0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
exp(itA) =
X (itA)k
k≥0
Hence
k!
1 0
0 1
= cos(t)
+ i sin(t)
0 1
1 0
1
cos(t)
exp(itA)
=
0
i sin(t)
t=
9
3π
10
0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
exp(itA) =
X (itA)k
k≥0
Hence
k!
1 0
0 1
= cos(t)
+ i sin(t)
0 1
1 0
1
cos(t)
exp(itA)
=
0
i sin(t)
t=
9
4π
10
0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
exp(itA) =
X (itA)k
k≥0
Hence
k!
1 0
0 1
= cos(t)
+ i sin(t)
0 1
1 0
1
cos(t)
exp(itA)
=
0
i sin(t)
t=
9
π
2
0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
exp(itA) =
X (itA)k
k≥0
Hence
k!
1 0
0 1
= cos(t)
+ i sin(t)
0 1
1 0
1
cos(t)
exp(itA)
=
0
i sin(t)
t=
9
6π
10
0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
exp(itA) =
X (itA)k
k≥0
Hence
k!
1 0
0 1
= cos(t)
+ i sin(t)
0 1
1 0
1
cos(t)
exp(itA)
=
0
i sin(t)
t=
9
7π
10
0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
exp(itA) =
X (itA)k
k≥0
Hence
k!
1 0
0 1
= cos(t)
+ i sin(t)
0 1
1 0
1
cos(t)
exp(itA)
=
0
i sin(t)
t=
9
8π
10
0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
exp(itA) =
X (itA)k
k≥0
Hence
k!
1 0
0 1
= cos(t)
+ i sin(t)
0 1
1 0
1
cos(t)
exp(itA)
=
0
i sin(t)
t=
9
9π
10
0 1
0 1
(odd)
A=
A
=
1 0
1 0
(even)
A
1 0
=
0 1
exp(itA) =
X (itA)k
k≥0
Hence
k!
1 0
0 1
= cos(t)
+ i sin(t)
0 1
1 0
1
cos(t)
exp(itA)
=
0
i sin(t)
t=π
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A taste of how things are in general....
Suppose A = A(X ) has distinct eigenvalues θ0 > ... > θd .
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Spectral decomposition:
A=
d
X
θr E r
r =0
Function on A ⇒ function on the eigenvalues.
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A=
d
X
θr E r
r =0
Function on A ⇒ function on the eigenvalues. Thus:
!
!
d
d
X
X
PST
itθr
exp(itA)eu =
e Er eu = λev = λ
Er ev
r =0
r =0
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A=
d
X
θr E r
r =0
Function on A ⇒ function on the eigenvalues. Thus:
!
!
d
d
X
X
PST
itθr
exp(itA)eu =
e Er eu = λev = λ
Er ev
r =0
r =0
Necessary condition for PST: Er eu = ±Er ev for all r .
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Known results:
Problem was introduced by Bose in 2003 and subsequently studied
in more detail by Christandl and others in 2005.
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Known results:
Perfect state transfer happens in:
Paths P2 and P3
Iterated cartesian powers - hypercubes.
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Known results:
Paths P2 and P3
Does not happen in:
Paths Pn , n > 3.
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Known results:
Paths P2 and P3
Does not happen in:
Paths Pn , n > 3.
It doesn’t happen in complete graphs Kn with n > 2.
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Known results:
Paths P2 and P3
Does not happen in:
Paths Pn , n > 3.
It doesn’t happen in complete graphs Kn with n > 2.
Theorem (Godsil ’12)
There are only finitely many graph with maximum valency k in
which perfect state transfer can occur.
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Perfect state transfer in circulants (Cayley graphs for Zn ) was
characterized by Bašić and others 2011 and 2013.
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Perfect state transfer in cubelike graphs (Cayley graphs for
Z2d ) was characterized by Godsil and others in 2008 and 2011.
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Perfect state transfer in cubelike graphs (Cayley graphs for
Z2d ) was characterized by Godsil and others in 2008 and 2011.
The effect of certain graph operations (joins, products) was
studied by Tamon and others 2009-2012.
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Cartesian product:
Consider graphs X and Y and their cartesian product X Y .
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Cartesian product:
We have
A(X Y ) = A(X ) ⊗ I + I ⊗ A(Y )
As a consequence (Christandl et al. ’05)
exp itA(X Y ) = exp itA(X ) ⊗ exp itA(Y )
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Cartesian product:
We have
A(X Y ) = A(X ) ⊗ I + I ⊗ A(Y )
As a consequence (Christandl et al. ’05)
exp itA(X Y ) = exp itA(X ) ⊗ exp itA(Y )
Thus X Y admits perfect state transfer at time τ if and only if
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X admits perfect state transfer at time τ , and
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Y admits perfect state transfer at time τ or contains a
periodic vertex at time τ .
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Example: P3 P3 admits perfect state transfer between the pairs of
π
vertices below at time √ .
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Similarly, P3 K1,8 admits perfect state transfer.
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π
2
14
π
2
14
π
2
14
π
2
14
π
2
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π
2
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π
2
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π
2
14
π
2
14
π
2
14
π
2
14
π
2
14
π
2
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Our goal: find more examples of perfect state transfer.
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Our method: use a general form of graph products.
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Suppose X is a graph such that
A(X ) = B1 ⊗ B2 + C1 ⊗ C2 ,
where B1 , B2 , C1 , C2 are 01-symmetric matrices (NEPS).
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Suppose X is a graph such that
A(X ) = B1 ⊗ B2 + C1 ⊗ C2 ,
where B1 , B2 , C1 , C2 are 01-symmetric matrices (NEPS).
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This generalizes: cartesian, direct, lexicographic products.
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Double covers can also be expressed in this way.
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Direct product:
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Defined as A(X × Y ) = A(X ) ⊗ A(Y ).
If perfect state transfer happens in the direct product X × Y ,
it must happen in at least one of the factors.
On the other hand, the conditions required in the other factor
are fairly weak.
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Direct product:
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Defined as A(X × Y ) = A(X ) ⊗ A(Y ).
If perfect state transfer happens in the direct product X × Y ,
it must happen in at least one of the factors.
On the other hand, the conditions required in the other factor
are fairly weak.
Theorem (Coutinho and Godsil ’13)
If
∗ Y admits perfect state transfer, and
∗ eigenvalues of X and Y are integers or integer multiples of a
square root, and
∗ integer parts of eigenvalues of X are all divisible by same
powers of 2,
then there is a k0 ∈ Z+ such that X × Y m.k0 admits perfect state
transfer for all m ∈ Z.
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Two families of examples:
If Y admits perfect state transfer at time
π
2,
then
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K1,n × Y 2m for all n and m.
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if X is one of the 32548 non-isomorphic strongly regular
graphs with parameters (36, 15, 6, 6) and eigenvalues
{15, 3, −3}, then X × Y 2m admits perfect state transfer.
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Double covers:
Given graphs X and Y on the same set of vertices, we define the
graph X n Y by
A(X ) A(Y )
A(X n Y ) =
A(Y ) A(X )
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If A(X ) ◦ A(Y ) = 0, this is a double cover of the graph with
adjacency matrix A(X ) + A(Y ).
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If Y = X , this is a double cover of the complete graph and is
known as the switching graph of X .
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If A(X ) and A(Y ) commute, then perfect state transfer happens in
X n Y if and only if both X and Y are periodic at the same vertex
at the same time with certain phase factors.
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If A(X ) and A(Y ) commute, then perfect state transfer happens in
X n Y if and only if both X and Y are periodic at the same vertex
at the same time with certain phase factors.
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If X = Kn Kn , then X n X admits perfect state transfer for
all n divisible by 4.
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Open questions:
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Find examples that do not depend on previous constructions.
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Characterize perfect state transfer in Cayley graphs for abelian
groups.
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Characterize perfect state transfer in trees (possibly the only
trees admitting pst are P2 and P3 ...)
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New Examples of Perfect State Transfer - IME-USP

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