Quantum Processes in Graph Computing

Quantum Processes in Graph Computing
Dr. Marko A. Rodriguez
DataStax and Apache TinkerPop
Rodriguez, M.A., Watkins, J.H., "Quantum Walks with Gremlin," GraphDay '16, pp. 1-16, Austin, Texas, January 2016.
http://arxiv.org/abs/1511.06278

Part 1
Wave Dynamics in Graphs

However, vertices don't "move."
The graph topology is the only space.
There is no "space" outside the graph.
"Then how do you model an oscillating vertex?"
By computing the amount of "energy"
at the vertex over time.
"But traversers are indivisible objects!
Energy must be able to be
divided indeﬁnitely and take negative values."
Lets look at Gremlin Sacks.

Gremlin Sacks
(a data structure local to the traverser)
(each traverser can carry an object along its way)
The traverser is not divisible, but that
double ﬂoating point number in his sack is!

1.0
0.0
-1.0
1.0
Don't worry about the algorithm, just watch the visualization for now.

-0.5 0.5
0.0
-1.0
1.0

-0.5 0.250.25
0.0
-1.0
1.0

-0.125 0.125-0.3750.375
0.0
-1.0
1.0

0.375-0.3125 -0.3125
0.0
-1.0
1.0

0.15625 -0.156250.34375-0.34375
0.0
-1.0
1.0

1 2 30 4 5 6
1.0
g.withSack(1.0,sum).V(3).
repeat(
union(
sack(mult).by(-0.5).out('left'),
sack(mult).by(0.5).out('right')
)
).times(3)
An initial energy of 1.0 perturbs the graph.

1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.5 0.51.0
out('left') out('right')
The energy diffuses left and right along the graph.

1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.5 0.5
1
tim
e
The total energy in the system is always 1.

Energy waves reverberate and thus, prior "left energy" can go right.
1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.5 0.250.25 -0.5 0.5
out('left') out('right') out('left') out('right')

1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.5 0.250.25 -0.5 0.5
out('left') out('right') out('left') out('right')
(-0.5 * 0.5) + (0.5 * -0.5)
-0.5
-0.25 + -0.25
Energy constructively and destructively interferes.

1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.5 0.250.25
2
tim
es

1 2 30 4 5 6
repeat(
union(
)
).times(3)
-0.125 0.125-0.3750.375 -0.5 0.250.25
out('left') out('left') out('left')out('right') out('right') out('right')

1 2 30 4 5 6
repeat(
union(
)
).times(3)
3
tim
es
-0.125 0.125-0.3750.375

A wave is a diffusion of energy within a space.
The space is the graph.
The energy is contained within the traversers (via their sacks).
When a traverser splits, its sack energy is divided amongst its children
traversers' sacks.
When traversers merge, their sack energy is summed to a
single traverser sack.
If two energy sacks with the same parity are merged,
constructive interference.
If a negative and a positive energy sack are merged,
destructive interference.

Part 2
Quantum Mechanics
via the
Copenhagen Interpretation

1
2
3
0 4 5
gremlin> g.V(0)
==>v[0]

1
2
3
0 4 5
gremlin> g.V(0).out()
==>v[1]

1
2
3
0 4 5
gremlin> g.V(0).out().out()
==>??
?
?
"Two outgoing adjacent vertices,
what should I do?"

1
2
3
0 4 5
==>v[2]
==>v[3]

1
2
3
0 4 5
==>v[2]
==>v[3]
2 particles are created from 1.
In graph computing: "There are two legal paths, so take both."
In physical computing: "Violates conservation of energy law (matter is created)."

1
2
3
0 4 5
gremlin> g.V(0).out().out().out()
==>v[4]
==>v[4]
In graph computing: "Two length 3 paths lead to vertex 4."
In physical computing: "The twin particles coexist at vertex 4."

1
2
3
0 4 5
gremlin> g.V(0).out().out().out().out()
==>v[5]
==>v[5]
In graph computing: "Two length 3 paths lead to vertex 5."
In physical computing: "The twin particles coexist at vertex 5."

1
2
3
0 4 5
What does a physical particle do when it is faced with a choice?
?
?

1
2
3
0 4 5
The particle takes both options!
The indivisible "particle" divides!

1
2
3
0 4 5
Two particles can't be "seen," but we know both options are taken.
When we try to see both "particles", they instantly become one particle.
"Why do you say the particle took both paths?
If you ﬁnd it only on a single path, then wasn't only one path taken?"
Wave interference!
Explanation coming soon...
"Wave Function Collapse"

The smallest "quanta" allowed is the particle.
Only one particle is ever directly observed.
However, prior to observation, the particle splits into pieces (a wave).
When we observe the system, the pieces becomes one again (a particle).
Other Interpretations of Quantum Mechanics
Multi-World Interpretation
Pilot Wave Interpretation
photon
electron
bacteria!
an isolated system

Part 3
The Quantum Wave Function

Classical Particle
(A single unique "thing" exists)
The Particle Diffuses over Space as a Wave
(The particle is in a quantum superposition)
The Wave Function is Perturbed
(e.g., light tries to "see" the particle)
The Wave Function Collapses to a Classical Particle
(The wave turns into a particle)

Classical Particle
(A single unique "thing" exists)
The Particle Diffuses over Space as a Wave
(The particle is in a quantum superposition)
The Wave Function is Perturbed
(e.g., light tries to "see" the particle)
The Wave Function Collapses to a Classical Particle
(The wave turns into a particle)
ObservedObservedInferred

48 49 5047 51 52 53
A 101 vertex line graph with the center being vertex 50.
Space

Classical Initial Step
The particle has a deﬁnite location and a spin/potential to the left.
48 49 5047 51 52 53
[1, 0]
[1,0]50
left-ness
right-ness
For each option (degree of freedom),
we need a vector component.
In the article, we call it "traversal superposition."

0.00.20.40.60.81.0
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53
The probability of seeing the particle at vertex 50 is 100%.
[1, 0]
12
+ 02
= 1
Wave Function to Probability Distribution

Coin Step
48 49 5047 51 52 53
H =
1
√
2
1 1
1 −1
The particle's spin is put into superposition.
[1, 0] →
1
√
2
,
1
√
2
Ithastheoptiontogo
leftorrightsoitgoesboth!

Coin Step
48 49 5047 51 52 53
H =
1
√
2
1 1
1 −1
[1, 0] →
1
√
2
,
1
√
2
"right particles" going right ﬂip their sign.
"left particles" going left or right keep their sign.

Coin Step
48 49 5047 51 52 53
H =
1
√
2
1 1
1 −1
[1, 0] →
1
√
2
,
1
√
2
1
√
2
1 1
1 −1
· [1, 0] =
1√
2
1√
2
1√
2
− 1√
2
· [1, 0] =
1
√
2
,
1
√
2

Coin Step
48 49 5047 51 52 53
H =
1
√
2
1 1
1 −1
1 ·
1
√
2
+ 0 ·
1
√
2
, 1 ·
1
√
2
+ 0 · −
1
√
2
=
1
√
2
,
1
√
2
[1, 0] →
1
√
2
,
1
√
2
1
√
2
1 1
1 −1
· [1, 0] =
1√
2
1√
2
1√
2
− 1√
2
· [1, 0] =
1
√
2
,
1
√
2

Coin Step
48 49 5047 51 52 53
1
√
2
,
1
√
2

The "particle" spinning left goes left and
the "particle" spinning right goes right.
Shift Step
48 49 5047 51 52 53
1
√
2
, 0 0,
1
√
2
1
√
2
,
1
√
2

48 49 5047 51 52 53
1
√
2
, 0 0,
1
√
2
Shift Step

P(v) = |c0|2
+ |c1|2
48 49 5047 51 52 53
1
√
2
, 0 0,
1
√
2
1
√
2
2
+ 02
= 0.5 02
+
1
√
2
2
= 0.5
The probability of locating the classical particle each vertex
is a function of the total wave amplitude at that vertex.
50%-chance 50%-chance

0.00.10.20.30.40.5
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53
1
√
2
, 0 0,
1
√
2
1
√
2
2
+ 02
= 0.5 02
+
1
√
2
2
= 0.5
The probability of seeing the particle at vertex 49 or 51 is 50%.

48 49 5047 51 52 53
H =
1
√
2
1 1
1 −1
Each "particles'" spin is put into superposition.
Coin Step
1
2
,
1
2
1
2
, −
1
2

48 49 5047 51 52 53
Shift Step
1
2
,
1
2
The "particles" spinning left go left and
the "particles" spinning right go right.
0, −
1
2
1
2
, 0
1
2
,
1
2
1
2
, −
1
2

48 49 5047 51 52 53
Shift Step
1
2
,
1
2
0, −
1
2
1
2
, 0
Constructive Interference

0.00.10.20.30.40.5
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53
1
2
,
1
2
0, −
1
2
1
2
, 0
1
2
2
+ 02
=
1
4
1
2
2
+
1
2
2
=
1
2
02
+ −
1
2
2
=
1
4

48 49 5047 51 52 53
Coin Step
H =
1
√
2
1 1
1 −1
The "particles'" spin is put into superposition.
1
2
√
2
,
1
2
√
2
−
1
2
√
2
,
1
2
√
2
1
√
2
, 0

48 49 5047 51 52 53
Shift Step
The "particles" spinning left go left and
the "particles" spinning right go right.
1
2
√
2
, 0
1
√
2
,
1
2
√
2
−
1
2
√
2
, 0 0,
1
2
√
2
Destructive InterferenceConstructive Interference

0.00.10.20.30.40.50.6
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53
1
2
√
2
, 0
1
√
2
,
1
2
√
2
−
1
2
√
2
, 0 0,
1
2
√
2
1
2
√
2
2
+ 02
=
1
8
02
+
1
2
√
2
2
=
1
8
−
1
2
√
2
2
+ 02
=
1
8
1
√
2
2
+
1
2
√
2
2
=
5
8

0.00.10.20.30.40.50.6
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.00.10.20.30.40.5
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.00.10.20.30.4
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.000.050.100.150.200.250.30
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.000.050.100.150.200.25
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.00.10.20.3
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
48 49 5047 51 52 53

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
13 14 1512 16 17 18

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
Wave Function is Perturbed by Light
13 14 1512 16 17 18

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
Wave Function Collapse
13 14 1512 16 17 18
sample

Classical Particle
13 14 1512 16 17 18
[1, 0]
0.00.20.40.60.81.0
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
13 14 1512 16 17 18
Could
have
been
thisvertex.
Again,itsa
probabilitydistribution.

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
13 14 1512 16 17 18
Could not be at this vertex because there
is no wave energy at that vertex.

g.withSack([1,0],sackSum).V(50).
repeat(
sack(hadamard).
union(
sack(project).by(constant([1,0])).out('left'),
sack(project).by(constant([0,1])).out('right')
)
).times(50).
group().by(id).by(sack().map(norm)).
unfold().sample(1).by(values)
48 49 5047 51 52 53

48 49 5047 51 52 53
[1, 0]
repeat(
sack(hadamard).
union(
)
).times(50).
Classical Initial Step
sackSum = { a,b -> [a[0] + b[0],a[1] + b[1]] }

hadamard = { a,b ->
[(1/Math.sqrt(2)) * (a[0] + a[1]), (1/Math.sqrt(2)) * (a[0] - a[1])]
}
repeat(
sack(hadamard).
union(
)
).times(50).
Coin Step
H =
1
√
2
1 1
1 −1
48 49 5047 51 52 53
1
√
2
,
1
√
2

project = { a,b -> [a[0] * b[0], a[1] * b[1]] }
repeat(
sack(hadamard).
union(
)
).times(50).
Shift Step
48 49 5047 51 52 53
1
√
2
, 0 0,
1
√
2
1
√
2
,
1
√
2

repeat(
sack(hadamard).
union(
)
).times(50).
Diffuse the wave function for 50 steps
48 49 5047 51 52 53

0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
norm = { Math.pow(it.get()[0],2) +
Math.pow(it.get()[1],2) }
repeat(
sack(hadamard).
union(
)
).times(50).
13 14 1512 16 17 18
P(v) = |c0|2
+ |c1|2

repeat(
sack(hadamard).
union(
)
).times(50).
Wave Function Collapse (Classical Particle Manifested)
13 14 1512 16 17 18
0.000.050.100.15
vertices
probability
0 4 8 12 17 22 27 32 37 42 47 52 57 62 67 72 77 82 87 92 97
sample

repeat(
sack(hadamard).
union(
)
).times(50).
A Quantum Walk on a Line with Gremlin

H =
1
√
2
1 1
1 −1
Y =
1
√
2
1 i
i 1
Balanced Unitary OperatorUnbalanced Unitary Operator
0.000.050.100.15
vertices
probability
0 6 12 20 28 36 44 52 60 68 76 84 92 100
0.000.020.040.060.080.10
vertices
probability
0 6 12 20 28 36 44 52 60 68 76 84 92 100
Any energy split operator can be used as long as:
1. The operation conserves the total energy in the system.
2. It is possible to reverse the operation.
In general, a unitary operator is used to diffuse energy.
complexnumbers
i2
=−1
U · U∗
= I
nofriction
noinformation
loss

The probability of seeing the particle at a particular vertex after
50 iterations for both a quantum walk and a classical walk.
Quantum walks can explore more graph in a shorter amount of time!
0.000.020.040.060.080.10
vertices
probability
0 6 12 20 28 36 44 52 60 68 76 84 92 100
√
50 = 7.07
50 = 50
√
n
n
Y =
1
√
2
1 i
i 1

Part 4
The Double Slit Experiment

Thomas Young (1773-1829)
Question
"Is light a wave or a particle?"
continuous light source double-slit screen photoelectric ﬁlm

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