Finding Braids for Nonabelian Particles

1
Finding Braids for Nonabelian Particles
Nick Bonesteel Layla Hormozi Georgios Zikos
Florida State University
Steven H. Simon
Lucent Technologies

• NEB, L. Hormozi, G. Zikos, S.H. Simon, Phys.
  Rev. Lett. 95 140503 (2005)
• L. Hormozi, G. Zikos, NEB, and S.H. Simon, Phys.
  Rev. B, In Press. (quant-ph/0610111)
• Some unpublished results

Support US DOE
2
Topological Quantum Computation
time
3
Topological Quantum Computation
time
Matrix depends only on the topology of the braid
swept out by quasiparticle world lines! Robust
quantum computation? (Kitaev 97 Freedman,
Larsen and Wang 01)
4
SU(2)k Nonabelian Particles
Describe quasiparticle excitations of the
Read-Rezayi Parafermion states at level k (up
to Abelian phases).
Read and Rezayi, 1999
Slingerland and Bais, 2001
Are universal for quantum computation for k3 and
k gt 4.
Freedman, Larsen, and Wang, 2001
But before SU(2)k, there was just plain old
SU(2).
5
Particles with Ordinary Spin SU(2)

• Particles have spin s 0, 1/2, 1, 3/2 ,

spin ½
2. Triangle Rule for adding angular momentum
For example
Two particles can have total spin 0 or 1.

a
b
0
1
Numbers label total spin of particles inside ovals
6
Hilbert Space
7
Hilbert Space
S
7/2
3
Paths are states
5/2
2
3/2
1
1/2
N
11
12
0
1
2
3
4
5
6
7
8
10
9
1
1/2
1
3/2
8
S
44
1
8
7/2
35
7
1
3
110
6
27
1
5/2
75
20
5
1
2
165
14
48
4
1
3/2
90
28
9
3
1
1
132
5
14
42
2
1
1/2
5
1
132
2
14
42
N
11
12
0
1
2
3
4
5
6
7
8
10
9
9
Exchange-Based Quantum Computing
Compute by turning on and off the exchange
interaction between neighboring spin-1/2
particles for some time t, splitting the energy
of the total spin 0 and total spin 1 sectors by e.
10
Encoded Universality
J. Kempe, D. Bacon, D.A. Lidar, and K.B. Whaley
(2001)
t
4 spin encoding
0
0
0
1
1/2
1
1
0
0
1
2
3
4
11
Encoded Qubit Space
S
7/2
3
5/2
2
3/2
1
1/2
N
11
12
0
1
2
3
4
5
6
7
8
10
9
12
Universal Set of Gates
M.Hsieh, J. Kempe, S. Myrgren and K.B. Whaley
(2003)
Single Qubit Gates

0
Controlled-NOT gate
0

0
13
Particles with Ordinary Spin SU(2)

• Particles have spin s 0, 1/2, 1, 3/2 ,

spin ½
2. Triangle Rule for adding angular momentum
For example
Two particles can have total spin 0 or 1.

a
b
0
1
14
Nonabelian Particles SU(2)k

• Particles have topological charge s 0, 1/2, 1,
  3/2 , , k/2

topological charge ½
2. Fusion Rule for adding topological charge
For example
Two particles can have total topological
charge 0 or 1.
a
b
0
1
15
S
44
1
8
7/2
35
7
1
3
110
6
27
1
5/2
75
20
5
1
2
165
14
48
4
1
3/2
90
28
9
3
1
1
132
5
14
42
2
1
1/2
5
1
132
2
14
42
N
11
12
0
1
2
3
4
5
6
7
8
10
9
16
k 4
17
k 3
Fib(N1)
18
k 2
S
7/2
3
5/2
2
3/2
32
16
8
4
2
1
1
32
4
8
16
2
1
1/2
4
1
32
2
8
16
N
11
12
0
1
2
3
4
5
6
7
8
10
9
2N/2-1
19
Exchange-Based Quantum Computing
Compute by turning on and off the exchange
interaction between neighboring spin-1/2
particles for some time t, splitting the energy
of the total spin 0 and total spin 1 sectors by e.
20
Topological Quantum Computing
Compute by braiding quasiparticles around one
another.
R matrix
0
0
1
1
21
4 particle encoding
22
Encoded Qubit Space
23
Lets focus on k3 first.
For k3, if we only have particles with
topological charge 1, the fusion rules are simply
The remaining k3 fusion rules, such as
3
3
1
3
3
1
3
3

Ä

Ä

Ä

Ä
0

1

1

0
2
2
2
2
2
2
2
2
imply that the 3/2 particle is effectively
abelian (it has non-branching fusion rules) and
the braiding properties of particles with
topological charges 1/2 and 1 are equivalent (up
to a phase).
topological charge 1 in SU(2)3
Fibonacci anyon
24
Braiding Matrices for 4 Fibonacci anyons
25
Braiding Matrices for 4 Fibonacci anyons
M T

s1-1 s2 s1-1 s2 M
26
Single Qubit Operations
General rule Braiding inside an oval does not
change the total topological charge of the
enclosed particles.
Important consequence As long as we braid
within a qubit, there is no leakage error.
0
0
Can we do arbitrary single qubit rotations this
way?
27
Single Qubit Operations are Rotations
2 p
The set of all single qubit rotations lives in a
solid sphere of radius 2p.
2 p
-2 p
-2 p
28
s12
2 p
s22
-2 p
-2 p
2 p
s2-2
-2 p
s1-2
29
N 1
30
N 2
31
N 3
32
N 4
33
N 5
34
N 6
35
N 7
36
N 8
37
N 9
38
N 10
39
N 11
40
Brute Force Search
Brute force searching rapidly becomes infeasible
as braids get longer.
Fortunately, a clever algorithm due to Solovay
and Kitaev allows for systematic improvement of
the braid given a sufficiently dense covering of
SU(2).
41
Solovay-Kitaev Construction
e
(Actual calculation)
Braid Length
,
42
Two Qubit Gates
a
?
a
?
0
?
b
?
b
?
0
?
Problems 1. We are pulling
quasiparticles out of qubits Leakage error!
2. 168 dimensional search space (as opposed
to 3 for three-particle braids).
Straightforward brute force search is not
feasible.
43
Two Qubit Gates
a
?
a
?
0
?
b
?
b
?
0
?
ab 00 01 10 11
4 x 4 block acting on the logical qubit space. In
general this matrix is not unitary due to leakage
error.
44
Two Qubit Controlled Gates
a
Control Qubit
a
0
b
Target Qubit
b
0
ab 00 01 10 11
45
Two Qubit Controlled Gates
a
Key Idea Weave this pair (the control pair)
of particles around the particles in the target
qubit.
a
0
b
b
0
ab 00 01 10 11
46
Two Qubit Controlled Gates
a
?
Key Idea Weave this pair (the control pair)
of particles around the particles in the target
qubit.
a
a
0
?
b
?
b
?
0
?
ab 00 01 10 11
47
Two Qubit Controlled Gates
a
a
a
a
0
?
b
?
b
?
0
?
ab 00 01 10 11
48
Two Qubit Controlled Gates
a
a
a
a
0
?
b
?
b
?
0
?
ab 00 01 10 11
Important Rule Braiding an object with
topological charge 0 does not induce any
transitions.
Only a1 sector is nontrivial.
49
Two Qubit Controlled Gates
1
1
1
1
0
?
b
b
b
b
0
?
ab 00 01 10 11
Another idea Weave control pair around pairs of
particles in the target qubit. If b 0 this
braid again produces no transition.
Only ab 11 sector is nontrivial.
50
Two Qubit Controlled Gates
1
1
1
1
0
?
1
1
1
1
0
?
ab 00 01 10 11
Another idea Weave control pair around pairs of
particles in the target qubit. If b 0 this
braid again produces no transition.
Only ab 11 sector is nontrivial.
51
Two Qubit Controlled Gates
?
0
?
0
0
0
For Fibonacci anyons this is equivalent to
finding a single qubit rotation!
52
Two Qubit Controlled Gates
?
0
?
0
0
0
a44

Result of brute force search
53
Two Qubit Controlled Gates
?
0
?
0
0
0
a44

Result of brute force search
54
Two Qubit Controlled Gates
a
a
a
a
0
0
b
b
b
b
0
0
ab 00 01 10 11
ù
é
0
0
0
1
ú
ê
0
0
1
0
ú
ê
Controlled-Phase Gate

U
ú
ê
-
qubit
two
0
1
0
0
ú
ê
-1O(10-3)
0
0
0
û
ë
55
Solovay-Kitaev Improved Controlled-phase gate
56
Universal Set of Gates
M.Hsieh, J. Kempe, S. Myrgren and K.B. Whaley
(2003)
Single Qubit Gates

0
Controlled-NOT gate
0

0
57
Universal Set of Gates
Single Qubit Gates

0
Controlled-NOT gate
0
0
58
What about k gt 3?
59
N 1
k2
k3
k4
k5
60
N 2
k2
k3
k4
k5
61
N 3
k2
k3
k4
k5
62
N 4
k2
k3
k4
k5
63
N 5
k2
k3
k4
k5
64
N 6
k2
k3
k4
k5
65
N 7
k2
k3
k4
k5
66
N 8
k3
k2
k5
k4
67
N 9
k2
k3
k4
k5
68
Does this construction work for kgt4?
a
a
a
a
0
?
b
b
b
b
0
?
ab 00 01 10 11
Only ab 11 sector is nontrivial.
69
In principle yes, but not efficient.
1
1
1
1
c
0
1
1
1
1
c
0
0
0
For kgt3, finding approximate gates requires
Solovay-Kitaev in SU(3).
c 0, 1, 2
This is feasible, but it is more efficient to
break the problem into smaller SU(2) problems.
New label occurs for k gt 3
70
OK, try weaving through top two particles
a
a
a
a
0
?
b
?
b
b
0
?
71
OK, try weaving through top two particles
1
1
1
1
0
?
b
?
b
b
0
?
Again, if a0, nothing happens, so only the a1
case is relevant.
72
OK, try weaving through top two particles
1
1
1
1
0
?
b
d
?
d
b
b
0
?
73
OK, try weaving through top two particles
1
1
b
d
?
d
db 01 10 11 21
These three matrix elements must be equal for
there to be no leakage error.
ù
é
a
0
0
0
ú
ê
c
b
0
0
ú
ê

U
ú
ê
e
d
0
0
ú
ê
f
0
0
0
û
ë
74
Try weaving through top two particles
1
1
b
d
?
d
Useful fact For braids with zero winding
determinant of each block must be 1
db 01 10 11 21
ù
é
1
0
0
0
ú
ê
0
1
0
0
ú
ê
Only braid which does not lead to leakage error
gives the identity operation.

U
ú
ê
1
0
0
0
ú
ê
1
0
0
0
û
ë
75
What if we dont bring the blue pair back?
1
?
b
d
1
d
db 01 10 11 21
Because the three b1 matrix elements are equal,
when b1 this operation simply swaps the blue
pair with the green pair.
ù
é
1
0
0
0
ú
ê
0
1
0
0
ú
ê

U
ú
ê
1
0
0
0
ú
ê
1
0
0
0
û
ë
76
1
1
1
0
0
1
0
1
0
0
0
1
If b0, the fusion rules imply that the overall
label of injected target must be 1
77
1
1
1
0
1
0
1
1
1
1
0
0
If b1, we simply swap the blue pair with the
green pair, and the overall label of injected
target remains 0.
78
Intermediate State
79
db 01 10 11 21
ù
é
1
0
0
0
b0
ú
ê
0
-1
0
0
ú
ê

U
b1
ú
ê
-1
0
0
0
ú
ê
1
0
0
0
û
ë
80
1
1
0
b
b
0
O(10-3)
81
k5 Controlled-Phase Gate
O(10-3)
82
Universal Set of Gates for k5
Single Qubit Gates

0
Controlled-NOT gate
0

0