fibonacci anyon with topological “charge” 1 fusion rules

−=

−−

−−

12/1

2/11

ϕϕϕϕ

FF Matrix:

F F

F

F

F

Pentagon

Fusion rules: 1x1 = 0+1; 1x0 = 0x1 = 1; 0x0 = 0

Fibonacci anyon with topological “charge” 1

aR=a a

=

1

0

00R

RR

The R Matrix

a a

=

1

0

00R

RR

The R Matrix

R

a

Equivalent to

1 a

a

Equivalent to R

a

1 a

a

Equivalent to R

a

R 1 a 1 a

F F

R

R

F F

R

F

R

R

F F

F

R

R

F F

R

F F

R

F

R

R

The Hexagon Equation

0 1 0 1


0 1 0 1


1

0 1 0 1


a 1

0 1 0 1


a 1 F0a

0 1 0 1


a 1 1 a F0a

0 1 0 1


a 1

δa,0 + R1 δa,1

1 a F0a

0 1 0 1


a 1 F0a

δa,0 + R1 δa,1

1 a Fa0

0 1 0 1


a 1 F0a

δa,0 + R1 δa,1

1 a Fa0

0 1

0 1 0 1


a 1 F0a

δa,0 + R1 δa,1

1 a Fa0

R0

0 1

0 1 0 1


a 1 F0a

δa,0 + R1 δa,1

1 a Fa0

R0

0 1 1

0 1 0 1


a 1 F0a

δa,0 + R1 δa,1

1 a Fa0

R0

0 1 1 0

0 1 0 1


a 1 F0a

δa,0 + R1 δa,1

1 a Fa0

R0

0 1 1 0 F00

0 1 0 1


a 1 F0a

δa,0 + R1 δa,1

1 a Fa0

R0

0 1 1 0 F00

R0

0 1 0 1


a 1 F0a

δa,0 + R1 δa,1

1 a Fa0

R0

0 1 1 0 F00

R0

000001,10,0 )( RFRFRF aaaa

a =+∑ δδTop path

Bottom path


Only two solutions:

=

−

5/3

5/4

00π

π

i

i

ee

R

=

− 5/3

5/4

00

π

π

i

i

ee

R

F F

R F

R

R

Basic Content of an Anyon Theory

−=

−−

−−

12/1

2/11

ϕϕϕϕ

F

=

−

5/3

5/4

00π

π

i

i

ee

R

F Matrix:

R Matrix:

F F

F

F

F

F F

R

F

R

R Pentagon Hexagon

“Charge” values: 0, 1

Fusion rules: 1x1 = 0+1; 1x0 = 0x1 = 1; 0x0 = 0

Qubit Encoding

0 1

Non-Computational State

State of qubit is determined by “charge” of two leftmost

particles

Qubit States

1 0

Transitions to this state are leakage errors

1 1 =1

=0

Initializing a Qubit

Pull two quasiparticle-quasihole pairs out of the “vacuum”.

0 0 0

These three particles have total q-spin 1

Initializing a Qubit

Pull two quasiparticle-quasihole pairs out of the “vacuum”.

0 1

0

Measuring a Qubit

Try to fuse the leftmost quasiparticle-quasihole pair.

10 βα +

? 1

Measuring a Qubit

If they fuse back into the “vacuum” the result of the measurement is 0.

1 0

0

Measuring a Qubit

If they cannot fuse back into the “vacuum” the result of the measurement is 1

1 1

1

1

Elementary Braid Matrices

a 1

time

?1 =σ

a 1

time

==

−

5/3

5/4

1 00π

π

σ i

i

ee

R


a = 0 1

a 1

time

==

−

5/3

5/4

1 00π

π

σ i

i

ee

R

a 1

?2 =σ


a 1

time

==

−

5/3

5/4

1 00π

π

σ i

i

ee

R

a 1

−−

==−

−−

ττττσ

π

ππ

5/3

5/35/

2 i

ii

eeeFRF


12/)15( −=−= ϕτ

a 1

−−

==−

−−

ττττσσ

π

ππ

5/3

5/35/

12 i

ii

eeeFF

σ1−1 σ2 σ1

−1 σ2 = M

i Ψ

f Ψ M T = i Ψ

=

−

5/3

5/4

1 00π

π

σ i

i

ee

a 1

1 1


A Quantum Bit: A Continuum of States

0

1

θ

φ 2

10 i−2

10 i+

1 2

sin 0 2

cos φ θ θ ψ i e − + =

Single Qubit Operations: Rotations

1 2

sin 0 2

cos φ θ θ ψ i e − + =

0

1

ψα

α = rotation vector

Direction of α is the rotation axis

Magnitude of α is the rotation angle


0

1

ψ

ψαU

α = rotation vector

α 2

exp σαα

⋅=

iU

ψ ψ α U α U


2 π

−2 π

2 π −2 π

The set of all single qubit

rotations lives in a solid sphere of

radius 2π.

α

Qubits are sensitive: Easy to make errors!

ψ ψ α U α U

2 π

−2 π

2 π −2 π

σ12

σ1-2

σ2-2

−2 π σ22

N = 10

N = 11

Brute Force Search

= )10(0

0 3−+

Oi

i

Braid Length

Brute Force Search

=

Braid Length |ln| εFor brute force search: |ln ε |

L. Hormozi, G. Zikos, NEB, S.H. Simon, PRB ‘07

ε “error”

)10(0

0 3−+

Oi

i

Brute Force Search

=

ε “error”

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).

)10(0

0 3−+

Oi

i

Solovay-Kitaev Construction

Braid Length c|ln| ε 4≈c,

ε “error”

)10(0

0 5−+

Oi

i

Single Qubit Operations

time

U ψ ψU

Two Qubit Gates

U

U

ψ ψ

ψ

0 0

1 1

ψU

X

X

0 0

0 1

0 0

1 1

Controlled-NOT Gate

X

X

1 1

1 0

0 0

1 1

=

0110

X

X

X

X

X U

U

Any unitary operation can be expressed as a quantum circuit

Universal Set of Gates

U X

Quantum Circuit

X

Two Qubit Gates (CNOT)

NEB, L. Hormozi, G. Zikos, & S. Simon, PRL 2005 L. Hormozi, G. Zikos, NEB, and S. H. Simon, PRB 2007

=

0110

X

SK Improved Controlled-NOT Gate

X

X X

X U

U

Quantum Circuit

Quantum Circuit

X

X X

X U

U

SU(2)k Nonabelian Particles Quasiparticle excitations of the ν=12/5 state (if it is a Read-Rezayi state) are described (up to abelian phases) by SU(2)3 Chern-Simons theory.

Universal for quantum computation for k=3 and k > 4.

Slingerland and Bais, 2001 Read and Rezyai, 1999

Freedman, Larsen, and Wang, 2001

But before SU(2)k, there was just plain old SU(2)….

In fact, Read and Rezayi proposed an infinite sequence of nonabelian states labeled by an index k with quasiparticle excitations described by SU(2)k Chern-Simons theory.

1. Particles have spin s = 0, 1/2, 1, 3/2 , …

spin = ½

2. “Triangle Rule” for adding angular momentum:

Particles with Ordinary Spin: SU(2)

Two particles can have total spin 0 or 1.

α + β 0 1

1021

21

⊕=⊗For example:

( ) 21212121 1 ssssssss +⊕⊕+−⊕−=⊗

Numbers label total spin of particles inside ovals

N 0

7/2 3

5/2 2

3/2 1

1/2

S

1 2 3 4 5 6 7 8 10 9 11 12

Hilbert Space

N 0

7/2 3

5/2 2

3/2 1

1/2

S

1 2 3 4 5 6 7 8 10 9 11 12

Hilbert Space

1 1/2 1 3/2

Paths are states

N 0

1

1

1

1

2

1

3

2

5

9

5

1

4

5

6

14

14

20

28

14

1

1 7

7/2 3

5/2 2

3/2 1

1/2

S

1 2 3 4 5 6 7 8 10 9 11 12

27

48

42

8

55

110

132

44

20

90

42

35

132

Hilbert Space

1. Particles have spin s = 0, 1/2, 1, 3/2 , …

spin = ½

2. “Triangle Rule” for adding angular momentum:

Particles with Ordinary Spin: SU(2)

Two particles can have total spin 0 or 1.

α + β 0 1

1021

21

⊕=⊗For example:

( ) 21212121 1 ssssssss +⊕⊕+−⊕−=⊗

1. Particles have topological charge s = 0, 1/2, 1, 3/2 , …, k/2

topological charge = ½

2. “Fusion Rule” for adding topological charge:

Nonabelian Particles: SU(2)k

Two particles can have total topological charge 0 or 1.

α + β 0 1

1021

21

⊕=⊗For example:

( ) [ ]2,min1 2121212121 kssssssssss −++⊕⊕+−⊕−=⊗

N 0

1

1

1

1

2

1

3

2

5

9

5

1

4

5

6

14

14

20

28

14

1

1 7

7/2 3

5/2 2

3/2 1

1/2

S

1 2 3 4 5 6 7 8 10 9 11 12

27

48

42

8

110

165

132

44

75

90

42

35

132

Dim(N) ~ 2N

N 0

1

1

1

1

2

1

3

2

9

5

4

5

13

14

27

14

7/2 3

5/2 2

3/2 1

1/2

S

1 2 3 4 5 6 7 8 10 9 11 12

40

41

121

122

81

41 122

4 13 40 121

k = 4

Dim(N) ~ 3N/2

N 0

1

1

1

1

2

3

2

8

5

3

5

8

13

21

13

7/2 3

5/2 2

3/2 1

1/2

S

1 2 3 4 5 6 7 8 10 9 11 12

21

34

55

89

55

34 89

k = 3

Dim(N) = Fib(N+1) ~ φΝ

N 0

1

1

1 2

2

2

4

4 4 8

8

8

7/2 3

5/2 2

3/2 1

1/2

S

1 2 3 4 5 6 7 8 10 9 11 12

16 32

16

16 32

32

k = 2

Dim(N) = 2N/2

The F Matrix

a c = b c

=

100

0]2[

1]2[]3[

0]2[]3[

]2[1

qq

q

q

q

q

−

0 1/2

1 1/2

1 3/2

0 1/2

1 1/2

1 3/2

∑a

cabF

2/12/1

2/2/

][ −

−

−−

=qqqqm

mm

qq-integers: )2/(2 += kieq π;

π4212

++

= kki

e

π42

1+= k

ie

=

+

++

π

π

421

4212

0

0k

i

kki

e

eR0

1

0

1

The R Matrix

R matrix

k=2

k=4

k=3

k=5

N = 1

http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Fibonacci.html

k=2

k=4

k=3

k=5

N = 2


k=2

k=4

k=3

k=5

N = 3


k=2

k=4

k=3

k=5

N = 4


k=2

k=4

k=3

k=5

N = 5


k=2

k=4

k=3

k=5

N = 6


k=2

k=4

k=3

k=5

N = 7


k=3

k=5

k=2

k=4

N = 8


k=2

k=4

k=3

k=5

N = 9


k=6

k=8

k=7

k=9

N = 1

k=6

k=8

k=7

k=9

N = 2

k=6

k=8

k=7

k=9

N = 3

k=6

k=8

k=7

k=9

N = 4

k=6

k=8

k=7

k=9

N = 5

k=6

k=8

k=7

k=9

N = 6

k=6

k=8

k=7

k=9

N = 7

k=6

k=8

k=7

k=9

N = 8

k=6

k=8

k=7

k=9

N = 9

X

CNOT gate for SU(2)5

L. Hormozi, NEB, & S.H. Simon, PRL 2009

“Surface Code” Approach

Key Idea: Use a quantum computer to simulate a theory of anyons.

More speculative idea: simulate Fibonacci anyons.

The simulation “inherits” the fault-tolerance of the anyon theory.

Most promising approach is based on “Abelian anyons.” High error thresholds, but can’t compute purely by braiding.

Bravyi & Kitaev, 2003 Raussendorf & Harrington, PRL 2007 Fowler, Stephens, and Groszkowski, PRA 2009

Konig, Kuperberg, Reichardt, Ann. Phys. 2010 NEB, D.P. DiVincenzo, PRB 2012

2D Array of Qubits

Trivalent Lattice

Trivalent Lattice

∑∑ −−=p

pv

v BQH

“Fibonacci” Levin-Wen Model Levin & Wen, PRB 2005

Trivalent Lattice

∑∑ −−=p

pv

v BQH

“Fibonacci” Levin-Wen Model

Vertex Operator

v

Qv = 0,1

Levin & Wen, PRB 2005

Trivalent Lattice

∑∑ −−=p

pv

v BQH


Vertex Operator

Plaquette Operator

v p

Qv = 0,1 Bp = 0,1

Levin & Wen, PRB 2005

Trivalent Lattice

∑∑ −−=p

pv

v BQH

Vertex Operator

Plaquette Operator

v p

Ground State Qv = 1 on each vertex Bp = 1 on each plaquette

Qv = 0,1 Bp = 0,1


Trivalent Lattice

∑∑ −−=p

pv

v BQH

Vertex Operator

Plaquette Operator

v p

Ground State Qv = 1 on each vertex Bp = 1 on each plaquette

Excited States are Fibonacci Anyons Qv = 0,1 Bp = 0,1


Vertex Operator: Qv

i

j

k v vQ ijkδ= i

j

k v 0001010100 === δδδ

All other 1=ijkδ

∑∑ −−=p

pv

v BQH


∑∑ −−=p

pv

v BQHVertex Operator: Qv

i

j

k v vQ ijkδ= i

j

k v

0=

1=

1v =Qon each vertex “branching” loop states.

0001010100 === δδδAll other 1=ijkδ


Plaquette Operator: Bp

Very Complicated 12-qubit Interaction!

∑∑ −−=p

pv

v BQH

1p =B on each plaquette superposition of loop states

( ) fmnmns

elmlms

dklkls

cjkjks

bijijs

aninmlkjisijklmn FFFFFFabcdefB ′′′′′′′′′′′′

′′′′′′ = nis,,p

sBp f

a

b

d

c

m

j

k

l

n

i

e

p ( )∑′′′′′′

′′′′′′=nmlkji

nmlkjisijklmn abcdefB ,

,p f

a

b

d

c

m’

j’

k’

l’

n’

i’

e

p

2

10

1 ϕϕ

+

+= pp

p

BBB

Plaquette Operator: Bp ∑∑ −−=p

pv

v BQH

1p =B on each plaquette superposition of loop states

Very Complicated 12-qubit Interaction!

( ) fmnmns

elmlms

dklkls

cjkjks

bijijs

aninmlkjisijklmn FFFFFFabcdefB ′′′′′′′′′′′′

′′′′′′ = nis,,p

sBp f

a

b

d

c

m

j

k

l

n

i

e

p ( )∑′′′′′′

′′′′′′=nmlkji

nmlkjisijklmn abcdefB ,

,p f

a

b

d

c

m’

j’

k’

l’

n’

i’

e

p

2

10

1 ϕϕ

+

+= pp

p

BBB

“Surface Code” Approach

v p

Use quantum computer to repeatedly measure Qv and Bp.

How hard is it to measure Qv and Bp with a quantum computer?

If Qv=1 on each vertex and Bp=1 on each plaquette, good. If not, treat as an “error.”

Konig, Kuperberg, Reichardt, Ann. Phys. 2010

Quantum Circuit for Measuring Qv

1

2

3

v

0

1

2

3

v1 Q−X X X X

NEB, D.P. DiVincenzo, PRB 2012


1

2

3

v

0

1

2

3

v1 Q−X X X X

Syndrome qubit



That was easy! What about Bp?

1

2

3

v

0

1

2

3

v1 Q−X X X X

Syndrome qubit


6-sided Plaquette: Bp is a 12-qubit Operator

Physical qubits are fixed in space



Trivalent lattice is abstract and can be “redrawn”





6-sided plaquette becomes 5-sided plaquette!

The F–Move

Locally redraw the lattice five qubits at a time.

The F–Move

a d

b c

e ∑′

′e

ebaedcF e’

a d

b c

Apply a unitary operation on these five qubits to stay in the Levin-Wen ground state.

−=

−−

−−

12/1

2/11

ϕϕϕϕ

F2

15 +=ϕ

a

c

e

b

d = F

X X

X

X X

a d

b c

e ∑′

′e

ebaedcF e’

a d

b c

F Quantum Circuit


a

c

e

b

d

1 2 3 4 5

a d

b c

e ∑′

′e

ebaedcF e’

a d

b c

F Quantum Circuit

1 5 4

3 2

1 5 4 3 2


Plaquette Reduction

Plaquette Reduction

2-qubit “Tadpole”

−+

=1

1

11

2 ϕϕ

ϕS

Measuring Bp for a Tadpole is Easy!

S Quantum Circuit

a

b a

b

0p1 B−

a

b

X

= a b S

X X

Plaquette Restoration

0

3 4 5 6

1 2

9 10 11 12

7 8

b

a e c

d

b

e c

d a

a e

b

c a b

b

c

d

a

b

a e c

d

b

e c

d a

a e

b

c a b

b

e c

d

a

b

e c

d

a

p1 B−

a

b

c

d

e e

X

6

1

2

4

3

12

9 10

11 7

8

5

p

Quantum Circuit for Measuring Bp


Gate Count

0

3 4 5 6

1 2

9 10 11 12

7 8

b

a e c

d

b

e c

d a

a e

b

c a b

b

c

d

a

b

a e c

d

b

e c

d a

a e

b

c a b

b

e c

d

a

b

e c

d

a

p1 B−

a

b

c

d

e e

X

371 CNOT Gates 392 Single Qubit Rotations

8 5-qubit Toffoli Gates 2 4-qubit Toffoli Gates 10 3-qubit Toffoli Gates 43 CNOT Gates 24 Single Qubit Gates

6

1

2

4

3

12

9 10

11 7

8

5

p


Creating Excited States: Fibonacci Anyons Konig, Kuperberg, Reichardt, Ann. Phys. 2010

Introduce “holes” (Stop measuring Qv and Bp

inside hole)


Fix boundary condition by modifying Qv operators

0=

1=


0=

1=

Fibonacci anyons


1 F-moves Konig, Kuperberg, Reichardt, Ann. Phys. 2010

56 F-moves Braid


1 2 3 4

5 6

7

1 2 3 4

5 6

7

5 6

7

1 2 3 4

5

5 6

7 1 2 3 4

6

7

1 2 3 4

1 2 3 4

5 6

7

6

5 6

7 1 2 3 4

5 6

7 1 2 3 4

5

7

1 2 3 4

1 2 3 4

5 6

7

5 6

1 2 3 4

5

6

7

1 2 3 4

7

5 6

7 1 2 3 4

5 6

7

1 2 3 4

1 2 3 4

5 6

7

6

1 2 3 4

7

1 2 3 4

5

7

1 2 3 4

5 6

7

5 6

7 1 2 3 4

5 6

7

1 2 3 4

1 2 3 4

5 6

7

6 5

6 6

1 2 3 4

6 5

7

1 2 3 4

5

7

1 2 3 4

5

7

5 6

7 1 2 3 4

5 6

7

1 2 3 4

1 2 3 4

5 6

7

6 6

1 2 3 4

6 5

7 SWAP

1 2 3 4

5

7

1 2 3 4

5

7

5 6

7 1 2 3 4

5 6

7

1 2 3 4

1 2 3 4

5 6

7

3

4

5

6

1

2

7

b

e a

d

c

d

e

c

b a

c

e

b

a d

a e

d

c

b

c

e

b

a

d

=

SWA

P


Pentagon Equation as a Quantum Circuit

1

1

1

1

1

3

4

5

6

1

2

7

b

e a

d

c

d

e

c

b a

c

e

b

a d

a e

d

c

b

c

e

b

a

d

=

SWA

P




−=

−−

−−

12/1

2/11

ϕϕϕϕ

F2

15 +=ϕ

Simplified Pentagon Circuit

=

SWA

P

F F

F F

F


Non-Abelian Anyons and Topological Quantum Computation, C. Nayak et al., Rev. Mod. Phys. 80, 1083 (2008). (arXiv:0707.1889v2)

Lectures on Topological Quantum Computation, J. Preskill, Available online at: www.theory.caltech.edu/~preskill/ph219/topological.pdf

Some excellent reviews:

Support: US DOE

NEB, L. Hormozi, G. Zikos, S.H. Simon, Phys. Rev. Lett. 95 140503 (2005). S.H. Simon, NEB, M.Freedman, N, Petrovic, L. Hormozi, Phys. Rev. Lett. 96, 070503 (2006). L. Hormozi, G. Zikos, NEB, and S.H. Simon, Phys. Rev. B 75, 165310 (2007). L. Hormozi, NEB, and S.H. Simon, Phys. Rev. Lett. 103, 160501 (2009). M. Baraban, NEB, and S.H. Simon, Phys. Rev. A 81, 062317 (2010). NEB and D.P. DiVincenzo, Phys. Rev. B 86, 165113 (2012).

fibonacci anyon with topological “charge” 1 fusion rules

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