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Adiabatic Gate Teleportation
     & Its Applications
                      Dave Bacon
     Department of Computer Science & Engineering
       University of Washington, Seattle, WA USA




        [joint work with Steve Flammia (Perimeter),
 Alice Neels (Washington), Andrew Landahl (Sandia Labs)]
Oops, I forgot to
strap myself in!
Look over there!
  A jackalope!
I’ll bet most states are
too entangled to be used as
 computational resources
ENIAC, 1946
        17,468 vacuum tubes
        7,200 crystal diodes
        1,500 relays
        70,000 resistors
        10,000 capacitors
        ~5,000,000 hand-soldered
      joints
ENIAC, 1946
                               17,468 vacuum tubes
                               7,200 crystal diodes
                               1,500 relays
                               70,000 resistors
                               10,000 capacitors
                               ~5,000,000 hand-soldered
                             joints
           ~ one vacuum tube failure / day
(equivalent to ~ one transistor failure /second today)
ENIAC, 1946
                               17,468 vacuum tubes
                               7,200 crystal diodes
                               1,500 relays
                               70,000 resistors
                               10,000 capacitors
                               ~5,000,000 hand-soldered
                             joints
           ~ one vacuum tube failure / day
(equivalent to ~ one transistor failure /second today)
Alternative History
Imagine: no transistors, no integrated circuits...
Alternative History
Imagine: no transistors, no integrated circuits...
Alternative History
Imagine: no transistors, no integrated circuits...
Alternative History
Imagine: no transistors, no integrated circuits...
Alternative History
     Imagine: no transistors, no integrated circuits...




Could we build a computer out of unreliable components?
    J. von Neumann, “Probabilistic Logics and the Synthesis of
    Reliable Organism from Unreliable Components” 1956
Von Neumann’s Solution
                           1x          1x
                      2x            1y      M
                 3x                1z
            4x                        2x                    M
       5x                           2y      M
                           1y
                                   2z                       M
                      2y              3x
              3
             My                     3y      M          π    M
            4y                     3z
       5y                             4x                    M
                                    4y      M
                           1z      4z
                      2z                                    M
                 3z                   5x
            4z                      5y      M
       5z                          5z



                                gate            error correction
Theorem (von Neumann): A circuit of g gates
can be made to fail with probability ε using
O(g log (g/ε) ) gates which fail with probability
p, assuming p<pth (vN estimated pth ~ 1%)
Engineering? We Don’t Need
  Your Stickin’ Engineering




   There are distinct physical reasons why robust
         classical computation is possible
Kitaev’s Idea
                   “Magnetism arise from spins of individual atoms. Each spin is
                   quite sensitive to thermal fluctuations. But the spins interact
                   with each other and tend to be oriented in the same
                   direction. If some spin flips to the opposite direction, the
                   interaction forces it to flip back to the direction of other
                   spins. This process is quite similar to the standard error
                   correction procedure for the repetition code. We may say
                   that errors are being corrected at the physical level. Can we
                   propose something similar in the quantum case? Yes, but
                   it is not so simple.”

Alexei Kitaev
 Kitaev, “Fault-tolerant quantum computation by anyons” (1997, published 2003)
Topological Quantum
             Computing
                              non-abelian Anyons used to build a
                              universal quantum computer




  Kitaev, “Fault-tolerant quantum computation by anyons” (1997, published 2003)
Freedman, “Quantum computation and the localization of modular functors” (2001)
Topological Quantum
             Computing
                              non-abelian Anyons used to build a
                              universal quantum computer

           √1     √1
   H=      √1
             2      2
                    1
                 − √2
                                                  quantum computing
             2
                                                  using representations
                                                  of braid group


  Kitaev, “Fault-tolerant quantum computation by anyons” (1997, published 2003)
Freedman, “Quantum computation and the localization of modular functors” (2001)
Kitaev’s Toric System
                     plaquette operator
                                      Sp =          Ze
                                             e∈δP
                     vertex operator
                                      Sv =          Xe
                                          e|v∈e
                     [Sp , Sv ] = 0    Sp = Sv
                                        2     2
                                                    =I
periodic boundary
conditions (torus)
Kitaev’s Toric System
                     plaquette operator
                                       Sp =           Ze
                                              e∈δP
                     vertex operator
                                       Sv =           Xe
                                              e|v∈e
                      [Sp , Sv ] = 0       Sp = Sv
                                            2     2
                                                       =I
periodic boundary    Kitaev Hamiltonian:
conditions (torus)         ∆
                       H=−                  Sp +       Sv
                           4           p           v
Controversy
Is topological quantum computing fault-tolerant?
                                   gate
                         2x
                              1x           1x
                                          1y    M
                                                    error correction
                    3x                1z
               4x                          2x                     M
          5x                              2y    M
                              1y
                                      2z                          M
                         2y                3x
                 3
                My                        3y    M            π    M
               4y                     3z
          5y                               4x                     M
                                          4y    M
                              1z      4z
                         2z                                       M
                    3z                     5x
               4z                      5y       M
          5z                          5z




   Theorem (von Neumann): A circuit of g gates
   can be made to fail with probability ε using
   O(g log (g/ε) ) gates which fail with probability
   p, assuming p<pth (vN estimated pth ~ 1%)
Controversy
Is topological quantum computing fault-tolerant?
                                   gate
                         2x
                              1x           1x
                                          1y    M
                                                    error correction
                    3x                1z

    If we use this definition of fault-tolerant,                   M
               4x                          2x
          5x                              2y    M
                              1y
                                      2z                          M

        then I believe the answer is NO!
                         2y                3x
                 3
                My                        3y    M            π    M
               4y                     3z
          5y                               4x                     M
                                          4y    M
                              1z      4z
                         2z                                       M
                    3z                     5x
               4z                      5y       M
          5z                          5z




   Theorem (von Neumann): A circuit of g gates
   can be made to fail with probability ε using
   O(g log (g/ε) ) gates which fail with probability
   p, assuming p<pth (vN estimated pth ~ 1%)
Controversy
Is topological quantum computing fault-tolerant?
                                   gate
                         2x
                              1x           1x
                                          1y    M
                                                    error correction
                    3x                1z

    If we use this definition of fault-tolerant,                   M
               4x                          2x
          5x                              2y    M
                              1y
                                      2z                          M

        then I believe the answer is NO!
                         2y                3x
                 3
                My                        3y    M            π    M
               4y                     3z
          5y                               4x                     M
                                          4y    M
                              1z      4z
                         2z                                       M
                    3z                     5x
               4z                      5y       M
          5z                          5z




   Theorem (von Neumann): A circuit of g gates
   can be made to fail with probability ε using
   But I also think this is the wrong definition
   O(g log (g/ε) ) gates which fail with probability
   p, assuming p<pth (vN estimated pth ~ 1%)
Cartoon Guide to Classical
 Physical Fault-Tolerance
                     Energy
                          E = −J                 si sj
                                        i,j
                                     neighbors

                     total magnetization
                          M=            si
                                 i
      +1
              spin    “0” as M > N0
      −1
  si ∈ {+1, −1}       “1” as M < −N0
Cartoon Guide to Classical
 Physical Fault-Tolerance
               Simple noise model
Cartoon Guide to Classical
 Physical Fault-Tolerance
                   Simple noise model
            1. Pick spin at random
Cartoon Guide to Classical
 Physical Fault-Tolerance
                   Simple noise model
            1. Pick spin at random
            2. If flipping spin would
            decrease energy
              a. flip spin
Cartoon Guide to Classical
 Physical Fault-Tolerance
                                Simple noise model
                         1. Pick spin at random
                         2. If flipping spin would
                         decrease energy
                           a. flip spin
                       3. If flipping spin would
                       increase energy:
“bare” noise time step
                         a. flip spin with probability
Nr runs of 1-3
                                 exp(−β∆E)      β=T   −1
β=T           −1

                                                     Memory
1D                                                      2D

                      m versus time
     1.0

     0.5

     0.0
m




     0.5

     1.0
        0.0   0.5     1.0   1.5    2.0   2.5   3.0
                            time

rate ~ exp(−cβJ)
                    m versus temperature
       1.0

       0.5

       0.0
 m




       0.5

       1.0
          0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
                    temperature
β=T           −1

                                                     Memory
1D                                                                         2D

                      m versus time                                      m versus time                                        m versus time
     1.0                                                  1.0                                                1.0

     0.5                                                  0.5                                                0.5

     0.0                                                  0.0                                                0.0

                                                      m
m




                                                                                                         m
     0.5                                                  0.5                                                0.5

     1.0                                                  1.0                                                1.0
        0.0   0.5     1.0   1.5    2.0   2.5   3.0           0.0   0.5   1.0    1.5 2.0    2.5   3.0            0.0     0.5   1.0   1.5    2.0   2.5   3.0
                            time                                               time                                                 time

rate ~ exp(−cβJ)                                                   T < TC                                             T > TC
                    m versus temperature
       1.0                                                                                       m versus temperature
                                                                                     1.0
       0.5
                                                                                     0.5
       0.0
 m




                                                                                     0.0
                                                                                 m




       0.5
                                                                                     0.5
       1.0
          0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5                                            1.0
                                                                                        0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
                    temperature
                                                                                                  temperature
Fault-Tolerance
                1D                                                2D

                      m versus time                                     m versus time
    1.0
                                                         1.0

    0.5                                                  0.5

                                                         0.0



                                                     m
    0.0
m




    0.5                                                  0.5

                                                         1.0
    1.0
          0.0   0.5   1.0          1.5   2.0   2.5
                                                            0.0   0.5    1.0      1.5   2.0   2.5
                            time                                               time


      Resilience to “gate” of flipping spins: self-correcting
Toy Model / Strawman
                     plaquette operator
                                      Sp =          Ze
                                             e∈δP
                     vertex operator
                                      Sv =          Xe
                                          e|v∈e
                     [Sp , Sv ] = 0    Sp = Sv
                                        2     2
                                                    =I
periodic boundary
conditions (torus)
Toy Model / Strawman
                      plaquette operator
                                        Sp =           Ze
                                               e∈δP
                      vertex operator
                                        Sv =           Xe
                                               e|v∈e
                       [Sp , Sv ] = 0       Sp = Sv
                                             2     2
                                                        =I
periodic boundary    Kitaev Toric Code Hamiltonian:
conditions (torus)          ∆
                        H=−                  Sp +       Sv
                            4           p           v
A Problem?
Toric code is like 1D Ising:
                         m versus time
          1.0                                           “On Thermalization in Kitaev’s 2D
          0.5
                                                        model” R. Alicki, M. Fannes, and M.
          0.0
                                                        Horodecki, arXiv:0810.4584
      m




          0.5

          1.0
             0.0   0.5   1.0   1.5    2.0   2.5   3.0
                               time

 Decay rate for information encoded into ground state is
 independent of system size, but proportional to
                          exp(−cβJ)
Contrast with 4D model
  “On thermal stability of topological qubit in Kitaev's 4D model”
  R. Alicki, (M.+P. +R.) Horodecki, arXiv:0811.0033
Not a Problem?
Factor a 1000 bit number:




Naive algorithm: ~109 gates    Gap to temp ratio: ~20

             Seems likely that exponential
               suppression as function of
             temp well worth working for
Not a Problem?
Factor a 1000 bit number:
      18819881292060796383869723946165043
      98071635633794173827007633564229888
      59715234665485319060606504743045317
      38801130339671619969232120573403187
      9550656996221305168759307650257059

Naive algorithm: ~109 gates    Gap to temp ratio: ~20

             Seems likely that exponential
               suppression as function of
             temp well worth working for
A Problem?
What about during computing?

                    Creation and manipulation of
                    excited states require
                    operators like:

                    How to do this?
                                 not
  How do you “create” and “move” anyons without
  creating errors = creating extra anyons
TQC: Gates
“We now introduce into the system’s Hamiltonian a scalar potential
composed of many local “traps,” each sufficient to capture exactly one
quasiparticle. These traps may be created by impurities, by small
gates, or by the potential created by tips of scanning microscopes.
The quasiparticle’s charge screens the potential introduced by the
trap and the quasiparticle-tip combination cannot be observed by
local measurements from far away. We denote the positions of these
traps by R1,...,Rk, and assume that these positions are well spaced
from each other compared to the microscopic length scales. A state
with quasiparticles at these positions can be viewed as an excited
state of the Hamiltonian without the trap potential or, alternatively, as
the ground state in the presence of the trap potential....”
C. Nayak, S. H. Simon, A. Stern, M. Freedman, and S. Das Sarma “Non-Abelian anyons
      and topological quantum computation” Rev. Mod. Phys. 80, 1083 (2008).
TQC Problems?
How does one guarantee that one can trap one and
only one anyon?

Every operation the previous paragraph needs to be
demonstrated that it can be done with
EXCEPTIONALLY small error if this is to be called
physical fault-tolerance.
Adiabatic Dragging
      Energy

Quantum                                        Quantum
  error                                          error
correcting                                     correcting
codespace                                      codespace

                                      Time
     Require: degenerate through entire evolution
             “Open loop” holonomic evolution
Adiabatic Teleportation

Initial Code    Final Code
Adiabatic Teleportation

  Initial Code        Final Code




Initial Stabilizer   Final Stabilizer
Adiabatic Teleportation
  Initial Code           Final Code




Initial Hamiltonian   Final Hamiltonian




     Evolution:
Adiabatic Teleportation


Initial Hamiltonian       Final Hamiltonian


                      Energy
                        1.0
                        0.5

                               0.2 0.4 0.6 0.8 1.0s
                        0.5
                        1.0
Adiabatic Teleportation


Initial Hamiltonian   Final Hamiltonian
Adiabatic Gate Teleportation

      Note: spectrum unchanged by unitary



Example (Hadamard gate):
Let’s Build a Computer
 One
 qubit
 gates

  Two
  qubit
  gates

Problem: initial Hamiltonian requires 3 qubit interactions
Example
       1    2    3


       4    5    6




To get 2 qubit interactions: perturbation theory gadgets
Gadgets
  1     2     3a    3b                  1      2    3a    3b


  4     5     6a    6b                  4      5    6a    6b




Replace 3,6 by encoded qubits:




 [S. D. Bartlett and T. Rudolph, Phys. Rev. A 74, 040302 (2006)]
Gadgets
 1    2   3a   3b             1     2   3a   3b


 4    5   6a   6b             4     5   6a   6b




                       Initial ground
Minimum gap:
                       state fidelity:
Variations on a Theme
Exchange interactions:


Adiabatically preparing rotated Hamiltonians:


   Not all U’s allow a gap



      gap:                        gap:
“Impossibility” Theorem
Is it possible to adiabatically SWAP with only one
adiabatic interpolation?




Diagonal in same basis so cannot transfer amplitude
to perform SWAP
Architectures



Sequence of Hamiltonians:
Sequence of single qubit gates applied:
Generalizing to more than 1 qubit:
  Universal QC by cycling between only 3 Hamiltonians
Architectures




         n qubit circuit

         4n or 5n qubits
Comparison
Holonomic quantum computing, but open loop
   open loop holomic QC:
   [D. Kult, J. Aberg, E. Sjoqvist, Phys. Rev. A 74, 022106 (2006)]
Adiabatic universal quantum computing, but with
piecewise evolution
  D. Aharonov et al., in 45th Annual IEEE FOCS (2004)]
Path robustness in holonomy is subsystem dependent
  compare:
  [O. Oreshkov, T. A. Brun, D. A. Lidar, Phys. Rev. Lett. 102, 070502 (2009)]
Piecewise keeps gap constant:
  compare spin-1 chain transmission of:
  [K. Eckert, O. Romero-Isart, and A. Sanpera, New J. Phys. 9, 155 (2007)]
                  Details: arXiv:0905.0901
Back to TQC!
Encoded Qubits




Smooth qubit   Rough qubit
Back to TQC!




 Smooth qubit
Prepare Smooth Puncture




       Turn off 2 plaquette operators
  adjacent to a qubit where X is turned on.
Prepare Smooth Puncture
Prepare Smooth Puncture


                                    commutes
                                     with H(t)

eigenspace of
                end up in +1 e.v. of smooth
                qubit encoded Z
Prepare Smooth Puncture




 Smooth qubit in +1 eigenvalue of encoded Z
Deform Smooth Puncture
                       Turn on


                       While turning off


                       + relevant vertex operators

 Rough qubits: Ditto         Rough qubit prepare +1
                             e.v. of encoded
Toolkit
1. Create small smooth qubit in +1 e.v. of Z
2. Create small rough qubit in +1 e.v. of X
3. Enlarge, move both punctures
4. Braid puncture: this give a CX target = rough

     These things all commute with each other, as
     of course these are Abelian anyons.
Universal Toolkit
1. Create small smooth qubit in +1 e.v. of Z
2. Create small rough qubit in +1 e.v. of X
3. Enlarge, move both punctures
4. Braid puncture: this give a CX target = rough

5. Measure encoded X,Z of qubits
6. Create magic states

                 Requires distillation.
Measurement
    Simple, right?

    But how to perform why
    Hamiltonian still on?
    Note: destructive Pauli
    measurements are tolerant to
    errors because (1)
    complementary errors are large
    and (2) we can remove region
    being measured
Universal Toolkit
1. Create small smooth qubit in +1 e.v. of Z
2. Create small rough qubit in +1 e.v. of X
3. Enlarge, move both punctures
4. Braid puncture: this give a CX target = rough

5. Measure encoded X,Z of qubits
6. Create magic states

       5+6 not tolerant to errors in this model
Summary
Adiabatically deforming between different
Hamiltonians which have stabilizer codewords as
energy eigenstates yields interesting protocols like
AGT

Using AGT as a motivator you can understand how
topological quantum computing can or cannot robustly
compute (and thus can provide rigorous error bounds.)
Adiabatic Gate Teleportation
                         Dave Bacon
          University of Washington, Seattle, WA USA




  [joint work with Steve Flammia (Perimeter), Alice Neels
        (Washington), Andrew Landahl (Sandia Labs)]
  8th Symposium on Topological Quantum Computing, Zurich 29th-31st August 2009
Open Questions
Like measurement based methods, but without having to
perform correction afterwards:
  Can we (1) do cluster state computation in this manner?
   (2) other measurement based models?
Self-Correcting
Calculate: free energy change to flip all spins
               = +1
              = −1

 1D                          2D
         E ∝ Constant

                                      E ∝ Perimeter
Self-Correcting
Calculate: free energy change to flip all spins
               = +1
              = −1

 1D                          2D
         E ∝ Constant

                                      E ∝ Perimeter


      entropy wins
Self-Correcting
Calculate: free energy change to flip all spins
               = +1
              = −1

 1D                          2D
         E ∝ Constant

                                      E ∝ Perimeter


      entropy wins                  T < TC energy wins
                                    T > TC entropy wins
Von Neumann’s Model
x               xy    f               full adder
y      S    f   00    1   c
                01    1
                          1       S
    NAND gate   10    1
                                                            s
                11    0   a               M             M


                xyz   f   1               S
 x
 y              000   0           M                         cout
       M    f             b
 z              001   0                       CS time
                010   0
majority gate   011   1       suppose each gate fails
                100   0       (outputs wrong answer)
                101   1
                110   1       independently and with
                111   1       probability p
Von Neumann’s Conundrum
Observation: If all gates fail with probability p, then
an output bit of any circuit most fail with at least
probability p

                     Rest of          M
                                           output
      input
                     Circuit
Von Neumann’s Conundrum
Observation: If all gates fail with probability p, then
an output bit of any circuit most fail with at least
probability p

                     Rest of          M
                                            output
      input
                     Circuit

“Loophole”: replace wire by “bundle” of N wires
                       ≤βN                0s decode as “1”
                     N
                       ≥(1-β)N            0s decode as “0”
                                                  0<β<1/2
Von Neumann’s Solution 1
                                                     x
Simulate the majority gate redundantly:              y             M               f
                                                     z
                                  1x        1x
                             2x            1y    M
  x                     3x             1z
                   4x                       2x
              5x                           2y    M
                                  1y   2z                                    1f

                                       M
                             2y             3x                          2f
  y                4y
                        3y
                                       3z
                                           3y    M            4f
                                                                   3f              f
              5y                            4x           5f
                                  1z   4z
                                           4y    M                                N=5
                             2z
  z                     3z                  5x
                   4z                      5y    M
              5z                       5z
Von Neumann’s Solution 1
                                                        x
Simulate the majority gate redundantly:                 y             M               f
                                                        z
                                  1x        1x
                             2x            1y    M
  x                     3x             1z
                   4x                       2x
              5x                           2y    M
                                  1y   2z                                       1f

                                       M
                             2y             3x                             2f
  y                4y
                        3y
                                       3z
                                           3y    M               4f
                                                                      3f              f
              5y                            4x              5f
                                  1z   4z
                                           4y    M                                   N=5
                             2z
  z                     3z                  5x
                   4z                      5y    M
              5z                       5z


                                                 1. M gates may fail
Von Neumann’s Solution 1
                                                        x
Simulate the majority gate redundantly:                 y             M               f
                                                        z
                                  1x        1x
                             2x            1y    M
  x                     3x             1z
                   4x                       2x
              5x                           2y    M
                                  1y   2z                                       1f

                                       M
                             2y             3x                             2f
  y                4y
                        3y
                                       3z
                                           3y    M               4f
                                                                      3f              f
              5y                            4x              5f
                                  1z   4z
                                           4y    M                                   N=5
                             2z
  z                     3z                  5x
                   4z                      5y    M
              5z                       5z


 2. incoming errors may add                      1. M gates may fail
Von Neumann’s Solution 1
                                                        x
Simulate the majority gate redundantly:                 y             M               f
                                                        z
                                  1x        1x
                             2x            1y    M
  x                     3x             1z
                   4x                       2x
              5x                           2y    M
                                  1y   2z                                       1f

                                       M
                             2y             3x                             2f
  y                4y
                        3y
                                       3z
                                           3y    M               4f
                                                                      3f              f
              5y                            4x              5f
                                  1z   4z
                                           4y    M                                   N=5
                             2z
  z                     3z                  5x
                   4z                      5y    M
              5z                       5z


 2. incoming errors may add                      1. M gates may fail
Von Neumann’s Solution 2
“Restorative organ”:
                   M
                                          Α versus Α
                        1.0
                   M    0.8



          π
                        0.6
                   M
                        0.4
                                                                       Α
                   M    0.2
                                                                       Α
                        0.0
                              0.0   0.2    0.4       0.6   0.8   1.0
                   M                             Α


      random
    permutation
αN “0s”        [3α2(1-α)+α3]N “0s” = α*N “0”s

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Perimeter Institute Talk 2009

  • 1. Adiabatic Gate Teleportation & Its Applications Dave Bacon Department of Computer Science & Engineering University of Washington, Seattle, WA USA [joint work with Steve Flammia (Perimeter), Alice Neels (Washington), Andrew Landahl (Sandia Labs)]
  • 2.
  • 3. Oops, I forgot to strap myself in!
  • 4. Look over there! A jackalope!
  • 5. I’ll bet most states are too entangled to be used as computational resources
  • 6. ENIAC, 1946 17,468 vacuum tubes 7,200 crystal diodes 1,500 relays 70,000 resistors 10,000 capacitors ~5,000,000 hand-soldered joints
  • 7. ENIAC, 1946 17,468 vacuum tubes 7,200 crystal diodes 1,500 relays 70,000 resistors 10,000 capacitors ~5,000,000 hand-soldered joints ~ one vacuum tube failure / day (equivalent to ~ one transistor failure /second today)
  • 8. ENIAC, 1946 17,468 vacuum tubes 7,200 crystal diodes 1,500 relays 70,000 resistors 10,000 capacitors ~5,000,000 hand-soldered joints ~ one vacuum tube failure / day (equivalent to ~ one transistor failure /second today)
  • 9. Alternative History Imagine: no transistors, no integrated circuits...
  • 10. Alternative History Imagine: no transistors, no integrated circuits...
  • 11. Alternative History Imagine: no transistors, no integrated circuits...
  • 12. Alternative History Imagine: no transistors, no integrated circuits...
  • 13. Alternative History Imagine: no transistors, no integrated circuits... Could we build a computer out of unreliable components? J. von Neumann, “Probabilistic Logics and the Synthesis of Reliable Organism from Unreliable Components” 1956
  • 14. Von Neumann’s Solution 1x 1x 2x 1y M 3x 1z 4x 2x M 5x 2y M 1y 2z M 2y 3x 3 My 3y M π M 4y 3z 5y 4x M 4y M 1z 4z 2z M 3z 5x 4z 5y M 5z 5z gate error correction Theorem (von Neumann): A circuit of g gates can be made to fail with probability ε using O(g log (g/ε) ) gates which fail with probability p, assuming p<pth (vN estimated pth ~ 1%)
  • 15. Engineering? We Don’t Need Your Stickin’ Engineering There are distinct physical reasons why robust classical computation is possible
  • 16. Kitaev’s Idea “Magnetism arise from spins of individual atoms. Each spin is quite sensitive to thermal fluctuations. But the spins interact with each other and tend to be oriented in the same direction. If some spin flips to the opposite direction, the interaction forces it to flip back to the direction of other spins. This process is quite similar to the standard error correction procedure for the repetition code. We may say that errors are being corrected at the physical level. Can we propose something similar in the quantum case? Yes, but it is not so simple.” Alexei Kitaev Kitaev, “Fault-tolerant quantum computation by anyons” (1997, published 2003)
  • 17. Topological Quantum Computing non-abelian Anyons used to build a universal quantum computer Kitaev, “Fault-tolerant quantum computation by anyons” (1997, published 2003) Freedman, “Quantum computation and the localization of modular functors” (2001)
  • 18. Topological Quantum Computing non-abelian Anyons used to build a universal quantum computer √1 √1 H= √1 2 2 1 − √2 quantum computing 2 using representations of braid group Kitaev, “Fault-tolerant quantum computation by anyons” (1997, published 2003) Freedman, “Quantum computation and the localization of modular functors” (2001)
  • 19. Kitaev’s Toric System plaquette operator Sp = Ze e∈δP vertex operator Sv = Xe e|v∈e [Sp , Sv ] = 0 Sp = Sv 2 2 =I periodic boundary conditions (torus)
  • 20. Kitaev’s Toric System plaquette operator Sp = Ze e∈δP vertex operator Sv = Xe e|v∈e [Sp , Sv ] = 0 Sp = Sv 2 2 =I periodic boundary Kitaev Hamiltonian: conditions (torus) ∆ H=− Sp + Sv 4 p v
  • 21. Controversy Is topological quantum computing fault-tolerant? gate 2x 1x 1x 1y M error correction 3x 1z 4x 2x M 5x 2y M 1y 2z M 2y 3x 3 My 3y M π M 4y 3z 5y 4x M 4y M 1z 4z 2z M 3z 5x 4z 5y M 5z 5z Theorem (von Neumann): A circuit of g gates can be made to fail with probability ε using O(g log (g/ε) ) gates which fail with probability p, assuming p<pth (vN estimated pth ~ 1%)
  • 22. Controversy Is topological quantum computing fault-tolerant? gate 2x 1x 1x 1y M error correction 3x 1z If we use this definition of fault-tolerant, M 4x 2x 5x 2y M 1y 2z M then I believe the answer is NO! 2y 3x 3 My 3y M π M 4y 3z 5y 4x M 4y M 1z 4z 2z M 3z 5x 4z 5y M 5z 5z Theorem (von Neumann): A circuit of g gates can be made to fail with probability ε using O(g log (g/ε) ) gates which fail with probability p, assuming p<pth (vN estimated pth ~ 1%)
  • 23. Controversy Is topological quantum computing fault-tolerant? gate 2x 1x 1x 1y M error correction 3x 1z If we use this definition of fault-tolerant, M 4x 2x 5x 2y M 1y 2z M then I believe the answer is NO! 2y 3x 3 My 3y M π M 4y 3z 5y 4x M 4y M 1z 4z 2z M 3z 5x 4z 5y M 5z 5z Theorem (von Neumann): A circuit of g gates can be made to fail with probability ε using But I also think this is the wrong definition O(g log (g/ε) ) gates which fail with probability p, assuming p<pth (vN estimated pth ~ 1%)
  • 24. Cartoon Guide to Classical Physical Fault-Tolerance Energy E = −J si sj i,j neighbors total magnetization M= si i +1 spin “0” as M > N0 −1 si ∈ {+1, −1} “1” as M < −N0
  • 25. Cartoon Guide to Classical Physical Fault-Tolerance Simple noise model
  • 26. Cartoon Guide to Classical Physical Fault-Tolerance Simple noise model 1. Pick spin at random
  • 27. Cartoon Guide to Classical Physical Fault-Tolerance Simple noise model 1. Pick spin at random 2. If flipping spin would decrease energy a. flip spin
  • 28. Cartoon Guide to Classical Physical Fault-Tolerance Simple noise model 1. Pick spin at random 2. If flipping spin would decrease energy a. flip spin 3. If flipping spin would increase energy: “bare” noise time step a. flip spin with probability Nr runs of 1-3 exp(−β∆E) β=T −1
  • 29. β=T −1 Memory 1D 2D m versus time 1.0 0.5 0.0 m 0.5 1.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 time rate ~ exp(−cβJ) m versus temperature 1.0 0.5 0.0 m 0.5 1.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 temperature
  • 30. β=T −1 Memory 1D 2D m versus time m versus time m versus time 1.0 1.0 1.0 0.5 0.5 0.5 0.0 0.0 0.0 m m m 0.5 0.5 0.5 1.0 1.0 1.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 time time time rate ~ exp(−cβJ) T < TC T > TC m versus temperature 1.0 m versus temperature 1.0 0.5 0.5 0.0 m 0.0 m 0.5 0.5 1.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 1.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 temperature temperature
  • 31. Fault-Tolerance 1D 2D m versus time m versus time 1.0 1.0 0.5 0.5 0.0 m 0.0 m 0.5 0.5 1.0 1.0 0.0 0.5 1.0 1.5 2.0 2.5 0.0 0.5 1.0 1.5 2.0 2.5 time time Resilience to “gate” of flipping spins: self-correcting
  • 32. Toy Model / Strawman plaquette operator Sp = Ze e∈δP vertex operator Sv = Xe e|v∈e [Sp , Sv ] = 0 Sp = Sv 2 2 =I periodic boundary conditions (torus)
  • 33. Toy Model / Strawman plaquette operator Sp = Ze e∈δP vertex operator Sv = Xe e|v∈e [Sp , Sv ] = 0 Sp = Sv 2 2 =I periodic boundary Kitaev Toric Code Hamiltonian: conditions (torus) ∆ H=− Sp + Sv 4 p v
  • 34. A Problem? Toric code is like 1D Ising: m versus time 1.0 “On Thermalization in Kitaev’s 2D 0.5 model” R. Alicki, M. Fannes, and M. 0.0 Horodecki, arXiv:0810.4584 m 0.5 1.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 time Decay rate for information encoded into ground state is independent of system size, but proportional to exp(−cβJ) Contrast with 4D model “On thermal stability of topological qubit in Kitaev's 4D model” R. Alicki, (M.+P. +R.) Horodecki, arXiv:0811.0033
  • 35. Not a Problem? Factor a 1000 bit number: Naive algorithm: ~109 gates Gap to temp ratio: ~20 Seems likely that exponential suppression as function of temp well worth working for
  • 36. Not a Problem? Factor a 1000 bit number: 18819881292060796383869723946165043 98071635633794173827007633564229888 59715234665485319060606504743045317 38801130339671619969232120573403187 9550656996221305168759307650257059 Naive algorithm: ~109 gates Gap to temp ratio: ~20 Seems likely that exponential suppression as function of temp well worth working for
  • 37. A Problem? What about during computing? Creation and manipulation of excited states require operators like: How to do this? not How do you “create” and “move” anyons without creating errors = creating extra anyons
  • 38. TQC: Gates “We now introduce into the system’s Hamiltonian a scalar potential composed of many local “traps,” each sufficient to capture exactly one quasiparticle. These traps may be created by impurities, by small gates, or by the potential created by tips of scanning microscopes. The quasiparticle’s charge screens the potential introduced by the trap and the quasiparticle-tip combination cannot be observed by local measurements from far away. We denote the positions of these traps by R1,...,Rk, and assume that these positions are well spaced from each other compared to the microscopic length scales. A state with quasiparticles at these positions can be viewed as an excited state of the Hamiltonian without the trap potential or, alternatively, as the ground state in the presence of the trap potential....” C. Nayak, S. H. Simon, A. Stern, M. Freedman, and S. Das Sarma “Non-Abelian anyons and topological quantum computation” Rev. Mod. Phys. 80, 1083 (2008).
  • 39. TQC Problems? How does one guarantee that one can trap one and only one anyon? Every operation the previous paragraph needs to be demonstrated that it can be done with EXCEPTIONALLY small error if this is to be called physical fault-tolerance.
  • 40. Adiabatic Dragging Energy Quantum Quantum error error correcting correcting codespace codespace Time Require: degenerate through entire evolution “Open loop” holonomic evolution
  • 42. Adiabatic Teleportation Initial Code Final Code Initial Stabilizer Final Stabilizer
  • 43. Adiabatic Teleportation Initial Code Final Code Initial Hamiltonian Final Hamiltonian Evolution:
  • 44. Adiabatic Teleportation Initial Hamiltonian Final Hamiltonian Energy 1.0 0.5 0.2 0.4 0.6 0.8 1.0s 0.5 1.0
  • 46. Adiabatic Gate Teleportation Note: spectrum unchanged by unitary Example (Hadamard gate):
  • 47. Let’s Build a Computer One qubit gates Two qubit gates Problem: initial Hamiltonian requires 3 qubit interactions
  • 48. Example 1 2 3 4 5 6 To get 2 qubit interactions: perturbation theory gadgets
  • 49. Gadgets 1 2 3a 3b 1 2 3a 3b 4 5 6a 6b 4 5 6a 6b Replace 3,6 by encoded qubits: [S. D. Bartlett and T. Rudolph, Phys. Rev. A 74, 040302 (2006)]
  • 50. Gadgets 1 2 3a 3b 1 2 3a 3b 4 5 6a 6b 4 5 6a 6b Initial ground Minimum gap: state fidelity:
  • 51. Variations on a Theme Exchange interactions: Adiabatically preparing rotated Hamiltonians: Not all U’s allow a gap gap: gap:
  • 52. “Impossibility” Theorem Is it possible to adiabatically SWAP with only one adiabatic interpolation? Diagonal in same basis so cannot transfer amplitude to perform SWAP
  • 53. Architectures Sequence of Hamiltonians: Sequence of single qubit gates applied: Generalizing to more than 1 qubit: Universal QC by cycling between only 3 Hamiltonians
  • 54. Architectures n qubit circuit 4n or 5n qubits
  • 55. Comparison Holonomic quantum computing, but open loop open loop holomic QC: [D. Kult, J. Aberg, E. Sjoqvist, Phys. Rev. A 74, 022106 (2006)] Adiabatic universal quantum computing, but with piecewise evolution D. Aharonov et al., in 45th Annual IEEE FOCS (2004)] Path robustness in holonomy is subsystem dependent compare: [O. Oreshkov, T. A. Brun, D. A. Lidar, Phys. Rev. Lett. 102, 070502 (2009)] Piecewise keeps gap constant: compare spin-1 chain transmission of: [K. Eckert, O. Romero-Isart, and A. Sanpera, New J. Phys. 9, 155 (2007)] Details: arXiv:0905.0901
  • 58. Back to TQC! Smooth qubit
  • 59. Prepare Smooth Puncture Turn off 2 plaquette operators adjacent to a qubit where X is turned on.
  • 61. Prepare Smooth Puncture commutes with H(t) eigenspace of end up in +1 e.v. of smooth qubit encoded Z
  • 62. Prepare Smooth Puncture Smooth qubit in +1 eigenvalue of encoded Z
  • 63. Deform Smooth Puncture Turn on While turning off + relevant vertex operators Rough qubits: Ditto Rough qubit prepare +1 e.v. of encoded
  • 64. Toolkit 1. Create small smooth qubit in +1 e.v. of Z 2. Create small rough qubit in +1 e.v. of X 3. Enlarge, move both punctures 4. Braid puncture: this give a CX target = rough These things all commute with each other, as of course these are Abelian anyons.
  • 65. Universal Toolkit 1. Create small smooth qubit in +1 e.v. of Z 2. Create small rough qubit in +1 e.v. of X 3. Enlarge, move both punctures 4. Braid puncture: this give a CX target = rough 5. Measure encoded X,Z of qubits 6. Create magic states Requires distillation.
  • 66. Measurement Simple, right? But how to perform why Hamiltonian still on? Note: destructive Pauli measurements are tolerant to errors because (1) complementary errors are large and (2) we can remove region being measured
  • 67. Universal Toolkit 1. Create small smooth qubit in +1 e.v. of Z 2. Create small rough qubit in +1 e.v. of X 3. Enlarge, move both punctures 4. Braid puncture: this give a CX target = rough 5. Measure encoded X,Z of qubits 6. Create magic states 5+6 not tolerant to errors in this model
  • 68. Summary Adiabatically deforming between different Hamiltonians which have stabilizer codewords as energy eigenstates yields interesting protocols like AGT Using AGT as a motivator you can understand how topological quantum computing can or cannot robustly compute (and thus can provide rigorous error bounds.)
  • 69. Adiabatic Gate Teleportation Dave Bacon University of Washington, Seattle, WA USA [joint work with Steve Flammia (Perimeter), Alice Neels (Washington), Andrew Landahl (Sandia Labs)] 8th Symposium on Topological Quantum Computing, Zurich 29th-31st August 2009
  • 70. Open Questions Like measurement based methods, but without having to perform correction afterwards: Can we (1) do cluster state computation in this manner? (2) other measurement based models?
  • 71. Self-Correcting Calculate: free energy change to flip all spins = +1 = −1 1D 2D E ∝ Constant E ∝ Perimeter
  • 72. Self-Correcting Calculate: free energy change to flip all spins = +1 = −1 1D 2D E ∝ Constant E ∝ Perimeter entropy wins
  • 73. Self-Correcting Calculate: free energy change to flip all spins = +1 = −1 1D 2D E ∝ Constant E ∝ Perimeter entropy wins T < TC energy wins T > TC entropy wins
  • 74. Von Neumann’s Model x xy f full adder y S f 00 1 c 01 1 1 S NAND gate 10 1 s 11 0 a M M xyz f 1 S x y 000 0 M cout M f b z 001 0 CS time 010 0 majority gate 011 1 suppose each gate fails 100 0 (outputs wrong answer) 101 1 110 1 independently and with 111 1 probability p
  • 75. Von Neumann’s Conundrum Observation: If all gates fail with probability p, then an output bit of any circuit most fail with at least probability p Rest of M output input Circuit
  • 76. Von Neumann’s Conundrum Observation: If all gates fail with probability p, then an output bit of any circuit most fail with at least probability p Rest of M output input Circuit “Loophole”: replace wire by “bundle” of N wires ≤βN 0s decode as “1” N ≥(1-β)N 0s decode as “0” 0<β<1/2
  • 77. Von Neumann’s Solution 1 x Simulate the majority gate redundantly: y M f z 1x 1x 2x 1y M x 3x 1z 4x 2x 5x 2y M 1y 2z 1f M 2y 3x 2f y 4y 3y 3z 3y M 4f 3f f 5y 4x 5f 1z 4z 4y M N=5 2z z 3z 5x 4z 5y M 5z 5z
  • 78. Von Neumann’s Solution 1 x Simulate the majority gate redundantly: y M f z 1x 1x 2x 1y M x 3x 1z 4x 2x 5x 2y M 1y 2z 1f M 2y 3x 2f y 4y 3y 3z 3y M 4f 3f f 5y 4x 5f 1z 4z 4y M N=5 2z z 3z 5x 4z 5y M 5z 5z 1. M gates may fail
  • 79. Von Neumann’s Solution 1 x Simulate the majority gate redundantly: y M f z 1x 1x 2x 1y M x 3x 1z 4x 2x 5x 2y M 1y 2z 1f M 2y 3x 2f y 4y 3y 3z 3y M 4f 3f f 5y 4x 5f 1z 4z 4y M N=5 2z z 3z 5x 4z 5y M 5z 5z 2. incoming errors may add 1. M gates may fail
  • 80. Von Neumann’s Solution 1 x Simulate the majority gate redundantly: y M f z 1x 1x 2x 1y M x 3x 1z 4x 2x 5x 2y M 1y 2z 1f M 2y 3x 2f y 4y 3y 3z 3y M 4f 3f f 5y 4x 5f 1z 4z 4y M N=5 2z z 3z 5x 4z 5y M 5z 5z 2. incoming errors may add 1. M gates may fail
  • 81. Von Neumann’s Solution 2 “Restorative organ”: M Α versus Α 1.0 M 0.8 π 0.6 M 0.4 Α M 0.2 Α 0.0 0.0 0.2 0.4 0.6 0.8 1.0 M Α random permutation αN “0s” [3α2(1-α)+α3]N “0s” = α*N “0”s

Editor's Notes