Three-qubit quantum error correction
with superconducting circuits
Matt Reed
Yale University
Boston, MA - February 28, 2012
Leo DiCarlo
Simon Nigg
Luyan Sun
Luigi Frunzio
Steven Girvin
Robert Schoelkopf
Outline
• Why is QEC necessary?
• Repetition codes
• Our architecture: cQED
• Adiabatic and sudden two-qubit phase gates
• GHZ states
• Efficient Toffoli gate using third-excited state
• Bit- and phase-flip error correction
Reed, et al. Nature 482, 382 (2012)
Why do we need to correct?
Classical bit
Quantum bit
State value
z
“1”
y
“0”
x
Control signal
Small control fluctuations do not
change the system state –
compressed phase space
Error probability ~ 10-15
Small control fluctuations do cause
a change in the system state!
p ~ 10-2 - 10-5
To get p~10-15 would need T1 ~ 1 year
Classical repetition code
0
0
Sent
p
p
1
1
1-p
Received
1-p
Probability p of having
a bit flipped
“Binary symmetric
channel”
Repetition code: send each bit three times, then vote
0
1
000
111
Reduces classical error rate to 3p2 – 2p3
Quadratic!
Can we do this for quantum computing? Some reasons to think no:
• No cloning theorem
• Measurements project qubits
• Errors are continuous
GHZ-like states
(
) (
) (
It is not possible to go from a 0 + b 1 ® a 0 + b 1 Ä a 0 + b 1 Ä a 0 + b 1
But we can make a 0 + b 1 ® a 000 + b 111
All ZiZj correlations are +1, independent of a and b
a 000 + b 111
Qubits 1 and 2 are either:
both 0 or both 1
Z1Z2 = a ( -1´ -1) + b (1´1) = a + b
2
2
2
2
= +1
“I don’t know where they are pointing, but I know they’re pointing in the same direction”
)
Flipping GHZs
What happens when we flip one of the qubits in a GHZ-like state?
a 000 + b 111 ® a 100 + b 011
Z1Z2 = +1
Z2Z3 = +1
®
Z1Z2 = -1
Z2Z3 = +1
Independent of
a and b
Flipped
State
Z1Z2
Z2Z3
None
a 000 + b 111
+1
+1
Q1
a 100 + b 011
-1
+1
Q2
a 010 + b 101
-1
-1
Q3
a 001 + b 110
+1
-1
Each error has a different observable! - The basis for the bit flip code
Four errors = two classical bits
Circuit quantum electrodynamics
Our system: superconducting qubits coupled to a microwave resonator
Transmon qubits
Transmission-line resonator bus
In analogy to cavity QED:
• Protection from spontaneous emission
• Qubit readout
• Multiplexed qubit drives (single-qubit gates)
• Mediate qubit coupling (multi-qubit gates)
cQED: Wallraff Nature 431, 162 (2004)
Bus: Majer Nature 449, 443 (2007)
Readout: Reed PRL 105, 173601 (2010)
Four-qubit cQED device
• Four transmon qubits coupled to single 2D microwave resonator
• Three qubits biased at 6, 7, and ~8 GHz
• Fourth qubit above cavity and unused
• T1 ~ 1 μs, T2 ~ 0.5 μs
Frequency (GHz)
cavity
Q3
Q2
• Flux bias lines to control frequency
• Nanosecond speed - two qubit gates
Q1
Flux bias on Qubit 1 (a.u.)
DiCarlo, et al. Nature 467 574 (2010)
Adiabatic multiqubit phase gates
A two qubit phase gate can be written:
Entanglement!
Interactions on two excitation manifold
give entangling two-qubit conditional phases
Top qubit flux bias (a.u.)
DiCarlo, et al. Nature 460, 240 (2009)
Adiabatic multiqubit phase gates
A two qubit phase gate can be written:
Entanglement!
Interactions on two excitation manifold
give entangling two-qubit conditional phases
Can give a universal “Conditional Phase Gate”
Top qubit flux bias (a.u.)
DiCarlo, et al. Nature 460, 240 (2009)
Sudden multiqubit phase gates
Suddenly move
11 into resonance with 02
11 ® y ( t = 0) = + + -
y (t ) = e
iDt 2
+ +e
-iDt 2
02
11
-
y ( t = 2p D) = - ( + + - ) ® - 11
Previously proposed:
Strauch et al., PRL 91, 167005 (2003)
-
11
02
Crossing measurement:
• Jump to a flux
• Wait some time
• Jump back
• Measure if in 11 (black)
or 02 (white)
+
t =12 ns
Or transfer to
02 in 6 ns!
Entangled states on demand
0
R y /2
R y /2
R y /2
State
Tomography
0
01
0
ZII
ABC = y A1B 2C 3 y
IXX
F   T   T  94%
DiCarlo, et al. Nature 467 574 (2010)
T
1
 0 
 00  11 
2
IZZ
IYY
ZXX
ZZZ
ZYY
GHZ states on demand
0
0
R y /2
R
 /2
y
R y /2
R y /2
State
Tomography
0
01
10
R y /2
ZZI
ZIZ
F  GHZ  GHZ  88%
y target =
IZZ
000 + 111 )
(
2
1
XXX
XYY
YXX
YYX
Can simply change the preparation of Q2 to encode any state
DiCarlo, et al. Nature 467 574 (2010)
Error correction with GHZ states
Measurements force finite
rotations to full flips
encode
0
error
diagnose
nose
fix
X
X
or
 0  1
X
X
0
 0  1
or
0
X
X
Z 2 Z3
Z1Z2
0
Logic
GHZ state for
Works for any
single error
Z1Z2 Z2 Z3
Nielsen & Chuang
NMR: Cory et al. PRL 81, 2152 (1998)
Ions: Chiaverini et al. Nature 432, 602 (2004)
Measurement-free QEC
Feed-forward measurement hard in this first expt
- Measurement-free version of the code
encode
diagnose
fix
Z1Z2
0
 0  1
Toffoli implements classical logic
• only acts on flipped subspace
 0  1
X
0
0
Z 2 Z3
Toffoli
(CCNot)
gate
0
Reset
(potentially)
Toffoli can be constructed with five two-qubit gates, but that’s expensive
How can we do better?
Nielsen & Chuang Cambridge Univ. Press
Ions: P. Schindler et al. Science 332, 1059 (2011)
Toffoli gate with noncomputational states
Two-qubit gate requires two excitations
Three-qubit interaction: third excited state
The essence!
This interaction is small, so use intermediary
Sudden transfer:
Identical for:
Adiabatic interaction:
Three-qubit phase here!
Classical truth table
How do we prove the gate works? First, measure classical action
Classically, a phase gate does nothing. So we dress it up to make it a CCNOT
F = 86%
(>50% the time
of an equivalent
construction)
Optics: Lanyon Nat. Phys. 5, 134 (2009)
Ions: Monz PRL 102, 040501 (2009)
SCQs: Mariantoni Science 334, 61 (2011)
Fedorov Nature 481, 170 (2012)
Quantum process tomography of CCPhase
Want to know the action on superpositions:
(but now with 64 basis states)
Invert to find
0.6
Theory
Experiment
0.3
0.0
4032 Pauli correlation measurements (90 minutes)
F=
Protection from single qubit bit-flip errors
“Error” rotation
by some angle
Prepare
Encode in
three-qubit
state
Correct subspace
with error
Decode
syndromes
Measure
single-qubit
state fidelity to
Ry (q )
Ry (q ) = cos2 (q / 2)I - isin2 (q / 2)Y
Ideally, there should be no dependence
of fidelity on the error rotation angle
Correction fidelity vs. bit-flip error rotation
Encode, single known error, decode, fix, and measure resulting state fidelity
Proteced qubit state fidelity
1.0
0.8
No correction
0.6
Error on Q2
Error on Q1
0.4
Error on Q3
0.2
0.0
-2p
--1p
0
0
p1
2p
Error rotation angle
(No-correction curve has finite fidelity because its duration is the same as the corrected curves)
Error syndromes
Is the algorithm really doing what we think?
Look at two-qubit density matrices of ancillas after a full flip
00
No error
0
0
y
y
0
0
Top flip
Bottom flip
01
1
0
-1
0
0
y
y
0
10
11
0
y
y
y
0
0
0
X
1
Protected flip
1
0
X
1
X
y
1
Phase-flip error correction code
Bit-flips are not common errors, but phase flips are – modify code
Differs from bit-flip code by single qubit rotations; e.g. change of coordinate system
More realistic error model:
Simultaneous flips on each
qubit happen with probability
p  sin ( / 2)
2
Rz ( )
Rz ( )
Rz ( )
Apply errors and measure fidelity to
the prepared state as a function of p
Code only works for single errors.
P(2 or 3 errors) = 3p2 – 2p3
Expect quadratic dependence on p
Simultaneous phase-flip errors
To measure the effect of the code on any state, test with four one-qubit basis states
State fidelity
State fidelity
1.0
0.8
Depends only quadratically
on error probability!
0.6
0.4
0.2
No correction
0.0 0.2 0.4 0.6 0.8 1.0
1.0 Phase flip probability
0.8
0.6
0.4
0.2
Corrected
0.0 0.2 0.4 0.6 0.8 1.0
Phase flip probability
For better coherence, see 3D Cavity session L39 (room 109B)
Summary
• Realized the simplest version of gate-based QEC
• Both bit- and phase-flip correction
• Not fault-tolerant (gate based, un-encoded)
• Based on new three-qubit phase gate
• Adiabatic interaction with transmon third excited state
• Works for any three nearest-neighbor qubits
• 86% classical fidelity and 78% quantum process fidelity
Reed, et al. Nature 482, 382 (2012)
Questions?
Reed, et al. Nature 482, 382 (2012)
CCNot gate pulse sequence
More than two times faster than equivalent two-qubit gate sequence
Three qubit state tomography
Rx0,
M  000 000
 ZII  IZI  IIZ  ZZI  ZIZ  IZZ  ZZZ
Rx0,
Rx0,
Example: extract
Joint
Readout
M
000 000
ZZZ
no pre-rotation:  ZII  IZI  IIZ  ZZI  ZIZ  IZZ  ZZZ
Rx ( ) on Q1 and Q2:  ZII  IZI  IIZ  ZZI  ZIZ  IZZ  ZZZ
Rx ( ) on Q1 and Q3:  ZII  IZI  IIZ  ZZI  ZIZ  IZZ  ZZZ
Rx ( ) on Q2 and Q3:  ZII  IZI  IIZ  ZZI  ZIZ  IZZ  ZZZ
4 ZZZ
DiCarlo, et al. Nature 467 574 (2010)