← All modules 09 Noise and quantum error correction

SESAME · Day 3 · Module 09

How can we detect an error without learning the qubit’s state?

Noise and quantum error correction

We want to preserve an unknown input:

\[|\psi\rangle=\alpha|0\rangle+\beta|1\rangle.\]
01 / Noise and responses02 / Encoding a qubit03 / Comparisons and recovery

SESAME · Day 3 · Module 09

We can correct errors without learning the quantum state we are protecting.

01Noise changes the results we are trying to estimate.

02We can encode one qubit's information across three qubits.

03Comparisons reveal an error while preserving the encoded state.

01 / Noise and responses · 1.1

Errors can enter while we store, manipulate or measure qubits.

During storage

The qubit interacts with its surroundings.

Energy can be lost; relative phase can drift.

During a gate

The control operation differs from the intended one.

For example, a rotation goes slightly too far.

At readout

The recorded bit can misreport the outcome.

For example, a 0 is recorded as 1.

Noise can affect the state, the operations and the final records.

01 / Noise and responses · 1.2

A noise model describes how an operation can differ from the intended one.

Our model: an unwanted X occurs with probability 10% during this interval.

The illustrative bit-flip noise model |0⟩Specified intervalNo change, or an unwanted X0 or 1
Modelled eventProbabilityOutput from input \(|0\rangle\)
No error90%0
X error10%1

A specified fault, location and probability. This 10% is illustrative; it is not a device calibration.

01 / Noise and responses · 1.2

More shots estimate the noisy distribution more precisely.

Ideal prediction: 0% onesOur noise model: 10% ones
Fresh runsRecorded onesObserved frequency
1001313%
10,0001,00410.04%

Sampling uncertainty shrinks. The noise model still predicts 10% ones.

Illustrative batches from the same model. Actual finite samples vary.

01 / Noise and responses · 1.3

We can reduce physical errors, improve estimates, or protect the quantum information.

Suppression

Make errors less likely

Remove unnecessary gates and shorten waiting times.

Mitigation

Improve estimates from counts

For readout errors: test known 0s and 1s, then use the error rates to adjust estimated probabilities.

Correction

Recover the state

For a single bit flip: use extra qubits to locate it, then apply X.

02 / Encoding a qubit · 2.1

The classical encodings 000 and 111 let us identify a single flipped bit.

Classical bitStored patternAfter a flip in the middle
0000010
1111101

With at most one flip, the two agreeing positions identify the original bit.

Only 000 and 111 are valid stored patterns. A single flip creates a recognizable mismatch.

02 / Encoding a qubit · 2.1

We cannot make identical copies of an arbitrary unknown qubit state.

Three independent copies

\[|\psi\rangle|\psi\rangle|\psi\rangle\]

We cannot copy an arbitrary unknown state this way.

Direct data readout

\[\alpha|0\rangle+\beta|1\rangle\]

Reading each qubit generally destroys the superposition we want to preserve.

The protection has to preserve the unknown amplitudes.

02 / Encoding a qubit · 2.2

The states 000 and 111 can serve as the two reference states of an encoded qubit.

Reference zero

\[|0\rangle\longmapsto|000\rangle\]

Reference one

\[|1\rangle\longmapsto|111\rangle\]
\[\alpha|0\rangle+\beta|1\rangle\longmapsto\alpha|000\rangle+\beta|111\rangle.\]

Three independent copies would be

\[(\alpha|0\rangle+\beta|1\rangle)(\alpha|0\rangle+\beta|1\rangle)(\alpha|0\rangle+\beta|1\rangle).\]

For a general superposition, three copies have eight terms; the encoding has two.

One qubit's information is carried by a joint state across three qubits.

02 / Encoding a qubit · 2.2

Two CNOTs create the encoding from the input and two qubits initially in zero.

The two CNOT encoding circuit q₀|ψ⟩q₁|0⟩q₂|0⟩ Jointstate
Initial q₀q₁q₂After CNOT q₀ → q₁After CNOT q₀ → q₂
\(|000\rangle\)\(|000\rangle\)\(|000\rangle\)
\(|100\rangle\)\(|110\rangle\)\(|111\rangle\)

Linearity gives \(\alpha|000\rangle+\beta|111\rangle\) for the unknown input.

The input is on q₀. The three data qubits are written in q₀, q₁, q₂ order.

02 / Encoding a qubit · 2.3

One bit flip creates the same mismatch in both encoded terms.

\[\alpha|000\rangle+\beta|111\rangle\xrightarrow{X\text{ on }q_1}\alpha|010\rangle+\beta|101\rangle.\]

Term with coefficient α

q₀0q₁1q₂0

Term with coefficient β

q₀1q₁0q₂1

The middle label differs from both neighbours in each term.

Worked model: at most one X after encoding; encoding, checks and recovery are ideal.

03 / Comparisons and recovery · 3.1

A helper qubit can record whether two data qubits agree.

Three data qubits and two helpers perform both comparisons Data3 qubits q₀q₁q₂Helpers2 qubits h₀₁|0⟩h₁₂|0⟩ s₀₁ s₁₂

Do q₀q₁ = 00 and 11 give different h₀₁ results?

q₀q₁ labels00110110
h₀₁ readout: s₀₁0011

0 means agreement; 1 means disagreement. Only h₀₁ and h₁₂ are measured.

03 / Comparisons and recovery · 3.1

These comparisons preserve the superposition because both terms give the same results.

h₀₁ compares q₀ with q₁; h₁₂ compares q₁ with q₂.

Data termq₀ compared with q₁q₁ compared with q₂Record s₀₁s₁₂
\(\alpha|010\rangle\)DifferentDifferent11
\(\beta|101\rangle\)DifferentDifferent11

Can this readout distinguish 010 from 101?

\[\underbrace{\bigl(\alpha|010\rangle+\beta|101\rangle\bigr)}_{\text{data }q_0q_1q_2}\;\underbrace{|11\rangle}_{\text{helpers }h_{01}\,h_{12}}\]

Reading the helpers leaves the data superposition intact.

03 / Comparisons and recovery · 3.2

Two comparison bits locate a single bit flip.

The comparison record is the syndrome, written in the order s₀₁, s₁₂.

s₀₁: q₀ with q₁s₁₂: q₁ with q₂Error location
00No bit flip
10q₀?
11q₁, the worked example
01q₂?

Only the first check reports disagreement. Which qubit could have flipped?

This table assumes at most one X error and ideal checks.

03 / Comparisons and recovery · 3.2

Applying X to the identified qubit restores the encoded input state.

Measured syndrome s₀₁s₁₂

11

Classical choice

X on q₁
\[\alpha|010\rangle+\beta|101\rangle\xrightarrow{X\text{ on }q_1}\alpha|000\rangle+\beta|111\rangle.\]

The original amplitudes survive encoding, checking and recovery.

If we undo the two encoding CNOTs

\[\alpha|000\rangle+\beta|111\rangle\longmapsto(\alpha|0\rangle+\beta|1\rangle)|00\rangle.\]

03 / Comparisons and recovery · 3.3

A phase error can change the encoded state while leaving both comparisons unchanged.

\[\frac{|000\rangle+|111\rangle}{\sqrt2}\xrightarrow{Z\text{ on }q_0}\frac{|000\rangle-|111\rangle}{\sqrt2}.\]
States₀₁ from h₀₁s₁₂ from h₁₂
Before Z00
After Z00

The labels still agree. These checks do not detect the changed relative sign.

Two bit flips can also imitate a different single flip. This code does not protect against every quantum error.

03 / Comparisons and recovery · 3.3

Real error correction must also tolerate faults in its checks and corrections.

Encoded data

Three data qubits

Checks

Two helpers and their measurements

Recovery

Classical decision and gates

A wrong comparison record can trigger a wrong correction.

Reliable designs account for faults throughout this process.

Extra qubits and operations are resources to account for. They can introduce errors of their own.

Noise and quantum error correction

We can correct errors without learning the quantum state we are protecting.

01Noise changes the results we are trying to estimate.

02We can encode one qubit's information across three qubits.

03Comparisons reveal an error while preserving the encoded state.

The checks identify an allowed error while preserving the unknown amplitudes.

Practice / The same encoding and comparisons

Which correction restores this encoded state?

A single bit flip leaves the data in:

\[\alpha|100\rangle+\beta|011\rangle.\]
01

What do the q₀–q₁ and q₁–q₂ comparisons report, in that order?

02

Which qubit receives X to restore the encoding?

03

Why does the helper readout preserve this superposition?

At most one X error; ideal comparisons and recovery. Only h₀₁ and h₁₂ are measured.