Qubit Repetition Code for Bit Flips

Classically, the repetition code is the simplest way of introducing redun-

dancy to protect information. Assume that noise in the channel is modelled

as a bit flip with probability p (and hence 1 − p for not flipping). This is

schematised in Figure 10.2. This is known as the binary symmetric channel.

0

1−p

//

p

0

1

1−p

//

p

77

1

FIGURE 10.2: Binary symmetric channel for bit flips.

To protect against errors, each logical bit, indicated by the tilde, is encoded

using three identical physical bits.

0 →

˜

0 = 000, 1 →

˜

1 = 111. (10.1)

If p is sufficiently small, then the majority value decides what the original bit

was. The total probability of error is the sum of probability that 2 bits flipped

and that 3 bits flipped which is 3p

(

1 −p) + p

3

= 3p

2

−2p

3

. If p < 1/2, this is

much smaller than the probability of error without encoding, which is p.

Now suppose we have a quantum channel that was susceptible to only

qubit flips. We can model such a channel by X acting with probability p on

a state passing through the channel. The quantum equivalent of the 3-bit

repetition code represents each basis state by 3 identical qubits in the same

basis state:

|0i → |

˜

0i = |000i; |1i → |

˜

1i = |111i (10.2)

Quantum Error Correction 199

so that an arbitrary state is encoded as

|ψi = α|0i + β|1i −→ |

˜

ψi = α|000i + β|111i. (10.3)

This encoding process is easily achieved by the circuit of Figure 10.3.

|ψi

• •

|0i

|

˜

ψi

|0i

FIGURE 10.3: Encoding circuit for the 3-qubit bit-flip code.

This does not make three copies of the original state, however, and neither

can we try to measure the state after it passes through the noise to check how

it has changed, as that would destroy the superposition. If we assume that

the channel is capable of flipping only one qubit, then the codeword state |

˜

ψi

could have changed into four possible output states:

S

0

: |

˜

ψi → |ψ

0

i = |

˜

ψi = α|000i + β|111i (10.4a)

S

1

: |

˜

ψi → |ψ

1

i = X

1

|

˜

ψi = α|001i + β|110i (10.4b)

S

2

: |

˜

ψi → |ψ

2

i = X

2

|

˜

ψi = α|010i + β|101i (10.4c)

S

3

: |

˜

ψi → |ψ

3

i = X

3

|

˜

ψi = α|100i + β|011i (10.4d)

These states are known as syndromes, since we can diagnose the affliction due

to noise by detecting which one occurred! But how do we detect the syndrome

without measuring or copying? The way out is to use ancillary qubits, with

controlled gates acting on them and to measure the ancillaries. We have to

ensure that we do not get any information about the original state by this

measurement, but still detect the syndrome. Note that the four syndrome

states are all mutually orthogonal. Therefore it is possible to distinguish them

by measuring a 2-qubit ancilla. Consider the circuit in Figure 10.4. Each qubit

of the input state is indicated by its label.

3

|

¯

ψi

2

• •

1

|0i

x

|0i

y

FIGURE 10.4: Syndrome measurement for the 3-qubit bit-flip code

200 Introduction to Quantum Physics and Information Processing

TABLE 10.1: Syndrome measurement: outcomes.

Syndrome xy

0

i 00

1

i 01

2

i 11

3

i 10

You can verify that the measured 2-bit number xy give you the syndrome

as in Table 10.1.

The information contained in the input state |

˜

ψi is not revealed by the

measurements. It is easy to see that for each syndrome |ψ

i

i, the error can be

corrected by applying X on the i

th

qubit. This action can be linked to the xy

values, which control the action of X on the corresponding qubit: X

x¯y

on the

first qubit, X

xy

on the second, and X

¯xy

on the last qubit of the codeword.

The nice thing about expressing it as this controlled action is that the

process can be automated, bypassing the need for measurement, by applying

suitable controlled gates as in Figure 10.5.

3

SM

|

¯

ψi

2

1

|0i

• •

|0i

• •

FIGURE 10.5: Error detection and correction for 3-qubit bit-flip code. Here

SM is the syndrome measurement circuit


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