5 Qubits Error 5 Qubits Error CorrectingCorrecting

Shor’s code uses 9 qubits to encode 1 qubit, but more efficient codes exist.

Given our error model where errors can be any of the Pauli matrices applied to any qubit.

To recover from 1 qubit errors, we need a minimum of 5 qubits to encode 1 qubit.

zyx , ,

How to calculate this number?

How to use syndrome bits to How to use syndrome bits to calculate the minimum lengthcalculate the minimum length

Argument:Supposing we encode 1 qubit using n qubits.We can have n-1 syndrome bits, the values of

which tells us the exact error that occurred. Hence 2n-1 errors can be represented by the

syndrome bits

We have n qubits, and so 3n possible errors. Consider also the case of no errors.

Hence,Least value of n solving this is 5.

1213 nnPauli rotations

Encoding

Use 5 qubits to encode 1 qubit :

111001001101111000008

10 encode

11010101010100100110

000110110010000111118

11 encode

00101010101011011001

From above, we calculate:

Encoding circuitEncoding circuit

10100010100110110011

110000011000001111118

b

01011101011001001100

001111100111110000008

10

54321543215432154321

54321543215432154321

54321543215432154321

54321543215432154321

aencode

Flip phase

Rules of flipping phase and bits If qubits 2,3 and 4 are ‘1’, flip

the phase

If qubits 2 and 4 is ‘0’ and qubit 3 is ‘1’, flip the phase

If qubit 1 is ‘1’, flip qubits 3 and 5

Signal after Hadamards:

Encoding circuit

Hadamards

Step-by-step analysis of the encoding circuit

flipping phase

flipping phase

Step-by-step analysis of the encoding circuitFlipping bits 3 and 5

Flipping bits 3 and 5 with “1” in bits 2 and 4

Step-by-step analysis of the encoding circuit

Flipping phase in data bit when bits 4 and 5 are “1”

Assuming at most 1 qubit error and the error is just as likely to affect any qubit.

The decoding circuit is the encoding circuit in reverse:

Step-by-step analysis of the decoding circuit

This is the decoding circuitThis is the decoding circuit

Error is phase and bit flip on 3Error is phase and bit flip on 3rdrd qubit qubit Example: Assume encoded qubit damaged such that:

Continuation of error analysis in decoderContinuation of error analysis in decoder

10000011100100110111

111000001000101110118

01111100011011001000

000111110111010001008

54321543215432154321

54321543215432154321

54321543215432154321

54321543215432154321

10000011100100110111

111000001000101110118

01111100011011001000

000111110111010001008

54321543215432154321

54321543215432154321

54321543215432154321

54321543215432154321

Phase and bit flip on 3rd qubit

Flip phase when bits 4 and 5 are “1”

10000110100110100111

110001001000101011118

11011100010011001100

100111100101110001008

54321543215432154321

54321543215432154321

54321543215432154321

54321543215432154321

10000110101100110011

110001001010001110118

11111101011011011100

101111110111110101008

54321543215432154321

54321543215432154321

54321543215432154321

54321543215432154321

Continuation of error analysis in decoderContinuation of error analysis in decoderInverting bits 3 and 5

Inverting bits 3 and 5

10000110101100110011

110001001010001110118

11111101011011011100

101111110111110101008

54321543215432154321

54321543215432154321

54321543215432154321

54321543215432154321

Continuation of error analysis in decoderContinuation of error analysis in decoderFlipping phase on bit 5

Re-express equation to prepare for Hadamard transform:

12

1001

2

10

2

10

2

110101

2

10

2

10

0010112

11110122

10

2

10

00011011001110014

110101111111010142

10

10000110101100110011

110001001010001110118

11111101011011011100

101111110111110101008

54321543215432154321

54321543215432154321

54321543215432154321

54321543215432154321

Continuation of error analysis in decoderContinuation of error analysis in decoder

This is on inputs to Hadamards

Qubits 1,2,4 and 5 are the syndrome bits which indicate the exact error that occurred and the current state of qubit 3.

544

332211 1

2

1001

2

10

2

10

543321 100111

Continuation of error analysis in decoderContinuation of error analysis in decoder

This syndrom on bits 1,2,4,5 will be now used to modify the bit 3

Syndromes Table

Syndromebits

q1q2q4q5

Erro r State o f q 3

0 0 0 0 No n e

0 0 0 1 B2

0 0 1 0 S4

0 0 1 1 B5

0 1 0 0 S2

0 1 0 1 BS2

0 1 1 0 B1

0 1 1 1 B3

1 0 0 0 S1

1 0 0 1 BS4

1 0 1 0 S3

1 0 1 1 B4

1 1 0 0 S5

1 1 0 1 BS3

1 1 1 0 BS1

1 1 1 1 BS5

10

01

10

10

10

10

10

10

10

10

10

01

01

01

01

01

* B n d e n o te s b it flip o n n th q u b it

* S n d e n o te s s ig n flip o n n th q u b it

* B S n d e n o te s s ig n a n d b it flip o n n th q u b it

From previous slide

Execution of correction based on syndromesExecution of correction based on syndromes

According to syndrome table, the 3rd qubit is in state .

So apply a phase flip and a bit flip to obtain the protected qubit .

544

332211 1

2

1001

2

10

2

10

543321 100111

1101 BS3 01

01

10

The 5 qubits error correcting circuit

Before transmission

After transmission

We did not show how to do the correction according to the syndrome. I leave it to you.

Concatenated CodeConcatenated Code 1 qubit can be encoded using 5 qubits. Each of the 5 qubits can be further

encoded using 5 qubits. Continue doing this until some number of

hierarchical levels is reached.

Illustration: We will use the 5 qubit encoding. Assume probability of single qubit error is e and that

errors are uncorrelated.

Example of Concatenated CodeExample of Concatenated Code

For 2 levels, number of qubits required is 52 = 25 This encoding will fail when 2 or more sub

blocks of 5 qubits cannot recover from errors. Hence probability of recovery failure is in order

of = (e2)2 = e4

e4 < e2. 2 levels encoding has better probability of error recovery than 1 level if e is small enough

Example of Concatenated CodeExample of Concatenated Code

For 3 levels, number of qubits required is 53 = 125 This encoding will fail when 2 or more sub blocks of

25 qubits cannot recover from errors. Hence probability of recovery failure is in order of =

(e4)2 = e8

e8 < e4. 3 levels encoding has better probabilitybetter probability of error recovery than 2 levels if e is small enough.

Example of Concatenated CodeExample of Concatenated Code

In general for L levels, Number of qubits required is 5L

Probability of recovery failure is in the order of

Advantages of concatenated code: If probability of individual qubit error, e, is pushed

below a certain threshold value, adding more levels will reduce probability of recovery failure.

i.e. we can increase the accuracy of our encoding indefinitely by adding more levels.

Error correction is simple using a divide and conquer strategy.

L

e2

Example of Concatenated CodeExample of Concatenated Code

Disadvantages of concatenated coding: If probability of individual qubit error, e, is

above the threshold value, adding more levels will make things worse.( i.e. probability of recovery failure will be higher)

Exponential number of qubits needed. Note:

Threshold value depends on Type of encoding used Types of errors that occurs. When the errors are likely to occur (during qubit

storage, or gate processing)

Concatenated CodeConcatenated Code