5 Qubits Error Correcting

1
5 Qubits Error Correcting

• Shors 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.
  

How to calculate this number?
2
How to use syndrome bits to calculate 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.

Pauli rotations
3
Encoding

• Use 5 qubits to encode 1 qubit

4
From above, we calculate

• Encoding circuit

Flip phase
5
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

6
Signal after Hadamards
Hadamards
Encoding circuit
7
Step-by-step analysis of the encoding circuit
flipping phase
flipping phase
8
Step-by-step analysis of the encoding circuit
Flipping bits 3 and 5
Flipping bits 3 and 5 with 1 in bits 2 and 4
9
Step-by-step analysis of the encoding circuit
Flipping phase in data bit when bits 4 and 5 are
1
10
Step-by-step analysis of the decoding circuit

• 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

This is the decoding circuit
11
Error is phase and bit flip on 3rd qubit

• Example Assume encoded qubit damaged such that

12
Continuation of error analysis in decoder
Phase and bit flip on 3rd qubit
Flip phase when bits 4 and 5 are 1
13
Continuation of error analysis in decoder
Inverting bits 3 and 5
Inverting bits 3 and 5
14
Continuation of error analysis in decoder
Flipping phase on bit 5
15
Continuation of error analysis in decoder

• Re-express equation to prepare for Hadamard
  transform

This is on inputs to Hadamards
16
Continuation of error analysis in decoder

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

This syndrom on bits 1,2,4,5 will be now used to
modify the bit 3
17
Syndromes Table
From previous slide
18
Execution 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 .

19
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.
20
Concatenated 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.

21
Example of Concatenated Code

• Illustration
• We will use the 5 qubit encoding.
• Assume probability of single qubit error is e and
  that errors are uncorrelated.

22
Example 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 lt e2. 2 levels encoding has better probability
  of error recovery than 1 level if e is small
  enough

23
Example 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 lt e4. 3 levels encoding has better probability
  of error recovery than 2 levels if e is small
  enough.

24
Example 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.

25
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)