 # 5 Qubits Error Correcting 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.

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5 Qubits Error Correcting 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. How to calculate this number?

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 2 n-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

Encoding Use 5 qubits to encode 1 qubit :  

From above, we calculate: Encoding circuit Encoding circuit 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

Step-by-step analysis of the encoding circuit flipping phase

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

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 circuit

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

Continuation of error analysis in decoder Phase and bit flip on 3 rd qubit Flip phase when bits 4 and 5 are “1”

Continuation of error analysis in decoder Inverting bits 3 and 5

Continuation of error analysis in decoder Flipping phase on bit 5

Re-express equation to prepare for Hadamard transform: Continuation 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. Continuation of error analysis in decoder This syndrom on bits 1,2,4,5 will be now used to modify the bit 3

Syndromes Table From previous slide

Execution of correction based on syndromes According to syndrome table, the 3 rd qubit is in state. So apply a phase flip and a bit flip to obtain the protected qubit.

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 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 Code

For 2 levels, number of qubits required is 5 2 = 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 = (e 2 ) 2 = e 4 e 4 < e 2. 2 levels encoding has better probability of error recovery than 1 level if e is small enough Example of Concatenated Code

For 3 levels, number of qubits required is 5 3 = 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 = (e 4 ) 2 = e 8 better probability e 8 < e 4. 3 levels encoding has better probability of error recovery than 2 levels if e is small enough. Example of Concatenated Code

In general for L levels,  Number of qubits required is 5 L  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. Example 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 Code Concatenated Code

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