A qubit is encoded using a 3-bit repetition code. If it is known that the qubits will only ever encounter noise that can
Posted: Mon Aug 14, 2023 5:57 am
Question - A qubit is encoded using a 3-bit repetition code. If it is known that the qubits will only ever encounter noise that can be modelled as independent, identically distributed bit-flips, with the probability of a bit flipping equal to p, then give the threshold of this code. State any assumptions made
Answer -
The threshold of a quantum error-correcting code represents the critical error rate beyond which the code becomes ineffective at correcting errors. In the context of the given 3-bit repetition code encoding a qubit with independent, identically distributed bit-flip noise, let's calculate the threshold for this code.
The 3-bit repetition code encodes a single logical qubit into three physical qubits. The encoding is as follows:
|Logical Qubit| |Physical Qubits|
|------------| |---------------|
| 0 |->| 000 |
| 1 |->| 111 |
In this encoding, the logical states |0⟩ and |1⟩ are mapped to the corresponding bit strings 000 and 111, respectively.
Assumptions made:
i) We assume that the errors are independent and identically distributed (i.i.d.) bit-flip errors with a probability
p of occurring on each physical qubit.
2) We assume that the errors occur with a bit-flip probability p on each of the three physical qubits.
Now, let's analyze the threshold of this code. The threshold is the error rate at which the code can still correct errors effectively. Beyond this threshold, the probability of successfully decoding the logical qubit approaches zero.
For the 3-bit repetition code, the code can correct errors as long as at most one out of the three physical qubits is flipped. This is because a majority vote will still recover the correct logical value.
Threshold calculation:
The code can correct an error if at most one bit out of three flips. We can calculate the threshold by considering the cases where exactly one bit flips (1 bit-flip) and where exactly two bits flip (2 bit-flips).
1 bit-flips case
• The probability of exactly one bit flipping out of three is 3p(1−p)2. (Three ways to choose which bit flips, and the other two bits remain unflipped.)
2 bit-flips case
The probability of exactly two bits flipping out of three is 3p2(1−p). (Three ways to choose which two bits flip, and the remaining bit remains unflipped.)
To find the threshold, we want the sum of the probabilities of these two cases to be less than or equal to 1 (meaning the code can correct errors):
3p(1−p)2+3p2(1−p)≤1
Solving this inequality for p, we can find the threshold of the code. This threshold represents the maximum error rate (p) at which the code can still effectively correct errors
Please note that solving this inequality may involve numerical methods, and the exact threshold value will depend on the specific details of the calculation
Answer -
The threshold of a quantum error-correcting code represents the critical error rate beyond which the code becomes ineffective at correcting errors. In the context of the given 3-bit repetition code encoding a qubit with independent, identically distributed bit-flip noise, let's calculate the threshold for this code.
The 3-bit repetition code encodes a single logical qubit into three physical qubits. The encoding is as follows:
|Logical Qubit| |Physical Qubits|
|------------| |---------------|
| 0 |->| 000 |
| 1 |->| 111 |
In this encoding, the logical states |0⟩ and |1⟩ are mapped to the corresponding bit strings 000 and 111, respectively.
Assumptions made:
i) We assume that the errors are independent and identically distributed (i.i.d.) bit-flip errors with a probability
p of occurring on each physical qubit.
2) We assume that the errors occur with a bit-flip probability p on each of the three physical qubits.
Now, let's analyze the threshold of this code. The threshold is the error rate at which the code can still correct errors effectively. Beyond this threshold, the probability of successfully decoding the logical qubit approaches zero.
For the 3-bit repetition code, the code can correct errors as long as at most one out of the three physical qubits is flipped. This is because a majority vote will still recover the correct logical value.
Threshold calculation:
The code can correct an error if at most one bit out of three flips. We can calculate the threshold by considering the cases where exactly one bit flips (1 bit-flip) and where exactly two bits flip (2 bit-flips).
1 bit-flips case
• The probability of exactly one bit flipping out of three is 3p(1−p)2. (Three ways to choose which bit flips, and the other two bits remain unflipped.)
2 bit-flips case
The probability of exactly two bits flipping out of three is 3p2(1−p). (Three ways to choose which two bits flip, and the remaining bit remains unflipped.)
To find the threshold, we want the sum of the probabilities of these two cases to be less than or equal to 1 (meaning the code can correct errors):
3p(1−p)2+3p2(1−p)≤1
Solving this inequality for p, we can find the threshold of the code. This threshold represents the maximum error rate (p) at which the code can still effectively correct errors
Please note that solving this inequality may involve numerical methods, and the exact threshold value will depend on the specific details of the calculation