I've been interested in gate fidelity lately.

In the meantime, I came up with a specific example. Let's think about the encoding of situation $\alpha_0 |0\rangle + \alpha_1 |1\rangle \rightarrow \alpha_0 |000\rangle + \alpha_1 |111\rangle$, and implement a circuit that can correct the case where there is an error in one bit out of three qubits. That is, consider a quantum error correction.

In that time, what is the threshold for gate fidelity at which error correction stops (or becomes meaningless)?

I think fidelity is simply a connection between the ideal and the realistic, but when I try to think of it as a concrete example, it is difficult to figure out how to calculate it. (I know the formula for calculating fidelity.)



1 Answer 1


First, take the case without encoding. Let there be an $X$ error with probability $p$, so the output state is $$ \rho=(1-p)|\psi\rangle\langle\psi|+pX|\psi\rangle\langle\psi|X. $$ Normally, I would just talk about the probability of an error (it lets you avoid some of the averaging I'm about to do), but we can talk about fidelity if you want. $$ F=\langle\psi|\rho|\psi\rangle=1-p+p|\langle\psi|X|\psi\rangle|^2 $$ To get just a single number, you might average over all possible initial states $|\psi\rangle$. You'll find that $$ \bar F=1-\gamma p $$ for some positive value $\gamma$ that is not so important to calculate right now.

By comparison, consider the encoded qubit going through the error channel. If $$ |\Psi\rangle=\alpha_0|000\rangle+\alpha_1|111\rangle, $$ then $$ \rho_2=(1-p)^3|\Psi\rangle\langle\Psi|+(1-p)^2p(X_1|\Psi\rangle\langle\Psi|X_1+X_2|\Psi\rangle\langle\Psi|X_2+X_3|\Psi\rangle\langle\Psi|X_3)+(1-p)p^2X_1X_2|\Psi\rangle\langle\Psi|X_1X_2+\ldots+p^3X_1X_2X_3|\Psi\rangle\langle\Psi|X_1X_2X_3. $$ If you error correct this, it will succeed so long as there's no more than one error $$ \rho_2\rightarrow (1-p)^2(1+2p)|\Psi\rangle\langle\Psi|+p^2(3-2p)X_1X_2X_3|\Psi\rangle\langle\Psi|X_1X_2X_3 $$ which will decode to $$ (1-p)^2(1+2p)|\psi\rangle\langle\psi|+p^2(3-2p)X|\psi\rangle\langle\psi|X $$ and hence have fidelity $$ \bar F=1-\gamma p^2(3-2p). $$

This fidelity is better than the fidelity from no encoding provided $$ p^2(3-2p)<p. $$ This is fulfilled for $0<p<\frac12$.

  • $\begingroup$ Thank you for your answer. But, using the assumption that there are no more than two errors, terms with two errors disappear, but why do you consider probabilities? $\endgroup$
    – pioneer
    Commented Oct 24, 2021 at 10:03
  • 1
    $\begingroup$ What do you understand by "threshold" if it's not a threshold probability? $\endgroup$
    – DaftWullie
    Commented Oct 24, 2021 at 11:31
  • $\begingroup$ I know that threshold is probability. I mean when calculating $\rho_2$, the assumption that there were no more than two errors was used. And after correcting the error, I thought that the probabilities of the term with two errors would disappear and be computed. Am I wrong? $\endgroup$
    – pioneer
    Commented Oct 25, 2021 at 1:00
  • $\begingroup$ No, that's not the assumption I used to calculate $\rho_2$. My assumption was that each qubit suffers a bit-flip with probability $p$. So, there's a probability $p^3$ that 3 errors happen. When you later apply error correction, all states are mapped to the closest logical state. In the case of 2 errors, the closest logical state is the one with 3 $X$ errors, so that's what they get mapped to. $\endgroup$
    – DaftWullie
    Commented Oct 25, 2021 at 9:18

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