5
$\begingroup$

What is a difference between error rates and qubit/gate fidelity? A bit of maths in the explanation is fine but I am an A Level student doing a research project so definitions would be preferred.

$\endgroup$
4
$\begingroup$

Usually, error rate for a qubit is defined as probability of undesired change in the qubit state (see for example this paper).

Then we have state fidelity, which is a measure of the difference between the state we have and the state we would like to have, for any (single or multi qubit) quantum system. Quantum state tomography is a means to characterize the actual state.

Finally, gate fidelity is a measure of the difference between the operation the gate actually performs and the operation we would like the gate to be performed. In this case, you need quantum process tomography.

$\endgroup$
2
  • $\begingroup$ If you just care about the fidelity or want just one figure of merit for your gate, I would argue that randomized benchmarking is much better and more widely used than full QPT. I would even go as far as calling the gate fidelity that which comes from randomized benchmarking, and the 'value' that comes from QPT as the process fidelity. $\endgroup$
    – JSdJ
    Jun 10 '20 at 7:32
  • $\begingroup$ Thank you so much! Really clear and helpful! $\endgroup$
    – 980emma
    Jun 10 '20 at 10:03
3
$\begingroup$

For a single operation, the notions of error rate and fidelity are clearly closely related. Error rate is the probability with which something goes wrong, while fidelity measures how accurately the actual output matches the desired one.

Let me illustrate this with an example. Let's say I have an initial state $|0\rangle$ which I'm just storing. So my expected output is $|\psi_\text{target}\rangle=|0\rangle$. However, what actually happens is that with probability $p$, there is an $X$ error (this flips the bit value). So my real output is a density matrix $$ \rho_\text{out}=(1-p)|0\rangle\langle 0|+p X|0\rangle\langle 0|X=(1-p)|0\rangle\langle 0|+p |1\rangle\langle 1|, $$ which you can interpret as "with probability $(1-p)$, I get the 0 state, and with probability 1, I get the 1 state".

The error rate in this case is simply $p$. (If you want to interpret it as a rate, it's a probability $p$ per operation.) The fidelity, on the other hand, is $$ F=\langle\psi_\text{target}|\rho|\psi_\text{target}\rangle=1-p. $$ So, in that sense, the error rate and fidelity are opposite sides of the same coin: one is how badly things go wrong, while the other is to what extent things go right.

However, it's not quite that simple, because while the error rate doesn't depend on the input state, the fidelity does. If I repeat my previous example with the input state $|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$, then the error rate is still $p$. However, $\rho_\text{out}=|+\rangle\langle +|$, and thus, the fidelity is 1. This particular input was unaffected by noise.

This is why you might prefer, instead, to evaluate the fidelity of the operation rather than the fidelity of the state. This might be the average of the state fidelity over all possible inputs, or it could be the worst-case fidelity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.