4
$\begingroup$

CTs / ZEFOZs: Energy level structures that include avoided crossings at accessible energies tend to be resilient to noise and therefore present high coherence times, at least in the case of spin qubits and magnetic noise: as the at first order effect of the magnetic field on the qubit energy vanishes, so does effectively most magnetic noise. Different people call these Atomic Clock Transitions (CTs or ACTs) or Zero First-Order-Zeeman (ZEFOZ) shift, but it's essentially the same phenomenon. This is experimentally expressed as high spin-spin $T_2$ relaxation times, even in presence of relatively high sources of noise.

However, in a quantum computing scenario, what we want is typically not a qubit surviving for long periods of inactivity (high $T_2$) but rather to obtain a high fidelity after a series of quantum gates, which can in general be rather complicated and involve entanglement with other qubits.

My question: Do CTs / ZEFOZs with their high relaxation times generally also translate into a high fidelity after a complicated series of quantum gates?

For context, the examples I have in mind are solid-state, mainly these two:

As well as this more refined variation, with electric field:

$\endgroup$
2
  • 1
    $\begingroup$ How is it possible for a longer coherence time to not increase fidelity? $\endgroup$ May 12, 2018 at 19:37
  • $\begingroup$ I clarified my thoughts in the form of a minimalistic answer, below; I can move the text to the question instead if you think that makes more sense. $\endgroup$ May 13, 2018 at 5:25

1 Answer 1

1
$\begingroup$

The best I have it's this generic answer, which I put here for clarity, hoping for improvements/corrections or even to be superseded by something better:

If the limiting factor for fidelity in a given architecture+algorithm are the single-qubit gates, or the two-qubit gates, or the measurement, and if this limiting factor is not optimized in a ZEFOZ point, then effectively quantum fidelity will not be optimal in the ZEFOZ point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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