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For transmon qubits, there is an always-on zz interaction between coupled qubits, thus introducing static zz crosstalk. The figures are from this paper. I know zz crosstalk can change the frequency of the spectator qubit (Q2 in the figure).

My question is: since zz crosstalk is always on, why is there a difference of frequency for Q2 when Q1/Q3 is set to 0 or 1 state? (second figure)

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the difference in frequency is precisely because the interaction is always on.

Let's start with a scenario where there is no ZZ interaction. In this case, the Hamiltonian during the free evolution time is simply

$$ H=\sum_{i=1,2,3}\frac{1}{2}\hbar\omega_{i}\sigma_{zi}. $$ Then the phase accumulated during the free evolution time for qubit 2 is $\omega_2 \Delta t$, where $\Delta t$ is the free evolution time, regadless of the state of qubits 1 and 3.

If, however, we include ZZ interaction (and assuming for simplicity we only have qubits 1 and 2), the Hamiltonian becomes: $$ H=\sum_{i=1,2}\frac{1}{2}\hbar\omega_{i}\sigma_{zi} + \frac{1}{2}J\sigma_{z1}\sigma_{z2}. $$ Now, if we set qubit 1 to the $|0\rangle$ state, the evolution seen by qubit 2 is effectively $$ \langle0|H|0\rangle=\frac{1}{2}(\hbar\omega_{2}+J)\sigma_{z2} $$ and so the phase accumulated during free evolution will be $(\omega_2 +J )\Delta t$. If, on the other hand, the state is $|1\rangle$, the accumulated phase will be $(\omega_2 -J )\Delta t$. This is the origin of the change in oscillation frequency in the figure above.

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  • $\begingroup$ Thanks! Could you explain why do we need to add a phase (Rz) in the circuit? $\endgroup$
    – peachnuts
    Apr 4, 2022 at 15:23
  • $\begingroup$ @peachnuts yes. They want to see the oscillations, so they add a $\theta \propto \tau$ signal. If they wouldn't have done that, they would be required to detune the signal which would result in a lower $\pi/2$ gate fidelity. see e.g. "Ramsey measurement" quantum-machines.co/solutions/superconducting-qubits here $\endgroup$
    – Lior
    Apr 4, 2022 at 16:28
  • $\begingroup$ Thanks! From my understanding, the detuning frequency is quite small. They add another phase rotation (f_offset) to make it easier to measure so that f_detunning = f_obtained - f_offset. Is it correct? If so, which value should we generally set for f_offset? $\endgroup$
    – peachnuts
    Apr 4, 2022 at 17:50
  • $\begingroup$ ideally you want to aim to have like 5-6 oscillations on your ramsey scan to get a good fit. This means that you want your f_offset to be about 5-6 times the expected $1/T_2^*$. Then, when your pulses are right on resonance, you'll see this number of oscillations. If you detune the control pulse, the oscillation frequency will grow or decay such that the oscillation frequency will always be |f_offset - f_detuning|. $\endgroup$
    – Lior
    Apr 4, 2022 at 18:24
  • $\begingroup$ for f_offset you mean the $Rz(\theta)$ where $\theta = 2\pi f_{offset} \times delay$? From the tutorial of T2 Ramsey Experiment in Qiskit (qiskit.org/documentation/experiments/tutorials/…), it writes: f_measure = f_detuning + f_offset (f_osc induced by user), why do you think f_measure = |f_offset - f_detuning|? I might misunderstand something... $\endgroup$
    – peachnuts
    Apr 5, 2022 at 8:09

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