When a single-qubit gate is operated by Rabi driving, the resulting operator is equivalent to
$$ \hat{O} = R_z(\omega t) R_x(\Omega t)\tag{1}$$ where $\omega$ is the frequency of the driving field, $\Omega$ is the Rabi frequency, and the phase of the driving field is assumed to be $\delta = 0$.

Thus, after the pulse, the qubit state will have rotated by an angle $\theta = \Omega t$ around the $x$-axis, but also of an angle $\gamma = \omega t$ around the $z$-axis. This is not a problem if only a single gate has to be applied, because the probability amplitudes for the $|0\rangle$ and $|1\rangle$ states will be unaffected by this last rotation.

However, if more than one gate has to be applied to the same qubit, these $z$-phases would accumulate and can completely change the operation of the circuit. Suppose $t_1 = \pi /\Omega$, and we wish to apply a rotation $R_x(2\pi)$ using two Rabi driving pulses. Then, after two $t_1$ pulses, the equivalent operator will be given by: \begin{equation} \hat{O} = R_z(\omega t_1)R_x(\pi)R_z(\omega t_1)R_x(\pi)\tag{2} \end{equation} which however, is completely different from: \begin{equation} \hat{O} = R_x(2\pi) = \hat{I}\tag{3} \end{equation}

To get rid of the spurious phase, I would guess one should either add a forced $R_z(-\omega t_1)$ rotation in between the pulses or wait for the qubit to "resynchronize" while precessing freely. However, none of the papers I've read mention this issue. What am I missing?


1 Answer 1


Good question! Your first equation mixes a couple of abstractions that we typically sweep under the rug, so let's start by taking a step back. When we apply a control pulse to a qubit, it's usually some sort of oscillating electric field $E(t)$ or magnetic field $B(t) = A \cos(\omega t + \delta)$ (I'm going to use $B$ throughout). Then our qubit's Hamiltonian is given by $H = \frac{\omega_0}{2}\sigma_z + \gamma B(t) \sigma_x$ where $\omega_0$ is the qubit frequency and $\gamma$ is a general coupling parameter between our drive field and qubit. The problem with working with this Hamiltonian is the $\sigma_z$ and $\sigma_x$ terms don't commute, one is time dependent and the other isn't, and the math gets ugly quickly when you want to calculate time evolutions.

The typical way to deal with this is to shift to something called the rotating frame or interaction picture. Rather than a time-independent state $|\psi\rangle = a|0\rangle + b|1\rangle$, we can instead consider a state/wavefunction/Bloch sphere that's rotating at the qubit frequency $\omega_0$ specifically $|\psi_I\rangle = e^{i\frac{\omega_0}{2} t\sigma_z} = a|0\rangle + be^{i\omega_0 t}|1\rangle$. Working through the math with the Schrodinger equation and using the rotating wave approximation, you can rewrite the Hamiltonian in this new interaction picture as $H = \frac{\omega-\omega_0}{2}\sigma_z + \frac{\Omega}{2} \sigma_x$, removing the time dependence.

Now why does this interaction picture help us? Well in most quantum experiments, we have some microwave (or laser) source that we're switching on and off to apply pulses to our qubit, but even when we have it switched off it's still oscillating at its drive frequency $\omega$ so when we apply the next pulse, it's still in-phase from the last pulse, basically taking care of the "resynchronizing" you mentioned for you.


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