Leakage of superconducting qubit: why it occurs with short driving pulses and not as well with long ones My question in very short:
Short pulses on superconducting qubit usually induce leakage. But a long pulse can be seen as a sequence of short pulses (imagine all of the short pulse as square pulses: combining them you obtain a longer square pulse). Thus how if short pulses induce leakage, long pulses induce less leakage ?
I am not necesseraly looking for a super precise model but I just want to get the basic intuition of the physics.

Let us consider a superconducting qubit. In practice such system does not consist into two well isolated energy level but is encoded in an anharmonic oscillator. Let me summarize it with only its three first levels with the associated energies:
$$E_0=0, E_1=\hbar \omega_{01}, E_2=\hbar (\omega_{01}+\omega_{12})$$
To drive such system, we usually aim to send resonnant signals at the frequency of the two first level: $\omega_{01}$ which define our qubit. A "good" driving is a driving that avoids to excite other transitions than $\omega_{01}$ in order to avoid the qubit to "leak" in the third excited level.
Allright.
Now, I remind that the frequencies composing a signal are entirely determined by the shape of the signal in time by Fourier transform:
$$s(\omega)=\int_{-\infty}^{+\infty} s(t) e^{-i \omega t} dt$$
The only signal perfectly resonnant with the frequency $\omega_{01}$ is (up to some phase):
$$s(t)=e^{i \omega_{01} t}$$
Let's assume two experiments. On the first one, experiment $A)$, I send a driving signals that starts at $t=t_0$ and is then cut at $t=t_1$. On the second one, experiment $B)$ I send a driving signals starting at $t=t_0$ but that is cut later on, at an instant $t=t_2$. When the signal is "on", it can be described by the function $s(t)$.

And I want to study the probability that my qubits is found in the highest energy state along time, thus that it has leaked.
Between $t_0$ and $t_1$, for both experiment everything happening is exactly the same. Thus, if the qubit has leaked (with a high probability) in experiment $A)$ it means that it must have leaked on experiment $B)$ during this interval of time.
But if on the experiment $B)$, we assume that the length of the signal is long enough, it will be a good approximation of the purely monochromatic signal $s(t)$ and then we would conclude that the qubit does not leak.
How to reconcile those visions ?
Is it that on experiment $B)$, if I measured the probability to find the qubit into the highest energy level I would find that it has indeed leaked but if I wait some time longer (after $t_1$) it will have come back in the two first energy levels ? Thus the dynamic is actually pretty complicated but this is what is happening in the end ?
Thus somehow, the qubit always leaks in the middle of the experiment, but what matters is that when my gate is done it ends up in the subspace it should belong to.
Are there some papers or book that discuss this kind of behavior in a simple manner ? I am not interested in a very detailed calculation but I just want to have the basic intuition to remove the kind of paradox I am confused by.
 A: There are lots of subtleties and tricks designed for 3-level systems, but the basic physics of the effect you are interested in can be understood from a basic time-dependent perturbation theory (any graduate level text would work, for instance Landau and Lifshitz, paragraphs 40-42).
Here is the short summary of things that relate directly to your example. Assuming your initial state $|\psi_0\rangle$ doesn't change much over the time of perturbation, you can integrate Shroedinger equation, and obtain that after the system started being driven with operator $\hat{V}e^{i\omega t}$ for some time $\tau$, the amplitude of the $n$-th state is going to be:
$$
c_n = \int_{0}^\tau V_{0n}e^{i\omega t}e^{-i\omega_nt}dt,\quad (1)
$$
where $\omega_n$ is the energy of the n-th state (for simplicity we assumed $\omega_0=0$).
Integrating yields the probability of the system to be in state $n$
$$
P(n)=|V_{0n}|^2 \frac{4\sin{(\frac{1}{2}(\omega_{n}-\omega)\tau)}^2}{(\omega_{n}-\omega)^2}.
$$
(Thats formula (42.3) in L&L)
As you can see, the probability oscillates as a function of $\tau$ between $0$ and $|V_{0n}|^2/{(\omega_{n}-\omega)^2}$, and never decays, which is easy to understand, since the argument in Eq. (1) is just an oscillating exponent.
Applying the analysis to your qubit, as one can see, longer drive does not help reducing the amplitude in the 3rd state, as the amplitude continues to bounce around with the magnitude of $2|V_{0n}/(\omega_{n}-\omega)|$.
What makes the situation drastically improve is the slow and smooth turn-up and turn-down of the driving amplitude. By multiplying the driving operator $\hat{V}e^{i\omega t}$ by a modulating function $M(t)=e^{-t^2/4T^2}$ ($T$ is the effective driving length) and performing integration (1) from $t=-\infty$ to $t=\infty$, we obtain that the amplitude of the n-th state at the end is
$$
c_n =V_{0n} e^{-(\omega_n-\omega)^2T^2}.
$$
In other words, for a fixed $\omega_n-\omega\neq 0$ the amplitude decays faster than exponentially (!!!) with $T$, and for fixed $T$ the amplitude drops as quickly with detuning $\omega_n-\omega\neq 0$. We can understand a general case of $M(t)$ by noticing that the integral $\int_{-\infty}^{\infty} V_{0n}M(t)e^{i(\omega-\omega_n) t}dt$ is just a Fourier transform of $M(t)$, and if k-th derivative of $M(t)$ is continuous, $c_n$ should decay at least as $O(|\omega-\omega_n|^{-(k+2)})$.
Summary. If you stop your drive abruptly, you will see leaking into other states of the magnitude $|V_{0n}/(\omega-\omega_n)|$, but if you wind down your drive slowly, the leak will disappear.
