Time-optimal control with finite bandwidth

Abstract

Time-optimal control theory provides recipes to achieve quantum operations with high fidelity and speed, as required in quantum technologies such as quantum sensing and computation. While technical advances have achieved the ultrastrong driving regime in many physical systems, these capabilities have yet to be fully exploited for the precise control of quantum systems, as other limitations, such as the generation of higher harmonics or the finite response time of the control apparatus, prevent the implementation of theoretical time-optimal control. Here we present a method to achieve time-optimal control of qubit systems that can take advantage of fast driving beyond the rotating wave approximation. We exploit results from time-optimal control theory to design driving protocols that can be implemented with realistic, finite-bandwidth control fields, and we find a relationship between bandwidth limitations and achievable control fidelity.

Appendices

Time-optimal bang–bang control

Bang–bang control has been shown to achieve time-optimal control of two-level systems. General bounds and prescriptions for the time-optimal bang–bang control have been provided [29, 32,33,34, 51, 68, 69]. These results can be used to numerically obtain a solution to the time-optimal problem with BB control for a general unitary. However, for some target unitaries and Hamiltonian parameters it is possible to find analytical solutions. In the main text we focused on these cases since they allow to more easily study trends in the fidelity and robustness of the FATO control strategy. Here we describe how, exploiting known results in BB control, we obtained the specific TO solutions for the two gates and two driving strength regimes considered.

The general goal is to find the optimal times such that a sequence of “bangs” under alternating Hamiltonian ${\mathcal {H}}_{\pm ,0}$ can achieve the desired unitary. Simple algebraic arguments [33] impose constraints on the middle times of any TO decomposition. Then the TO problem reduces to finding three times, $t_i, t_m$ and $t_f$, such that concatenating the unitaries $U_{0,\pm }(t)=e^{-it{\mathcal {H}}_{0,\pm }}$ achieves the two desired gates, $\sigma _x$ and $\sigma _y$. Since we focused on achieving $\pi$ rotations, we can apply (in addition to results found in [33,34,35]) the results of [32] relating to a north-to-south pole transformation only. Their constraints still need to be valid sufficient conditions (although not necessary) for the TO unitary.

1.1 Weak driving

For weak driving, it was found in [32] that there should be no singular bangs in the TO solution. In addition, for $\alpha =\pi /2n$ the solution $U_{NS}$ for the north-to-south transition is obtained as

$$\begin{aligned} U_{NS}=[U_+(\pi )U_-(\pi )]^n \end{aligned}$$

(7)

We find that if n is even, $U_{NS}=\sigma _y$, while if n is odd $U_{NS}=\sigma _x$. Ref. [32] allowed for a second type of solution, with $n+1$ bangs. These are candidates solutions for $\sigma _y$ if n is odd and $\sigma _x$ if n is even. While it is possible to find analytical solutions for the optimal times in these cases as well, the expressions become more involved and thus we limited our extended analysis in the main text to the simplest case.

We note that even for X rotations we consider an even number $n+1$ of bangs to build the FATO approximation. This ensures that the function f(t) is odd, yielding a sine Fourier series which is zero at $t=0$ for any bandwidth.

1.2 Ultrastrong Driving

The strong driving occurs when $\vartheta >\pi /4$. In this case, following [32] and [34], we find that the TO solution is composed of at most three bangs. It is then easy to find analytical solutions for the times $\{t_i, t_m, t_f\}$, for example by following the construction described in [70].

We can first verify that two bangs are not enough to generate the desired rotations. Defining $\mathbf {v}_{\pm ,0}$ the vectors corresponding to the Hamiltonians ${\mathcal {H}}_{\pm ,0}$ and $R=X,Y$ the rotations in SO(3) corresponding to $\sigma _x, \sigma _y$, we need to verify whether $\mathbf {v}_i\cdot \mathbf {v}_j=\mathbf {v}_i\cdot R\cdot \mathbf {v}_j$. However $\mathbf {v}_+\cdot \mathbf {v}_{-}=\cos (2\vartheta )$, $\mathbf {v}_\pm \cdot \mathbf {v}_0=\cos (\vartheta )$, while $\mathbf {v}_0\cdot R\cdot \mathbf {v}_\pm =-\cos (\vartheta )$, $\mathbf {v}_+\cdot Y\cdot \mathbf {v}_-=-\cos (2\vartheta )$ and $\mathbf {v}_+\cdot X\cdot \mathbf {v}_-=-1$.

Similarly, we can easily identify all the allowed three-bang constructions that achieve the desired unitaries. We find that the central bang must be singular ($\varOmega =0$) for the $\sigma _y$ gate. Then, since the desired $\sigma _y$ gate cannot have a component along $\sigma _x$ we have to set

$$\begin{aligned} \frac{i}{2}\mathrm {Tr}\{(U_\pm U_0U_\mp )\sigma _x\}\!=\!\sin (\vartheta )\sin \!\left( \!\frac{\omega _0t_2}{2}\!\right) \sin \!\left[ \frac{\omega _0(t_1\!-\!t_3)}{2}\right] \!=\!0 \end{aligned}$$

by selecting $t_1^y=t_3^y$ (since $t_2=0$ has already been excluded). Finally, $t_1^y$ and $t_2^y$ can be easily found algebraically, yielding Eq. (3) in the main text. For the $\sigma _x$ gate, solutions with and without a singular bang are allowed. In both cases the outer rotations are about the same axis, e.g., $U_+U_0U_+$ or $U_+U_-U_+$, and of the same duration, $t_1^x=t_3^x$. We can then calculate explicitly the times and compare the two possible solutions to select the time-optimal one. We find that the shortest evolution is obtained by discarding the singular solution, resulting in the times in Eq. (3) in the main text.

1.3 Time-optimal control of two qubits

Extending the results of BB TO control to more than one qubit is generally difficult, but results have been found for some particular cases [51, 63, 65]. In particular, it has been found [51] that for two qubits with opposite drift terms and under the same control,

$$\begin{aligned} {\mathcal {H}}_\pm =\frac{\omega _0}{2}(\sigma _z^1-\sigma _z^2)\pm \frac{\varOmega }{2}(\sigma _x^1+\sigma _x^2) \end{aligned}$$

the TO problem can be solved simultaneously. In particular, for $\pi$ rotations, we recover the same solutions found for one qubit. Then we could repeat the analysis performed for FATO control of one qubit; the fidelity is simply the square of the fidelity found for one qubit.

We note that while these analytical results are restricted to particular cases, they could be at the basis of numerical searches in more complex situations. For example, knowing the control function and required bandwidth for two non-interacting qubits could be used as initial guess for numerical searches of the control profile for two interacting qubits.

Bang–bang control can as well be used to achieve two-qubit gates. Time-optimal solutions for the steering of two qubits with Hamiltonian ${\mathcal {H}}=J\sigma _z^1\sigma _z^2/2$ were indeed found under the assumption of delta pulses (zero-duration pulses at infinite driving power). While the control solutions obtained when relaxing these assumptions might not be time-optimal, we can still aim to preserve the optimal time and look for the ensuing drop in fidelity. In the main text, we considered an exemplary gate, the SWAP gate between two qubits (swapping their states):

$$\begin{aligned} U_{swap}=\left[ \begin{array}{cccc} 1&{}0&{}0&{}0\\ 0&{}0&{}1&{}0\\ 0&{}1&{}0&{}0\\ 0&{}0&{}0&{}1 \end{array}\right] \end{aligned}$$

As shown in [52], this gate can be obtained in a time $t=3\pi /2J$ by applying instantaneous $\pi /2$ pulses about x and y. This ideal control can be in practice replaced by finite-duration (rectangular) pulses, to account for finite control strength. In turn rectangular pulses can be replaced by FATO driving to take into account the control finite bandwidth. Figure 7 shows the control sequence we implemented to analyze the fidelity behavior of FATO for two-qubit control, as shown in Fig. 6 of the main text.

Fig. 7

SWAP Gate. We show the control profile for realizing a SWAP gate with the FATO control. In red is the driving along the x axis and in blue along the y axis. For the bandwidth considered here ($\varDelta \omega =400J$) the FATO control completely masks the BB control (solid black lines) (Color figure online)

Fidelity

Here we provide further details on the calculations of the fidelity. Due to FATO control, the system evolves under the Hamiltonian ${\mathcal {H}}\!=\!{\mathcal {H}}_{id}(t)-{\mathcal {H}}_R(t)$, with

$$\begin{aligned} \begin{array}{l} {\mathcal {H}}_{id}(t)\!=\!\omega _0/2[\sigma _z+f(t)\tan (\vartheta )\sigma _x]\\ {\mathcal {H}}_R(t)\!=\!\omega _0/2\tan (\vartheta ) R_K\!(t)\sigma _x, \end{array} \end{aligned}$$

Here f(t) is the BB control function $f(t)=\varOmega (t)/{\overline{\varOmega }}$ and $R_K(t)$ the remainder of the truncated Fourier series of f(t),

$$\begin{aligned} R_K(t)=\sum _{k=K+1}^\infty [s_k\sin ({2\pi k t}/T)+ c_k\cos ({2\pi k t}/T)]. \end{aligned}$$

(8)

In order to calculate the infidelity, we consider the propagator $U_K$ achieved by implementing a control field according to FATO to order K, satisfying

$$\begin{aligned} i\dot{U}_K(t)\!=[{\mathcal {H}}_{id}(t)-{\mathcal {H}}_R(t)]U_K(t) \end{aligned}$$

(9)

and the propagator due to the ideal BB evolution, $U_{id}(t)$ defined by

$$\begin{aligned} i\dot{U}_{id}(t)={\mathcal {H}}_{id}(t)U_{id}(t) \end{aligned}$$

(10)

The infidelity of the truncated FATO control can be evaluated using the entanglement fidelity [48]

$$\begin{aligned}F=|\mathrm {Tr}[U_{id}^\dag (T)U_K(T)]|/2\end{aligned}$$

We can decompose the total propagator $U_K(T)$ as $U_K(T)=U_{id}(T)U_R(T)$, by defining the error propagator $U_R(t)=U_{id}^\dag (t)U_K(t)$. Then the fidelity is simply defined as $F=|\mathrm {Tr}[U_{R}(T)]|/2$.

The error propagator can be evaluated by moving to the toggling frame defined by the ideal control Hamiltonian. In this frame, the Hamiltonian becomes

$$\begin{aligned}{\widetilde{{\mathcal {H}}}}_R(t) = U^\dag _{id}(t){\mathcal {H}}_R(t)U_{id}(t)\end{aligned}$$

and the error propagator satisfies the Schr$\ddot{\text {o}}$dinger equation

$$\begin{aligned}i\dot{{\widetilde{U}}}_R(t)={\widetilde{{\mathcal {H}}}}_R(t){\widetilde{U}}_R(t)\end{aligned}$$

We can approximate $U_R$ with a first-order Magnus expansion given by the effective Hamiltonian

$$\begin{aligned}{\overline{{\mathcal {H}}}}_R=\int _0^T {\widetilde{{\mathcal {H}}}}_R(t')dt'\end{aligned}$$

Consider for example the weak driving regime. The contribution to ${\overline{{\mathcal {H}}}}_R$ from each bang is given by the integral over the interval $[t_j,t_{j+1}]$ of

$$\begin{aligned} {\widetilde{{\mathcal {H}}}}_j\!=\!\omega _0\tan \!(\vartheta )R_K\!(t)U_{id}^\dag (t_j) [e^{i{\mathcal {H}}_\pm t}\sigma _xe^{-i{\mathcal {H}}_\pm t}]U_{id}(t_j) \end{aligned}$$

(11)

Each pair of ideal propagators $U_-U_+$ creates a rotation $e^{-2i\sigma _y\vartheta }$. Since the angle $\vartheta$ is small for weak driving, we can approximate this expression by ignoring the time evolution due to ${\mathcal {H}}_{id}$ during the $j^{th}$ time interval and only considering its effects stroboscopically. Then Eq. (11) reduces to

$$\begin{aligned}&{\widetilde{{\mathcal {H}}}}_j=\frac{\omega _0}{2}\tan (\vartheta )R_K(t)U_{id}(t_{j+1})\sigma _xU_{id}(t_{j+1})\\&\qquad = \frac{\omega _0}{2}\tan (\vartheta )R_K(t)[(-1)^j\cos (2j\vartheta )\sigma _x+\sin (2j\vartheta )\sigma _z]\nonumber \end{aligned}$$

(12)

Note that the sign of the $\sigma _x$ terms follows the same pattern as the BB function f(t). Setting $j(t)=\lceil t/T\rceil$, we then need to evaluate the integrals

$$\begin{aligned}\frac{2}{T}\int _0^T f(t)\cos [2j(t)\vartheta ]\sin (2\pi kt/T)dt=s_k\frac{\vartheta }{\pi }\end{aligned}$$

and

$$\begin{aligned}\frac{2}{T}\int _0^T \sin [2j(t)\vartheta ]\sin (2\pi kt/T)dt=-s_k\frac{\vartheta }{\pi }\tan (\vartheta )\end{aligned}$$

We thus obtain the average Hamiltonian

$$\begin{aligned} {\overline{{\mathcal {H}}}}_R=\frac{\omega _0}{2}\tan (\vartheta )\frac{\vartheta }{\pi }\sum _{k=K+1}^\infty \!\!\!s_k^2\ [\sigma _x-\tan (\vartheta )\sigma _z], \end{aligned}$$

(13)

where we recognize the mean error of the truncated Fourier series, ${\mathcal {E}}_K=\frac{1}{2}\sum _{k=K+1}^\infty \!s_k^2$. The fidelity is then given by $F=\cos (\Vert {\overline{{\mathcal {H}}}}_R\Vert T)$:

$$\begin{aligned}&F=\cos \left( \frac{\vartheta }{\pi }\frac{\omega _0T}{2\cos (\vartheta )} \tan (\vartheta ){\mathcal {E}}_K \Vert \cos (\vartheta )\sigma _x-\sin (\vartheta )\sigma _z\Vert \right) \nonumber \\&\quad =\cos \left( \frac{\pi }{2}\tan (\vartheta ){\mathcal {E}}_K\right) . \end{aligned}$$

(14)

A similar calculation can be done for the ultrastrong driving case. However, in that case each “bang” has a long duration, and thus we need to start from Eq. (11) to find the average Hamiltonian and only approximate or numerical solutions can be found. Still, we find that the solutions depend on the mean Fourier error in a simple way, $F_s^x\approx \cos [2/\pi \sin (\vartheta ){\mathcal {E}}_K]$ and $F_s^y\approx \cos [2/\pi \tan (\vartheta ){\mathcal {E}}_K]$.