# Pulse control of affine systems with applications to quantum control

Ivan Beschastnyi, Lamberto Dell'Elce, Jean-Baptiste Pomet, Ludovic Sacchelli, David Tinoco

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Ivan Beschastnyi, Lamberto Dell'Elce, Jean-Baptiste Pomet, Ludovic Sacchelli, David Tinoco. Pulse control of affine systems with applications to quantum control. 64th IEEE Conf. on Decision and Control (CDC), Dec 2025, Rio De Janeiro, Brazil. hal-05042176

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Submitted on 22 Apr 2025

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Distributed under a Creative Commons Attribution 4.0 International License
Pulse control of affine systems with applications to quantum control

Ivan Beschastnyi<fnref n="1" />*, Lamberto Dell'Elce<fnref n="2" />*, Jean-Baptiste Pomet<fnref n="3" />*,
Ludovic Sacchelli<fnref n="4" />* and David Tinoco<fnref n="5" />*

**Abstract**—Motivated by problems in quantum control, we introduce a novel method for studying time optimal control problems of affine systems with unbounded controls. Several examples, including generation of noiseless and noisy qubit gates, as well as population transfer of open two-level systems, are discussed.

I. INTRODUCTION

Quantum technology has the potential to revolutionize
numerous fields, from metrology to computation. However,
the development of physical devices is severely constrained
by the phenomenon of decoherence, and one of the primary
challenges remains achieving effective control over open
quantum systems [8]. Since decoherence gradually degrades
the purity of a quantum state, it is reasonable to seek control
strategies that reach the desired state as accurately and as
quickly as possible. This naturally leads to an optimal control
problem.

Pulse-based engineering has become ubiquitous in quan-
tum information for effective qubit control [2], [9], [15].
However, from the optimization point of view, if we assume
that control functions can be unbounded, we encounter the
issue that such an optimal control problem becomes ill-
posed. Indeed, by increasing the control magnitude indef-
initely, we could theoretically steer the system from some
state to another arbitrarily fast (or at least, as close as possible
to the theoretical limit [6]). In this case, the minimal control
might take the form of a pulse causing an instantaneous jump
in the system's state. Consequently, the minimal control may
not belong to the usual L∞-class, where standard existence
theorems apply. The issue of the existence of a minimum
is not unique to quantum control problems but arises more
generally in any time-optimal control problem for a system
of the form:

$$
\dot{q} = f_0(q) + \sum_{i=1}^{k} u_i f_i(q), \quad (I.1)
$$

where $q \in M$ are points in a smooth manifold $M$, $f_0, \dots, f_k$
are vector fields on $M$ and $u = (u_1, \dots, u_k) \in \mathbb{R}^k$,
where $k < \dim M$. In this article we study the minimum

time problem for (I.1) with unbounded controls and give
applications to particular quantum systems.

System (I.1) is typically analyzed using a reduction pro-
cedure. Consider first the simplest case of a single control,
k = 1. Intuitively, when |u₁| becomes very large, the
dynamics along the vector field f₀ + u₁f₁ will converge to
the dynamics of the vector field u₁f₁. This suggests that, in
the limit, motion occurs instantaneously along the integral
curves of f₁.

One can define an equivalence relation on M as follows.
Two points q₁ ~ q₂ are equivalent if and only if they lie on
the same integral curve of f₁. If the quotient M/~ is well-
behaved, a reduced control system can be defined on M/~,
preserving all the essential information of the original control
problem while ensuring a well-posed time-minimum control
problem [1, Chapter 22]. For k > 1, the same principle
applies when the vector fields f₁, ..., fₖ commute and are
analytic. In this case, the Stefan-Sussmann theorem (see,
for example, [1, Proposition 5.3]) guarantees that f₁, ..., fₖ
generate a singular foliation, and the reduced system is
defined on the corresponding leaf space.

The reduction method has been successfully applied to
controlled quantum systems in several articles [1], [7], [10],
[12]. However, it comes with several notable drawbacks.
First, the leaf space M/~ may not always be a smooth
manifold. For instance, in the case of two-level open quantum
systems with a single control function, the resulting space is
a semidisc,

$$
D_{+}^{2} = \{(x, y) \in \mathbb{R}^{2} : x^{2} + y^{2} = 1, y \geq 0\},
$$

as shown in [10]. Second, the calculation of the reduced
system requires calculating pushforwards of the vector fields
f₀ under the flows of a linear combination of f₁, ..., fₖ,
which can be a difficult task on its own.

In this article, we propose an alternative approach to
solving the minimum time problem for the system (I.1) that
does not rely on the reduction procedure at all. Instead, we
introduce a time rescaling technique. As the control vector
u increases in magnitude, we proportionally dilate the time
variable, so that in the limit, we obtain a new artificial time in
which the instantaneous jumps along the leaves generated by
f₁, ..., fₖ are no longer instantaneous. This leads to a natural
extension of the original system, ensuring the existence of
solutions to the minimum time problem.

The advantage of our approach is that it eliminates the
need for complex calculations to explicitly describe the
extended system. It remains on the original manifold M
and can be analyzed using standard tools of optimal control

Ivan Beschastnyi was supported by the French government through the
France 2030 investment plan managed by the National Research Agency
(ANR), as part of the Initiative of Excellence of Université Côte d'Azur
under reference number ANR-15-IDEX-01

<footnote>*Université Côte d'Azur, Inria, CNRS, LJAD, France.</footnote>

<footnote n="1">¹ivan.beschastnyi@inria.fr</footnote>

<footnote n="2">²lamberto.dell-elce@inria.fr</footnote>

<footnote n="3">³jean-baptiste.pomet@inria.fr</footnote>

<footnote n="4">⁴ludovic.sacchelli@inria.fr</footnote>

<footnote n="5">⁵david.tinoco@inria.fr</footnote>
theory. In fact, we will provide simplified proofs of the results from [7], [10] regarding the structure of optimal solutions and offer additional insights into minimal trajectories. Moreover, this technique can also be applied in certain non-autonomous cases.

Our method has one drawback: the optimal control problem for the extended system is highly degenerate due to a large symmetry group, making the minimum non-unique. On one hand, when we revert to the original time, the entire family of solutions collapses into a single trajectory, meaning that from a theoretical perspective, this degeneracy does not introduce additional difficulties. On the other hand, it poses a challenge in numerical computations, leading to instabilities when solving the extended problem. We address this issue in a specific application: quantum gate generation for a model of a qubit interacting with its environment.

The main result of this paper is the relaxation procedure, described in Section II, which provides a mathematically rigorous framework for handling impulsive controls. Additionally, we present several theoretical and numerical applications, including quantum gate generation for both closed and open two-level systems, as well as optimal state transfer for an open two-level system.

Various other approaches for quick control of open quantum systems and generation of gates have been proposed in the literature [13], [14]. For example, limiting the control amplitudes leads to a well-defined optimal control problem. For two-level systems specifically, the time optimal gate generation problem for isolated systems was studied in [5], while population transfer for open systems was studied in [3], [17]. Although this approach is physically more realistic, it can result in more complex control strategies compared to our approach. Another way is to look for the minimum among strategies that use jumps and zero controls (pure impulse controls). This was studied, for example, in [16].

The structure of the paper is the following. In Section II we discuss the change-of-time method that we use, give the description of the extended system, and apply Pontryagin's Maximum Principle (PMP). In Section III we find optimal solutions in the problem of gate generation for a qubit, in Section IV we discuss the structure of optimal trajectories for an open two-level system. In the final Section V we discuss numerical solutions to the problem of generating gates for a noisy qubit.

## II. GENERAL SETTING

### A. Formulation and relaxation of the problem

Consider a time-minimum problem for the control system of the form (I.1)

$$ \dot{q} = f_0(q) + \sum_{i=1}^{k} u_i f_i(q) $$

with unbounded controls and fixed endpoints. The minimum does not exist for $L^\infty$ control, for which Pontryagin's Maximum Principle (PMP) is usually formulated. Hence, we need to relax the problem to allow for a larger class of controls.

The relaxation procedure goes as follows. Consider a positive bounded from below continuous function $g: \mathbb{R}_+ \to \mathbb{R}_+$ with linear growth at infinity. Notable examples include

$$ g_1(x) = \max\{1, x\}, \quad g_2(x) = \sqrt{1 + x^2}. \quad (II.1) $$

Let $u \in L^\infty$ be a fixed control function and define

$$ \tilde{u}(t) = g(\|u\|), $$

where $\|u\|$ is the Euclidean norm of $u \in \mathbb{R}^k$. We can now rewrite (I.1) as

$$ \dot{q} = \tilde{u}(t) \left( \frac{1}{\tilde{u}(t)} f_0(q) + \sum_{i=1}^{k} \frac{u_i(t)}{\tilde{u}(t)} f_i(q) \right). \quad (II.2) $$

Next, we introduce a new time variable $\tau$. System (II.2) in the new time will be of the form

$$ \frac{dq}{d\tau} = \frac{dq}{dt} \frac{dt}{d\tau} = \frac{dt}{d\tau} \tilde{u}(t) \left( \frac{1}{\tilde{u}(t)} f_0(q) + \sum_{i=1}^{k} \frac{u_i(t)}{\tilde{u}(t)} f_i(q) \right). \quad (II.3) $$

Hence we choose

$$ \frac{dt}{d\tau} \tilde{u}(t(\tau)) \equiv 1 \Rightarrow \tau(t) = \int_{0}^{t} \tilde{u}(s)ds. $$

This is a normalization condition that slows the time when the norm of the control function is large. Note that for a fixed $u \in L^\infty$, the integrand is a locally integrable function and $\tau$ is well-defined.

Now we can consider the system

$$ \frac{dq}{d\tau} = v_0 f_0(q) + \sum_{i=1}^{k} v_i f_i(q), \quad (II.4) $$

essentially taking $v_0 = 1/\tilde{u}$ and $v_i = u_i/\tilde{u}$ in (II.3). We need to understand what values $v_0, v_i$ can take. It will be a continuous surface without boundary $\Gamma \subset \mathbb{R}_+^{k+1}$ where

$$ \mathbb{R}_+^{k+1} = \{(v_0, v) \in \mathbb{R}^{k+1} : v_0 \ge 0\}. $$

Note that $\Gamma \cap \{v_0 = 0\} = \emptyset$, because $v_0 = 0$ would mean that $u$ is infinite. However, by taking the closure $\bar{\Gamma}$ we add exactly the possible infinite controls, and we get

$$ \bar{\Gamma} \setminus \Gamma \subset \{v_0 = 0\}. $$

The set $\bar{\Gamma}$ is not convex and needs to be convexified to ensure existence of a minimum.

The original time variable is recovered by taking

$$ t(\tau) = \int_{0}^{\tau} v_{0}(s)ds. \quad (II.5) $$

When $v_0 \equiv 0$ on some closed interval of time $[\tau_1, \tau_2]$, in original time this will correspond to instantaneous jumps to the endpoint at $\tau_2$. On the intervals where $v_0 \ne 0$ the original control is recovered via

$$ u_i = v_i/v_0. $$
Summing everything up, we get the relaxed optimal control problem:

$$ 
\begin{gathered}
\frac{d q}{d \tau}=v_{0} f_{0}(q)+\sum_{i=1}^{k} v_{i} f_{i}(q), \quad(v_{0}, v) \in \operatorname{conv} \bar{\Gamma}, \\
q \in M, \quad q(0)=q_{0}, \quad q(T)=q_{T}, \quad T \text { is free, } \quad(II.6) \\
\int_{0}^{T} v_{0}(\tau) \mathrm{d} \tau \rightarrow \min .
\end{gathered}
$$

Let us consider some examples of possible sets $\operatorname{conv} \bar{\Gamma}$.

**Example 2.1:** Choose the function $g$ to be $g_1$ defined in (II.1). Controls that satisfy $\|u(t)\| \le 1$ will correspond to points inside the set

$$ \{(v_0, v) \in \mathbb{R}^{k+1} : v_0 = 1, \|v\| \le 1\}, $$

while controls that satisfy $\|u(t)\| \ge 1$ will correspond to points inside the set

$$ \{(v_0, v) \in \mathbb{R}^{k+1} : v_0 \in (0, 1], \|v\| = 1\}. $$

Then $\Gamma$ is the union of those two sets.

As discussed, we extend to non-bounded controls by taking the closure of $\Gamma$. It corresponds to adding points $(v_0, v)$, where $v_0 = 0$ and at least one $\|v\| = 1$, i.e. a sphere. Now when we take the convexification which gives us a cylinder

$$ \operatorname{conv} \bar{\Gamma}=\{\left(v_{0}, v\right) \in \mathbb{R}^{k+1} : 0 \le v_{0} \le 1, \|v\| \le 1\}. $$

**Example 2.2:** Choose $g$ to be $g_2$ from (II.1). Then the new control

$$ (v_0, v) = \left( \frac{1}{\sqrt{1 + \|u\|^2}}, \frac{u_1}{\sqrt{1 + \|u\|^2}}, \dots, \frac{u_k}{\sqrt{1 + \|u\|^2}} \right) $$

The set $\bar{\Gamma}$ is a hemisphere

$$ S_+^k = \{(v_0, v) \in \mathbb{R}^{k+1} : v_0^2 + \|v\|^2 = 1, v_0 \ge 0\}, $$

and its convexification is one half of a Euclidean ball:

$$ \operatorname{conv} \bar{\Gamma}=\{\left(v_{0}, v\right) \in \mathbb{R}^{k+1} : v_{0}^{2}+\|v\|^{2} \leq 1, v_{0} \geq 0\}. $$

To justify this procedure we state and prove two lemmas.

**Lemma 2.3:** The control system (II.4) with controls either in the set $\Gamma$ or $\operatorname{conv} \bar{\Gamma}$ have the same closures of reachable sets.

**Proof:** We only need to prove that the closure stays the same as we pass from $\Gamma$ to $\operatorname{conv} \bar{\Gamma}$. The passage from $\bar{\Gamma}$ to $\operatorname{conv} \bar{\Gamma}$ is the standard Fillipov's relaxation [1, Theorem 8.2]. Therefore, it remains to check that $\Gamma$ and $\bar{\Gamma}$ will produce the same closure of reachable sets. Fix a time $T$. Since simple functions are dense in $L^\infty[0, T]$, it is sufficient to prove that the closure does not change after adding constant controls $(0, v)$ to $\Gamma$. Consider a sequence of constants controls $(v_0^i, v^i)$ converging to a constant control $(0, v)$. Then, by the continuous dependence of solutions of ODEs on parameters, the corresponding endpoints $q(T, v_0^i(T), v^i(T))$ of the trajectories of (II.4) that all start at some $q_0 \in M$ will converge to $q(T, 0, v(T))$. Therefore, $q(T, 0, v(T))$ is in the

closure of the reachable set of the point $q_0$ at time less than $T$ of (II.4) with control functions taking values in $\Gamma$. ■

**Lemma 2.4:** Any two choices of functions $g_1, g_2$ will result in equivalent control problems.

**Proof:** Let $g_1, g_2: \mathbb{R}_+ \to \mathbb{R}_+$ be two positive bounded from below continuous functions with linear growth at infinity. Consider the corresponding controls $(v_0^i, v^i)$, $i = 1, 2$ as functions of $u$. Then we have

$$ v_0^1 = \frac{g_2(\|u\|)}{g_1(\|u\|)} v_0^2, \quad v^1 = \frac{g_2(\|u\|)}{g_1(\|u\|)} v^2. \quad (II.7) $$

We consider the system (II.4) with controls replaced by $(v_0^i, v^i)$. From (II.7) we see that trajectories with controls in $(v_0^i, v_1)$, $\Gamma_0^i$ and time $\tau_1$ will be just time rescaled trajectories of (II.4) with controls $(v_0^2, v_2) \in \Gamma_0^i$, where the time is

$$ \int_{0}^{\tau_1} \frac{g_2(\|u(s)\|)}{g_1(\|u(s)\|)} \mathrm{d}s. $$

But this is just the passage between time $\tau_1$ and time $\tau_2$. Moreover, by assumptions $g_2(\|u(s)\|)/g_1(\|u(s)\|)$ is continuous bounded and positive, which allows us to pass freely between the new times.

The extension procedure of taking the closure and convexification is now performed on two equivalent control systems. In addition, the functionals are preserved under the passage from $(v_0^1, v^1, \tau_1)$ to $(v_0^2, v^2, \tau_2)$. ■

**Remark 2.5:** This Lemma shows that it does not matter which function $g$ we choose for the relaxation. In this article we will use only $g_1$ from (II.1).

The relaxation procedure will guarantee that the optimal control problem has minimum, but it is nevertheless quite degenerate, because there is a big symmetry group present. For example, it is immediately evident that all trajectories with $v_0 \equiv 0$ are automatically optimal. To avoid this issue we prove the following lemma.

**Lemma 2.6:** Without any loss of generality, we can assume that the minimal control takes values in $\bar{\Gamma}$.

**Proof:** Let $(v_0(t), v_1(t))$ be a non-zero control. Then we can consider a new control $(r(t)v_0(t), r(t)v(t))$, where $r(t)$ is chosen in such a way that this new control lies on $\Gamma$. Consequently, $r(t) \ge 1$ for all times. Due to the linearity of relaxed control system will follow the same trajectory but arrive at the same point instead of time $T$ at time equal to

$$ F(T) = \int_{0}^{T} \frac{d\tau}{r(\tau)}. $$

But then, by a change of variables, the functional is equal to

$$ \int_{0}^{F(T)} r(\tau)v_{0}(\tau) \mathrm{d}\tau=\int_{0}^{T} \frac{r(\tau)}{r(\tau)} v_{0}(\tau) \mathrm{d}\tau=\int_{0}^{T} v_{0}(\tau) \mathrm{d}\tau. $$

In particular, if $(v_0(t), v_1(t))$ is minimal, then $(r(t)v_0(t), r(t)v(t))$ is minimal as well. ■

A less formal, but more intuitive way of understanding this lemma is the following. If $v_0 \ne 0$ on an arc, then in the original time any control rescaled by a positive function would correspond to the same control $u$. If $v_0 \equiv 0$ along an arc, then in the original time there is a jump to the end-point
of the trajectory in the new time. It does not matter how slow or fast we go in the new time, since in the old time the movement is instantaneous. Hence, we can assume that we go with maximum speed, which corresponds to $||v|| = 1$.

## B. Pontryagin's Maximum Principle

Let us write down the PMP for the optimal control problem (II.6). We will make an additional assumption specific to our applications.

**Assumption.** The vector fields $f_1, \dots, f_k$ commute with respect to the Lie bracket of vector fields

$$0 = [f_i, f_j] := f_i \circ f_j - f_j \circ f_i, \quad \forall i, j = 1, \dots, k. \quad (II.8)$$

The PMP Hamiltonian is given by

$$H_v = v_0 h_0 + \sum_{i=1}^{k} v_i h_i - \nu v_0,$$

where $h_i(\lambda) = \langle \lambda, X_i(\pi(\lambda)) \rangle$ for $\lambda \in T^*M$. Defining $\tilde{h}_0 = h_0 - \nu$ this gives us

$$H_v = v_0 \tilde{h}_0 + \sum_{i=1}^{k} v_i h_i.$$

For the relaxation we use function $g_1$ from (II.1). The maximized hamiltonian over $\text{conv } \bar{\Gamma}$ has the form

$$H = \max\{\tilde{h}_0, 0\} + \|h\|,$$

where $\|h\| = \sqrt{h_1^2 + \dots + h_k^2}$. For further calculations we will use the Lie-Poisson bracket on TM defined on the basis hamiltonians as

$$\{h_i, h_j\}(\lambda) = \langle \lambda, [f_i, f_j](\lambda) \rangle.$$

To shorten the notations we will denote

$$h_{ij} = \{h_i, h_j\}, h_{ijk} = \{h_i, \{h_j, h_k\}\}.$$

If $h_0 > 0$ or there at least one non-zero $h_i \neq 0$, then the extremals are not minimal, because $H \neq 0$. This gives us two possibilities:

1) $\tilde{h}_0 = 0$ and $h_i = 0$ for $i = 1, \dots, k$. Then none of the controls are determined. Differentiating all of them along the flow of $H_v$ gives due to (II.8) $h_{0i} = 0$. Differentiating once more gives

$$v_0 h_{00i} + \sum_{j=1}^{k} v_j h_{j0i} = 0. \quad (II.9)$$

which is a homogeneous linear system of $k$ equations with $k+1$ variables. Recall that by Lemma 2.6 we can always normalize the control to take values in $\bar{\Gamma}$. For all applications in this article $k=1$, and we get just one equation:

$$v_0 h_{001} + v_1 h_{101} = 0. \quad (II.10)$$

2) $\tilde{h}_0 < 0$ and $\|h\| = 0$. In this case $v_0 = 0$ and $v_1, \dots, v_k$ are to be determined. Differentiating $h_i \equiv 0$ gives $v_0 h_{0i} \equiv 0$. If $h_{0i} \neq 0$, then differentiating these equations gives exactly the singular control equation (II.9). If $v_0 \neq 0$, then we don't get any information. So $v$ can be any control inside the unit

disc. Note that $v_0 \equiv 0$ means that we move instantaneously along the integral curve of a combination of $f_i$. multiplying $v$ by a non-zero function will neither change the extremal trajectory nor the functional. Hence, we can assume without any loss of generality that $\|v\| = 1$. We call such arcs pulse arcs.

## III. GATE GENERATION FOR NOISELESS QUBITS

We consider the problem of gate generation for a two-level quantum system. This problem was previously considered in [7]. Here we give a simplified proof of the optimal synthesis.

The time minimization problem is of the form

$$i\dot{U} = \frac{1}{2}(\omega\sigma_z + u\kappa\sigma_x)U, \\ U(0) = \text{id}, U(T) = U_T.$$

where $U$ is a unitary matrix, $\sigma_x, \sigma_y, \sigma_z$ are Pauli matrices and $\omega, \kappa$ are two positive numbers. We can define

$$A_i = -\frac{i\sigma_i}{2},$$

where $i$ takes the values in the set $\{x, y, z\}$, which gives the following commutation relations

$$[A_x, A_y] = A_z, \quad [A_y, A_z] = A_x, \quad [A_z, A_x] = A_y.$$

which are generators of the Lie algebra $\mathfrak{su}(2)$. Denote $H_i(\lambda) := \langle \lambda, A_i\pi(\lambda) \rangle$. In order to write down the Hamiltonian system for the relaxed problem, we use the Poisson bracket induced by the Lie bracket on $\mathfrak{su}(2)$:

$$\{H_i, H_j\}(\lambda) := \langle \lambda, [A_i, A_j]\pi(\lambda) \rangle.$$

The Hamiltonian of the relaxed problem is given by

$$H = v_0\omega H_z + v_1\kappa H_x.$$

The Hamiltonian system takes the following form in the $H_x, H_y, H_z$ basis:

$$\begin{aligned} \dot{H}_x &= -v_0\omega H_y, \\ \dot{H}_y &= v_0\omega H_x - v_1\kappa H_z, \\ \dot{H}_z &= v_1\kappa H_y. \end{aligned}$$

**Lemma 3.1:** Every optimal trajectory must be of the form pulse-singular-pulse.

*Proof:* Consider a pulse arc. Integral curves $\sigma_x$ are closed circles. Therefore, for a pulse arc we have $v_0 = 0$, and can choose $v_1 = +1$. This implies that the Hamiltonian system becomes

$$\begin{aligned} \dot{H}_x &= 0, \\ \dot{H}_y &= -\kappa H_z, \\ \dot{H}_z &= \kappa H_y. \end{aligned}$$

which is just a constant-speed rotation in the $y, z$ plane. Hence, we introduce polar coordinates

$$\begin{aligned} H_y &= \rho \cos \varphi, \\ H_z &= \rho \sin \varphi. \end{aligned}$$
The Hamiltonian along this trajectory is equal to $\kappa H_x$. Since it must be zero we find that $H_x \equiv 0$.

Now, assume that a pulse arc is connected to a singular arc. Then at the junction time we must have

$$ 
\begin{aligned}
0 &= \tilde{h}_0(t) = \omega\rho\sin\varphi - \nu, \\
0 &= h_{01}(t) = -\omega\kappa\rho\cos\varphi.
\end{aligned}
 $$

Since $\omega\kappa \neq 0$, we must have from the second equation that either $\rho = 0$ or $\varphi \in \{\pi/2, 3\pi/2\}$. If $\rho(t) = 0$, then $\rho \equiv 0$ and as consequence $\nu = 0$. But then we get a trivial extremal which is not optimal by PMP. Hence, $\rho \neq 0$ and $\varphi \in \{\pi/2, 3\pi/2\}$. This implies that $\pm\omega\rho = -\nu$, but since $\nu$ is non-positive, this leaves us with only possible choice of $\varphi$ based on the sign of $\omega$. Therefore, there is a unique $\varphi$ allowing for a junction between a pulse and a singular arc. This means that in the relaxed problem we can have a singular arc, a full-circle rotation, singular arc and so on. But all of them in the original time correspond to just moving along the singular trajectory, since we are jumping from the junction point to itself. Therefore, no non-trivial pulse arc can join to singular curves.

Now we consider the singular arcs. We have

$$ h_{001} = -\omega^2 \kappa H_x, \quad h_{101} = \omega \kappa^2 H_z. $$

Along the singular trajectory we have $h_1 \equiv 0$, which in this notation reads as $H_x \equiv 0$. Therefore,

$$ v_0 h_{001} + v_1 h_{101} = v_1 \omega \kappa^2 H_z = 0. $$

If $H_z \equiv 0$, then as a consequence $\nu = 0$ and in combination with $0 \equiv H_{01} \simeq H_y$ this would give a trivial extremal. Hence, we can only have $v_1 \equiv 0$. In addition, we can assume that $v_0 \neq 0$, because else we don't move. Passing to the original time we see that it just means $u \equiv 0$. Hence, we have the following result on the structure of optimal trajectories:

**Proposition 3.1:** All optimal trajectories must be of the form: pulse-wait-pulse.

This proposition allows us for a complete optimal synthesis. Indeed, jump-wait-jump strategy will give us a unitary operator of the form

$$ U = e^{-\frac{it_3 \kappa \sigma_x}{2}} e^{-\frac{it_2 \omega \sigma_z}{2}} e^{-\frac{it_1 \kappa \sigma_x}{2}}, \quad (\text{III.1}) $$

which is just the usual parametrization of special unitary matrices using Euler angles. Hence, we can solve the above equations in the range $\kappa t_1, \kappa t_3 \in [0, 2\pi)$ and $\omega t_2 \in [0, \pi)$, which shows that original control time is bounded by $\pi/\omega$.

## IV. COHERENT CONTROL OF A SINGLE PARTICLE

### A. Formulation of the problem

We consider the model of an open two-level quantum system studied in [10], which models interaction of a qubit with zero-temperature environment. The corresponding Lindblad equation reads as

$$ i\dot{\rho} = \frac{1}{2}[\omega\sigma_z + \kappa u\sigma_x, \rho] + \gamma \left( \sigma^- \rho\sigma^+ - \frac{1}{2}\{\sigma^+\sigma^-, \rho\} \right), \quad (\text{IV.1}) $$

where $\rho(t)$ is density matrix, $\gamma \ge 0$ is a parameter that describes the intensity of interaction between the qubit and the environment, $\{\cdot, \cdot\}$ is the anti-commutator of matrices defined as

$$ \{A, B\} = AB + BA $$

and $\sigma^{\pm} = (\sigma_x \pm \sigma_y)/2$.

Since density matrices are Hermitian, we can write them down in the basis of Pauli matrices

$$ \rho = \frac{1}{2}(\tau \text{id} + x\sigma_x + y\sigma_y + z\sigma_z) $$

Plugging this expression into the Lindblad equation gives us

$$ \frac{d}{dt} \begin{pmatrix} x \\ y \\ z \\ \tau \end{pmatrix} = \begin{pmatrix} -\omega y - \gamma x/2 \\ \omega x - \gamma y/2 \\ \gamma \tau - \gamma z \\ 0 \end{pmatrix} + u \begin{pmatrix} 0 \\ -\kappa z \\ \kappa y \\ 0 \end{pmatrix}. \quad (\text{IV.2}) $$

We further simplify this system by noting that the trace variable $\tau$ should be equal to one for density matrix, and also that one can set $\omega = 1$ by appropriately choosing the time scale and $\kappa = 1$ by absorbing into a new control $\kappa u$. This gives us the final control system:

$$ \frac{d}{dt} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -y - \gamma x/2 \\ x - \gamma y/2 \\ \gamma - \gamma z \end{pmatrix} + u \begin{pmatrix} 0 \\ -z \\ y \end{pmatrix}. \quad (\text{IV.3}) $$

We are interested in solving the time minimum problem for (IV.3) with fixed end-points $\rho_0, \rho_T$. However, the system (IV.1) is not controllable due to decoherence [11]. Therefore, to guarantee the existence of the minimum, we must minimize a functional of the form

$$ (1 - \alpha)d(\rho(T), \rho_T) + T \to \min. $$

### B. Structure of optimal trajectories of the relaxed problem

We consider the relaxed problem corresponding to (IV.3). The main result is the following theorem:

**Theorem 4.1:** Any extremal must be of the form pulse-singular-pulse arc. More precisely, two singular arcs can not be connected by a pulse arc.

Applying the PMP to this concrete case using the canonical Poisson bracket on $T^*\mathbb{R}^3$ gives us the following Hamiltonian system:

$$ 
\begin{aligned}
\dot{x} &= -v_0 \left(y + \frac{\gamma x}{2}\right) & \dot{p}_x &= v_0 \left(\frac{\gamma}{2} p_x - p_y\right) \\
\dot{y} &= v_0 \left(x - \frac{\gamma y}{2}\right) - v_1 z & \dot{p}_y &= v_0 \left(p_x + p_y \frac{\gamma}{2}\right) - v_1 p_z \\
\dot{z} &= v_0 \gamma (1 - z) + v_1 y & \dot{p}_z &= v_0 \gamma p_z + v_1 p_y
\end{aligned}
 $$

At a pulse arc we have $v_0 \equiv 0$ and $h_1 = 0$. This reduces to the equations $\dot{x} = 0, \dot{p}_x = 0$ and a system

$$ 
\begin{aligned}
\dot{y} &= -v_1 z \\
\dot{z} &= v_1 y
\end{aligned}
 \qquad \begin{aligned}
\dot{p}_y &= -v_1 p_z \\
\dot{p}_z &= v_1 p_y
\end{aligned}
 $$

Hence, we see that solutions are simultaneous rotations in the $(y, z)$ and $(p_y, p_z)$ planes with the same speed. In addition, $0 \equiv h_1 = -p_y z + p_z y$ implies that $(y(t), z(t))$
and $(p_y(t), p_z(t))$ are linearly dependent for all times. We can introduce polar coordinates, then the solutions are of the form

$$ 
\begin{aligned}
y &= \rho_1 \cos \phi(t) & p_y &= \sigma \rho_2 \cos \phi(t) \\
z &= \rho_1 \cos \phi(t) & p_z &= \sigma \rho_2 \cos \phi(t)
\end{aligned}
$$

where $\sigma = \pm 1$ and $\rho_i$ are non-negative.

If a pulse arc is connected to a singular set, then at the junction by continuity we must have $h_0(t) - \nu = h_{01} = 0$. Let us write the two sides explicitly. First we have that both of them are trigonometric polynomials. Moreover, if $\tilde{h}_0(t) = P(\phi(t))$, then $h_{10}(t) = P'(\phi(t))$. Indeed, $f_1 = \partial_\phi$ in cylindrical coordinates. Thus we get

$$ P(\phi) = a_1 \cos \phi + a_2 \sin \phi + b \cos 2\phi + c \quad (IV.4) $$

where

$$ 
\begin{aligned}
a_1 &= x\sigma\rho_2 - p_x\rho_1, & a_2 &= \gamma\rho_2\sigma, \\
b &= \gamma\rho_1\rho_2\sigma/4, & c &= -\nu - \gamma(2xp_x + 3\rho_1\rho_2\sigma)/4.
\end{aligned}
$$

Assume by contradiction that two singular arcs are connected by a pulse arc. Let the switching times be $t_1, t_2$. Then the polynomial (IV.4) must have two distinct modulo $2\pi$ double roots at $\phi(t_1)$ and $\phi(t_2)$. We claim that this impossible for our system, which is a consequence of the following lemma.

**Lemma 4.2:** If a polynomial $P$ of the form (IV.4) has two distinct double zeroes modulo $2\pi$, then one of the following two conditions must hold:

1) $b \neq 0$, $|a_1| < 4|b|$, $a_2 = 0$, $c = b + \frac{a_1^2}{8b}$ and $\phi_1, \phi_2$ are two distinct roots of $\cos \phi = -a_1/4b$;

2) $b \neq 0$, $a_1 = 0$, $|a_2| < 4|b|$, $c = -b - \frac{a_1^2}{8b}$ and $\phi_1, \phi_2$ are two distinct roots of $\sin \phi = a_2/4b$;

*Proof:* We have $P'(\phi) = -a_1 \sin \phi + a_2 \cos \phi - 2b \sin 2\phi$. Then the system

$$ P(\phi_1) = P(\phi_2) = P'(\phi_1) = P'(\phi_2) = 0 $$

can be rewritten in the form

$$ 
\begin{pmatrix} \cos \phi_1 & \sin \phi_1 & \cos 2\phi_1 & 1 \\ -\sin \phi_1 & \cos \phi_1 & -2 \sin 2\phi_1 & 0 \\ \cos \phi_2 & \sin \phi_2 & \cos 2\phi_2 & 1 \\ -\sin \phi_2 & \cos \phi_2 & -2 \sin 2\phi_2 & 0 \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \\ b \\ c \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.
$$

In order for this system to be solve, the determinant of the square matrix must be zero, which after many simplifications is equal to:

$$ -8 \sin^4 \left( \frac{\phi_1 - \phi_2}{2} \right) \sin(\phi_1 + \phi_2) = 0. $$

The first function is zero if and only if $\phi_1 - \phi_2$ is a multiple of $2\pi$, while the other sinus vanishes when $\phi_1 = -\phi_2$ or $\phi_1 = \pi - \phi_2$. Plugging this into the systems allows us to determine the roots and the coefficient $c$. Note that the inequalities are strict, because in the case of equalities the two roots are separated by a multiple of $2\pi$. The same happens if $b = 0$.

Now to finish the proof we notice that $b = \frac{1}{4}\rho_1 a_2$. If $a_2 = 0$, then $b = 0$, which gives a contradiction. If $a_2 \neq 0$, then we must have

$$ \left| \frac{a_2}{4\beta} \right| = \frac{1}{\rho_1} < 1. $$

But this is impossible, because we are inside a ball and hence $\rho_1 \le 1$.

### C. Singular arcs and feedback control

Now that we have the structure of optimal solutions we can even try to construct an optimal synthesis. For this let us analyze the singular arcs. First, let us write down the vector fields and possible dependencies between them:

$$ 
\begin{aligned}
f_0 &= (-y - \gamma x/2)\partial_x + (x - \gamma y/2)\partial_y + \gamma(1 - z)\partial_z, \\
f_1 &= -z\partial_y + y\partial_z, \\
f_{01} &= -z\partial_x + \frac{\gamma}{2}(z - 2)\partial_y + \left(\frac{\gamma}{2}y + x\right)\partial_z, \\
f_{001} &= \gamma(z - 2)\partial_x + \left(1 - \frac{\gamma^2}{4}\right)z\partial_y \\
&\quad + \left(\gamma x + \left(\frac{\gamma^2}{4} - 1\right)y\right)\partial_z, \\
f_{101} &= -y\partial_x + (x + \gamma y)\partial_y + \gamma(1 - z)\partial_z.
\end{aligned}
$$

We have

$$ 
\begin{aligned}
4 \det(f_0, f_1, f_{01}) &= S(x, y, z) \\
&= y^2 z + \gamma(z - 1)(y^2 + z^2) \\
&\quad + xy \left( z \left( \frac{\gamma^2}{4} + \frac{\gamma}{2} - 1 \right) - \frac{\gamma^2}{2} \right).
\end{aligned}
$$

In particular, the surface $\{S = 0\}$ contains the horizontal line $L = \{y = z = 0\}$ and the north pole $N = \{z = 1, x = y = 0\}$. Note also that on $\{S = 0\} \setminus L$ the vector fields $f_1, f_{01}$ are linearly independent.

Next we have a linear dependence relation

$$ c_1 f_1 + c_{01} f_{01} + c_{001} f_{001} + c_{101} f_{101} = 0, \quad (IV.5) $$

where $c_I$ are polynomials and all vanish at the north pole. One can take

$$ 
\begin{aligned}
c_{001} &= y^2 + z^2(1 - z), \\
c_{101} &= (1 - z)(2xz + \gamma y(z - 2)).
\end{aligned}
$$

Here it is important note that outside $L \cup \{N\}$ we have $c_{001} \neq 0$ and even more $f_1, f_{01}, f_{101}$ form a basis because

$$ \det(f_1, f_{01}, f_{101}) = -16(y^2 + z^2(1 - z)). $$

So let us assume that we have a singular arc that passes through point $q \in \mathbb{R}^3 \setminus (L \cup \{N\})$. From (II.10) and relation (IV.5) it follows that along a singular arc that does not cross $L \cup \{N\}$ we have

$$ 
\begin{aligned}
v_0 h_{001} + v_1 h_{101} &= 0, \\
c_{001} h_{001} + c_{101} h_{101} &= 0.
\end{aligned}
$$

And therefore

$$ v_1 c_{001} - v_0 c_{101} = 0, $$
which defines $v_0, v_1$ up to a multiplicative constant. By Lemma 2.6, we can assume that our control takes values in $\bar{\Gamma}$. This gives us a control in a feed-back form

$$ (v_0, v_1) = r(q)(c_{001}(q), c_{101}(q)), $$

where

$$ r(q) = \begin{cases} 1/c_{001}(q), & |c_{001}(q)| \ge |c_{101}(q)|, \\ 1/|c_{101}(q)|, & |c_{001}(q)| \le |c_{101}(q)| \end{cases} $$

## V. COHERENT CONTROL GATE GENERATION

### A. Formulation of the problem

We now consider the problem of gate generation for the system described by (IV.1). When interpreted as an equation over the space of Hermitian matrices, this becomes a linear system, explicitly given by (IV.2). Equivalently, we define the matrices:

$$ M_0 = \begin{pmatrix} -\frac{\gamma}{2} & -\omega & 0 & 0 \\ \omega & -\frac{\gamma}{2} & 0 & 0 \\ 0 & 0 & -\gamma & \gamma \\ 0 & 0 & 0 & 0 \end{pmatrix}, M_1 = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -\kappa & 0 \\ 0 & \kappa & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}. $$

Abusing notation slightly, we write the state as a vector $\rho = (x, y, z, \tau)^T$, which allows us to express the system in compact form as:

$$ \dot{\rho} = (M_0 + uM_1)\rho. $$

Since we are interested in gate generation, we need to control all possible density matrices. This means that we need to pass to the equation for the fundamental matrix

$$ \dot{P} = (M_0 + uM_1)P, \quad P(0) = \text{id} $$

where $P$ is a $4 \times 4$ real matrix. Similar to the population transfer problem, this system is not controllable due to decoherence. Therefore to make it well-posed, we need to minimize

$$ (1 - \alpha)d(P(T), P_U)^2 + \alpha T \to \min $$

where $\alpha$ is a real parameter and $P_U = Ad_U$ is the operator that corresponds to the gate $U$ that we want to generate.

For example, for the Hadamard gate $H$, we have

$$ Ad_H = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

By applying our small-time variable change technique, we reformulate the dynamics as a control problem of the form:

$$ \begin{gathered} \dot{P} = (v_0 M_0 + v_1 M_1)P, \quad P(0) = \text{id}, \\ 0 \le v_0 \le 1, \quad |v_1| \le 1, \\ (1 - \alpha)d(P(T), P_U)^2 + \alpha \int_0^T v_0(s)ds \to \min \end{gathered} \quad (V.1) $$

The problem is now well-posed and amenable to numerical optimization. In the next section, we describe how this formulation is implemented computationally and present numerical results for the Hadamard gate.

### B. Numerical Results

In this section, we present some results obtained using the OptimalControl.jl package, which is the core package of the control-toolbox.org ecosystem [4]. This ecosystem includes several Julia packages designed for mathematical control and its applications. Its main objective is to provide tools for modeling and solving optimal control problems involving ordinary differential equations, using both direct and indirect methods.

To solve the problem numerically, we formulate the problem (V.1) in Julia code. We must choose a quantum gate $U$ to be generated. We use the Euler angle representation (III.1) and construct the corresponding operator $Ad_U$. In the case with decoherence, i.e. when $\gamma > 0$, we set the parameter $\alpha$ to 0. This is because prolonged exposure to decoherence will steer us away from the perfect gates and hence it is enough to minimize the distance between $P(T)$ and $P_U$. We provide an initial guess to the solver to assist with convergence by using the already known optimal solution for the decoherence-free case $\gamma = 0$ given by the pulse-wait-pulse pattern from Proposition 3.1:

$$ \left\{ \begin{array}{ll} v_0 = 0, & v_1 = 1, \quad \text{if } 0 \le \tau < t_1, \\ v_0 = 1, & v_1 = 0, \quad \text{if } t_1 \le \tau < t_1 + t_2, \\ v_0 = 0, & v_1 = 1, \quad \text{if } t_1 + t_2 \le \tau \le t_1 + t_2 + t_3. \end{array} \right. \quad (V.2) $$

In addition, for numerical stability we impose the additional constraint

$$ v_0 + |v_1| \ge 1. $$

In other words, we avoid situations where the control input is weak.

Consider an example of the Hadamard gate generation $Ad_H$ for parameters $\omega = \kappa = 1$ and the coherence constant $\gamma = 0.1$. The resulting numerically found controls are depicted in 1.

Note that when $v_0$ is equal to zero over an interval $[\tau_1, \tau_2]$, this implies that, in the original time $t$, the system undergoes an instantaneous jump to the point corresponding to $\tau_2$. In such cases, the control $v_1$ must be maximal in order to perform the rotation as quickly as possible. By observing Figure 1, this phenomenon can be clearly seen in two instances: one at the beginning and another at the end of the process. In other words, the trajectory follows a pattern close to pulse-wait-pulse: a jump occurs when $v_0 = 0$ and $v_1 = 1$, while the system follows closely its drift when $v_0 = 1$ and $v_1 = 0$. Another way to visualize the gate evolution is shown in Figure 2. The three trajectories inside the Bloch ball correspond to the evolution of the operators $(id + \sigma_x)/2$ (red), $(id + \sigma_y)/2$ (green), and $(id + \sigma_z)/2$ (blue), projected onto the Bloch sphere. The bullet markers indicate the initial points of each trajectory. The purity decay of individual trajectories is shown in Figure 3 with the same color scheme. The red and blue graphs overlap.

To interpret the evolution in real time $t$, rather than the artificial time variable $\tau$ used in the relaxed problem formulation, we compute the integral (II.5). Through numerical
<figure><img src="image_2.png" /><figcaption>Fig. 1: $\tau$ time evolution of the controls $v_0$ and $v_1$ to generate the gate $Ad_H$.</figcaption></figure>

<figure><img src="image_3.png" /><figcaption>Fig. 2: Optimal gate generation of $Ad_H$. The three lines indicate the corresponding evolution of $(id+\sigma_x)/2$ (red), $(id+\sigma_y)/2$ (green), and $(id+\sigma_z)/2$ (blue).</figcaption></figure>

integration, we find that the final real time is approximately 1.55.

**Acknowledgments:** we thank J.-B. Caillau for many helpful discussions.

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