A Reduced Complexity Min-Plus Solution Method to the Optimal Control of Closed Quantum Systems

Abstract

The process of obtaining solutions to optimal control problems via mesh based techniques suffers from the well known curse of dimensionality. This issue is especially severe in quantum systems whose dimensions grow exponentially with the number of interacting elements (qubits) that they contain. In this article we develop a min-plus curse-of-dimensionality-free framework suitable to a new class of problems that arise in the control of certain quantum systems. This method yields a much more manageable complexity growth that is related to the cardinality of the control set. The growth is attenuated through $\max$-plus projection at each propagation step. The method’s efficacy is demonstrated by obtaining an approximate solution to a previously intractable problem on a two qubit system.

Appendices

Appendix 1: Proofs of Proposition 3.1 and Lemma 3.10

Proof

[Proposition 3.1] We prove two inequalities which when combined yield the result. Let $T\ge T_0$. From the definition of $C_{0}(\cdot )$ in Eq. (2.5) and Eq. (3.5), we have that for all ${U_0}\in \mathbf {G}$ and all $\delta >0$, there exists $\hat{v}=\hat{v}(U_0,\delta ) \in \mathcal {V}^e_g$ such that

$$\begin{aligned} U(T;v,0,{U_0}) = I \quad \text{ and } \quad \int \limits _0^T{\sqrt{{\hat{v}}^{T} R\,{\hat{v}}}\, dt} < C_{0}({U_0}) + \delta . \end{aligned}$$

Combining this with the definition of $C^{\epsilon }_0$, one sees that for all $\epsilon >0$, $C^{\epsilon }_0({U_0})< C_{0}({U_0}) + \delta$. Since this holds for all $\delta >0$, we have that

$$\begin{aligned} C^{\epsilon }_0({U_0}) \le C_{0}({U_0})\quad \forall \,\epsilon >0. \end{aligned}$$

(9.1)

To prove the reverse inequality to this, we proceed as follows. Let $\epsilon \in (0,1]$. From the definition of $C^{\epsilon }_0$, $\forall {U_0}\in \mathbf {G}, \forall \delta >0, \exists \hat{v}=\hat{v}(U_0,\delta ) \in \mathcal {V}_{g}$, such that

$$\begin{aligned} C^{\epsilon }_0({U_0})+ \delta \ge \int \limits _0^{T}{\sqrt{{\hat{v}(t)}^{T} R\,{\hat{v}(t)}}\,\,dt} + \frac{1}{\epsilon } \hat{\varphi }\Big (U(T;\hat{v},0,{U_0})\Big ). \end{aligned}$$

(9.2)

From the assumption that $\hat{\varphi }(\cdot ) \ge C_{0}(\cdot )$ we have that $\forall \epsilon \in (0,1]$,

$$\begin{aligned}&\int \limits _0^{T}{\sqrt{{\hat{v}(t)}^{T} R\,{\hat{v}(t)}}\,\,dt} + \frac{1}{\epsilon } \hat{\varphi }\Big (U(T;\hat{v},0,{U_0})\Big )\\&\quad \ge \int \limits _0^{T}{\sqrt{{\hat{v}(t)}^{T} R\,{\hat{v}(t)}}\,\,dt} + C_{0}\Big (U(T;\hat{v},0,{U_0})\Big ). \end{aligned}$$

Using the dynamic programming equation for $C_{0}$ in the right hand side of the equation above it follows that

$$\begin{aligned} \int \limits _0^{T}{\sqrt{{\hat{v}(t)}^{T} R\,{\hat{v}(t)}}\,\,dt} + \frac{1}{\epsilon } \hat{\varphi }\Big (U(T;\hat{v},0,{U_0})\Big ) \ge C_{0}(U_{0}). \end{aligned}$$

(9.3)

From Eqs. (9.2) and (9.3) we see that $\forall {U_0}\in \mathbf {G}, \forall \delta >0$, $\forall \epsilon \in (0,1]$, one has $C^{\epsilon }_0({U_0})+ \delta \ge C_{0}(U_{0})$. Since this holds for all $\delta >0$ we have that

$$\begin{aligned} C^{\epsilon }_0({U_0}) \ge C_{0}({U_0})\quad \forall \,\epsilon \in (0,1]. \end{aligned}$$

(9.4)

Combining (9.1) and (9.4), one has the desired result. $\square$

Proof

[Theorem 3.10 for the dynamic programming equation for the relaxation of the optimal cost function]

Let ${U_0}\in \mathbf {G}$ and $\tau \in [0, T-s]$. Consider any particular control $\alpha (t), t \in [s, s+\tau ]$. By the definition of $C^{\epsilon }_r$ in Eqs. (3.1) and (3.2), $\forall {U_0}\in \mathbf {G}, \forall \delta >0, \exists$ control $\beta (\cdot )$ such that

$$\begin{aligned} J^{\epsilon }_{s+\tau } (U(s+\tau ; \alpha ,s,{U_0}), \beta ) \le C^\epsilon _{s+\tau } (U(s+\tau ;\alpha ,s,{U_0})) + \delta . \end{aligned}$$

(9.5)

Constructing a composite control

$$\begin{aligned} u(t) = \left\{ \begin{array}{lr} \alpha (t) &{}t \in [s, s+\tau ) \\ \beta (t) &{} t \in [s+\tau , T) \end{array} \right. , \end{aligned}$$

we have from the dynamic programming principle that

$$\begin{aligned} C^{\epsilon }_s({U_0}) \le \int \limits _s^{s+\tau } \sqrt{{\alpha (t)}^{T} R\,{\alpha (t)}}\, dt + J^{\epsilon }_{s+\tau }(U(s+\tau ;\alpha ,s,{U_0}),\beta ). \end{aligned}$$

Using Eq. (9.5) we obtain

$$\begin{aligned} C^{\epsilon }_s({U_0})&\le \int \limits _s^{s+\tau } \sqrt{{\alpha (t)}^{T} R\,{\alpha (t)}}\, dt + C^\epsilon _s (U(s+\tau ;\alpha ,s,{U_0})) + \delta . \end{aligned}$$

(9.6)

Hence taking the infimum among possible controls $\alpha (\cdot )$ and then letting $\delta \downarrow 0$,

$$\begin{aligned} C^{\epsilon }_s({U_0})&\le \mathop {\inf }\limits _{\alpha \in \mathcal {V}^{e}_{g}} \left\{ \int \limits _s^{s+\tau } \sqrt{{\alpha (t)}^{T} R\,{\alpha (t)}}\, dt + C^\epsilon _s (U(s+\tau ;\alpha ,s,{U_0}))\right\} . \end{aligned}$$

(9.7)

Now to prove the other side of the inequality: from the definition of $C^{\epsilon }_s (\cdot )$ in Eqs. (3.1) and (3.2), it follows that $\forall \delta >0$, $\exists \bar{v}$ s.t

$$\begin{aligned} C^{\epsilon }_s({U_0}) + \delta \ge J^\epsilon _s({U_0},\bar{v}). \end{aligned}$$

(9.8)

The control $\bar{v}$ is partitioned as

$$\begin{aligned} \bar{v}(t) = \left\{ \begin{array}{lr} \alpha (t) &{}t \in [s, s+\tau ) \\ \beta (t) &{} t \in [s+\tau , T) \end{array} \right. . \end{aligned}$$

Hence from Eq. (9.8) and the definition of $J^{\epsilon }_s(\cdot )$ we have

$$\begin{aligned} C^{\epsilon }_s({U_0}) + \delta&\ge \int \limits _s^{s+\tau } {\sqrt{{\alpha (t)}^{T} R\,{\alpha (t)}}\, dt}+ \int \limits _{s+\tau }^T{\sqrt{{\beta (t)}^{T} R\,{\beta (t)}}\,}\nonumber \\&+\,\frac{1}{\epsilon } \varphi \Big (U\big (T;\beta ,s+\tau ,U(s+\tau ;\alpha ,s,{U_0})\big )\Big ). \end{aligned}$$

(9.9)

Now, the definition of $C^{\epsilon }_s(\cdot )$ implies that

$$\begin{aligned} C^{\epsilon }_{s+\tau }(U(\tau ;\alpha ,s,{U_0}))&\le \int \limits _{s+\tau }^T {\sqrt{{\beta (t)}^{T} R\,{\beta (t)}}\, dt}\nonumber \\&+\, \frac{1}{\epsilon } \varphi \Big (U\big (T;\beta , s+\tau ,U(s+\tau ;\alpha ,s,{U_0})\big )\Big ).\qquad \end{aligned}$$

(9.10)

Therefore from Eqs. (9.9) and (9.10) we have

$$\begin{aligned} C^{\epsilon }_s({U_0}) + \delta \ge \int \limits _s^{s+\tau }{\sqrt{{\alpha (t)}^{T} R\,{\alpha (t)}}\, dt} + C^{\epsilon }_{s+\tau }\big (U(\tau ;\alpha ,s,{U_0})\big ). \end{aligned}$$

Taking the infimum over all $\alpha (\cdot )$ and then letting $\delta \rightarrow 0$

$$\begin{aligned} C^{\epsilon }_s({U_0}) \ge \mathop {\inf }\limits _{\alpha \in \mathcal {V}^{e}_{g}} \left\{ \int \limits _s^{s+\tau }{\sqrt{{\alpha (t)}^{T} R\,{\alpha (t)}}\, dt} + C^{\epsilon }_{s+\tau }\big (U(\tau ;\alpha ,s,{U_0}) \big ) \right\} . \end{aligned}$$

(9.11)

Thus from Eqs. (9.7) and (9.11) the result follows. $\square$

Appendix 2: Proof of Upper Bound on the Time to Reach the Identity

In this section we prove various results that lead up to the proof of Theorem 3.2. The intuition behind this consists of two main steps: (1) determining how to generate a desired control action via a sequence of propagation steps using the basis elements of the Lie algebra, and (2) obtaining the sequence of propagations from the available control set that would produce the desired propagation steps.

We also note that the following proof is for general $\mathbf {G}\doteq SU(2^n)$ for $n\in \mathbf{N}$, and in particular, ${\mathfrak {g}}\doteq \mathfrak {su}(2^n)$. We begin by introducing some notation: Let $M={\tilde{N}}^2-1=2^{2n}-1$, the dimension of ${\mathfrak {g}}$. Let $\mathcal{M}=]1,M[$, where we remind the reader that $]a,b[$ is used throughout to denote $\{a,a+1,a+2,\ldots , b\}$ for integer pairs $a<b$. Let $\overline{\mathcal{H}}\doteq \{H_m\,\vert \, m\in \mathcal{M}\}$ be a set of orthonormal basis vectors for ${\mathfrak {g}}$. For some $M_0\in ]1,M[$, $\mathcal{M}^0=]1,M_0[$ and

$$\begin{aligned} \mathcal{H}^0=\{H_m\,\vert \, m\in \mathcal{M}^0\}, \end{aligned}$$

(10.1)

where $\mathcal{H}^0$ is the set of basis directions for the available control. Let $\mathcal{M}^1=]M_0+1,M[$ and

$$\begin{aligned} \mathcal{H}^1=\{H_m\,\vert \, m\not \in \mathcal{M}^0\}=\overline{\mathcal{H}}\setminus \mathcal{H}^0. \end{aligned}$$

We suppose that for each $k\in \mathcal{M}^1$, there exist $m^1(k),m^2(k)\in \mathcal{M}^0$ and $\alpha _k\in \mathbb {R}$ such that

$$\begin{aligned} H_k&=\alpha _k[H_{m^1(k)},H_{m^2(k)}]\end{aligned}$$

(10.2)

$$\begin{aligned}&=\alpha _k\bigl (H_{m^1(k)}H_{m^2(k)}-H_{m^2(k)}H_{m^1(k)}\bigr ). \end{aligned}$$

(10.3)

As the $\alpha _k$ only modify the result by a fixed, multiplicative constant, for simplicity of the already-technical proof, we suppose $\alpha _k=1$ for all $k\in \mathcal{M}^1$. Similarly, for simplicity, we are also assuming that the $H_k$ are orthogonal.

Let ${\overline{D}_\beta }<\infty$ represent the maximum magnitude of the control and $\overline{B}_{\overline{D}_\beta }= \overline{B}_{\overline{D}_\beta }(0)\subset I\!R^{M_0}$, denote the closure of the ball of radius ${\overline{D}_\beta }$ centered at the origin. Let

$$\begin{aligned} \mathcal{V}^e_{0,{\overline{D}_\beta }}\doteq \bigl \{v:\,[0,\infty )\rightarrow I\!R^{M_0}\,\bigl \vert \, \Vert v(t)\Vert \in \{0,{\overline{D}_\beta }\}\,\forall t\in [0,\infty ), \text{ measurable }\bigr \} \end{aligned}$$

The controlled dynamics are

$$\begin{aligned} \dot{U}_t=\left[ \sum _{m\in \mathcal{M}_0}[v_t]_m H_m \right] U_t. \end{aligned}$$

(10.4)

We remind the reader that we use the norm on $\mathbf {G}$ given by $\Vert U\Vert =\Bigl [\sum _{i,j}|U_{i,j}|^2\Bigr ]^{1/2} = \Bigl \{\text{ tr }[UU^\dag ]\Bigr \}^{1/2}$.

Lemma 10.1

There exists $\mathcal{D}_{\tilde{N}}<\infty$ such that $\Vert \log (U)\Vert \le \mathcal{D}_{\tilde{N}}\Vert U-I\Vert$ for all $U\in \mathbf {G}$.

Proof

This is a slight variant of a standard calculation. For $U\in \mathbf {G}$ satisfying $\Vert U-I\Vert \le {\frac{1}{2}}$, one has (c.f., [24])

$$\begin{aligned} \Vert \log (U)-(U-I)\Vert \le \Vert U-I\Vert ^2\sum _{k=2}^\infty \frac{\left( \frac{1}{2}\right) ^{k-2}}{k}. \end{aligned}$$

which implies

$$\begin{aligned} \Vert \log (U)\Vert \le \Vert U-I\Vert +\Vert U-I\Vert ^2\le \frac{3}{2}\Vert U-I\Vert . \end{aligned}$$

(10.5)

More generally, there exists $\overline{\mathcal{D}}_{\tilde{N}}$ such that $\Vert \log (U)\Vert \le \overline{\mathcal{D}}_{\tilde{N}}$ for all $U\in \mathbf {G}$. Consequently, for $U\in \mathbf {G}$ satisfying $\Vert U-I\Vert \ge {\frac{1}{2}}$, one has

$$\begin{aligned} \Vert \log (U)\Vert \le 2\overline{\mathcal{D}}_{\tilde{N}}\Vert U-I\Vert . \end{aligned}$$

(10.6)

Combining (10.5) and (10.6), and taking $\mathcal{D}_{\tilde{N}}=\max \{3/2,2 \overline{\mathcal{D}}_{\tilde{N}}\}$ yields the result. $\square$

The following is a standard calculation obtained using Taylor expansions; c.f., [27].

Lemma 10.2

Suppose $G,H\in \mathfrak {su}({\tilde{N}})$ with ${\tilde{N}}\in \mathbf{N}$ and $\Vert G\Vert ,\Vert H\Vert \le D<\infty$. Suppose $\delta \le 1$. Then, there exists $c_0=c_0({\tilde{N}},D)$ such that

$$\begin{aligned} e^{{\sqrt{\delta }}G}e^{{\sqrt{\delta }}H}&e^{-{\sqrt{\delta }}G}e^{-{\sqrt{\delta }}H} =I+\delta [G,H]+\delta ^{\frac{3}{2}} F_0(G,H), \end{aligned}$$

(10.7)

where $\Vert F_0(G,H)\Vert \le c_0$.

We are looking for an upper bound on the minimal time to travel from $U^1$ to $U^2$ in $\mathbf {G}$ by application of a bounded control in directions spanned by $\mathcal{H}^0$. By rotation of coordinates, this problem is equivalent to the problem of finding an upper bound on the minimal time to travel from $U^0\in \mathbf {G}$ to $I$; (see Theorem 2.1). We will obtain an upper bound for a slightly more constrained problem, which will then provide an upper bound for this problem.

Specifically we obtain the upper bound by considering the more restrictive case where, for all $t$, $[v_t]_m=\alpha _t\hat{\delta }_{j_t,m}$ for some $\alpha _t\in \{0,-{\overline{D}_\beta },{\overline{D}_\beta }\}$, $j_t\in \mathcal{M}_0$. Here $\hat{\delta }$ denotes the Dirac delta function. We let

$$\begin{aligned} \tau ^{opt}\!=\!\tau ^{opt}(U^0) \doteq \inf _{v\in \mathcal{V}^e_{0,{\overline{D}_\beta }}}\{ t\ge 0\,\vert \, U_t=I,\, U, \text { satisfying Eq.}~(10.4) \text { with }\, U_0=U^0\}. \end{aligned}$$

(10.8)

By the nature of $\mathbf {G}$, there exists $\overline{H}^0\in {\mathfrak {g}}$ such that

$$\begin{aligned} \overline{H}^0\in argmin\{ \Vert H_{0}\Vert \vert \,H^0\in {\mathfrak {g}}\,, U^0=e^{H^0}\}. \end{aligned}$$

Let $\overline{H}=\overline{H}^0/\Vert \overline{H}^0\Vert$. Let ${D_\beta }=\min \bigl \{{\overline{D}_\beta },{\overline{D}_\beta }^2\bigr \}$, so that $\max \bigl \{{D_\beta },\sqrt{{D_\beta }}\bigr \}={\overline{D}_\beta }$, and let $T~=~\Vert \overline{H}^0\Vert /{D_\beta }$. Then, $U^0=e^{{D_\beta }\overline{H}T}$. As ${\overline{\mathcal{H}}}$ is an orthonormal basis for ${\mathfrak {g}}$, it follows that we can generate a linear combination satisfying

$$\begin{aligned} {D_\beta }\overline{H}=\sum _{m\in \mathcal{M}}\beta ^0_m H_m, \end{aligned}$$

where $\sum _{m\in \mathcal{M}}(\beta ^0_m)^2={D_\beta }^2$. We have

$$\begin{aligned} U^0=e^{\sum _{m\in \mathcal{M}}\beta ^0_m H_m T}. \end{aligned}$$

Remark 10.3

Note that we have chosen ${D_\beta }$ such that when we apply control for motion in the $\mathcal{H}^1$ directions via bracket operations (such as indicated in Lemma 10.2), the actual applied control will not exceed ${\overline{D}_\beta }$ in magnitude.

We will choose a sufficiently small $\delta >0$ (where this size will be clarified below), such that certain estimates will hold. If $T$ is sufficiently small we will simply take $\delta =T$. Otherwise, we will take $T=\hat{N}\delta$ where $\hat{N}$ is sufficiently large so that $\delta$ satisfies the estimates. Then,

$$\begin{aligned} U^0=\left[ e^{\sum _{m\in \mathcal{M}}\beta ^0_m H_m \delta }\right] ^{\hat{N}}. \end{aligned}$$

Let $U^\kappa \doteq \left[ e^{\sum _{m\in \mathcal{M}}\beta ^0_m H_m \delta }\right] ^{\hat{N}-\kappa }$ for $\kappa \in ]0,\hat{N}[$. We will drive the system from $U_0=U^0$ to $I$ by repeatedly driving the system from $U^\kappa$ to $U^{\kappa +1}$, that is, by repeatedly applying $e^{-\sum _{m\in \mathcal{M}}\beta ^0_m H_m \delta }$. Our goal is to obtain transition matrix $e^{-\sum _{m\in \mathcal{M}}\beta ^0_m H_m \delta }$ via some control in $\mathcal{V}^e_{0,{\overline{D}_\beta }}$; this will lead to our upper bound on the travel time to the identity element. We begin with another Taylor series based calculation.

Lemma 10.4

Suppose $\widetilde{H}_1,\widetilde{H}_2\in {\mathfrak {g}}$ with $\Vert \widetilde{H}_1\Vert ,\Vert \widetilde{H}_2\Vert =1$, $\langle \widetilde{H}_1,\widetilde{H}_2\rangle =0$ (where we note that the appropriate inner product is given by $\langle \widetilde{H}_1,\widetilde{H}_2\rangle \doteq \sum _{i,j}[\widetilde{H}_1]_{i,j}[\bar{\widetilde{H}}_2]_{i,j}$). Suppose $\tilde{\beta }_1^2+\tilde{\beta }_2^2\le {D_\beta }^2$. Then, there exists $c_1=c_1({\tilde{N}},{D_\beta })<\infty$, $\widehat{H}_1 \in {\mathfrak {g}}$ such that for any $\delta >0$,

$$\begin{aligned} {e^{-\delta \tilde{\beta }_1\widetilde{H}_1}}e^{\delta (\tilde{\beta }_1\widetilde{H}_1+\tilde{\beta }_2\widetilde{H}_2)} = {e^{\delta \tilde{\beta }_2\widetilde{H}_2}}e^{\widehat{H}_1}, \end{aligned}$$

where $\Vert \widehat{H}_1\Vert \le \delta ^2 c_1$.

Proof

We simply expand the exponentials in their Taylor series. First, note that

$$\begin{aligned} {e^{-\delta \tilde{\beta }_1\widetilde{H}_1}}e^{\delta {(\tilde{\beta }_1\widetilde{H}_1+\tilde{\beta }_2 \widetilde{H}_2)}}&= \left[ I-\delta \tilde{\beta }_1 \widetilde{H}_1+\delta ^2\sum _{k=2}^\infty \frac{(\tilde{\beta }_1\widetilde{H}_1)^k}{k!}\delta ^{k-2} \right] \nonumber \\&\quad \times \left[ I+\delta {(\tilde{\beta }_1\widetilde{H}_1+\tilde{\beta }_2 \widetilde{H}_2)}+ \delta ^2\sum _{k=2}^\infty \frac{{(\tilde{\beta }_1\widetilde{H}_1+\tilde{\beta }_2 \widetilde{H}_2)}^k}{k!}\delta ^{k-2} \right] \nonumber \\&=I+\delta \tilde{\beta }_2\widetilde{H}_2+\delta ^2\overline{F}_1, \end{aligned}$$

(10.9)

where $\Vert \overline{F}_1\Vert \le \bar{c}_1({\tilde{N}},{D_\beta })$. Combining this with another expansion for ${e^{-\delta \tilde{\beta }_2\widetilde{H}_2}}$ yields

$$\begin{aligned} {e^{-\delta \tilde{\beta }_2\widetilde{H}_2}}{e^{-\delta \tilde{\beta }_1\widetilde{H}_1}}e^{\delta {(\tilde{\beta }_1\widetilde{H}_1+\tilde{\beta }_2 \widetilde{H}_2)}}&= \left[ I-\delta \tilde{\beta }_2 \widetilde{H}_2+\delta ^2\sum _{k=2}^\infty \frac{(\tilde{\beta }_2\widetilde{H}_2)^k}{k!}\delta ^{k-2} \right] \nonumber \\&\times \left[ I+\delta \tilde{\beta }_2\widetilde{H}_2+\delta ^2\overline{F}_1 \right] \doteq I+\delta ^2 F_1,\qquad \qquad \end{aligned}$$

(10.10)

where $\Vert F_1\Vert \le \tilde{c}_1({\tilde{N}},{D_\beta })<\infty$. Then, by Lemma 10.1,

$$\begin{aligned} \Vert \log (I+\delta ^2 F_1)\Vert \le \mathcal{D}_{\tilde{N}}\Vert \delta ^2 F_1\Vert \le \delta ^2\tilde{c}_1\doteq \delta ^2 c_1. \end{aligned}$$

Consequently, letting $\widehat{H}_1\doteq \log (I+\delta ^2 F_1)$, one has the desired result. $\square$

We define the ordered product notation

$$\begin{aligned} \prod _{k=1}^K A_k\doteq A_K A_{K-1}\cdots A_2 A_1. \end{aligned}$$

Lemma 10.5

Suppose $\sum _{m\in \mathcal{M}}(\beta ^0_m)^2 \le {D_\beta }^2$. There exists $c_1=c_1({\tilde{N}},{D_\beta })>0$ such that for any $\delta >0$,

$$\begin{aligned} \left[ \prod _{m=1}^{M_0} e^{-\delta \beta ^0_m H_m} \right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}= e^{\delta \sum _{m=M_0+1}^M\beta ^0_m H_m}\prod _{m=1}^{M_0} e^{\widehat{H}_m},\qquad \qquad \qquad \end{aligned}$$

(10.11)

where $\Vert \widehat{H}_m\Vert \le \delta ^2 c_1$ for all $m\in \mathcal{M}_0$, and $c_{1}$ is as indicated in Lemma 10.4.

Proof

Suppose $M_0>1$ (the case of $M_0 = 1$ has been dealt with in Lemma. 10.4). Let

$$\begin{aligned} \widetilde{H}_1=H_1,\,\, \tilde{\beta }_1=\beta ^0_1, \quad \widetilde{H}_2=\frac{\sum _{m=2}^M\beta ^0_m H_m}{\Vert \sum _{m=2}^M\beta ^0_m H_m\Vert } =\frac{\sum _{m=2}^M\beta ^0_m H_m}{\bigl [\sum _{m=2}^M(\beta ^0_m)^2\bigr ]^{\frac{1}{2}}}\qquad \qquad \end{aligned}$$

(10.12)

and $\tilde{\beta }_2=\bigl [\sum _{m=2}^M(\beta ^0_m)^2\bigr ]^{\frac{1}{2}}$. Then,

$$\begin{aligned} \sum _{m\in \mathcal{M}}\beta ^0_m H_m= \tilde{\beta }_1 \widetilde{H}_1 + \tilde{\beta }_2 \widetilde{H}_2, \end{aligned}$$

with $\Vert \widetilde{H}_2\Vert =1$. We have

$$\begin{aligned} e^{-\delta \beta ^0_1 H_1}{e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}&={e^{-\delta \tilde{\beta }_1\widetilde{H}_1}}e^{\delta {(\tilde{\beta }_1\widetilde{H}_1+\tilde{\beta }_2 \widetilde{H}_2)}}, \end{aligned}$$

which, by Lemma 10.4 yields

$$\begin{aligned} e^{-\delta \beta ^0_1 H_1}{e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}&= {e^{\delta \tilde{\beta }_2\widetilde{H}_2}}e^{\widehat{H}_1}\nonumber \\&= e^{\delta \sum _{m=2}^M\beta ^0_m H_m}e^{\widehat{H}_1}, \end{aligned}$$

(10.13)

with $\Vert \widehat{H}_1\Vert \le \delta ^2 c_1$.

Suppose that for some ${\widehat{M}}\in ]1,M_0-1[$,

$$\begin{aligned} \left[ \prod _{m=1}^{{\widehat{M}}} e^{-\delta \beta ^0_m H_m} \right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}=e^{\delta \sum _{m={\widehat{M}}+1}^M\beta ^0_m H_m}\prod _{m=1}^{\widehat{M}}e^{\widehat{H}_m}, \end{aligned}$$

(10.14)

where $\Vert \widehat{H}_m\Vert \le \delta ^2 c_1$ for all $m\in ]1,{\widehat{M}}[$, and note that by Eq. (10.13) this is true for ${\widehat{M}}=1$. Now let $\widetilde{H}_1=H_{{\widehat{M}}+1}$ and $\tilde{\beta }_1=\beta ^0_{{\widehat{M}}+1}$. Let

$$\begin{aligned} \widetilde{H}_2=\frac{\sum _{m={\widehat{M}}+2}^M\beta ^0_m H_m}{\Vert \sum _{m={\widehat{M}}+2}^M\beta ^0_m H_m\Vert } =\frac{\sum _{m={\widehat{M}}+2}^M\beta ^0_m H_m}{\bigl [\sum _{m={\widehat{M}}+2}^M(\beta ^0_m)^2\bigr ]^{\frac{1}{2}}} \end{aligned}$$

(10.15)

and $\tilde{\beta }_2=\bigl [\sum _{m={\widehat{M}}+2}^M(\beta ^0_m)^2\bigr ]^{\frac{1}{2}}$, where we are ignoring the trivial case $M_0=M$, ${\widehat{M}}=M_0-1$. Then,

$$\begin{aligned} e^{-\delta \beta ^0_{{\widehat{M}}+1} H_{{\widehat{M}}+1}} e^{\delta \sum _{m={\widehat{M}}+1}^M\beta ^0_m H_m} = {e^{-\delta \tilde{\beta }_1\widetilde{H}_1}}e^{\delta {(\tilde{\beta }_1\widetilde{H}_1+\tilde{\beta }_2 \widetilde{H}_2)}} \end{aligned}$$

(10.16)

which, by Lemma 10.4, gives

$$\begin{aligned} e^{-\delta \beta ^0_{{\widehat{M}}+1} H_{{\widehat{M}}+1}} e^{\delta \sum _{m={\widehat{M}}+1}^M\beta ^0_m H_m}&={e^{\delta \tilde{\beta }_2\widetilde{H}_2}}e^{\widehat{H}_{{\widehat{M}}+1}}\end{aligned}$$

(10.17)

$$\begin{aligned}&=e^{\delta \sum _{m={\widehat{M}}+2}^M\beta ^0_m H_m}e^{\widehat{H}_{{\widehat{M}}+1}}, \end{aligned}$$

(10.18)

where $\Vert \widehat{H}_{{\widehat{M}}+1}\Vert \le \delta ^2 c_1$. Rearranging, we have

$$\begin{aligned} e^{\delta \sum _{m={\widehat{M}}+1}^M\beta ^0_m H_m}&= e^{\delta \beta ^0_{{\widehat{M}}+1} H_{{\widehat{M}}+1}} e^{\delta \sum _{m={\widehat{M}}+2}^M\beta ^0_m H_m}e^{\widehat{H}_{{\widehat{M}}+1}}. \end{aligned}$$

(10.19)

Substituting (10.19) into (10.14), and rearranging, yields

$$\begin{aligned} \left[ \prod _{m=1}^{{\widehat{M}}+1} e^{-\delta \beta ^0_m H_m} \right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}= e^{\delta \sum _{m={\widehat{M}}+2}^M\beta ^0_m H_m}\prod _{m=1}^{{\widehat{M}}+1} e^{\widehat{H}_m}, \end{aligned}$$

(10.20)

with $\Vert \widehat{H}_m\Vert \le \delta ^2 c_1$ for all $m\in ]1,{\widehat{M}}+1[$. By induction, one has the desired result. $\square$

We have now dealt with all the $m\in \mathcal{M}_0$, and must consider the slightly more difficult case of $m\in \mathcal{M}_1$. For $m\in \mathcal{M}_1$, we define

$$\begin{aligned} {\phi ^\delta _m}={\phi ^\delta _m}(\beta ^0_m)=&\bigg \{e^{\sqrt{\delta |\beta ^0_{m}|}\,{\hbox {sgn}}(\beta ^0_{m}) H_{m^1(m)}}\cdot e^{-\sqrt{\delta |\beta ^0_{m}|}\, H_{m^2(m)}} \bigg \}\nonumber \\&\quad \times \bigg \{ e^{-\sqrt{\delta |\beta ^0_{m}|}\,{\hbox {sgn}}(\beta ^0_{m}) H_{m^1(m)}}\cdot e^{\sqrt{\delta |\beta ^0_{m}|}\,H_{m^2(m)}}\bigg \}, \end{aligned}$$

(10.21)

where $m^1$ and $m^2$ are given in (10.2). We recall that for all $m$, $|\beta ^0_m|\le {D_\beta }$, where we chose ${D_\beta }$ such that $\sqrt{{D_\beta }}\le {\overline{D}_\beta }$ where ${\overline{D}_\beta }$ is the maximum control magnitude.

Lemma 10.6

There exists $c_2=c_2({\tilde{N}},{D_\beta })<\infty$ such that for any $\delta \in (0,1]$ and any $m\in \mathcal{M}_1$, there exists $H^\prime _m$ such that

$$\begin{aligned} {\phi ^\delta _m}=e^{H^\prime _m}e^{-\delta {\beta ^0_m}H_m}, \end{aligned}$$

where $\Vert H^\prime _m\Vert \le \delta ^{3/2} c_2$.

Proof

Let $m\in \mathcal{M}_1$. Then, using Lemma 10.2 in the case

$$\begin{aligned} G=\sqrt{|\beta ^0_{m}|}\,{\hbox {sgn}}(\beta ^0_{m}) H_{m^1(m)} \quad \text{ and } \quad H=-\sqrt{|\beta ^0_{m}|}\, H_{m^2(m)}, \end{aligned}$$

we obtain

$$\begin{aligned} {\phi ^\delta _m}= I-\delta {\beta ^0_m}\left[ H_{m^1(m)},H_{m^2(m)}\right] +\delta ^{3/2} \bar{F}_0\left( H_{m^1(m)},H_{m^2(m)},\beta ^0_{m},{\tilde{N}}\right) \end{aligned}$$

for $\delta \in (0,1]$ where $\Vert \bar{F}_0(H_{m^1(m)},H_{m^2(m)},\beta ^0_{m},{\tilde{N}})\Vert \le c_0({\tilde{N}},{D_\beta })$, which by the definition of $m^1,m^2$, yields

$$\begin{aligned} {\phi ^\delta _m}=I-\delta {\beta ^0_m}H_m+\delta ^{3/2} \bar{F}_0\left( H_{m^1(m)},H_{m^2(m)},\beta ^0_{m},{\tilde{N}}\right) . \end{aligned}$$

(10.22)

Combining this with the Taylor expansion for $e^{\delta {\beta ^0_m}H_m}$, one has

$$\begin{aligned} {\phi ^\delta _m}e^{\delta {\beta ^0_m}H_m}&= \left[ I-\delta {\beta ^0_m}H_m+\delta ^{3/2}\bar{F}_0\right] \left[ I+\delta {\beta ^0_m}H_m+\delta ^2\sum _{k=2}^\infty \frac{({\beta ^0_m}H_m)^k\delta ^{k-2}}{k!}\right] \\&= I+\delta ^{3/2}F_2\left( H_{m^1(m)},H_{m^2(m)},\beta ^0_{m},{\tilde{N}}\right) , \end{aligned}$$

where $\Vert F_2(H_{m^1(m)},H_{m^2(m)},\beta ^0_{m},{\tilde{N}})\Vert \le c'_2=c'_2({\tilde{N}},{D_\beta })$. Then, by Lemma 10.1,

$$\begin{aligned} \Vert \log (I+\delta ^{3/2} F_2)\Vert \le \mathcal{D}_{\tilde{N}}\delta ^{3/2} c_2^\prime \doteq \delta ^{3/2} c_2. \end{aligned}$$

Letting $H^\prime _m\doteq \log (I+\delta ^{3/2} F_2)$, we have

$$\begin{aligned} {\phi ^\delta _m}e^{\delta {\beta ^0_m}H_m}=e^{H^\prime _m} \end{aligned}$$

with $\Vert H^\prime _m\Vert \le \delta ^{3/2} c_2$. $\square$

The proof of the following lemma is similar to the above calculations, and we do not include it. We only note that the fact that $F_3\in {\mathfrak {g}}$ below follows immediately from the fact that the left-hand side of (10.23) is in $\mathbf {G}$.

Lemma 10.7

Let $\delta \le 1$, $D^1<\infty$ and $H,G\in {\mathfrak {g}}$ with $\Vert H\Vert ,\Vert G\Vert \le D^1$. Then, there exists $c_3=c_3({\tilde{N}},D^1)$ such that for any $k\in \mathbf{N}$ and $m,n\in [0,\infty )$,

$$\begin{aligned} e^{-\delta ^{m+n}H}e^{-\delta ^m G} e^{\delta ^{m+n}H}e^{\delta ^m G} =e^{F_3}, \end{aligned}$$

(10.23)

where $\Vert F_3\Vert \le \delta ^{2m} c_3$ and $F_{3} \in {\mathfrak {g}}$.

Theorem 10.8

There exist $c_1({\tilde{N}},{D_\beta }),c_2({\tilde{N}},{D_\beta }),\bar{c}_3({\tilde{N}},{D_\beta })<\infty$ such that for all $\delta \in (0,1]$,

$$\begin{aligned}&\left[ \prod _{m=M_0+1}^M{\phi ^\delta _m}\!\! \right] \left[ \prod _{m=1}^{M_0} e^{-\delta {\beta ^0_m}H_m} \right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}&\!=\!e^{H^\prime _M} \left[ \prod _{m=M_0+1}^{M-1}\! \! e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m}\!\! \right] \prod _{m=1}^{M_0} e^{\widehat{H}_m}, \end{aligned}$$

(10.24)

where $H^\prime _m, H^{\prime \prime }_m, \widehat{H}_m \in {\mathfrak {g}}$, $\Vert H^\prime _m\Vert \le \delta ^{3/2} c_2$ for all $m\in \mathcal{M}_1$, $\Vert H^{\prime \prime }_m\Vert \le \delta ^2 c_3$ for all $m\in \mathcal{M}_1$, and $\Vert \widehat{H}_m\Vert \le \delta ^2 c_1$ for all $m\in \mathcal{M}$.

Proof

Suppose there exists

$$\begin{aligned} M^\prime \in ]M_0,M-1[ \end{aligned}$$

such that

$$\begin{aligned}&\left[ \prod _{m=M_0+1}^{M^\prime }{\phi ^\delta _m}\right] \left[ \prod _{m=1}^{M_0} e^{-\delta {\beta ^0_m}H_m} \right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}\nonumber \\&\quad =e^{\delta \sum _{M^\prime +1}^M{\beta ^0_m}H_m} \left[ \prod _{m=M_0+1}^{M^\prime } e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m} \right] \prod _{m=1}^{M_0} e^{\widehat{H}_m}, \end{aligned}$$

(10.25)

where the $\widehat{H}_m$, $H^\prime _m$, $H^{\prime \prime }_m$ satisfy the conditions in the theorem statement. Note that by Lemma 10.5, this is true for $M^\prime =M_0$. If, by an induction argument, we show that (10.25) holds for $M^\prime +1$, we will be done up to $M^\prime +1=M-1$.

Let $\tilde{\beta }_1=\beta ^0_{M^\prime +1}$, $\widetilde{H}_1=H_{M^\prime +1}$,

$$\begin{aligned} \widetilde{H}_2=\frac{\sum _{m=M^\prime +2}^M {\beta ^0_m}H_m}{\Vert \sum _{m=M^\prime +2}^M {\beta ^0_m}H_m\Vert } =\frac{\sum _{m=M^\prime +2}^M {\beta ^0_m}H_m}{[\sum _{m=M^\prime +2}^M ({\beta ^0_m})^2]^{\frac{1}{2}}} \end{aligned}$$

and $\tilde{\beta }_2=\left[ \sum _{m=M^\prime +2}^M ({\beta ^0_m})^2\right] ^{\frac{1}{2}}$. Then,

$$\begin{aligned} e^{-\delta \beta ^0_{M^\prime +1} H_{M^\prime +1}} e^{\delta \sum _{M^\prime +1}^M{\beta ^0_m}H_m} ={e^{-\delta \tilde{\beta }_1\widetilde{H}_1}}e^{\delta {(\tilde{\beta }_1\widetilde{H}_1+\tilde{\beta }_2 \widetilde{H}_2)}}, \end{aligned}$$

which by Lemma 10.4,

$$\begin{aligned}&={e^{-\delta \tilde{\beta }_2\widetilde{H}_2}}e^{\widehat{H}_{M^\prime +1}} =e^{-\delta \sum _{M^\prime +2}^M{\beta ^0_m}H_m}e^{\widehat{H}_{M^\prime +1}}, \end{aligned}$$

(10.26)

where $\Vert \widehat{H}_{M^\prime +1}\Vert \le \delta ^2 c_1=\delta ^2 c_1({\tilde{N}},{D_\beta })$. Also, assuming that $\delta \le 1$, one has from Lemma 10.6 that

$$\begin{aligned} \phi ^\delta _{M^\prime +1}= e^{H^\prime _{M\prime +1}} e^{-\delta \beta ^0_{M\prime +1} H_{M\prime +1}}, \end{aligned}$$

(10.27)

where $\Vert H^\prime _{M^\prime +2}\Vert \le \delta ^{3/2} c_2$. Now, using (10.26) to substitute on the right-hand side in Eq. (10.25) yields

$$\begin{aligned}&e^{-\delta \beta ^0_{M\prime +1} H_{M\prime +1}} \left[ \prod _{m=M_0+1}^{M^\prime }{\phi ^\delta _m}\right] \left[ \prod _{m=1}^{M_0} e^{-\delta {\beta ^0_m}H_m} \right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}\nonumber \\&\qquad = e^{\delta \sum _{M^\prime +2}^M{\beta ^0_m}H_m}e^{\widehat{H}_{M^\prime +1}} \left[ \prod _{m=M_0+1}^{M^\prime } e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m} \right] \prod _{m=1}^{M_0} e^{\widehat{H}_m}, \end{aligned}$$

(10.28)

and substituting for the first term on the left-hand side of this using (10.27) yields

$$\begin{aligned}&\left[ \prod _{m=M_0+1}^{M^\prime +1}{\phi ^\delta _m}\right] \left[ \prod _{m=1}^{M_0} e^{-\delta {\beta ^0_m}H_m} \right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}\nonumber \\&\qquad = e^{H^\prime _{M^\prime +1}} e^{\delta \sum _{M^\prime +2}^M{\beta ^0_m}H_m}e^{\widehat{H}_{M^\prime +1}} \cdot \left[ \prod _{m=M_0+1}^{M^\prime } e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m} \right] \cdot \prod _{m=1}^{M_0} e^{\widehat{H}_m}.\qquad \qquad \end{aligned}$$

(10.29)

Finally, using Lemma 10.7 to reorder the first two terms on the right-hand side of Eq. (10.29) yields

$$\begin{aligned} \left[ \prod _{m=M_0+1}^{M^\prime +1}{\phi ^\delta _m}\right] \left[ \prod _{m=1}^{M_0} e^{-\delta {\beta ^0_m}H_m} \right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}= e^{\delta \sum _{M^\prime +2}^M{\beta ^0_m}H_m} \left[ \prod _{m=M_0+1}^{M^\prime +1} e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m} \right] \prod _{m=1}^{M_0} e^{\widehat{H}_m},\nonumber \\ \end{aligned}$$

where $\Vert H^{\prime \prime }_{M^\prime +1}\Vert \le \delta ^2 \bar{c}_3({\tilde{N}},{D_\beta })<\infty$. By induction, we obtain

$$\begin{aligned} \left[ \prod _{m=M_0+1}^{M-1}{\phi ^\delta _m}\right] \left[ \prod _{m=1}^{M_0} e^{-\delta {\beta ^0_m}H_m} \right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}=e^{\delta \beta ^0_M H_M} \left[ \prod _{m=M_0+1}^{M-1} e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m} \right] \prod _{m=1}^{M_0} e^{\widehat{H}_m},\nonumber \\ \end{aligned}$$

(10.30)

where the $\widehat{H}_m$, $H^\prime _m$, $H^{\prime \prime }_m$ satisfy the conditions in the theorem statement.

It only remains to work on the term $e^{\delta \beta ^0_M H_M}$. Note that Eq. (10.30) is

$$\begin{aligned} e^{-\delta \beta ^0_M H_M} \left[ \prod _{m=M_0+1}^{M-1}{\phi ^\delta _m}\right] \left[ \prod _{m=1}^{M_0} e^{-\delta {\beta ^0_m}H_m} \right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}= \left[ \prod _{m=M_0+1}^{M-1} e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m} \right] \prod _{m=1}^{M_0} e^{\widehat{H}_m}, \end{aligned}$$

which, by Lemma 10.6, yields

$$\begin{aligned}&\left[ \prod _{m=M_0+1}^{M}{\phi ^\delta _m}\!\right] \left[ \prod _{m=1}^{M_0} e^{-\delta {\beta ^0_m}H_m} \!\right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}\!=\! e^{H^\prime _M} \left[ \prod _{m=M_0+1}^{M-1} e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m} \!\right] \prod _{m=1}^{M_0} e^{\widehat{H}_m}. \end{aligned}$$

Theorem 10.9

There exists $c_4({\tilde{N}},{D_\beta })<\infty$ and $\check{H}^0 \in {\mathfrak {g}}$ such that for any $\delta \in (0,1]$,

$$\begin{aligned} \left[ \prod _{m=M_0+1}^M{\phi ^\delta _m}\right] \left[ \prod _{m=1}^{M_0} e^{-\delta {\beta ^0_m}H_m} \right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}=e^{\check{H}^0}, \end{aligned}$$

where $\Vert \check{H}^0\Vert \le \delta ^{3/2} c_4$.

Proof

We need only make a substitution on the right-hand side of (10.24). By Lemma 10.1,

$$\begin{aligned}&\left\| \log \left\{ e^{H^\prime _M} \left[ \prod _{m=M_0+1}^{M-1} e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m} \right] \left[ \prod _{m=1}^{M_0} e^{\widehat{H}_m} \right] \right\} \right\| \nonumber \\&\quad \le \mathcal{D}_{\tilde{N}}\left\| e^{H^\prime _M} \left[ \prod _{m=M_0+1}^{M-1} e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m} \right] \left[ \prod _{m=1}^{M_0} e^{\widehat{H}_m} \right] - I \right\| . \end{aligned}$$

(10.31)

Let

$$\begin{aligned} \Delta _m\doteq {\left\{ \begin{array}{ll} e^{H^\prime _m}&{}\quad \text {if}\;m=M\\ e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m}&{}\quad \text {if}\;m\in ]M_0+1,M-1[\\ e^{\widehat{H}_m} &{} \quad \text {if}\;m\in \mathcal{M}_0\\ I &{} \quad \text {if}\;m=0. \end{array}\right. } \end{aligned}$$

(10.32)

With this notation definition, Eq. (10.31) becomes

$$\begin{aligned} \left\| \log \left[ \prod _{m=0}^{M} \Delta _m \right] \right\| \le \mathcal{D}_{\tilde{N}}\left\| \left[ \prod _{m=0}^{M} \Delta _m\right] -I \right\| , \end{aligned}$$

which is

$$\begin{aligned} \le \mathcal{D}_{\tilde{N}}\left[ \sum _{m=1}^M \left\| \prod _{k=0}^{m} \Delta _m-\prod _{k=0}^{m-1} \Delta _m \right\| \right] \le \mathcal{D}_{\tilde{N}}\sum _{m=1}^M \Vert \Delta _m-I\Vert \prod _{k=0}^{m-1} \Vert \Delta _m\Vert .\qquad \qquad \end{aligned}$$

(10.33)

For $m\in \mathcal{M}_0$, using Theorem 10.8,

$$\begin{aligned} \Vert \Delta _m\Vert =\Vert e^{\widehat{H}_m}\Vert \le e^{\Vert \widehat{H}_m\Vert }\le e^{\delta ^2 c_1({\tilde{N}},{D_\beta })}. \end{aligned}$$

(10.34)

Also, for all $m\in \mathcal{M}$, using Theorem 10.8 again,

$$\begin{aligned}&\Vert e^{\widehat{H}_m}-I\Vert \le \sum _{k=1}^\infty \frac{\Vert \widehat{H}_m\Vert ^k}{k!} \le \sum _{k=1}^\infty \frac{\delta ^{2k}c_1^k}{k!} \le \delta ^2 c_5({\tilde{N}},{D_\beta }), \end{aligned}$$

(10.35)

and

$$\begin{aligned}&\Vert e^{H^\prime _m}-I\Vert \le \sum _{k=1}^\infty \frac{\Vert H^\prime _m\Vert ^k}{k!} \le \sum _{k=1}^\infty \frac{\delta ^{2k}c_2^k}{k!} \le \delta ^2 c_5({\tilde{N}},{D_\beta }), \end{aligned}$$

(10.36)

for proper choice of $c_5({\tilde{N}},{D_\beta })<\infty$. Combining Eqs. (10.32)–(10.35) yields

$$\begin{aligned} \left\| \log \left[ \prod _{m=0}^{M} \Delta _m \right] \right\|&\le \mathcal{D}_{\tilde{N}}\left\{ \sum _{m=M_0+1}^M \Vert \Delta _m-I\Vert \left[ \prod _{k=M_0+1}^{m-1} \Vert \Delta _m\Vert \right] \cdot e^{\delta ^2 c_1 M_0} +\delta ^2 c_5\sum _{m=1}^{M_0} e^{\delta ^2 c_1(m-1)} \right\} \nonumber \\&\le c_6 \left\{ \sum _{m=M_0+1}^M \Vert \Delta _m-I\Vert \left[ \prod _{k=M_0+1}^{m -1} \Vert \Delta _m\Vert \right] +\delta ^2 \right\} , \end{aligned}$$

(10.37)

for a proper choice of $c_6=c_6({\tilde{N}},{D_\beta })<\infty$.

Now, for $m\in ]M_0+1,M-1[$, using Theorem 10.8,

$$\begin{aligned} \Vert \Delta _m\Vert&= \Vert e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m}\Vert \le e^{\Vert H^\prime _m\Vert }e^{\Vert H^{\prime \prime }_m\Vert }e^{\Vert \widehat{H}_m\Vert } \nonumber \\&\le e^{\delta ^2 c_1+\delta ^{3/2} c_2+\delta ^2 \bar{c}_3}, \end{aligned}$$

(10.38)

which implies that for $m\in ]M_0+1,M[$,

$$\begin{aligned} \prod _{k=M_0+1}^{m-1}\Vert \Delta _k\Vert \le c_7=c_7({\tilde{N}},{D_\beta })<\infty . \end{aligned}$$

(10.39)

Applying Eq. (10.39) in Eq. (10.37) yields

$$\begin{aligned} \left\| \log \left[ \prod _{m=0}^{M} \Delta _m \right] \right\| \le c_6 \left\{ c_7\Vert e^{H^\prime _M}-I\Vert + c_7\sum _{m=M_0+1}^{M-1} \Vert e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m}-I\Vert +\delta ^2 \right\} ,\nonumber \\ \end{aligned}$$

(10.40)

which by Eqs. (10.35) and (10.36), yields

$$\begin{aligned} \le c_6 \bigg \{ \delta ^2(c_7c_5+1) + c_7\sum _{m=M_0+1}^{M-1} \Vert e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m}-I\Vert \bigg \}. \end{aligned}$$

(10.41)

Now, as above,

$$\begin{aligned} \Vert e^{H^\prime _m}e^{H^{\prime \prime }_m}e^{\widehat{H}_m}-I\Vert&\le \Vert e^{H^\prime _m}-I\Vert e^{\Vert H^{\prime \prime }_m\Vert }e^{\Vert \widehat{H}_m\Vert } +\Vert e^{H^{\prime \prime }_m}-I\Vert e^{\Vert \widehat{H}_m\Vert } +\Vert e^{\widehat{H}_m}-I\Vert \end{aligned}$$

which, using (10.35) again,

$$\begin{aligned}&\le \Vert e^{H^\prime _m}-I\Vert e^{\delta ^2(c_1+c_3)} +\Vert e^{H^{\prime \prime }_m}-I\Vert e^{\delta ^2 c_1} +c_5\delta ^2, \end{aligned}$$

which, in a similar fashion to (10.35), (10.36),

$$\begin{aligned}&\le c_8\left[ \delta ^{3/2}e^{\delta ^2(c_1+c_3)}+\delta ^2e^{\delta ^2 c_1} \right] +c_5\delta ^2, \end{aligned}$$

(10.42)

where $c_8=c_8({\tilde{N}},{D_\beta })<\infty$. Substituting (10.42) into (10.41) yields

$$\begin{aligned} \left\| \log \left[ \prod _{m=0}^{M} \Delta _m\right] \right\| \le \delta ^{3/2} c_4, \end{aligned}$$

where $c_4=c_4({\tilde{N}},{D_\beta })<\infty$. Letting $\check{H}^0=\log \left[ \prod _{m=1}^{M} \Delta _m\right]$ yields the result. $\square$

We now describe a result regarding the generation of a control signal via a series of propagation steps. For $m\in \mathcal{M}$, let

$$\begin{aligned} {\tilde{\phi }^\delta _m}\doteq {\left\{ \begin{array}{ll} e^{-\delta {\beta ^0_m}H_m} &{}\quad \text {if}\;m\in \mathcal{M}_0,\\ {\phi ^\delta _m}&{}\quad \text {if}\;m\in \mathcal{M}_1. \end{array}\right. } \end{aligned}$$

We have

$$\begin{aligned} \left[ \prod _{m=1}^M{\tilde{\phi }^\delta _m}\right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}=e^{\check{H}^0}, \end{aligned}$$

(10.43)

where $\Vert \check{H}^0\Vert \le \delta ^{3/2} c_4$. Let

$$\begin{aligned} \sqrt{\delta }\le \frac{{D_\beta }}{2 c_4}. \end{aligned}$$

(10.44)

One has $\Vert \check{H}^0\Vert \le \frac{\delta }{2}{D_\beta }$. Consequently, $\check{H}^0$ may be written as a linear combination of the basis vectors of the algebra ${\mathfrak {g}}$ as follows

$$\begin{aligned} \check{H}^0=\frac{-\delta }{2}\sum _{m\in \mathcal{M}}{\beta ^1_m}H_m \end{aligned}$$

where $\sum _{m\in \mathcal{M}}({\beta ^1_m})^2 \le 1$. Using this in (10.43), we have

$$\begin{aligned} e^{\frac{\delta }{2}\sum _{m\in \mathcal{M}}{\beta ^1_m}H_m} \left[ \prod _{m=1}^M{\tilde{\phi }^\delta _m}\right] {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}=I. \end{aligned}$$

(10.45)

For $m\in \mathcal{M}$, let

$$\begin{aligned} {\tilde{\phi }^{\delta /2}_m}={\tilde{\phi }^{\delta /2}_m}({\beta ^1_m}) \doteq {\left\{ \begin{array}{ll} e^{\frac{-\delta }{2}{\beta ^1_m}H_m} &{}\quad \text {if}\;m\in \mathcal{M}_0,\\ \phi ^{\delta /2}_m(-{\beta ^1_m}) &{}\quad \text {if}\;m\in \mathcal{M}_1, \end{array}\right. } \end{aligned}$$

where $\phi ^\cdot _m(\cdot )$ is given by (10.21). By Theorem 10.9 with $\delta /2$ replacing $\delta$ and ${\beta ^1_m}$ replacing ${\beta ^0_m}$, we have

$$\begin{aligned} \left[ \prod _{m\in \mathcal{M}}{\tilde{\phi }^{\delta /2}_m}\right] e^{\frac{-\delta }{2}\sum _{m\in \mathcal{M}}{\beta ^1_m}H_m} =e^{\check{H}^1}, \end{aligned}$$

or

$$\begin{aligned} e^{-\check{H}^1}\left[ \prod _{m\in \mathcal{M}}{\tilde{\phi }^{\delta /2}_m}\right] =e^{\frac{\delta }{2}\sum _{m\in \mathcal{M}}{\beta ^1_m}H_m}, \end{aligned}$$

(10.46)

where $\Vert \check{H}^1\Vert \le c_4\left( \frac{\delta }{2}\right) ^{3/2}$. From inequality (10.44), we have

$$\begin{aligned} \Vert \check{H}^1\Vert \le \frac{\delta }{4\sqrt{2}}{D_\beta }<\frac{\delta }{2^2}{D_\beta }. \end{aligned}$$

(10.47)

Substituting (10.46) into (10.45) yields

$$\begin{aligned} \left[ \prod _{m\in \mathcal{M}}{\tilde{\phi }^{\delta /2}_m}\right] \left[ \prod _{m=1}^M{\tilde{\phi }^\delta _m}\right] e^{\delta \sum _{m\in \mathcal{M}}{\beta ^0_m}H_m} =e^{\check{H}^1}, \end{aligned}$$

where, from above, $\Vert \check{H}^1\Vert$ is bounded by the right-hand side of (10.47). We see that by repeating this process, one obtains the following.

Theorem 10.10

Let $\delta _5=\min \{1,{D_\beta }/(2c_4)\}$ (where $c_4$ was indicated in Theorem 10.9). Then,

$$\begin{aligned} \prod _{k=0}^\infty \left[ \prod _{m\in \mathcal{M}}{\tilde{\phi }}^{\delta /2^k}_m \right] =e^{-\delta \sum _{m\in \mathcal{M}}{\beta ^0_m}H_m}. \end{aligned}$$

(10.48)

for all $\delta \in (0,\delta _5]$ and all $\{{\beta ^0_m}\}_{m\in \mathcal{M}}$ such that $\sum _{m\in \mathcal{M}}({\beta ^0_m})^2\le 1$, where the ${\tilde{\phi }}^{\delta /2^k}_m$ are constructed as above.

The above result indicates that one may achieve control $e^{-\delta \sum _{m\in \mathcal{M}}{\beta ^0_m}H_m}$ by applying the sequence of sequences of propagation mappings

$$\begin{aligned} \left\{ \bigl \{{\tilde{\phi }}^{\delta /2^l}_m \bigr \}_{m\in \mathcal{M}} \right\} _{l=0}^\infty . \end{aligned}$$

(10.49)

The associated controls drive the system from $U^\kappa$ to $U^{\kappa +1}$, where $U^\kappa \doteq \left[ {e^{\delta \sum _{m\in \mathcal{M}}\beta ^0_m H_m}}\right] ^{\hat{N}-\kappa }$ for any $\kappa \in ]0,\hat{N}[$. Recall that the goal is to drive the system from $U^0$ to $U^{\hat{N}}=I$.

We now consider the time required to drive the system in this manner from $U^0$ to $I$. Note that $U^0=e^{\sum _{m\in \mathcal{M}}{\beta ^0_m}H_m T}$, and that we can expect the required time to be greater than $T$ when $M_0<M$. There are two cases to be considered: $T<\delta _5$ and $T\ge \delta _5$. First, we compute the time required for the control sequence corresponding to (10.48), (10.49). For $m\in \mathcal{M}_0$, the time required for transition ${\tilde{\phi }}^{\delta /2^k}_m$ is $\delta /2^k$. (Note that the actual control applied during this interval is ${\beta ^0_m}H_m$.) For $m\in \mathcal{M}_1$, the time required for transition ${\tilde{\phi }}^{\delta /2^k}_m$ is $4\sqrt{\delta /2^k}$. Therefore, the total time required for (10.48), (10.49) is

$$\begin{aligned} \tau ^\delta&\doteq \sum _{k=0}^\infty \frac{4\sqrt{\delta }}{(\sqrt{2})^k} +\sum _{k=0}^\infty \frac{\delta }{2^k} =\frac{4\sqrt{\delta }}{1-1/\sqrt{2}}+\frac{\delta }{1-1/2}\nonumber \\&= \frac{4\sqrt{2}}{\sqrt{2}-1}\sqrt{\delta }+2\delta . \end{aligned}$$

(10.50)

Consequently, in the case $T<\delta _5$, the minimal required time to drive the system from $U^0$ to $I$ is bounded above by

$$\begin{aligned} \overline{\tau }=\frac{4\sqrt{2}}{\sqrt{2}-1}\sqrt{T}+2T \le \left[ \frac{4\sqrt{2}}{\sqrt{2}-1}+2 \right] \sqrt{T}. \end{aligned}$$

(10.51)

In the case where $T\ge \delta _5$, we let $\bar{N}=\lceil (T/\delta _5)$ and $\delta =T/\bar{N}$. That is, we will have $\bar{N}$ steps of (10.48), (10.49), where the $\kappa$ step drives the system from $U^\kappa$ to $U^{\kappa +1}$. The total required time bound is

$$\begin{aligned} \overline{\tau }&= \frac{4\sqrt{2}}{\sqrt{2}-1}\bar{N}\sqrt{\delta }+2\bar{N}\delta \end{aligned}$$

(10.52)

$$\begin{aligned}&= \frac{4\sqrt{2}}{\sqrt{2}-1}\frac{T}{\sqrt{\delta }}+2T \le \left[ \frac{4\sqrt{2}}{\sqrt{2}-1}+2 \right] \frac{T}{\sqrt{\delta }}. \end{aligned}$$

(10.53)

Summarizing, we have now obtained Theorem 3.2.

Lastly, note that from Lemma 10.1, one has $\Vert \log U^0\Vert \le \frac{3}{2}\Vert U^0-I\Vert$. Using this, one may obtain Corollary 3.3.