How fast and robust is the quantum adiabatic passage?

In order to derive the optimal solution of the time-dependent Hamiltonian H(t) and state |ψ(t)〉, we define the action

$$\begin{equation} S = \int _0^T {\rm d}t (L_{\rm T}+L_{\rm S}+L_{\rm C}), \end{equation} \tag{ 1 }$$

as an integral between the initial and final time [25]. Each term is expressed as

$$\begin{equation} L_{\rm T} = \frac{\sqrt{\langle \dot{\psi }|(1-P)|\dot{\psi }\rangle }}{\Delta E}, \end{equation} \tag{ 2 }$$

$$\begin{equation} L_{\rm S} = {\rm i}\langle \phi |\dot{\psi }\rangle -\langle \phi |H|\psi \rangle +({\rm h.c.}), \end{equation} \tag{ 3 }$$

$$\begin{equation} L_{\rm C} = \sum _a \lambda _a f_a(H). \end{equation} \tag{ 4 }$$

We use the notations P = |ψ〉〈ψ|, ΔE² = 〈ψ|H²|ψ〉 − 〈ψ|H|ψ〉² and a dotted symbol for the time derivative. L_T dt represents the Fubini–Study distance 〈dψ|(1 − P)|dψ〉 divided by ΔE. The energy variance ΔE plays the role of the velocity in quantum systems [28], which means that ∫L_T dt represents the time to be taken for the evolution of the quantum state. The other two terms are introduced to impose constraints which must hold throughout the evolution. The Schrödinger equation ${\rm i}|\dot{\psi }\rangle =H|\psi \rangle$ is implemented by L_S and other constraints f_a(H) = 0 by L_C. |ϕ〉 and λ_a represent the Lagrange multipliers. They are determined in course of the calculation.

For the fixed initial state |ψ(0)〉 and the final one |ψ(T)〉, the action is extremized by variations H → H + δH and |ψ〉 → |ψ〉 + |δψ〉. Setting the linear term to be zero, we obtain

$$\begin{equation} {\rm i}\frac{{\rm d}}{{\rm d}t}\left(\frac{H-\langle H\rangle }{2\Delta E^2}\right)|\psi \rangle +\left({\rm i}\frac{{\rm d}}{{\rm d}t}-H\right)|\phi \rangle =0, \end{equation} \tag{ 5 }$$

$$\begin{equation} {-}\frac{1}{2\Delta E^2}(HP+PH-2P\langle H\rangle ) -(|\psi \rangle \langle \phi |+|\phi \rangle \langle \psi |) +F =0, \end{equation} \tag{ 6 }$$

where

$$\begin{equation} F = \sum _a \lambda _a f_a^{\prime }(H). \end{equation} \tag{ 7 }$$

The prime denotes the derivative with respect to H. After some calculations, we obtain

$$\begin{equation} F(t)= U(t)[|\tilde{\phi }(0)\rangle \langle \psi (0)| +|\psi (0)\rangle \langle \tilde{\phi }(0)|]U^\dag (t), \end{equation} \tag{ 8 }$$

where U(t) is the time-evolution operator for the Hamiltonian H(t) and

$$\begin{equation} |\tilde{\phi }(0)\rangle =|\phi (0)\rangle +\frac{H(0)-\langle \psi (0)|H(0)|\psi (0)\rangle }{2\Delta E^2(0)}|\psi (0)\rangle . \end{equation} \tag{ 9 }$$

Thus, F satisfies the QB equation [25]

$$\begin{equation} {\rm i}\dot{F}(t)=[H(t),F(t)], \end{equation} \tag{ 10 }$$

with the initial condition

$$\begin{equation} (1-P(0))F(0)(1-P(0))=0. \end{equation} \tag{ 11 }$$

We note that condition (11) is unchanged throughout the evolution. These equations are solved under fixed T. Then, we minimize the passage time T to accomplish our purpose to obtain the optimal solution.

For convenience of the calculation, we rewrite the QB equation by using the basis operators $\lbrace X_\mu \rbrace _{\mu =0,1,\ldots ,N^2-1}$ where N is the dimension of the Hilbert space. These satisfy the orthonormal relation

$$\begin{equation} \frac{1}{N}\mathop {\mathrm{Tr}}\nolimits X_\mu X_\nu = \delta _{\mu \nu }, \end{equation} \tag{ 12 }$$

and the completeness relation

$$\begin{equation} \frac{1}{N}\sum _{\mu =0}^{N^2-1}(X_\mu )_{ij}(X_\mu )_{kl}=\delta _{il}\delta _{jk}. \end{equation} \tag{ 13 }$$

The structure constant f_μνρ is defined by the commutation relation

$$\begin{equation} [X_\mu ,X_\nu ]={\rm i}\sum _\rho f_{\mu \nu \rho }X_\rho . \end{equation} \tag{ 14 }$$

We set X₀ to be the identity operator and use the vector representation X for other components $\lbrace X_a\rbrace _{a=1,\ldots ,N^2-1}$. Then, the Hamiltonian is represented as

$$\begin{equation} H(t)=h_0(t)+\boldsymbol {h}(t)\cdot \boldsymbol {X}. \end{equation} \tag{ 15 }$$

We note that the first term does not bring any quantum effect and can be eliminated by a proper gauge transformation |ψ(t)〉 → e^iφ(t)|ψ(t)〉.

Using this representation, we can parametrize P and F as

$$\begin{equation} P(t)=\frac{1}{N}+\frac{\sqrt{N-1}}{N}\boldsymbol {e}(t)\cdot \boldsymbol {X}, \end{equation} \tag{ 16 }$$

$$\begin{equation} F(t)= \lambda \left( \frac{\boldsymbol {l}(0)\cdot \boldsymbol {e}(0)}{\sqrt{N-1}} +\boldsymbol {l}(t)\cdot \boldsymbol {X}\right), \end{equation} \tag{ 17 }$$

where e and l are normalized vectors with the angle between them fixed: l(t) · e(t) = l(0) · e(0). These representations are derived from Tr P = Tr P² = 1 and Tr F = Tr F P. In the following analysis, the overall constant λ is not important and is left undetermined. We also note that the first term in (17) is derived from the condition that the first term in (15) is fixed and is not important as well. From equation (10), we can write the equations of motion as

$$\begin{equation} \dot{\boldsymbol {l}}(t) = \boldsymbol {h}(t)\times \boldsymbol {l}(t), \end{equation} \tag{ 18 }$$

where the vector product is defined by $(\boldsymbol {h}\times \boldsymbol {l})_a=\sum _{b,c}f_{abc}h_bl_c$. Since the state |ψ(t)〉 satisfies the Schrödinger equation, P follows the same equation as (10) and, as a result, $\dot{\boldsymbol {e}}(t)=\boldsymbol {h}(t)\times \boldsymbol {e}(t)$.

The solution of the QB equation strongly depends on the constraints imposed. We consider the case where components a = 1, 2, ..., k are fixed as

$$\begin{equation} H_{\rm C}(t)= \sum _{a=1}^k h_a^{(0)}(t)X_a. \end{equation} \tag{ 19 }$$

Correspondingly, f_a and F are given by

$$\begin{equation} f_a = {\mathrm{Tr}}\, H(t)X_a -Nh_a^{(0)}(t) \qquad (a=1,2,\ldots ,k), \end{equation} \tag{ 20 }$$

$$\begin{equation} F(t) = \lambda \sum _{a=1}^k l_a(t)X_a. \end{equation} \tag{ 21 }$$

The QB equation is solved to give other components:

$$\begin{equation} H_{\rm QB}(t)= \sum _{p=k+1}^{N^2-1}h_p^{(1)}(t)X_p. \end{equation} \tag{ 22 }$$

The total Hamiltonian is given by H(t) = H_C(t) + H_QB(t). For a given $h_a^{(0)}(t)$, time-dependences of l_a(t) and $h^{(1)}_p(t)$ are obtained by solving equation (18) and the initial condition is determined by equation (11).

The QB equation is written in terms of the operator $F=\sum _a\lambda _af_a^{\prime }(H)$ coming from the constraints f_a(H) = 0. It is crucial in the following analysis to notice that this quantity F is nothing but the Lewis–Riesenfeld invariant [29]. It satisfies equation (10) and, as a result, all eigenvalues of F become independent of time. We can write

$$\begin{equation} F(t) = \sum _{n}\lambda _n |n(t)\rangle \langle n(t)|, \end{equation} \tag{ 23 }$$

where the time-independent eigenvalue is denoted by λ_n and |n(t)〉 is the eigenstate defined at each time. By using the eigenstates, we can write the state as

$$\begin{equation} |\psi (t)\rangle = \sum _n c_n\, {\rm e}^{{\rm i}\alpha _n(t)}|n(t)\rangle , \end{equation} \tag{ 24 }$$

with a phase factor

$$\begin{equation} \alpha _n(t)=\int _0^t {\rm d}t^{\prime } \langle n(t^{\prime })| \left({\rm i}\frac{{\rm d}}{{\rm d}t^{\prime }}-H(t^{\prime })\right) |n(t^{\prime })\rangle . \end{equation} \tag{ 25 }$$

The real constant c_n is shown to be independent of time, which means that if we start the evolution from |n(0)〉, the state remains the eigenstate |n(t)〉 at arbitrary t.

In order to represent the optimal Hamiltonian in the basis of {|n(t)〉}, we multiply 〈m(t)| from the left and |n(t)〉 from the right to equation (10). We assume that there is no degeneracy in the spectrum for simplicity. Then, the off-diagonal element 〈m(t)|H(t)|n(t)〉 with m ≠ n is calculated as

$$\begin{equation} \langle m(t)|H(t)|n(t)\rangle = {\rm i}\langle m(t)|\dot{n}(t)\rangle . \end{equation} \tag{ 26 }$$

The Hamiltonian is thus written as H(t) = H₀(t) + H₁(t) where

$$\begin{equation} H_0(t) = \sum _{n}E_n(t)|n(t)\rangle \langle n(t)|, \end{equation} \tag{ 27 }$$

$$\begin{equation} H_1(t) = {\rm i}\sum _{m\ne n} |m(t)\rangle \langle m(t)|\dot{n}(t)\rangle \langle n(t)|. \end{equation} \tag{ 28 }$$

The diagonal part H₀(t) commutes with the invariant F(t). Its eigenvalue E_n(t) cannot be specified from equation (10). We see that this Hamiltonian represents the formula of the transitionless quantum driving [3]. In that method, the adiabatic state (24), with H(t) replaced by H₀(t), is constructed for a given Hamiltonian H₀(t). This is not the solution of the Schrödinger equation and we apply the counter-diabatic Hamiltonian H₁(t) to make the system transitionless. Then, the adiabatic state becomes the exact solution of the equation.

The result here shows that the assisted AP, transitionless quantum driving and shortcuts to adiabaticity are essentially equivalent and they can be derived from the QB equation.

Although we have shown that the solution of the QB equation can be interpreted as the AP, it is not clear from the beginning what kind of adiabatic state is obtained. It will be practically useful if the adiabatic state is determined by the constrained part H_C(t) so that we can control the system at will. That is, we want to know the cases where the relations H₀(t) = H_C and H₁(t) = H_QB hold. In the following, we set the initial state |ψ(0)〉 to be an eigenstate of F(0). In this case, we have |ϕ(t)〉∝|ψ(t)〉 and F(t)∝P(t).

The condition that H_C(t) determines the adiabatic state is given by the commutation relation

$$\begin{equation} [H_{\rm C}(t),F(t)]=0. \end{equation} \tag{ 29 }$$

This can be written as

$$\begin{equation} \boldsymbol {h}^{(0)}(t)\times \boldsymbol {l}(t) = 0. \end{equation} \tag{ 30 }$$

We easily see that a possible solution is given by l(t) ∝ h⁽⁰⁾(t). It is hard to imagine that the equation has other nontrivial solutions since the commutation relation (29) must hold for arbitrary t. Although it is an interesting problem to find such a solution in a specific example, we discuss the solution l(t) ∝ h⁽⁰⁾(t) in the following general analysis. Thus, F is equivalent to H_C(t) as

$$\begin{equation} F(t)=\frac{1}{h^{(0)}(t)}\sum _{a=1}^k h_a^{(0)}(t)X_a =\frac{1}{h^{(0)}(t)}H_{\rm C}(t), \end{equation} \tag{ 31 }$$

where $h^{(0)}(t)=|\boldsymbol {h}^{(0)}(t)|=\sqrt{\sum _ah_a^{(0)}(t)h_a^{(0)}(t)}$. We neglected the unimportant constant term and overall factor. The QB equation is given by

$$\begin{equation} \dot{\boldsymbol {h}}^{(0)}(t)-\frac{\dot{h}^{(0)}(t)}{h^{(0)}(t)}\boldsymbol {h}^{(0)}(t) =\boldsymbol {h}^{(1)}(t)\times \boldsymbol {h}^{(0)}(t). \end{equation} \tag{ 32 }$$

We note that this equation does not necessarily have the solution. If no solution exists, H₀(t) ≠ H_C(t) is implied. When the solution exists, each part of the Hamiltonian is given by

$$\begin{equation} H_{\rm C}(t)=H_0(t), \end{equation} \tag{ 33 }$$

$$\begin{equation} H_{\rm QB}(t)=H_1(t)+\delta H_1(t), \end{equation} \tag{ 34 }$$

where

$$\begin{equation} [H_0(t),\delta H_1(t)]=0. \end{equation} \tag{ 35 }$$

Generally, there exists the Hamiltonian δH₁(t) in H_QB(t) such that the adiabatic state is not disturbed. Since the QB equation is the most general equation, the result depends strongly on the constraints imposed. If the constraints are too loose, the solution has many ambiguities and is not determined uniquely. On the other hand, tight constraints will give trivial and ineffective solutions. It is important to choose proper constraints to obtain nontrivial results. Since it is hard to study general conditions, we examine the QB equation explicitly in the next section.

First, we consider the simplest example of N = 2. The Hamiltonian is written by the Pauli matrices σ as

$$\begin{equation} H(t) = \frac{1}{2}\boldsymbol {h}(t)\cdot \boldsymbol {\sigma } = \frac{1}{2}\left(\begin{array}{@{}cc@{}}h_3(t) & h_1(t)-{\rm i}h_2(t) \\ h_1(t)+{\rm i}h_2(t) & -h_3(t) \end{array}\right). \end{equation} \tag{ 36 }$$

The basis operator is given by X = (1, σ₁, σ₂, σ₃) and the structure constant by f_abc = 2_abc. Equation (32) is rewritten as

$$\begin{equation} \dot{\boldsymbol {h}}^{(0)}(t)-\frac{\dot{h}^{(0)}(t)}{h^{(0)}(t)}\boldsymbol {h}^{(0)}(t) =\boldsymbol {h}^{(1)}(t)\times \boldsymbol {h}^{(0)}(t), \end{equation} \tag{ 37 }$$

where the definition of the vector product is changed as $(\boldsymbol {h}^{(1)}\times \boldsymbol {h}^{(0)})_a=\sum _{bc}\epsilon _{abc}h_b^{(1)}h_c^{(0)}$. Considering the vector product with h⁽⁰⁾(t), we obtain

$$\begin{equation} \boldsymbol {h}^{(1)}(t)= \frac{\boldsymbol {h}^{(0)}(t)\times \dot{\boldsymbol {h}}^{(0)}(t)}{|\boldsymbol {h}^{(0)}(t)|^2}. \end{equation} \tag{ 38 }$$

Here, we used the property of the antisymmetric tensor ∑_aabcade = δ_bdδ_ce − δ_beδ_cd. This result gives the counter-diabatic part H₁(t) derived in the method of the transitionless quantum driving [3]. That is, we conclude that H_C(t) = H₀(t) and H_QB(t) = H₁(t) in the present case.

We note that the nontrivial result is obtained only in the case where two components of h⁽⁰⁾ are constrained. In that case, the third component is determined by equation (38). When one component of h⁽⁰⁾ is constrained, no quantum effect appears and we do not have any interesting results. For three components fixed, the Hamiltonian is completely specified and there is no room for finding the optimal Hamiltonian.

Next, we consider the case of N = 3. In this case, we use the Gell–Mann matrices

$$\begin{eqnarray} & & \lambda _1 = \left(\begin{array}{@{}ccc@{}}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right), \quad \lambda _2 = \left(\begin{array}{@{}ccc@{}}0 & -{\rm i} & 0 \\ {\rm i} & 0 & 0 \\ 0 & 0 & 0 \end{array}\right), \quad \lambda _3 = \left(\begin{array}{@{}ccc@{}}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right), \nonumber \\ & & \lambda _4 = \left(\begin{array}{@{}ccc@{}}0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array}\right), \quad \lambda _5 = \left(\begin{array}{@{}ccc@{}}0 & 0 & -{\rm i} \\ 0 & 0 & 0 \\ {\rm i} & 0 & 0 \end{array}\right), \quad \lambda _6 = \left(\begin{array}{@{}ccc@{}}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right), \nonumber \\ & & \lambda _7 = \left(\begin{array}{@{}ccc@{}}0 & 0 & 0 \\ 0 & 0 & -{\rm i} \\ 0 & {\rm i} & 0 \end{array}\right), \quad \lambda _8 = \frac{1}{\sqrt{3}} \left(\begin{array}{@{}ccc@{}}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{array}\right), \end{eqnarray} \tag{ 39 }$$

as the basis operators. Normalizing the operators, we have $X_{a}=\frac{\sqrt{6}}{2}\lambda _a$ and the structure constant is given in table 1. Since we do not have useful relations on the structure constant, it is hard to consider the general properties as we did in the case of N = 2. In the following, we study the properties of the solution explicitly.

Table 1. The structure constant f_abc for the basis operators $X_a=\frac{\sqrt{6}}{2}\lambda _a$ (a = 1, ...8) at N = 3. λ_a represent the Gell–Mann matrices in equation (39). f_abc is antisymmetric with respect to all pairs of indices. The other components not shown in the table are zero.

a	b	c	fabc	a	b	c	fabc
1	2	3	$\sqrt{6}$	3	4	5	$\frac{1}{2}\sqrt{6}$
1	4	7	$\frac{1}{2}\sqrt{6}$	3	6	7	−$\frac{1}{2}\sqrt{6}$
1	5	6	−$\frac{1}{2}\sqrt{6}$	4	5	8	$\frac{\sqrt{3}}{2}\sqrt{6}$
2	4	6	$\frac{1}{2}\sqrt{6}$	6	7	8	$\frac{\sqrt{3}}{2}\sqrt{6}$
2	5	7	$\frac{1}{2}\sqrt{6}$

3.2.1. k = 2

3.2.2. k = 3

The advantage of describing the AP as the QB is that one can study the stability of the driving. It is shown in several examples that the AP is robust against parameter variations [13, 19]. Here, we show that it is generally correct. We also find that the instability arises when we consider perturbation by different kinds of operators.

The QB equation is derived by expanding the action up to first order in δH and |δψ〉. The stability of the extremized solution is found by examining the second order. As a possible situation, we consider the case when the Hamiltonian is changed as H → H + δH. H and |ψ〉 are determined from the QB equation and we see what happens if the counter-diabatic Hamiltonian deviates from the ideal form.

The second order of the action in δH is calculated as

$$\begin{eqnarray} I(t) &=& -\frac{1}{2\Delta E^2(t)}( \langle \delta H^2(t)\rangle - \langle \delta H(t)\rangle ^2) \nonumber \\ && +\frac{3}{8}\frac{1}{\Delta E^4(t)}( \langle H(t)\delta H(t)+\delta H(t) H(t)\rangle -2\langle \delta H(t)\rangle \langle H(t)\rangle )^2, \end{eqnarray} \tag{ 67 }$$

where 〈(⋅⋅⋅)〉 = 〈ψ(t)|(⋅⋅⋅)|ψ(t)〉. The solution of the QB equation is stable when I(t) > 0. This is the general result applied to any QB trajectories. We are interested in the case where H(t) = H₀(t) + H₁(t) with H₀(t) is the Hamiltonian giving the adiabatic state and H₁(t) is the counter-diabatic Hamiltonian. In this case, the state |ψ〉 is the eigenstate of H₀(t) and the diagonal elements of H₁(t) are zero in this representation. Then, we can write

$$\begin{equation} I(t) = -\frac{\langle \delta H^2(t)\rangle - \langle \delta H(t)\rangle ^2}{2\langle H_1^2(t)\rangle } +\frac{3}{8} \frac{\langle H_1(t)\delta H(t)+\delta H(t) H_1(t)\rangle ^2}{\langle H_1^2(t)\rangle ^2}. \end{equation} \tag{ 68 }$$

This is the main result in this section. The stability of the AP is given by the condition I(t) > 0.

When the variation is proportional to the counter-diabatic Hamiltonian as δH(t) = c(t)H₁(t) with an arbitrary scalar function c(t), we can show that I(t) = c²(t), which means that the AP is stable against any parameter variations in the counter-diabatic Hamiltonian. This result is consistent with previous examples showing stable driving [13, 19]. It is also possible to consider the unstable perturbation in principle. Such a perturbation is realized when the second term of equation (68) becomes smaller compared with the first term. It is accomplished when H₁(t) and δH(t) anticommute with each other: H₁(t)δH(t) + δH(t)H₁(t) = 0. We can generally say that the AP is unstable against perturbations by operators anticommuting with the counter-diabatic Hamiltonian.

Equation I(t) > 0 represents a necessary condition of the stability. We examine this condition by using an example to see whether the expected behavior is obtained. Although the simplest example of the quantum system is the two-level system N = 2, unstable perturbations cannot be considered in this case. If we fix two components of the magnetic field, the third component is determined by the QB equation and no other components exist. For this reason, we treat the three-level system N = 3 in this subsection.

As an adiabatic Hamiltonian, we consider the magnetic field in the xy plane

$$\begin{equation} H_0(t) = h_1(t)S_1+h_2(t)S_2, \end{equation} \tag{ 69 }$$

$$\begin{equation} h_1(t)=h_0\cos \omega t, \end{equation} \tag{ 70 }$$

$$\begin{equation} h_2(t)=h_0\sin \omega t, \end{equation} \tag{ 71 }$$

where S₁ and S₂ are spin operators,

$$\begin{equation} S_1 = \frac{1}{\sqrt{2}} \left(\begin{array}{@{}ccc@{}}0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right), \end{equation} \tag{ 72 }$$

$$\begin{equation} S_2 = \frac{1}{\sqrt{2}} \left(\begin{array}{@{}ccc@{}}0 & -{\rm i} & 0 \\ {\rm i} & 0 & -{\rm i} \\ 0 & {\rm i} & 0 \end{array}\right), \end{equation} \tag{ 73 }$$

and can be expressed by linear combinations of the Gell–Mann matrices. The adiabatic state is given by

$$\begin{equation} |\psi _{\rm ad}(t)\rangle = \frac{1}{2} \left(\begin{array}{@{}cc@{}}\rm e^{-{\rm i}\omega t} \\ \sqrt{2} \\ {\rm e}^{{\rm i}\omega t} \end{array}\right). \end{equation} \tag{ 74 }$$

By using the formulas (27) and (28) of the transitionless quantum driving, we obtain the counter-diabatic Hamiltonian

$$\begin{equation} H_1(t)=\omega S_3 =\omega \left(\begin{array}{@{}ccc@{}}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{array}\right). \end{equation} \tag{ 75 }$$

From the general consideration, the driving should be stable against the variation ω → ω + δω(t) where δω(t) is an arbitrary function. It is also possible to consider perturbations inducing instabilities. For example, as operators which anticommute with S₃, we can use the Gell–Mann matrices λ₄ and λ₅.

We numerically solve the Schrödinger equation to calculate the fidelity

$$\begin{equation} f = |\langle \psi _{\rm ad}(t)|\psi (t)\rangle |^2, \end{equation} \tag{ 76 }$$

where |ψ_ad(t)〉 denotes the adiabatic state of H₀(t) in equation (74) and |ψ(t)〉 the solution of the Schrödinger equation with the Hamiltonian H₀(t) + H₁(t) + δH(t). We set the initial condition at t = 0 to be the eigenstate of S₁ with the eigenvalue +1 and calculate the probability of observing the eigenstate of S₂ with the eigenvalue +1.

We set parameters h₀ = 1 and ω = π/20. As possible perturbations, we consider the following four cases:

$$\begin{equation} \delta H(t) = \delta h(t)\times \left\lbrace \begin{array}{@{}l@{}}S_3 \\\ms 2\sqrt{\frac{2}{3}}\lambda _4 \\\ms 2\sqrt{\frac{2}{3}}\lambda _5 \\\ms \frac{4}{3}\lambda _8 \end{array} \right.. \end{equation} \tag{ 77 }$$

We set each coefficient so that the instability I(t), calculated respectively as

$$\begin{equation} I(t) = \frac{\delta h^2(t)}{\omega ^2}\times \left\lbrace \begin{array}{@{}l@{}}1 \\\ms -\frac{2}{3}[1+\sin ^2(2\omega t)] \\\ms -\frac{2}{3}[1+\cos ^2(2\omega t)] \\\ms -1 \end{array} \right., \end{equation} \tag{ 78 }$$

gives the same magnitude in average. We set δh(t) = 0.5.

The results are plotted in figures 1–4. We see in the stable case that the trajectory follows that of the ideal driving case while oscillating uniformly. In unstable cases, we see more nonuniform behavior and large deviations for λ₄ and λ₅. The deviation is relatively small for λ₈ but the oscillations are nonuniform compared with the stable case of S₃. We note that λ₈ commutes with the driving term S₃ and does not disturb the system significantly. Such a difference cannot be seen in the quantity I(t). We consider only the local instability of the solution of the QB equation. Although it is interesting to see nonlinear effects, such an analysis is beyond the scope of this study. The result here shows that the condition I(t) > 0 can be one of the criteria for the stable driving.

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