Interacting Stochastic Schrödinger Equation

Abstract

Being the annihilation and creation operators on the space $\mathfrak{h}$ of square integrable Bernoulli functionals, quantum Bernoulli noises (QBN) satisfy the canonical anti-commutation relation (CAR) in equal time. Let $\mathcal{K}$ be the Hilbert space of an open quantum system interacting with QBN (the environment). Then $\mathcal{K}\otimes \mathfrak{h}$ just describes the coupled quantum system. In this paper, we introduce and investigate an interacting stochastic Schrödinger equation (SSE) in the framework $\mathcal{K}\otimes \mathfrak{h}$, which might play a role in describing the evolution of the open quantum system interacting with QBN (the environment). We first prove some technical propositions about operators in $\mathcal{K}\otimes \mathfrak{h}$. In particular, we obtain the spectral decomposition of the tensor operator $I_{\mathcal{K}}\otimes N$, where $I_{\mathcal{K}}$ means the identity operator on $\mathcal{K}$ and N is the number operator in $\mathfrak{h}$, and give a representation of $I_{\mathcal{K}}\otimes N$ in terms of operators $I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k$, $k\ge 0$, where ${\partial }_k$ and ${\partial }_k^{\ast }$ are the annihilation and creation operators on $\mathfrak{h}$, respectively. Based on these technical propositions as well as Mora and Rebolledo’s results on a general SSE, we show that under some mild conditions, our interacting SSE has a unique solution admitting some regularity properties. Some other results are also proven.

1. Introduction

In quantum theory, quantum systems are divided into two categories: closed quantum systems and open quantum systems. A closed quantum system is isolated and has no interaction with the outside world. However, not all quantum systems are isolated. In practice, most quantum systems inevitably interact with their environments which influence them in a non-negligible way. Such quantum systems are known as open quantum systems [1,2,3].

Usually, an open quantum system interacts with another huge quantum system, namely, its environment is a huge quantum system. In that case, by joining them together, one gets a bigger quantum system, which is referred to as the coupled quantum system. The coupled quantum system can be regarded as a closed one, hence is subject to Hamiltonian dynamics.

Let ${\mathcal{H}}_S$ be the Hilbert space describing an open quantum system and ${\mathcal{H}}_E$ the Hilbert space describing its environment. Then, the coupled quantum system corresponds to the tensor space ${\mathcal{H}}_S\otimes {\mathcal{H}}_E$ and its Hamiltonian is given by

$$H=H_S\otimes I_E+H_I+I_S\otimes H_E,$$

where $I_S$ are $I_E$ are the identity operators on ${\mathcal{H}}_S$ and ${\mathcal{H}}_E$ respectively, $H_S$ stands for the Hamiltonian of the open quantum system, $H_E$ means the free Hamiltonian of the environment, and $H_I$ is the Hamiltonian describing the interaction between the open quantum system and its environment [1,3,4].

Quantum Bernoulli noises (QBN, for short) refer to the annihilation and creation operators $\{{\partial }_k^{\ast },{\partial }_k\mid k\ge 0\}$ acting on the space $\mathfrak{h}$ of square integrable Bernoulli functionals, which satisfy the canonical anti-commutation relation (CAR) in equal time [5,6]. Recent years have seen many applications of QBN in developing a discrete-time stochastic calculus in infinite dimensions. Indeed, in 2008, Privault [7] used the annihilation operators to define the gradients for Bernoulli functionals. Two years later, Nourdin et al. [8] investigated a normal approximation of Rademacher functionals (a special case of Bernoulli functionals) with the help of the annihilation operators. Recently, it has been shown that QBN can play an active role in the study of quantum Markov semigroups and quantum walks [9,10,11,12].

As is known, $\mathfrak{h}$ is a symmetric Hilbert space of infinite dimension and has an orthonormal basis $\{Z_{\sigma }\mid \sigma \in \Gamma \}$ indexed by the finite power set $\Gamma$ of the nonnegative integer set $\mathbb{N}$. And more importantly, as the annihilation and creation operators on $\mathfrak{h}$, QBN satisfies the CAR in equal time. Thus, from the perspective of mathematical physics, $\mathfrak{h}$ together with QBN is quite suitable for describing the environment of an open quantum system [13].

It is of great significance to develop effective tools dealing with the dynamics of open quantum systems. In the past few years, remarkable attention has been paid to the approach to open quantum systems provided by stochastic Schrödinger equations (SSE, for short), which are a class of stochastic differential equations dominated by Hamiltonian operators on complex Hilbert spaces. Barchielli, et al. [14,15] first investigated some linear SSEs with bounded operators as the coefficients and established the existence and uniqueness of strong solutions to such equations. Later, Holevo [16] obtained an existence and uniqueness result about weak topology solution for a general linear SSE with unbounded operators as the coefficients. In 2007, Mora and Rebolledo [17] further studied a more general class of SSEs and obtained the corresponding results. Now SSEs are widely used in different fields such as measurement theory, quantum optics, quantum chaos, solid states, etc, wherever quantum irreversibility matters [18,19]. They do not only serve as a fruitful theoretical concept but also as a practical method for computations in the form of quantum trajectories (see, e.g., [4,17,20,21] and references therein).

Let $\mathcal{K}$ be the Hilbert space of an open quantum system interacting with QBN (the environment). Then, the tensor space $\mathcal{K}\otimes \mathfrak{h}$ just represents the coupled quantum system. In this paper, we introduce in the framework $\mathcal{K}\otimes \mathfrak{h}$ an interacting SSE of the following type

$$X_t=\xi +{\int }_0^t\left\{-\mathrm{i}[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g\right)]-\frac{1}{2}{I}_{\mathcal{K}}\otimes N\right\}{X}_{s}ds+\sum _{k=0}^{\infty }{\int }_0^t{I}_{\mathcal{K}}\otimes \left({\partial }_k^{\ast }{\partial }_k\right)X_{s}d{W}_s^k,$$

(1)

with

$$\Phi \left(g\right)=\sum _{k=0}^{\infty }\left(\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast }\right),$$

(2)

where H and B are self-adjoint operators in $\mathcal{K}$, N is the number operator in $\mathfrak{h}$, g is a suitable function defined on the nonnegative integer set $\mathbb{N}$, and ${\left(W^k\right)}_{k\ge 0}$ is a sequence of independent real-valued Wiener processes on a filtered complete probability space $(\Omega ,\mathcal{F},{\left({\mathcal{F}}_t\right)}_{t\ge 0},\mathbb{P})$.

The physical meaning of Equation (1) lies in the following observations. As usual, the self-adjoint operator H can be viewed as the Hamiltonian of the open quantum system interacting with QBN (the environment), while the number operator N in $\mathfrak{h}$ represents the free Hamiltonian of QBN (the environment). Since $\Phi \left(g\right)={\sum }_{k=0}^{\infty }(\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast })$ is an operator in $\mathfrak{h}$, the tensor product operator $B\otimes \Phi \left(g\right)$ just reflects the interaction between the open quantum system and QBN (the environment), while the parameter $\lambda \ge 0$ means the coupling strength. The equation itself, then, describes a type of time evolution of the open quantum system interacting with QBN (the environment).

We mention that ref. [20] actually considers in the framework $\mathfrak{h}$ (the space of square integrable Bernoulli functionals as mentioned above) an SSE of the form

$$X_t=\xi +{\int }_0^t\left(-\mathrm{i}H-\frac{1}{2}N\right)X_{s}ds+\sum _{k=0}^{\infty }{\int }_0^t{\partial }_k^{\ast }{\partial }_k{X}_{s}d{W}_s^k,$$

(3)

where H is a self-adjoint operator in $\mathfrak{h}$ and N is the number operator in $\mathfrak{h}$. Recently, in the same framework, ref. [21] investigates a SSE of exclusion type, which reads

$$X_t=\xi +{\int }_0^{t}G{X}_{s}ds+\sum _{j,k=0}^{\infty }\sqrt{w(j,k)}{\int }_0^t{\partial }_j^{\ast }{\partial }_k{X}_{s}d{W}_s^{j,k},$$

(4)

where G is a suitable operator in $\mathfrak{h}$, w is a nonnegative function defined on ${\mathbb{N}}^2$, and $W^{j,k}$, j, $k\ge 0$ are independent real-valued Wiener processes. Clearly, as SSEs in the framework $\mathfrak{h}$, both Equations (3) and (4) do not belong to the category of interacting SSEs. Thus, from a perspective of mathematical physics, our interacting SSE (1) essentially differs from those SSEs considered in refs. [20,21]. Additionally, as far as we know, interacting SSEs like ours have not been considered yet in the literature. Finally, we mention that there are a lot of researches on stochastic equations of fractional order in the setting of Banach spaces (see, e.g., [22,23] and references therein).

Let us now briefly describe our main work in this paper as follows. We first prove in Section 2 several technical propositions about operators in $\mathcal{K}\otimes \mathfrak{h}$. In particular, we obtain the spectral decomposition of the tensor operator $I_{\mathcal{K}}\otimes N$ (see Proposition 3), where $I_{\mathcal{K}}$ denotes the identity operator on $\mathcal{K}$ and N is the number operator in $\mathfrak{h}$, and give a representation of $I_{\mathcal{K}}\otimes N$ in terms of operators $I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k$, $k\ge 0$ (see Proposition 4). And then, in Section 3, by using these technical propositions as well as Mora and Rebolledo’s results on a general SSE, we show that, under some mild conditions, our interacting SSE Equation (1) has a unique solution admitting some regularity properties (see Theorem 4 and Theorem 6). Some other results are also proven therein.

Throughout, $\mathbb{N}$, $\mathbb{R}$ and $\mathbb{C}$ stand for the set of nonnegative integers, the set of real numbers and the set of complex numbers, respectively. If z is a complex number, then $\Re z$ and $\Im z$ denote its real and imaginary parts, respectively. For a mapping A, we denote by $\mathrm{Dom}A$ its domain. If A is a densely defined operator in a Hilbert space, then $A^{\ast }$ means its adjoint operator.

2. Technical Preparations

In this section, we make some technical preparations, which will play an important role in explaining the precise meaning of Equation (1) and in proving our main theorems on Equation (1). We refer to Appendix A.2 for main notions, facts and notation about the space $\mathfrak{h}$ of square-integrable Bernoulli functionals and quantum Bernoulli noises (QBN).

Recall that $\mathcal{K}$ is the Hilbert space of an open quantum system interacting with QBN (the environment), while $\mathfrak{h}$ is the space where QBN lives. Thus $\mathcal{K}\otimes \mathfrak{h}$ just acts as the Hilbert space of the coupled system. We fix an ONB ${\left\{e_j\right\}}_{j\ge 1}$ for $\mathcal{K}$. Then the system $\{e_j\otimes Z_{\sigma }\mid j\ge 1,\sigma \in \Gamma \}$ forms an ONB for $\mathcal{K}\otimes \mathfrak{h}$, where $\{Z_{\sigma }\mid \sigma \in \Gamma \}$ is the canonical ONB for $\mathfrak{h}$. We use $\langle ·,·\rangle$ and $\parallel ·\parallel$ to mean the inner product and norm in $\mathcal{K}\otimes \mathfrak{h}$ respectively, while we write ${\langle ·,·\rangle }_{\mathcal{K}}$ and ${\parallel ·\parallel }_{\mathcal{K}}$ for the inner product and norm in $\mathcal{K}$ respectively.

Let S and T be densely-defined symmetric operators in $\mathcal{K}$ and $\mathfrak{h}$, respectively. Then, according to the general theory of operators in Hilbert spaces (see, e.g., [24,25,26]), their tensor product $S\otimes T$ is defined as the closure of the densely-defined symmetric (hence, closable) operator $S{\otimes }_{0}T$ in $\mathcal{K}\otimes \mathfrak{h}$ given by

$$\left(S{\otimes }_{0}T\right)(x\otimes \xi )=\left(Sx\right)\otimes \left(T\xi \right),x\in \mathrm{Dom}S,\xi \in \mathrm{Dom}T$$

(5)

and $\mathrm{Dom}\left(S{\otimes }_{0}T\right)=\mathrm{span}\{x\otimes \xi \mid x\in \mathrm{Dom}S,\xi \in \mathrm{Dom}T\}$, where the symbol $\mathrm{span}\{\star \}$ means the linear subspace spanned by a vector set $\{\star \}$. It is known that $S\otimes T$ remains symmetric, hence a closed symmetric operator in $\mathcal{K}\otimes \mathfrak{h}$. Further, if S and T are bounded symmetric operators, then $S\otimes T$ coincides with the usual tensor product of S and T as bounded operators.

Lemma 1.

[26] Let S and T be self-adjoint operators in $\mathcal{K}$ and $\mathfrak{h}$, respectively. Then, $S\otimes T$ is a self-adjoint operator in $\mathcal{K}\otimes \mathfrak{h}$. In particular, both $S\otimes I_{\mathfrak{h}}$ and $I_{\mathcal{K}}\otimes T$ are self-adjoint operators in $\mathcal{K}\otimes \mathfrak{h}$, where $I_{\mathfrak{h}}$ and $I_{\mathcal{K}}$ are the identity operators on $\mathfrak{h}$ and $\mathcal{K}$, respectively.

For a function $g:\mathbb{N}\to \mathbb{C}$, the operator $\Phi \left(g\right)={\sum }_{k=0}^{\infty }(\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast })$ in $\mathfrak{h}$ is naturally defined as

$$\Phi \left(g\right)\xi =\sum _{k=0}^{\infty }\left(\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast }\right)\xi ,\xi \in \mathrm{Dom}\Phi \left(g\right),$$

(6)

where $\mathrm{Dom}\Phi \left(g\right)$ consists of vectors $\xi \in \mathfrak{h}$ such that the series ${\sum }_{k=0}^{\infty }(\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast })\xi$ converges in norm. In what follows, for $p\ge 1$, we denote by ${\ell }^p\left(\mathbb{N}\right)$ the space of all complex-valued functions g on $\mathbb{N}$ satisfying that

$$\sum _{k=0}^{\infty }{|g\left(k\right)|}^p<\infty ,$$

(7)

and write ${\parallel ·\parallel }_{{\ell }^p}$ for the norm in ${\ell }^p\left(\mathbb{N}\right)$.

Proposition 1.

For all $g\in {\ell }^2\left(\mathbb{N}\right)$, $\Phi \left(g\right)$ is a densely-defined symmetric operator in $\mathfrak{h}$. Moreover, if $g\in {\ell }^1\left(\mathbb{N}\right)$, then $\Phi \left(g\right)$ is a self-adjoint bounded operator on $\mathfrak{h}$ with $\parallel \Phi \left(g\right)\parallel \le {2\parallel g\parallel }_{{\ell }^1}$.

Proof. 

Let $g\in {\ell }^2\left(\mathbb{N}\right)$ be given. Then, for all $\sigma \in \Gamma$ and all nonnegative integers m and n with $m\le n$, we have

$$\begin{array}{cc}\hfill \parallel \sum _{k=m}^n(\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast })Z_{\sigma }{\parallel }_{\mathfrak{h}}^2& \le 2\left[\parallel \sum _{k=m}^n\overline{g\left(k\right)}{\partial }_k{Z}_{\sigma }{\parallel }_{\mathfrak{h}}^2+\parallel \sum _{k=m}^{n}g\left(k\right){\partial }_k^{\ast }{Z}_{\sigma }{\parallel }_{\mathfrak{h}}^2\right]\hfill \\ \hfill & =2\left[\parallel \sum _{k=m}^n\overline{g\left(k\right)}{\mathbf{1}}_{\sigma }\left(k\right)Z_{\sigma \setminus k}{\parallel }_{\mathfrak{h}}^2+\parallel \sum _{k=m}^{n}g\left(k\right)[1-{\mathbf{1}}_{\sigma }\left(k\right)]Z_{\sigma \cup k}{\parallel }_{\mathfrak{h}}^2\right]\hfill \\ \hfill & =2\left[\sum _{k=m}^n|\overline{g\left(k\right)}{|}^2{\mathbf{1}}_{\sigma }\left(k\right)+\sum _{k=m}^n{|g\left(k\right)|}^2[1-{\mathbf{1}}_{\sigma }\left(k\right)]\right]\hfill \\ \hfill & \le 4\sum _{k=m}^n{|g\left(k\right)|}^2,\hfill \end{array}$$

which, together with ${\sum }_{k=0}^{\infty }{|g\left(k\right)|}^2<\infty$, implies that the series ${\sum }_{k=0}^{\infty }(\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast })Z_{\sigma }$ converges in norm, hence $Z_{\sigma }\in \mathrm{Dom}\Phi \left(g\right)$. Therefore, $\{Z_{\sigma }\mid \sigma \in \Gamma \}\subset \mathrm{Dom}\Phi \left(g\right)$, which implies that $\mathrm{Dom}\Phi \left(g\right)$ is a dense in $\mathfrak{h}$, namely, $\Phi \left(g\right)$ is a densely-defined operator in $\mathfrak{h}$.

It is easy to show that $\Phi \left(g\right)$ is symmetric. Now suppose that $g\in {\ell }^1\left(\mathbb{N}\right)$. Then, in view of the fact that $\parallel {\partial }_k\parallel =\parallel {\partial }_k^{\ast }\parallel =1$ for all $k\ge 0$, we find

$$\sum _{k=1}^{\infty }\parallel (\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast })\parallel \le \sum _{k=1}^{\infty }\left(|\overline{g\left(k\right)}|\parallel {\partial }_k\parallel +|g\left(k\right)|\parallel {\partial }_k^{\ast }\parallel \right)=2\sum _{k=1}^{\infty }|g\left(k\right)|={2\parallel g\parallel }_{{\ell }^1},$$

which implies that the operator series ${\sum }_{k=1}^{\infty }(\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast })$ converges in operator norm. Thus, $\mathrm{Dom}\Phi \left(g\right)=\mathfrak{h}$ and $\Phi \left(g\right)$ is a bounded operator on $\mathfrak{h}$ with $\parallel \Phi \left(g\right)\parallel \le {2\parallel g\parallel }_{{\ell }^1}$. Finally, using the symmetric property of $\Phi \left(g\right)$, we know that $\Phi \left(g\right)$ is self-adjoint. □

Proposition 2.

Let B be a self-adjoint operator in $\mathcal{K}$. Then, for each $g\in {\ell }^2\left(\mathbb{N}\right)$, $B\otimes \Phi \left(g\right)$ is a densely-defined closed symmetric operator in $\mathcal{K}\otimes \mathfrak{h}$.

Proof. 

B is a densely-defined symmetric operator in $\mathcal{K}$ since it is self-adjoint. On the other hand, using Proposition 1, we know that $\Phi \left(g\right)$ is also a densely-defined symmetric operator in $\mathfrak{h}$. Thus, by definition, $B\otimes \Phi \left(g\right)$ is a densely-defined closed symmetric operator in $\mathcal{K}\otimes \mathfrak{h}$. □

Recall that the number operator N in $\mathfrak{h}$ is a self-adjoint one. Thus, by Lemma 1, $I_{\mathcal{K}}\otimes N$ is a self-adjoint operator in $\mathcal{K}\otimes \mathfrak{h}$. The next proposition actually gives its spectral decomposition.

Proposition 3.

It holds true that $I_{\mathcal{K}}\otimes N={\sum }_{\sigma \in \Gamma }(\#\sigma )I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|$, where $|Z_{\sigma }\rangle \langle Z_{\sigma }|$ means the Dirac operator on $\mathfrak{h}$ associated with the basis vector $Z_{\sigma }$.

Proof. 

Write $\tilde{N}={\sum }_{\sigma \in \Gamma }(\#\sigma )I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|$. Then, by the general theory of spectral integrals [27], $\tilde{N}$ is the self-adjoint operator in $\mathcal{K}\otimes \mathfrak{h}$ given by

$$\tilde{N}u=\sum _{\sigma \in \Gamma }(\#\sigma )I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|u,u\in \mathrm{Dom}\tilde{N}$$

with

$$\mathrm{Dom}\tilde{N}=\left\{u\in \mathcal{K}\otimes \mathfrak{h}|\sum _{\sigma \in \Gamma }{(\#\sigma )}^2\parallel I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }{|u\parallel }^2<\infty \right\}.$$

Next, we prove that $I_{\mathcal{K}}\otimes N=\tilde{N}$. To this end, we consider the restriction $I_{\mathcal{K}}{\otimes }_{0}N$ of $I_{\mathcal{K}}\otimes N$ to

$$\mathcal{D}=\mathrm{span}\{x\otimes \xi \mid x\in \mathcal{K},\xi \in \mathrm{Dom}N\},$$

which is a dense subspace of $\mathcal{K}\otimes \mathfrak{h}$. Then, by the definition of $I_{\mathcal{K}}\otimes N$, we know that $I_{\mathcal{K}}\otimes N$ is the closure of $I_{\mathcal{K}}{\otimes }_{0}N$, equivalently $I_{\mathcal{K}}\otimes N={\left(I_{\mathcal{K}}{\otimes }_{0}N\right)}^{\ast \ast }$.

Let $x\in \mathcal{K}$ and $\xi \in \mathrm{Dom}N$ be given. A straightforward calculation gives

$$\sum _{\sigma \in \Gamma }{(\#\sigma )}^2\parallel (I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|)(x\otimes \xi ){\parallel }^2={\parallel x\parallel }^2\sum _{\sigma \in \Gamma }{(\#\sigma )}^2{|{\langle Z_{\sigma },\xi \rangle }_{\mathfrak{h}}|}^2<\infty ,$$

which implies that $x\otimes \xi \in \mathrm{Dom}\tilde{N}$. Additionally, by the definitions of $\tilde{N}$ and N, we have

$$\tilde{N}(x\otimes \xi )=x\otimes \sum _{\sigma \in \Gamma }(\#\sigma ){\langle Z_{\sigma },\xi \rangle }_{\mathfrak{h}}{Z}_{\sigma }=x\otimes N\xi =(I_{\mathcal{K}}\otimes N)(x\otimes \xi )=\left(I_{\mathcal{K}}{\otimes }_{0}N\right)(x\otimes \xi ).$$

Thus $\mathcal{D}\subset \mathrm{Dom}\tilde{N}$ and $\tilde{N}\phi =\left(I_{\mathcal{K}}{\otimes }_{0}N\right)\phi$, $\forall \phi \in \mathcal{D}$, namely $I_{\mathcal{K}}{\otimes }_{0}N\subset \tilde{N}$, which together with $\tilde{N}={\tilde{N}}^{\ast }$ implies that $\tilde{N}\subset {\left(I_{\mathcal{K}}{\otimes }_{0}N\right)}^{\ast }$. Thus, ${\left(I_{\mathcal{K}}{\otimes }_{0}N\right)}^{\ast \ast }\subset {\tilde{N}}^{\ast }=\tilde{N}$, which together with $I_{\mathcal{K}}\otimes N={\left(I_{\mathcal{K}}{\otimes }_{0}N\right)}^{\ast \ast }$ yields $I_{\mathcal{K}}\otimes N\subset \tilde{N}$. Since both $I_{\mathcal{K}}\otimes N$ and $\tilde{N}$ are self-adjoint operators in $\mathcal{K}\otimes \mathfrak{h}$, we finally know that $I_{\mathcal{K}}\otimes N=\tilde{N}$. □

The next proposition further shows that the operator $I_{\mathcal{K}}\otimes N$ has a representation in terms of the identity operator $I_{\mathcal{K}}$ on $\mathcal{K}$ as well as the product operators ${\partial }_k^{\ast }{\partial }_k$, $k\ge 0$, on $\mathfrak{h}$.

Proposition 4.

Let $u\in \mathcal{K}\otimes \mathfrak{h}$ be a vector in $\mathcal{K}\otimes \mathfrak{h}$. Then $u\in \mathrm{Dom}(I_{\mathcal{K}}\otimes \mathrm{N})$ if and only if the vector series ${\sum }_{k=0}^{\infty }(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u$ converges in norm. In that case, one has

$$\begin{array}{c}\hfill (I_{\mathcal{K}}\otimes \mathrm{N})u=\sum _{k=0}^{\infty }(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u.\end{array}$$

(8)

Proof. 

The “if” part. It follows from the norm convergence of the vector series ${\sum }_{k=0}^{\infty }(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u$ that there exist a finite constant $c\ge 0$ such that

$$\parallel \sum _{k=0}^n(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u\parallel \le c,\forall n\ge 0.$$

On the other hand, by using the continuity of operator $I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k$, we have

$$\begin{array}{cc}\hfill \sum _{k=0}^n(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u& =\sum _{k=0}^n(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)\sum _{j=1}^{\infty }\sum _{\sigma \in \Gamma }\langle e_j\otimes Z_{\sigma },u\rangle (e_j\otimes Z_{\sigma })\hfill \\ \hfill & =\sum _{j=1}^{\infty }\sum _{\sigma \in \Gamma }\left(\sum _{k=0}^n{\mathbf{1}}_{\sigma }\left(k\right)\right)\langle e_j\otimes Z_{\sigma },u\rangle (e_j\otimes Z_{\sigma }).\hfill \end{array}$$

Thus,

$$\parallel \sum _{k=0}^n(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u{\parallel }^2=\sum _{j=1}^{\infty }\sum _{\sigma \in \Gamma }{\left(\sum _{k=0}^n{\mathbf{1}}_{\sigma }\left(k\right)\right)}^2{|\langle e_j\otimes Z_{\sigma },u\rangle |}^2\le c^2<\infty ,$$

which, together with ${\sum }_{k=0}^{\infty }{\mathbf{1}}_{\sigma }\left(k\right)=\#\sigma$ as well as the Fatou’s theorem, gives

$$\sum _{\sigma \in \Gamma }{(\#\sigma )}^2\parallel (I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|){u\parallel }^2=\sum _{j=1}^{\infty }\sum _{\sigma \in \Gamma }{(\#\sigma )}^2{|\langle e_j\otimes Z_{\sigma },u\rangle |}^2\le c^2<\infty ,$$

which together with Proposition 3 implies that $u\in \mathrm{Dom}(I_{\mathcal{K}}\otimes N)$.

The “only if” part. Let $u\in \mathrm{Dom}(I_{\mathcal{K}}\otimes N)$ be given. Then, by Proposition 3, we know

$$\sum _{j=1}^{\infty }\sum _{\sigma \in \Gamma }{(\#\sigma )}^2|\langle e_j\otimes Z_{\sigma },u\rangle {|}^2=\sum _{\sigma \in \Gamma }{(\#\sigma )}^2{\parallel (I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|)u\parallel }^2<\infty .$$

On the other hand, we have

$${\left(\sum _{k=m}^{\infty }{\mathbf{1}}_{\sigma }\left(k\right)\right)}^2|\langle e_j\otimes Z_{\sigma },u\rangle {|}^2\le {(\#\sigma )}^2{|\langle e_j\otimes Z_{\sigma },u\rangle |}^2,\forall \sigma \in \Gamma ,j\ge 1,m\ge 0,$$

and for all $\sigma \in \Gamma ,j\in \mathbb{N}$,

$${\left(\sum _{k=m}^{\infty }{\mathbf{1}}_{\sigma }\left(k\right)\right)}^2{|\langle e_j\otimes Z_{\sigma },u\rangle |}^2\to 0(m\to \infty ).$$

Thus, by the well known dominated convergence theorem, we find

$$\sum _{j=1}^{\infty }\sum _{\sigma \in \Gamma }{\left(\sum _{k=m}^{\infty }{\mathbf{1}}_{\sigma }\left(k\right)\right)}^2{|\langle e_j\otimes Z_{\sigma },u\rangle |}^2\to \sum _{j=1}^{\infty }\sum _{\sigma \in \Gamma }0=0(m\to \infty ).$$

Now, for m, $n\ge 0$ with $m\le n$, by using properties of operator ${\partial }_k^{\ast }{\partial }_k$, we can get

$$\parallel \sum _{k=m}^n(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u{\parallel }^2=\sum _{j=1}^{\infty }\sum _{\sigma \in \Gamma }{\left(\sum _{k=m}^n{\mathbf{1}}_{\sigma }\left(k\right)\right)}^2|\langle e_j\otimes Z_{\sigma },u\rangle {|}^2\le \sum _{j=1}^{\infty }\sum _{\sigma \in \Gamma }{\left(\sum _{k=m}^{\infty }{\mathbf{1}}_{\sigma }\left(k\right)\right)}^2{|\langle e_j\otimes Z_{\sigma },u\rangle |}^2.$$

Thus

$$\parallel \sum _{k=m}^n(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u{\parallel }^2\to 0(m,n\to \infty ),$$

which implies that the series ${\sum }_{k=0}^{\infty }(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u$ converges in norm.

Finally, we verify Equality (8). To do so, we take $u\in \mathrm{Dom}(I_{\mathcal{K}}\otimes N)$. Then, by Proposition 3 as well as the dominated convergence theorem, we have

$$\begin{array}{c}\parallel (I_{\mathcal{K}}\otimes N)u-\sum _{k=0}^n(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u{\parallel }^2\hfill \\ =\parallel \sum _{\sigma \in \Gamma }(\#\sigma )(I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|)u-\sum _{k=0}^n(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u{\parallel }^2\hfill \\ =\parallel \sum _{j=1}^{\infty }\sum _{\sigma \in \Gamma }\#\sigma \langle e_j\otimes Z_{\sigma },u\rangle (e_j\otimes Z_{\sigma })-\sum _{j=1}^{\infty }\sum _{\sigma \in \Gamma }\left(\sum _{k=0}^n{\mathbf{1}}_{\sigma }\left(k\right)\right)\langle e_j\otimes Z_{\sigma },u\rangle (e_j\otimes Z_{\sigma }){\parallel }^2\hfill \\ =\sum _{\sigma \in \Gamma }\sum _{j=1}^{\infty }{\left(\sum _{k=n+1}^{\infty }{\mathbf{1}}_{\sigma }\left(k\right)\right)}^2{|\langle e_j\otimes Z_{\sigma },u\rangle |}^2\to 0(n\to \infty ).\hfill \end{array}$$

Therefore Equality (8) holds. □

Remark 1.

According to Proposition 4, the operator $I_{\mathcal{K}}\otimes N$ can actually be represented as

$$I_{\mathcal{K}}\otimes N=\sum _{k=0}^{\infty }{I}_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k.$$

This shows the close links between $I_{\mathcal{K}}\otimes N$ and the family $\{I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k\mid k\ge 0\}$ of bounded operators on $\mathcal{K}\otimes \mathfrak{h}$.

Recall that $I_{\mathcal{K}}\otimes N$ is a self-adjoint operator in $\mathcal{K}\otimes \mathfrak{h}$. Hence, for any $r\ge 0$, ${(I_{\mathcal{K}}\otimes N)}^r$ makes sense as a self-adjoint operator in $\mathcal{K}\otimes \mathfrak{h}$. In fact, using Proposition 3, ${(I_{\mathcal{K}}\otimes N)}^r$ can be directly defined as follows.

$${(I_{\mathcal{K}}\otimes N)}^{r}u=\sum _{\sigma \in \Gamma }{(\#\sigma )}^r(I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|)u,u\in \mathrm{Dom}{(I_{\mathcal{K}}\otimes N)}^r,$$

(9)

where

$$\mathrm{Dom}{(I_{\mathcal{K}}\otimes N)}^r=\left\{u\in \mathcal{K}\otimes \mathfrak{h}|\sum _{\sigma \in \Gamma }{(\#\sigma )}^{2r}{\parallel (I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|)u\parallel }^2<\infty \right\}.$$

(10)

According to Lemma 1, $I_{\mathcal{K}}\otimes N^r$ makes sense as a self-adjoint operator in $\mathcal{K}\otimes \mathfrak{h}$ for each real number $r\ge 0$. Using the same method as in the proof of Proposition 3, we can prove the following useful result.

Proposition 5.

Let $r\ge 0$ be a real number. Then it holds true that ${(I_{\mathcal{K}}\otimes N)}^r=I_{\mathcal{K}}\otimes N^r$.

3. Solutions to Interacting SSE

In this section, we consider the existence and uniqueness of a regular solution to Equation (1) in the framework of $\mathcal{K}\otimes \mathfrak{h}$, where, as shown above, $\mathcal{K}$ is the Hilbert space of an open quantum system interacting with QBN (the environment), and $\mathfrak{h}$ is the space of square integrable Bernoulli functionals, which describes QBN (the environment). We will freely use notions and known results about a general SSE, which are collected in Appendix A.1.

Recall that ${\left(W^k\right)}_{k\ge 0}$ is a sequence of independent real-valued Wiener processes on a filtered complete probability space $(\Omega ,\mathcal{F},{\left({\mathcal{F}}_t\right)}_{t\ge 0},\mathbb{P})$. In what follows, we use $\mathbb{E}$ to mean the expectation with respect to $\mathbb{P}$, and by “a.s.” we mean “almost surely with respect to $\mathbb{P}$”.

We note that H and B appearing in Equation (1) are self-adjoint operators in $\mathcal{K}$. Hence, by Lemma 1, $H\otimes I_{\mathfrak{h}}$ is a self-adjoint operator in $\mathcal{K}\otimes \mathfrak{h}$ and, by Proposition 2, $B\otimes \Phi \left(g\right)$ is a densely-defined closed symmetric operator in $\mathcal{K}\otimes \mathfrak{h}$ for each $g\in {\ell }^2\left(\mathbb{N}\right)$. Additionally, we always assume that the parameter $\lambda \ge 0$ in Equation (1) is given.

Theorem 1.

Suppose that $g\in {\ell }^2\left(\mathbb{N}\right)$ and $\mathrm{Dom}{I}_{\mathcal{K}}\otimes N\subset \mathrm{Dom}H\otimes I_{\mathfrak{h}}\cap \mathrm{Dom}B\otimes \Phi \left(g\right)$. Then, Equation (1) satisfies the fundamental hypothesis as indicated in Definition A1 of the Appendix A.

Proof. 

Let $G=-i[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g\right)]-\frac{1}{2}{I}_{\mathcal{K}}\otimes N$ and $L_k=I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k$ for $k\ge 0$. Then, by the assumptions, $\mathrm{Dom}G=\mathrm{Dom}{I}_{\mathcal{K}}\otimes N$. Clearly, $\mathrm{Dom}{L}_k=\mathcal{K}\otimes \mathfrak{h}$, $k\ge 0$, which means that

$$\mathrm{Dom}G\subset \mathrm{Dom}{L}_k,k\ge 0.$$

Now let $u\in \mathrm{Dom}G$ and write $\tilde{N}=I_{\mathcal{K}}\otimes N$. Then, by the symmetric property of $H\otimes I_{\mathfrak{h}}$ and $B\otimes \Phi \left(g\right)$, we have

$$2\Re \langle u,Gu\rangle =2\Re [-i\langle u,(H\otimes I_{\mathfrak{h}})u\rangle -i\lambda \langle u,(B\otimes \Phi \left(g\right))u\rangle -\frac{1}{2}\langle u,\tilde{N}u\rangle ]=-\langle u,\tilde{N}u\rangle .$$

On the other hand, for each $k\ge 0$, $I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k$ is a projection operator on $\mathcal{K}\otimes \mathfrak{h}$ since ${\partial }_k^{\ast }{\partial }_k$ is a projection operator on $\mathfrak{h}$ (see Appendix A.2). This, together with Proposition 4, gives

$$\sum _{k=0}^{\infty }\parallel L_k{u\parallel }^2=\sum _{k=0}^{\infty }{\parallel (I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u\parallel }^2=\sum _{k=0}^{\infty }\langle u,(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u\rangle =\langle u,\tilde{N}u\rangle .$$

Thus

$$2\Re \langle u,Gu\rangle +\sum _{k=0}^{\infty }{\parallel L_{k}u\parallel }^2=0,u\in \mathrm{Dom}G.$$

This means that Equation (1) satisfies the fundamental hypothesis as indicated in Definition A1 of the Appendix A.1. □

Recall that ${\Gamma }_{n]}$ denotes the collection of all subsets of $\{0,1,\cdots ,n\}$. In what follows, for $n\ge 1$, we set ${\mathcal{H}}_n=\mathrm{span}\{e_j\otimes Z_{\sigma }\mid \sigma \in {\Gamma }_{n]},1\le j\le n\}$, which is a finite-dimensional subspace of $\mathcal{K}\otimes \mathfrak{h}$. we denote by $P_n$ the projection operator from $\mathcal{K}\otimes \mathfrak{h}$ onto ${\mathcal{H}}_n$.

Theorem 2.

Let $r\ge 0$ be a nonnegative real number. Then, for all $n\ge 1$, ${(I_{\mathcal{K}}\otimes N)}^r{P}_n$ makes sense, and moreover, ${(I_{\mathcal{K}}\otimes N)}^r{P}_n=P_n{(I_{\mathcal{K}}\otimes N)}^r$ on $\mathrm{Dom}{(I_{\mathcal{K}}\otimes N)}^r$.

Proof. 

For brevity, we write $\tilde{N}=I_{\mathcal{K}}\otimes N$. Let $n\ge 1$. Clearly, ${\mathcal{H}}_n\subset \mathrm{Dom}{\tilde{N}}^r$, which, together with the fact that ${\mathcal{H}}_n$ is the range of $P_n$, implies that ${\tilde{N}}^r{P}_n$ makes sense. Now suppose that $u\in \mathrm{Dom}{\tilde{N}}^r$. Then, it follows from the definitions of ${\tilde{N}}^r$ and $P_n$ as well as the equality that

$$\begin{array}{cc}\hfill {\tilde{N}}^r{P}_{n}u& =\sum _{\sigma \in \Gamma }{(\#\sigma )}^r(I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|)P_{n}u\hfill \\ \hfill & =\sum _{\sigma \in \Gamma }{(\#\sigma )}^r(I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|)\sum _{j=1}^n\sum _{\tau \in {\Gamma }_{n]}}\langle e_j\otimes Z_{\tau },u\rangle (e_j\otimes Z_{\tau })\hfill \\ \hfill & =\sum _{\sigma \in {\Gamma }_{n]}}{(\#\sigma )}^r\sum _{j=1}^n\langle e_j\otimes Z_{\sigma },u\rangle (e_j\otimes Z_{\sigma })\hfill \\ \hfill & =\sum _{j=1}^n\sum _{\sigma \in {\Gamma }_{n]}}{(\#\sigma )}^r\langle e_j\otimes Z_{\sigma },u\rangle (e_j\otimes Z_{\sigma }).\hfill \end{array}$$

By using the equality $\langle e_j\otimes Z_{\tau },{\tilde{N}}^{r}u\rangle ={(\#\tau )}^r\langle e_j\otimes Z_{\tau },u\rangle$, $j\ge 1$, $\tau \in \Gamma$, we can similarly get

$$P_n{\tilde{N}}^{r}u=\sum _{j=1}^n\sum _{\tau \in {\Gamma }_{n]}}\langle e_j\otimes Z_{\tau },{\tilde{N}}^{r}u\rangle (e_j\otimes Z_{\tau })=\sum _{j=1}^n\sum _{\tau \in {\Gamma }_{n]}}{(\#\tau )}^r\langle e_j\otimes Z_{\tau },u\rangle (e_j\otimes Z_{\tau }).$$

Thus ${\tilde{N}}^r{P}_n=P_n{\tilde{N}}^r$. This completes the proof. □

Theorem 3.

Let $r\ge 1$ be a real number and $g\in {\ell }^2\left(\mathbb{N}\right)$ be a given function. Suppose that

$$\mathrm{Dom}{I}_{\mathcal{K}}\otimes N\subset \mathrm{Dom}H\otimes I_{\mathfrak{h}}\cap \mathrm{Dom}B\otimes \Phi \left(g\right)$$

and there exist two finite constants a, $b\ge 0$ such that

$$\Im 〈{(I_{\mathcal{K}}\otimes N)}^{2r}u,[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g\right)]u〉\le a\left(\parallel {(I_{\mathcal{K}}\otimes N)}^r{u\parallel }^2+{\parallel u\parallel }^2+b\right),$$

where $u\in \mathrm{Dom}{(I_{\mathcal{K}}\otimes N)}^{2r}$. Then ${(I_{\mathcal{K}}\otimes N)}^r$ is a reference operator of Equation (1).

Proof. 

As before, we write $\tilde{N}=I_{\mathcal{K}}\otimes N$, $G=-i[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g\right)]-\frac{1}{2}\tilde{N}$ and $L_k=I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k$ for $k\ge 0$. Obviously, $\mathrm{Dom}{\tilde{N}}^r\subset \mathrm{Dom}\tilde{N}$. Additionally, by the assumptions as well as Proposition 3, we have $\mathrm{Dom}G=\mathrm{Dom}\tilde{N}$ and $\mathrm{Dom}G\subset \mathrm{Dom}{G}^{\ast }$. Thus $\mathrm{Dom}{\tilde{N}}^r\subset \mathrm{Dom}G\cap \mathrm{Dom}{G}^{\ast }$.

Recall that the system $\{e_j\otimes Z_{\sigma }\mid j\ge 1,\sigma \in \Gamma \}$ is an ONB for $\mathcal{K}\otimes \mathfrak{h}$. For $j\ge 1$, $\tau \in \Gamma$, by a simple calculation, we find

$$\sum _{\sigma \in \Gamma }{(\#\sigma )}^{2r}\parallel (I_{\mathcal{K}}\otimes |Z_{\sigma }\rangle \langle Z_{\sigma }|)(e_j\otimes Z_{\tau }){\parallel }^2=\sum _{\sigma \in \Gamma }{(\#\sigma )}^{2r}{\parallel e_j\otimes \langle Z_{\sigma },Z_{\tau }\rangle Z_{\sigma }\parallel }^2={(\#\tau )}^{2r}<\infty .$$

Thus $\{e_j\otimes Z_{\sigma }\mid j\ge 1,\sigma \in \Gamma \}\subset \mathrm{Dom}{\tilde{N}}^r$.

Also for $j\ge 1$, $\tau \in \Gamma$, by using $L_k^{\ast }=L_k$, we have

$$\sum _{k=0}^{\infty }\parallel L_k^{\ast }(e_j\otimes Z_{\tau }){\parallel }^2=\sum _{k=0}^{\infty }\parallel (I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)(e_j\otimes Z_{\tau }){\parallel }^2=\sum _{k=0}^{\infty }{\parallel e_j\otimes \left({\mathbf{1}}_{\tau }\left(k\right)Z_{\tau }\right)\parallel }^2=\#\tau <\infty .$$

For all $n\ge 1$ and $u\in {\mathcal{H}}_n$, by direct computations we have

$${\tilde{N}}^{m}u\in {\mathcal{H}}_n\subset \mathrm{Dom}{\tilde{N}}^m,P_n{\tilde{N}}^{m}u={\tilde{N}}^{m}u,\forall m\ge 1$$

and

$$P_n(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u=(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u,{\tilde{N}}^r(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u=(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k){\tilde{N}}^{r}u,\forall k\ge 0,$$

which implies that

$$2\Re \langle {\tilde{N}}^{r}u,{\tilde{N}}^r{P}_{n}Gu\rangle =2\Re \langle P_n{\tilde{N}}^{2r}u,Gu\rangle =2\Re \langle {\tilde{N}}^{2r}u,Gu\rangle$$

and

$$\sum _{k=0}^{\infty }\parallel {\tilde{N}}^r{P}_n(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k){u\parallel }^2=\sum _{k=0}^{\infty }{\parallel (I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k){\tilde{N}}^{r}u\parallel }^2=\langle {\tilde{N}}^{2r}u,\tilde{N}u\rangle .$$

Thus

$$\begin{array}{cc}\hfill 2\Re & \langle {\tilde{N}}^{r}u,{\tilde{N}}^r{P}_{n}Gu\rangle +\sum _{k=0}^{\infty }{\parallel {\tilde{N}}^r{P}_n(I_{\mathcal{K}}\otimes {\partial }_k^{\ast }{\partial }_k)u\parallel }^2\hfill \\ \hfill & =2\Re \langle {\tilde{N}}^{2r}u,Gu\rangle +\langle {\tilde{N}}^{2r}u,\tilde{N}u\rangle \hfill \\ \hfill & =2\Re \langle {\tilde{N}}^{2r}u,-i(H\otimes I_{\mathfrak{h}})u\rangle +2\Re \langle {\tilde{N}}^{2r}u,-\lambda i(B\otimes \Phi \left(g\right))u\rangle \hfill \\ \hfill & =2\Im \langle {\tilde{N}}^{2r}u,[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g\right)]u\rangle \hfill \\ \hfill & \le a(\parallel {\tilde{N}}^{r}u{\parallel }^2+\parallel u{\parallel }^2+b).\hfill \end{array}$$

Finally, for all $u\in \mathrm{Dom}{\tilde{N}}^r$, using Theorem 2 yields

$$\underset{n\ge 1}{sup}\parallel {\tilde{N}}^r{P}_{n}u\parallel =\underset{n\ge 1}{sup}\parallel P_n{\tilde{N}}^{r}u\parallel \le \parallel {\tilde{N}}^{r}u\parallel .$$

Therefore, ${(I_{\mathcal{K}}\otimes N)}^r={\tilde{N}}^r$ is a reference operator of Equation (1). □

Theorem 4.

Let $r\ge 1$ be a real number and $g\in {\ell }^2\left(\mathbb{N}\right)$ a given function. Suppose further that $\mathrm{Dom}{I}_{\mathcal{K}}\otimes N\subset \mathrm{Dom}H\otimes I_{\mathfrak{h}}\cap \mathrm{Dom}B\otimes \Phi \left(g\right)$ and there exist two finite constants $a,b\ge 0$ such that

$$\Im 〈{(I_{\mathcal{K}}\otimes N)}^{2r}u,[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g\right)]u〉\le a\left(\parallel {(I_{\mathcal{K}}\otimes N)}^r{u\parallel }^2+{\parallel u\parallel }^2+b\right),$$

where $u\in \mathrm{Dom}{(I_{\mathcal{K}}\otimes N)}^{2r}$. Then, for each ${\mathcal{F}}_0$-measurable $\mathcal{K}\otimes \mathfrak{h}$-value random variable ξ with $\xi \in \mathrm{Dom}{(I_{\mathcal{K}}\otimes N)}^r$ a.s. and $\mathbb{E}(\parallel {(I_{\mathcal{K}}\otimes N)}^r\xi {\parallel }^2+\parallel \xi {\parallel }^2)<\infty$, there exists a unique ${(I_{\mathcal{K}}\otimes N)}^r$-strong solution ${\left(X_t\right)}_{t\ge 0}$ to Equation (1) such that $X_0=\xi$. And moreover, the solution ${\left(X_t\right)}_{t\ge 0}$ satisfies that

$$\mathbb{E}(\parallel X_t{\parallel }^2{)=\mathbb{E}(\parallel \xi \parallel }^2),t\ge 0$$

and

$$\mathbb{E}\parallel {(I_{\mathcal{K}}\otimes N)}^r{X}_t{\parallel }^2\le {\mathrm{e}}^{at}\left[\mathbb{E}\parallel {(I_{\mathcal{K}}\otimes N)}^r{\xi \parallel }^2+{at(\mathbb{E}\parallel \xi \parallel }^2+b)\right],t\ge 0.$$

(11)

Proof. 

Using Lemma A1 in the Appendix A and Theorem 3, we can come straightforward to the conclusion. □

As is seen, Theorem 4 offers conditions for Equation (1) to have a unique regular solution. From a viewpoint of application, however, it seems those conditions are not so easy to check. Next, we would like to find out easily-checking conditions.

For a function $g:\mathbb{N}\to \mathbb{C}$, we write $\Psi \left(g\right)={\sum }_{k=0}^{\infty }(g\left(k\right){\partial }_k^{\ast }-\overline{g\left(k\right)}{\partial }_k)$, which is naturally defined as an operator in $\mathfrak{h}$. Moreover, if $g\in {\ell }^1\left(\mathbb{N}\right)$, then $\Psi \left(g\right)$ is a bounded operator on $\mathfrak{h}$ with $\parallel \Psi \left(g\right)\parallel \le {2\parallel g\parallel }_{{\ell }^1}$. Cf the definition of the operator $\Phi \left(g\right)$ in (6) and Proposition 1.

Theorem 5.

Let H and B be self-adjoint bounded operators on $\mathcal{K}$, $\lambda \ge 0$ a real number and $g\in {\ell }^1\left(\mathbb{N}\right)$. Then, it holds true that

$$|\langle {(I_{\mathcal{K}}\otimes N)}^{2}u,[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g\right)]u\rangle |\le \left[\parallel H\parallel +{3\lambda \parallel B\parallel \parallel g\parallel }_{{\ell }^1}\right]\left[\parallel (I_{\mathcal{K}}\otimes N){u\parallel }^2+{\parallel u\parallel }^2\right],$$

(12)

where $u\in \mathrm{Dom}{(I_{\mathcal{K}}\otimes N)}^2$.

Proof. 

For $r\ge 1$, we set ${\mathcal{D}}^{\left(r\right)}=\mathrm{span}\{x\otimes \xi \mid x\in \mathcal{K},\xi \in \mathrm{Dom}{N}^r\}$. Then, by the definition of $I_{\mathcal{K}}\otimes N^r$, we know that ${\mathcal{D}}^{\left(r\right)}$ is a core of $I_{\mathcal{K}}\otimes N^r$, namely $\overline{(I_{\mathcal{K}}\otimes N^r){|}_{{\mathcal{D}}^{\left(r\right)}}}=I_{\mathcal{K}}\otimes N^r$. Additionally, it can be verified that $(I_{\mathcal{K}}\otimes N){\mathcal{D}}^{\left(2\right)}\subset {\mathcal{D}}^{\left(1\right)}$ and

$$(I_{\mathcal{K}}\otimes N^2)u=(I_{\mathcal{K}}\otimes N)\left[(I_{\mathcal{K}}\otimes N)u\right],\forall u\in {\mathcal{D}}^{\left(2\right)}.$$

Now, let us show some useful claims as follows:

Claim A: $(H\otimes I_{\mathfrak{h}}){\mathcal{D}}^{\left(2\right)}\subset {\mathcal{D}}^{\left(1\right)}$ and $(I_{\mathcal{K}}\otimes N)(H\otimes I_{\mathfrak{h}})=(H\otimes I_{\mathfrak{h}})(I_{\mathcal{K}}\otimes N)$ on ${\mathcal{D}}^{\left(2\right)}$. The proof of this claim is quite straightforward, hence is omitted here.

Claim B: For all $n\ge 0$, $(B\otimes \Phi \left(g_n\right)){\mathcal{D}}^{\left(2\right)}\subset {\mathcal{D}}^{\left(1\right)}$ and it holds on ${\mathcal{D}}^{\left(2\right)}$ that

$$(I_{\mathcal{K}}\otimes N)(B\otimes \Phi \left(g_n\right))=(B\otimes \Phi \left(g_n\right))(I_{\mathcal{K}}\otimes N)+B\otimes \Psi \left(g_n\right),$$

where $g_n$ is the function on $\mathbb{N}$ given by $g_n\left(k\right)=g\left(k\right)$ when $k\le n$ and $g_n\left(k\right)=0$ when $k>n$.

In fact, for all $x\in \mathcal{K}$ and $\xi \in \mathrm{Dom}{N}^2$, by the inclusion relations $\mathrm{Dom}{N}^2\subset \mathrm{Dom}N$ and $N\left(\mathrm{Dom}{N}^2\right)\subset \mathrm{Dom}N$ as well as Theorems 4.7 and 4.8 of [11], we have

$$\left(\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast }\right)\xi \in \mathrm{Dom}N,0\le k\le n$$

and

$$N\left(\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast }\right)\xi =\left(\overline{g\left(k\right)}{\partial }_k+g\left(k\right){\partial }_k^{\ast }\right)N\xi +\left(g\left(k\right){\partial }_k^{\ast }-\overline{g\left(k\right)}{\partial }_k\right)\xi ,0\le k\le n,$$

which implies that $(B\otimes \Phi \left(g_n\right))(x\otimes \xi )=Bx\otimes \Phi \left(g_n\right)\xi \in {\mathcal{D}}^{\left(1\right)}$ and

$$\begin{array}{cc}\hfill (I_{\mathcal{K}}\otimes N)(B\otimes \Phi \left(g_n\right))(x\otimes \xi )& =Bx\otimes N\Phi \left(g_n\right)\xi \hfill \\ \hfill & =Bx\otimes [\Phi \left(g_n\right)N\xi +\Psi \left(g_n\right)\xi ]\hfill \\ \hfill & =(B\otimes \Phi \left(g_n\right))(I_{\mathcal{K}}\otimes N)(x\otimes \xi )+(B\otimes \Psi \left(g_n\right))(x\otimes \xi ).\hfill \end{array}$$

Thus, Claim B is true.

Claim C: For all $u\in {\mathcal{D}}^{\left(2\right)}$, it holds true that

$$|〈(I_{\mathcal{K}}\otimes N^2)u,[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g\right)]u〉|\le \left[\parallel H\parallel +{3\lambda \parallel B\parallel \parallel g\parallel }_{{\ell }^1}\right]\left[\langle (I_{\mathcal{K}}\otimes N^2)u,u\rangle +{\parallel u\parallel }^2\right].$$

Indeed, for all $n\ge 0$, by Claims A and B, we have $[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g_n\right)]u\in {\mathcal{D}}^{\left(1\right)}$ and

$$\begin{array}{cc}\hfill (I_{\mathcal{K}}\otimes N)& [H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g_n\right)]u\hfill \\ \hfill & =(H\otimes I_{\mathfrak{h}})(I_{\mathcal{K}}\otimes N)u+\lambda (B\otimes \Phi \left(g_n\right))(I_{\mathcal{K}}\otimes N)u+\lambda (B\otimes \Psi \left(g_n\right))u,\hfill \end{array}$$

which together with $(I_{\mathcal{K}}\otimes N^2)u=(I_{\mathcal{K}}\otimes N)\left[(I_{\mathcal{K}}\otimes N)u\right]$ implies that

$$\begin{array}{c}|\langle (I_{\mathcal{K}}\otimes N^2)u,[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g_n\right)]u\rangle |\hfill \\ =|〈(I_{\mathcal{K}}\otimes N)u,(I_{\mathcal{K}}\otimes N)[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g_n\right)]u〉|\hfill \\ =|〈(I_{\mathcal{K}}\otimes N)u,(H\otimes I_{\mathfrak{h}})(I_{\mathcal{K}}\otimes N)u〉+\lambda 〈(I_{\mathcal{K}}\otimes N)u,(B\otimes \Phi \left(g_n\right))(I_{\mathcal{K}}\otimes N)u〉\hfill \\ +\lambda 〈(I_{\mathcal{K}}\otimes N)u,(B\otimes \Psi \left(g_n\right))u〉|\hfill \\ \le (\parallel H\parallel +\lambda \parallel B\otimes \Phi \left(g_n\right)\parallel )\parallel (I_{\mathcal{K}}\otimes N){u\parallel }^2+\lambda \parallel B\otimes \Psi \left(g_n\right)\parallel \parallel (I_{\mathcal{K}}\otimes N)u\parallel \parallel u\parallel \hfill \\ =(\parallel H\parallel +\lambda \parallel B\parallel \parallel \Phi \left(g_n\right)\parallel )\parallel (I_{\mathcal{K}}\otimes N){u\parallel }^2+\lambda \parallel B\parallel \parallel \Psi \left(g_n\right)\parallel \parallel (I_{\mathcal{K}}\otimes N)u\parallel \parallel u\parallel \hfill \\ \le [\parallel H\parallel +\lambda \parallel B\parallel \parallel \Phi \left(g_n\right)\parallel +\frac{1}{2}\lambda \parallel B\parallel \parallel \Psi \left(g_n\right)\parallel ]\parallel (I_{\mathcal{K}}\otimes N){u\parallel }^2+\frac{1}{2}\lambda \parallel B\parallel \parallel \Psi \left(g_n\right){\parallel \parallel u\parallel }^2\hfill \\ \le [\parallel H\parallel +\lambda \parallel B\parallel \parallel \Phi \left(g_n\right)\parallel +\frac{1}{2}\lambda \parallel B\parallel \parallel \Psi \left(g_n\right)\parallel ][\parallel (I_{\mathcal{K}}\otimes N){u\parallel }^2+{\parallel u\parallel }^2],\hfill \end{array}$$

which together with $\parallel (I_{\mathcal{K}}\otimes N){u\parallel }^2=\langle (I_{\mathcal{K}}\otimes N^2)u,u\rangle$ gives

$$\begin{array}{cc}\hfill |\langle (I_{\mathcal{K}}\otimes N^2)u,& [H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g_n\right)]u\rangle |\hfill \\ \hfill & \le [\parallel H\parallel +\lambda \parallel B\parallel \parallel \Phi \left(g_n\right)\parallel +\frac{1}{2}\lambda \parallel B\parallel \parallel \Psi \left(g_n\right)\parallel ][\langle (I_{\mathcal{K}}\otimes N^2)u,u\rangle +{\parallel u\parallel }^2],\hfill \end{array}$$

which, together with $\Phi \left(g_n\right)\to \Phi \left(g\right)$, $\Psi \left(g_n\right)\to \Psi \left(g\right)$ (both in operator norm), $\parallel \Phi \left(g\right)\parallel \le {2\parallel g\parallel }_{{\ell }^1}$ as well as $\parallel \Psi \left(g\right)\parallel \le {2\parallel g\parallel }_{{\ell }^1}$, implies that

$$|〈(I_{\mathcal{K}}\otimes N^2)u,[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g\right)]u〉|\le \left[\parallel H\parallel +{3\lambda \parallel B\parallel \parallel g\parallel }_{{\ell }^1}\right]\left[\langle (I_{\mathcal{K}}\otimes N^2)u,u\rangle +{\parallel u\parallel }^2\right].$$

Finally, we verify inequality (12). Let $u\in \mathrm{Dom}{(I_{\mathcal{K}}\otimes N)}^2=\mathrm{Dom}(I_{\mathcal{K}}\otimes N^2)$ be given. Then, by the self-adjoint property of $I_{\mathcal{K}}\otimes N^2$ as well as the fact that ${\mathcal{D}}^{\left(2\right)}$ is a core of $I_{\mathcal{K}}\otimes N^2$, there exists a sequence $\left\{u_n\right\}\subset {\mathcal{D}}^{\left(2\right)}$ such that $u_n\to u$ and $(I_{\mathcal{K}}\otimes N^2)u_n\to (I_{\mathcal{K}}\otimes N^2)u$, which together with Claim C implies that

$$|〈(I_{\mathcal{K}}\otimes N^2)u,[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g\right)]u〉|\le \left[\parallel H\parallel +{3\lambda \parallel B\parallel \parallel g\parallel }_{{\ell }^1}\right]\left[\langle (I_{\mathcal{K}}\otimes N^2)u,u\rangle +{\parallel u\parallel }^2\right],$$

which, together with ${(I_{\mathcal{K}}\otimes N)}^2=I_{\mathcal{K}}\otimes N^2$ and $\langle (I_{\mathcal{K}}\otimes N^2)u,u\rangle ={\parallel (I_{\mathcal{K}}\otimes N)u\parallel }^2$, yields

$$|\langle {(I_{\mathcal{K}}\otimes N)}^{2}u,[H\otimes I_{\mathfrak{h}}+\lambda B\otimes \Phi \left(g\right)]u\rangle |\le \left[\parallel H\parallel +{3\lambda \parallel B\parallel \parallel g\parallel }_{{\ell }^1}\right][\parallel (I_{\mathcal{K}}\otimes N){u\parallel }^2+{\parallel u\parallel }^2].$$

This completes the proof. □

Combining Theorem 4 with Theorem 5, we come to the next theorem, which provides easily-checking conditions for Equation (1) to have a regular solution.

Theorem 6.

Let H, B be self-adjoint bounded operators on $\mathcal{K}$, $\lambda \ge 0$ a real number and $g\in {\ell }^1\left(\mathbb{N}\right)$. Then, for each ${\mathcal{F}}_0$-measurable $\mathcal{K}\otimes \mathfrak{h}$-value random variable ξ with $\xi \in \mathrm{Dom}{I}_{\mathcal{K}}\otimes N$ a.s. and $\mathbb{E}\left(\parallel (I_{\mathcal{K}}\otimes N){\xi \parallel }^2{)+\parallel \xi \parallel }^2\right)<\infty$, there exists a unique $(I_{\mathcal{K}}\otimes N)$-strong solution ${\left(X_t\right)}_{t\ge 0}$ to Equation (1) such that $X_0=\xi$. And moreover, the solution ${\left(X_t\right)}_{t\ge 0}$ satisfies that

$$\mathbb{E}(\parallel X_t{\parallel }^2{)=\mathbb{E}(\parallel \xi \parallel }^2),t\ge 0$$

and

$$\mathbb{E}\parallel (I_{\mathcal{K}}\otimes N)X_t{\parallel }^2\le {\mathrm{e}}^{at}\left[\mathbb{E}\parallel (I_{\mathcal{K}}\otimes N){\xi \parallel }^2+at\mathbb{E}{\parallel \xi \parallel }^2\right],t\ge 0,$$

(13)

where $a=\parallel H\parallel +{3\lambda \parallel B\parallel \parallel g\parallel }_{{\ell }^1}$.

Usually, an open quantum system is quite small compared to its environment. Even in many cases, the open quantum system of interest can be described by a finite-dimensional Hilbert space. Thus, from a perspective of mathematical physics, the conditions provided in Theorem 6 are reasonable and mild.

Example 1.

Let $x={\left(x_k\right)}_{k\ge 1}$ and $y={\left(y_k\right)}_{k\ge 1}$ be bounded sequences of real numbers. Consider the function $g_0\left(k\right)=\frac{1}{2^k}$, $k\ge 0$ and the operators $H_0$, $B_0$ defined respectively by

$$H_0=\sum _{j=1}^{\infty }{x}_j|e_j\rangle \langle e_j|,B_0=\sum _{j=1}^{\infty }{y}_j|e_j\rangle \langle e_j|,$$

(14)

where ${\left\{e_j\right\}}_{j\ge 1}$ is the ONB for $\mathcal{K}$ and $|e_j\rangle \langle e_j|$ the Dirac operator associated with j. Clearly, both $H_0$ and $B_0$ are self-adjoint bounded operators on $\mathcal{K}$, and moreover $g_0\in {\ell }^1\left(\mathbb{N}\right)$. Now by letting $H=H_0$, $B=B_0$ and $g=g_0$, one can immediately see that H, B and g satisfy the conditions required in Theorem 6. This validates Theorem 6.

4. Concluding Remarks

SSEs are widely used in many research fields such as measurement theory, quantum optics, quantum chaos, solid states, etc, wherever quantum irreversibility matters. As is seen, in this paper, we introduce and investigate a SSE in terms of quantum Bernoulli noises (QBN), which might serve as a model describing the evolution of an open quantum system interacting with QBN. The main features of our SSE lie in that it contains an interacting term that reflects the interactions between the system and the environment described by QBN.

As can be seen, a SSE is usually considered in the setting of a Hilbert space. This is because a quantum system is usually described by a Hilbert space. From a mathematical point of view, however, one may also consider SSEs in the setting of Banach spaces, and even SSEs of fractal order.