---
abstract: 'We show that any arbitrary time-dependent density operator of an open system can always be described in terms of an operator-sum representation (Kraus representation) regardless of its initial condition and the path of its evolution in the state space, and we provide a general expression of Kraus operators for arbitrary time-dependent density operator of an $N$-dimensional system. Moreover, applications of our result are illustrated through several examples.'
address: |
    $^1$Department of Physics, National University of Singapore, 10 Kent Ridge Crescent, Singapore 119260, Singapore\
    $^2$Department of Physics, Shandong Normal University, Jinan 250014, People’s Republic of China\
    $^3$ National Institute of Education, Nanyang Technological University, 1 Nanyang Walk, Singapore 639798, Singapore 
author:
- 'D. M. Tong$^{1,2}$, L. C. Kwek$^{1,3}$, C. H. Oh$^1$, Jing-Ling Chen$^1$, and L. Ma$^1$'
title: 'Operator-sum representation of time-dependent density operators and its applications'
---

Arbitrary states of any quantum system, be it open or closed, can always be described by density matrices $\rho(t)$. If the system is closed, its time evolution is unitary, and there exists a unitary operator $U(t)$, such that $\rho(t)=U(t)\rho(0)U(t)^\dagger$, where $\rho(0)$ is the initial state. However, if the system is open, the evolution is not necessarily unitary and the above relation between $\rho(t)$ and $\rho(0)$ is in general not valid. To describe the evolution of an open system, one usually employs the Kraus representation[@Kraus]. If there exist operators $M_\mu(t)$, called Kraus operators, that satisfy $$\begin{aligned}
\rho(t)=\sum\limits_\mu M_\mu(t)\rho(0)M_\mu(t)^\dagger,
\label{rhot}\end{aligned}$$and $$\begin{aligned}
\sum\limits_\mu M_\mu(t)^\dagger M_\mu(t)=I, \label{mm}\end{aligned}$$ then the evolution of $\rho(t)$ is said to have operator-sum representation or the Kraus representation.

It has been known that a completely positive map always possesses the operator-sum representation while a general positive map may not do[@Kraus; @Preskill; @Pechukas; @Pomero; @Bouda; @Philip; @Gen; @Peixoto; @Mika; @Sonja]. The papers[@Peter; @Salgado; @Hayashi] furthermore pursued the existence of the Kraus representation by investigating the role of initial correlations between the open system and its environment, and showed that a map based on the reduced dynamics, in general, cannot be described as the Kraus representation in the presence of the initial correlations because an additional inhomogeneous part appears. On the other hand, we know that for a qubit system, although a non-completely positive map $\$:\rho(0)\rightarrow\rho(t)$ does not possess the Kraus representation with the Kraus operators independent of $\rho(0)$, the state $\rho(t)=\$(\rho(0))$ can still be expressed in the form of Eq. (\[rhot\]) with the operators $M_\mu(t)$ selected for $\rho(0)$[@Philip; @Gen; @Tong1]. Arbitrary density matrices $\rho(t)$ of a qubit system can always be described in terms of the operator-sum representation. This start us to wonder whether the same property is valid for $N$-dimensional systems. That is, for any arbitrary time-dependent state $\rho(t)$ and initial state $\rho(0)$ of an $N$-dimensional system, is it true that the operators $M_\mu(t) $ can always be found so that (\[rhot\]) and (\[mm\]) are fulfilled? This is a interesting issue as it extends the notion of the Kraus representation associated with a map to that associated with an evolution of the state and we find that it is useful to some physical problem.

In this paper, we prove that the time-dependent density operator of an arbitrary $N$-dimensional quantum system can always be described in terms of the Kraus representation and provide a general expression of Kraus operators for the $N$-dimensional system. Some applications of our result are also discussed.

Firstly, we prove that any arbitrary two density matrices $\rho_A$ and $\rho_B$ of an $N$-dimensional open system can always be connected by Kraus operators.

Suppose $$\begin{aligned}
\rho_A=\sum\limits_{i=1}^N p^A_i |A_i\rangle\langle A_i|=U_A\rho^d_AU_A^\dagger,\label{rhoab1}\\
\rho_B=\sum\limits_{i=1}^N p^B_i |B_i\rangle\langle
B_i|=U_B\rho^d_BU_B^\dagger, \label{rhoab2}\end{aligned}$$ where $p^A_i$ ($p^B_i$) and $|A_i\rangle$ ($|B_i\rangle$) are the eigenvalues and the orthonormal eigenvectors of the density matrix $\rho_A$ ($\rho_B$) respectively, $\rho^d_A$ ($\rho^d_B$) is the diagonal matrix with the entries $p^A_i$ ($p^B_i$), and $U_A$ ($U_B$) is the unitary transformation matrix, the $i$-th column of which is just the vector $|A_i\rangle $ ($|B_i\rangle$). We want to prove that there exist the operators $M_\mu$, satisfying $$\begin{aligned}
&&\rho_B=\sum\limits_\mu M_\mu\rho_AM_\mu^\dagger,\label{rhoa}\\
&&\sum\limits_\mu M_\mu^\dagger M_\mu=I. \label{mm2}\end{aligned}$$

To find the required Kraus operators, one may write $M_\mu$ as $N\times N$ matrices with unknown elements and then directly solve (\[rhoa\]) and (\[mm2\]) to determine the matrices. However, this method is not feasible for high dimensional systems due to the computational complexity. To overcome the problem, we first seek diagonal matrices, which are generally easier to manipulate. To this end, we look for operators $M'_i$ such that $$\begin{aligned}
&&\rho^d_B=\sum\limits_\mu M'_\mu \rho^d_A {M'_\mu}^\dagger,\label{rhod}\\
&&\sum\limits_\mu {M'_\mu}^\dagger M'_\mu=I. \label{mm3}\end{aligned}$$ To solve the above equations, we obtain the matrix operators $M'_\mu $ ($\mu=0,1,2,...N-1$) with the entries $$\begin{aligned}
 (M'_\mu)_{ij}= \sqrt{p^B_i}\delta_{i,j-\mu}+\sqrt{p^B_i}\delta_{i,j-\mu+N}~~,
\label{m}\end{aligned}$$ where $i,j=1,2,...N$[@Tong2].

With the matrices $M'_\mu $, we can construct the Kraus operators $M_\mu $ satisfying (\[rhoa\]) and (\[mm2\]). To see this, we use the relations, $$\begin{aligned}
\rho^d_A=U_A^\dagger\rho_AU_A,\label{rhoab3}\\
\rho^d_B=U_B^\dagger\rho_BU_B. \label{rhoab4}\end{aligned}$$ Substituting (\[rhoab3\]) and (\[rhoab4\]) into (\[rhod\]), we have $$\begin{aligned}
U_B^\dagger\rho_BU_B=\sum\limits_\mu M'_\mu U_A^\dagger\rho_A
U_A{M'_\mu}^\dagger,\end{aligned}$$ that is, $$\begin{aligned}
\rho_B=\sum\limits_\mu U_B M'_\mu U_A^\dagger\rho_A (U_BM'_\mu
U_A^\dagger)^\dagger.\end{aligned}$$ Let $$\begin{aligned}
M_\mu=U_BM'_\mu U_A^\dagger, \label{mmu}\end{aligned}$$ it is obvious that $M_\mu $ defined by (\[mmu\]) satisfy (\[rhoa\]) and (\[mm2\]). They are the Kraus operators connecting $\rho_A$ with $\rho_B$. Since $U_A$, $U_B$ and $M'_\mu $ can be deduced directly from $\rho_A$ and $\rho_B$, the explicit expression of $M_\mu$ is obtained for arbitrary density matrices.

Secondly, having proved that any arbitrary two density matrices $\rho_A$ and $\rho_B$ of an $N$-dimensional open system can always be connected by Kraus operators, we now revert to our discussion concerning the time evolution of open systems. We show that the time-dependent state of an arbitrary open system can always be described in terms of the Kraus representation.

We replace $\rho_A$ and $\rho_B$ in the above demonstration by $\rho(0)$ and $\rho(t)$ respectively. Note that for any evolution of an $N$-dimensional open system, the state $\rho(t)$ and the initial state $\rho(0)$ can always be expressed as $$\begin{aligned}
\rho(t)=\sum\limits_{i=1}^N p_i(t) |\psi_i(t)\rangle\langle
\psi_i(t)|
,\label{rhot1}\\
\rho(0)=\sum\limits_{i=1}^N p_i(0) |\psi_i(0)\rangle\langle
\psi_i(0)|.
\label{rhot2}\end{aligned}$$ By comparing (\[rhot1\]) (\[rhot2\]) with (\[rhoab1\]) (\[rhoab2\]) and using (\[mmu\]), one can immediately write down the Kraus operators $M_\mu (t)$ as $$\begin{aligned}
M_\mu(t)=U(t)M'_\mu(t) U(0)^\dagger, \label{mmt}\end{aligned}$$ where $U(t)$ ($U(0)$) is the unitary transformation matrix that diagonalizes $\rho(t)$ ($\rho(0)$) given explicitly by $$\begin{aligned}
&U(t)&=\left(\begin{array}{ccccc}\psi_1(t)~&\psi_2(t)~&...&...&~\psi_N(t)
\end{array}\right)\label{ut},\\
&U(0)&=\left(\begin{array}{ccccc}\psi_1(0)~&\psi_2(0)~&...&...&~\psi_N(0)
\end{array}\right),
\label{u0}\end{aligned}$$ and $M'_\mu(t) $ ($\mu=0,1,...N-1$) are given by $$\begin{aligned}
&M'_0(t)&=\left(\begin{array}{ccccc}
\sqrt{p_1(t)}&0&0&...&0\\
0&\sqrt{p_2(t)}&0&...&0\\
0&0&\sqrt{p_3(t)}&...&0\\
 &...&...&...&\\
0&0&0&0&\sqrt{p_N(t)}
\end{array}\right),
\nonumber\end{aligned}$$ $$\begin{aligned}
&M'_1(t)&=\left(\begin{array}{ccccc}
0&\sqrt{p_1(t)}&0&...&0\\
0&0&\sqrt{p_2(t)}&...&0\\
 &...&...&...&\\
0&0&0&...&\sqrt{p_{N-1}(t)}\\
\sqrt{p_N(t)}&0&0&0&0
\end{array}\right),\nonumber\end{aligned}$$ $$\begin{aligned}
...~~~~~~~...~~~~~~~...~~~~~~~\nonumber\end{aligned}$$

$$\begin{aligned}
&M'_{N-1}(t)&=\left(\begin{array}{ccccc}
0&0&...&0&\sqrt{p_1(t)}\\
\sqrt{p_2(t)}&0&0&...&0\\
0&\sqrt{p_3(t)}&0&...&0\\
 &...&...&...&\\
0&0&0&\sqrt{p_N(t)}&0
\end{array}\right).
\label{m0m1}\end{aligned}$$

The operators $M_\mu (t)$ defined by (\[mmt\])-(\[m0m1\]) satisfy (\[rhot\]) and (\[mm\]), giving the Kraus representation of $\rho(t)$. Generally speaking, one cannot assert that any arbitrary time-dependent density operator possesses the Kraus representation simply from the result that any two states can be connected by Kraus operators. However, here we can make the assertion because $p_i(t)$ are always positive and (\[m0m1\]) is always valid independent of time $t$.

So far, we have shown that the time-dependent state of an open system can always be described in terms of the Kraus representation. A general expression of Kraus operators for any arbitrary $N$-dimensional system is given. When the evolution of the open system is given by $\rho(t)$ and $\rho(0)$, the Kraus operators can be written immediately as $M_\mu(t)=U(t)M'_\mu(t)
U(0)^\dagger$, where $U(t)$, $U(0)$ and $M'_\mu(t)$ are explicitly given by (\[ut\]), (\[u0\]) and (\[m0m1\]). Since the Kraus operators are not unique, the expression (\[mmt\]) is only one set of them. The other equivalent expressions of the Kraus operators can be obtained by $\tilde{M}_\mu(t)=\sum\limits_\nu M_\nu(t) V_{\mu\nu}$, where $V_{\mu\nu}$ are the elements of an arbitrary unitary matrix $V$.

Thirdly, in the following paragraphs, we will illustrate some applications of our result.

The result can help to clarify some ambiguous concepts. For example, the [*[Kraus representation theorem]{}*]{}, recalling the well-known [*[representation theorem]{}*]{}, which states that a map has the Kraus representation if and only if it is linear, completely positive (CP) and trace preserving[@Kraus], one may thought that the expression $\rho'=\$(\rho)$ has no Kraus representation if the map $\$$ is not CP. However, our result shows that any two states $\rho$ and $\rho'$ can always be connected by the Kraus operators irrespective of the form of the map acting on $\rho$. This shows that although non-CP map does not possess the Kraus representation, the expression $\rho'=\$(\rho)$ can still be cast into the Kraus representation. For instance, the transposition operator $T: \rho \rightarrow
\rho^T$ is positive but not completely positive. According to the [*[Kraus representation theorem]{}*]{}, there does not exist the Kraus representation of the map $T$. However, the expression $
\rho^T=T(\rho)$ can still be described by Kraus operators. Let $\rho_A=\rho$ and $\rho_B=\rho^T$, the Kraus operators can be obtained directly using (\[mmu\]). The important key for clarifying the ambiguity is to note that there is a difference between the Kraus representation of a map and the Kraus representation for a state under the action of a map. The Kraus operators describing the Kraus representation of a map are independent of the state $\rho$ while the Kraus operators describing the Kraus representation of a state $\rho$ under the action of a map may be dependent on the state. A non-CP map has no Kraus representation, but $\rho'=\$(\rho)$, be $\$$ a CP map or not, always has the Kraus representation.

An important corollary of our result is that there always exists a CP map between any two quantum states and the map can be represented through $N$ Kraus operators. This corollary is obvious, because any two states can be connected by $N$ Kraus operators $M_\mu $ and the CP map can then be defined by the $N$ Kraus operators. The corollary shows that even for such two states $\rho$ and $\rho'$, where $\rho'$ is obtained from $\rho$ by a non-CP map $\$:\rho\rightarrow\rho'$, one can still find an alternative map ${\tilde{\$}}$ that is completely positive, satisfying ${\tilde{\$}}:\rho\rightarrow\rho'$.

The conclusion that an arbitrary time-dependent state $\rho(t)$ can always be described by the Kraus representation could have other deep applications in the study of open systems, especially in the field of quantum information. For example, it is useful for the study on geometric phase. As we know, geometric phases of both pure state and mixed state under unitary evolutions have been clarified[@Berry; @Aharonov; @Samuel; @Mukunda; @Pati; @Sjoqvistm; @Tong]. A new issue is on the geometric phases for open systems under nonunitary evolutions. Some papers[@Peixoto; @Ericsson; @Carollo] just use the Kraus operators $M_\mu$ to define and calculate the geometric phases of open systems. As described in ref. [@Ericsson], the relative phases are defined by $\alpha_\mu=\arg tr[M_\mu(\tau)\rho(0)]$, and the geometric phases can be calculated by making polar decomposition of $M_{\mu}(t)$, such that $M_{\mu}(t)=h_{\mu}(t)u_{\mu}(t)$, where $h_\mu(t)$ are Hermitian and positive, and $u_\mu(t)$ are unitary. The relative phases $\alpha_\mu$ will then lead to the geometric phase when $u_{\mu}(t)$ satisfy the $N$ parallel transport conditions $\langle \psi_i(0)|u_{\mu}(t)^+\dot
{u}_{\mu}(t)|\psi_i(0)\rangle =0,~~(i=1,2,...,N)$. With the scheme showed in the present paper, any time-dependent density matrix can be easily written as the Kraus representation. So one can transplant the notion of geometric phases defined by using Kraus operators to arbitrary evolutions of quantum systems by writing $\rho(t)$ as the Kraus representation.

Another example of its applications concerns the inverse problem of the evolutions of open systems. Suppose that the density matrix of a system is given as a time-dependent function $\rho(t)$, one wants inversely to deduce the evolutional operators or Hamiltonians that evolve the initial state $\rho(0)$ to the state $\rho(t)$. This is an important issue because physicists sometimes need to prepare experimentally a quantum system that is expected to evolve along a given path in the projected Hilbert space. The approach for solving the problem is to constitute a closed system by combining the open system with an ancilla. The open system will undergo nonunitary evolution while the combined system evolves unitarily as $\varrho(t)=U_{sa}(t)\varrho(0)U_{sa}(t)^+$, where $U_{sa}(t)$ are unitary operators acting on the combined system and $\varrho(0)=\rho(0)\otimes|0_a\rangle\langle 0_a|$ is the initial state. The issue becomes to finding the unitary operators $U_{sa}(t)$ so that $tr_e\varrho(t)=\rho(t)$. It is quite a difficult problem in general. However, the present paper can provide an effective approach to obtain the unitary operators. As concluded above, evolutions of an open system always have the Kraus representation and the Kraus operators can be directly deduced by the given $\rho(t)$. With the Kraus operators $M_\mu(t)$, one can easily obtain the unitary operator $U_{sa}(t)$. In fact, in order to satisfy $tr_e\varrho(t)=\rho(t)$, the elements of $U_{sa}(t)$ in the bases $|\Psi_i(0)\rangle \otimes|j_a\rangle$ are only required to be $[U_{sa}(t)]_{ij,k0}=[M_{j}(t)]_{ik}$ while $[U_{sa}(t)]_{ij,kl}~(l\neq 0)$ are arbitrary but keeping $U_{sa}(t)$ to be unitary and $U_{sa}(t)|_{t=0}=I$, where $|j_a\rangle~ (j=0,1,...K-1)$ are the bases of the ancilla. Obviously, there are infinitely many such unitary operators $U_{sa}(t)$, so do the Hamiltonians $H=i\dot U(t)U(t)^+$. One can choose the suitable ones which can be easily performed in the laboratory.

It is also interesting to note that the Kraus operators provided in the present paper have some intriguing properties. Since these Kraus operators are dependent on the eigenvectors of “input state" $\rho_A$ but independent of its eigenvalues, all the input states with distinct eigenvalues but the same eigenvectors will yield the same “output" state $\rho_B$. That is, the map defined by the Kraus operators can transform all the input states whose Bloch vectors lie on the same diameter of the Bloch sphere to the same output state. Specially, one may let $\rho_A=\frac{1}{N}I$, and then has $\rho_B=\frac{1}{N}\sum M_\mu
M_\mu^\dagger$. We suppose that these properties of the Kraus operators may be useful in the study of open systems.

The work was supported by NUS Research Project Grant: Quantum Entanglement with WBS R-144-000-089-112. J.L.C. acknowledges financial support from Singapore Millennium.

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