# The Franke–Gorini–Kossakowski–Lindblad–Sudarshan (FGKLS) Equation for Two-Dimensional Systems

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## Abstract

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## 1. Introduction

## 2. Pointers for the FGKLS Equation in Two Dimensions

- $\rho $ is Hermitian: $\rho ={\rho}^{\u2020}$, i.e.,$${f}_{11},{f}_{22}\in \mathbb{R};\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{f}_{21}={f}_{12}^{*}$$
- $Tr\rho =1$:$${f}_{11}+{f}_{22}=1$$
- $\rho $ is non-negative.

- The Lindblad operator is diagonal;
- The Lindblad operator is a Jordan block.

#### 2.1. Diagonal Lindblad Operator

- Determinant of the matrix in (24) does not vanish:$$|{e}_{21}{|}^{2}{|{\lambda}_{1}-{\lambda}_{2}|}^{2}\ne 0,$$$${\rho}^{(p)}=\left(\begin{array}{cc}\frac{1}{2}& 0\\ 0& \frac{1}{2}\end{array}\right).$$This pointer does not depend on the parameters $c,{\lambda}_{1},{\lambda}_{2}$ in $L,$ being a maximally mixed state.
- Determinant vanishes due to ${e}_{21}=0.$ Two options in this case exist:
- If, additionally, b vanishes, i.e., $-i({e}_{11}-{e}_{22})+{\lambda}_{1}{\lambda}_{2}^{*}-\frac{1}{2}|{\lambda}_{1}{|}^{2}-\frac{1}{2}{|{\lambda}_{2}|}^{2}=0,$ the pointer is an arbitrary matrix with unit trace:$${\rho}^{(p)}=\left(\begin{array}{cc}{f}_{11}^{(p)}& {f}_{12}^{(p)}\\ {f}_{21}^{(p)}& 1-{f}_{11}^{(p)}\end{array}\right).$$
- If, additionally, b does not vanish, i.e., $-i({e}_{11}-{e}_{22})+{\lambda}_{1}{\lambda}_{2}^{*}-\frac{1}{2}|{\lambda}_{1}{|}^{2}-\frac{1}{2}{|{\lambda}_{2}|}^{2}\ne 0,$ the pointer is an arbitrary diagonal matrix with unit trace:$${\rho}^{(p)}=\left(\begin{array}{cc}{f}_{11}^{(p)}& 0\\ 0& 1-{f}_{11}^{(p)}\end{array}\right)$$

- Determinant vanishes due to ${\lambda}_{1}={\lambda}_{2}.$ Then, the pointer is expressed via one arbitrary real parameter $x\equiv {f}_{11}^{(p)}:$$${\rho}^{(p)}=\left(\begin{array}{cc}x& {e}_{12}\frac{2x-1}{{e}_{11}-{e}_{22}}\\ {e}_{21}\frac{2x-1}{{e}_{11}-{e}_{22}}& 1-x\end{array}\right).$$

#### 2.2. Lindblad Operator of a Jordan Block Form

- For the special case of degenerate $H,$ then ${\epsilon}_{1}={\epsilon}_{2}$, $H=\epsilon \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$, the pointer is$${\rho}_{p,\mathrm{deg}}=\frac{1}{{2|\lambda |}^{2}+1}\left(\begin{array}{cc}{|\lambda |}^{2}+1& -{\lambda}^{*}\\ -\lambda & {|\lambda |}^{2}\end{array}\right)$$

## 3. The General Solution of the FGKLS Equation in Two-Dimensional Hilbert Space

#### 3.1. The Case of Diagonal Lindblad Operator

- ${\mathit{e}}_{\mathbf{12}}={\mathit{e}}_{\mathbf{21}}=\mathbf{0}$ or ${\mathit{\lambda}}_{1}={\mathit{\lambda}}_{2}$:Solving Equation (46), we obtain$${s}_{1}=0$$$${s}_{2,3}=-\frac{1}{2}{|{\lambda}_{1}-{\lambda}_{2}|}^{2}\pm i\sqrt{4|{e}_{12}{|}^{2}+{\left|{e}_{11}-{e}_{22}+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}}$$We see that, when ${\lambda}_{1}\ne {\lambda}_{2}$, according to (42), last two exponents decrease, while the first one reduces to a constant, which merges with a pointer. Therefore, we conclude that, during the evolution, such a system approaches a constant that is not constrained by interaction with an environment, parameters in Hamiltonian, etc.When ${\lambda}_{1}={\lambda}_{2}$, the real part of ${s}_{2,3}$ vanishes, and we have neverending oscillations of a solution.
- $|{\mathit{e}}_{\mathbf{12}}{|}^{\mathbf{2}}{|{\mathit{\lambda}}_{\mathbf{1}}-{\mathit{\lambda}}_{\mathbf{2}}|}^{\mathbf{2}}\ne \mathbf{0}$:Below, we will prove that, for this case $Res<0$ for each root s of this equation, i.e, according to (42), $\rho (t)$ converges to the pointer ${\rho}_{p}$ for $t\to \infty $. In general, two options are possible for this case.
- (a)
- All three roots of (46), ${s}_{1},{s}_{2},{s}_{3}$, are realThe form of l.h.s. of (46) is such that it is strictly positive for $s\ge 0$, and therefore, ${s}_{1},{s}_{2},{s}_{3}$ have to be negative.
- (b)
- To start with, we write Vieta’s formulas for Equation (46):$$2Re(s)+t=-|{\lambda}_{1}-{\lambda}_{2}{|}^{2}$$$${|s|}^{2}+2Re(s)t=4|{e}_{12}{|}^{2}+{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}+\frac{1}{4}{|{\lambda}_{1}-{\lambda}_{2}|}^{4}$$$${|s|}^{2}t=-2|{e}_{12}{|}^{2}{|{\lambda}_{1}-{\lambda}_{2}|}^{2}$$We are left to prove that $Re(s)<0$. Expressing ${|s|}^{2}$ and t from the system of Equations (49)–(51), we get the equation for $Res$:$$\begin{array}{c}Re{(s)}^{3}+{|{\lambda}_{1}-{\lambda}_{2}|}^{2}Re{(s)}^{2}\\ +\left(|{e}_{12}{|}^{2}+\frac{1}{4}{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}+\frac{5}{16}{|{\lambda}_{1}-{\lambda}_{2}|}^{4}\right)Re(s)\\ +\left(\frac{1}{4}|{e}_{12}{|}^{2}+\frac{1}{8}{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}+\frac{1}{32}{|{\lambda}_{1}-{\lambda}_{2}|}^{4}\right){|{\lambda}_{1}-{\lambda}_{2}|}^{2}=0\end{array}$$As before, we notice that $Res\ge 0$ cannot be a solution of this equation; therefore, $Res<0$.

The fact that $Re{s}_{1},Re{s}_{2},Re{s}_{3}<0$ is very important. It signifies the vanishing of the exponents in the general solution of the Lindblad equation (42) over time (according to the definition, ${\mathsf{\Lambda}}_{1}={c}^{2}s$, ${\mathsf{\Lambda}}_{2}={c}^{2}{s}^{*}$, ${\mathsf{\Lambda}}_{3}={c}^{2}t$), which, finally, gives that the density matrix for any values of the parameters in $H,L$ (of the form (29)) converges to the pointer (34).

#### 3.2. The Case of the Lindblad Operator of the Jordan Block Form

- If Equation (55) has two complex roots $s,{s}^{*}$, and one real root $t,$ all of them have negative real parts. This important fact is derived by means of the well-known Vieta’s formulas.To start with, we write Vieta’s formulas for Equation (55):$$2Re(s)+t=-2$$$${|s|}^{2}+2Re(s)t=\frac{5}{4}+{({e}_{11}-{e}_{22})}^{2}+4{\left|\frac{1}{2}\lambda +i{e}_{21}\right|}^{2}$$$${|s|}^{2}t=-\left(\frac{1}{4}+{({e}_{11}-{e}_{22})}^{2}+2{\left|\frac{1}{2}\lambda +i{e}_{21}\right|}^{2}\right)$$${|s|}^{2}\ne 0$, since $s=0$ cannot be the root of (55). Therefore, from the last equation, it is obvious that $t<0$.We are left to prove that $Re(s)<0$. Expressing ${|s|}^{2}$ and t from the system of Equations (56)–(58), we get the cubic equation for $Res$:$$\begin{array}{c}Re{(s)}^{3}+2Re{(s)}^{2}+Re(s)\left(\frac{21}{16}+\frac{1}{4}{({e}_{11}-{e}_{22})}^{2}+{\left|\frac{1}{2}\lambda +i{e}_{21}\right|}^{2}\right)+\\ \left(\frac{9}{32}+\frac{1}{8}{({e}_{11}-{e}_{22})}^{2}+\frac{3}{4}{\left|\frac{1}{2}\lambda +i{e}_{21}\right|}^{2}\right)=0.\end{array}$$As before, we conclude that $Res\ge 0$ cannot be a solution of this equation; therefore, $Res<0$.

## 4. Positivity of the Solution of FGKLS Equation

- For the case of two complex roots and one real root (${s}_{1},{s}_{2}={s}_{1}^{*}\in \mathbb{C},{s}_{3}\in \mathbb{R}$):$$\begin{array}{c}\rho (t)=\left(\begin{array}{cc}{f}_{p11}& {f}_{p12}\\ {f}_{p21}& {f}_{p22}\end{array}\right)+u{e}^{-i\varphi}{e}^{{s}_{1}{c}^{2}t}\left(\begin{array}{cc}d{e}^{-i\delta}& b{e}^{-i\gamma}\\ a{e}^{-i\alpha}& -d{e}^{-i\delta}\end{array}\right)\\ +u{e}^{i\varphi}{e}^{{s}_{1}^{*}{c}^{2}t}\left(\begin{array}{cc}d{e}^{i\delta}& a{e}^{i\alpha}\\ b{e}^{i\gamma}& -d{e}^{i\delta}\end{array}\right)\pm w{e}^{{s}_{3}{c}^{2}t}\left(\begin{array}{cc}h& p{e}^{i\frac{\beta}{2}}\\ p{e}^{-i\frac{\beta}{2}}& -h\end{array}\right)\end{array}$$Since $({v}_{1}^{k},{v}_{2}^{k},{v}_{3}^{k})$ is an eigenvector of the matrix A from (40), $\{{v}_{l}^{(k)}\}$ are completely determined by the particular form of the Hamiltonian H and Lindblad operator L, i.e., by the values of parameters ${\epsilon}_{1},{\epsilon}_{2},c,\lambda $. Therefore, $\{a,b,d,p,h,\alpha ,\beta ,\gamma ,\delta \}$ are also completely determined by H and L. As for $\{u,w\in {R}_{+},\varphi \in R$,“±” sign}, these are free parameters. They depend on the choice of the initial state $\rho (0)$.
- For the case of three real roots (${s}_{1},{s}_{2},{s}_{3}\in \mathbb{R}$):$$\begin{array}{c}\rho (t)=\left(\begin{array}{cc}{f}_{p11}& {f}_{p12}\\ {f}_{p21}& {f}_{p22}\end{array}\right)\pm \\ {w}^{(1)}{e}^{{s}_{1}{c}^{2}t}\left(\begin{array}{cc}{h}^{(1)}& {p}^{(1)}{e}^{i\frac{{\beta}^{(1)}}{2}}\\ {p}^{(1)}{e}^{-i\frac{{\beta}^{(1)}}{2}}& -{h}^{(1)}\end{array}\right)\pm {w}^{(2)}{e}^{{s}_{2}{c}^{2}t}\left(\begin{array}{cc}{h}^{(2)}& {p}^{(2)}{e}^{i\frac{{\beta}^{(2)}}{2}}\\ {p}^{(2)}{e}^{-i\frac{{\beta}^{(2)}}{2}}& -{h}^{(2)}\end{array}\right)\pm \\ {w}^{(3)}{e}^{{s}_{3}{c}^{2}t}\left(\begin{array}{cc}{h}^{(3)}& {p}^{(3)}{e}^{i\frac{{\beta}^{(3)}}{2}}\\ {p}^{(3)}{e}^{-i\frac{{\beta}^{(3)}}{2}}& -{h}^{(3)}\end{array}\right)\end{array}$$Interestingly, the initial state has just exactly three real parameters: $x,y,z\in \mathbb{R}$,$$\rho (0)=\left(\begin{array}{cc}z& x+iy\\ x-iy& 1-z\end{array}\right)$$

- $w=0$It is a trivial case. $\rho (t)$ is just a pointer in any moment of time t. It is positive, as we checked earlier, and the inequality (66) is obviously correct.
- “+” sign, $w>0$The inequalities (66) are reduced to:$${s}_{3}{c}^{2}t\le ln{x}_{2}-lnw.$$Thus, the solution $\rho (t)$ (64) of the FGKLS equation has physical sense only for$$t\ge \frac{lnw-ln{x}_{2}}{-{s}_{3}{c}^{2}}$$
- “−” sign, $w>0$In a similar way, we obtain:$${s}_{3}{c}^{2}t\le ln(-{x}_{1})-lnw,$$$$t\ge \frac{lnw-ln(-{x}_{1})}{-{s}_{3}{c}^{2}}.$$

## 5. The Behavior of Solutions for Weak Interaction with Environment

- ${a}_{0}=0$ and ${a}_{1}=0;$
- ${a}_{0}=i({\epsilon}_{11}-{\epsilon}_{22})$ and ${a}_{1}=-\frac{1}{2}(|{\lambda}_{1}{|}^{2}+|{\lambda}_{2}{|}^{2}-2{\lambda}_{1}^{*}{\lambda}_{2});$
- ${a}_{0}=-i({\epsilon}_{11}-{\epsilon}_{22})$ and ${a}_{1}=-\frac{1}{2}(|{\lambda}_{1}{|}^{2}+|{\lambda}_{2}{|}^{2}-2{\lambda}_{1}{\lambda}_{2}^{*}).$

- ${a}_{0}=0;\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{a}_{1}=-1;$
- ${a}_{0}=i({\epsilon}_{11}-{\epsilon}_{22});\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{a}_{1}=-\frac{1}{2};$
- ${a}_{0}=-i({\epsilon}_{11}-{\epsilon}_{22});\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}{a}_{1}=-\frac{1}{2}$

## 6. Unitons

- If the determinant does not vanish, the only solution is ${f}_{mn}^{(u)}=0\phantom{\rule{0.277778em}{0ex}}\forall m,n$, and no unitons exist in this case.
- If the determinant vanishes, three options appear:
- -
- The solution ${f}_{mn}^{(u)}\}$ has one free parameter.This parameter is eliminated by the trace condition ${\sum}_{k}{f}_{kk}^{(u)}=1$. We have a constant density matrix that has to satisfy (82). Thus, unitons exist only if they commute with Hamiltonian: $[H,{\rho}_{u}]=0$. As a result, the only uniton does not depend on time: ${\rho}_{u}(t)={\rho}_{u}(0)$. Actually, it can be considered as a pointer.
- -
- The solution $\{{u}_{mn}\}$ has two parameters.One of the parameters is eliminated by the trace condition ${\sum}_{k}{u}_{kk}=1$, and the dependence of the other on time is found from Equation (82). We have a solution with no free parameters, depending on time in some way.
- -
- The solution $\{{u}_{mn}\}$ has three or more parameters.The same, as in the previous case, but the solution, besides depending on time in some way, also contains one or more parameters. Their dependence on time can be chosen in any manner.

- Diagonal Lindblad operator (23)
- (a)
- ${\lambda}_{1}={\lambda}_{2}$, then$${\rho}_{u}=\left(\begin{array}{cc}{f}_{u11}& {f}_{u12}\\ {f}_{u21}& 1-{f}_{u11}\end{array}\right)$$If this condition for L is satisfied, then every density matrix turns out to be a uniton, obeying Equation (82).
- (b)
- ${\lambda}_{1}\ne {\lambda}_{2}$, then$${\rho}_{u}=\left(\begin{array}{cc}{f}_{u11}& 0\\ 0& 1-{f}_{u11}\end{array}\right)$$

- Lindblad operator of the Jordan block form (29)The uniton is found from the previous result for the pointer (34):$${\rho}_{u}=\left(\begin{array}{cc}\frac{{|\lambda |}^{2}+1}{{2|\lambda |}^{2}+1}& -\frac{{\lambda}^{*}}{{2|\lambda |}^{2}+1}\\ -\frac{\lambda}{{2|\lambda |}^{2}+1}& \frac{{|\lambda |}^{2}}{{2|\lambda |}^{2}+1}\end{array}\right)$$

## 7. Conclusions

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Conflicts of Interest

## Appendix A. Expression for a Pointer for Weak Interaction with Environment

## Appendix B. Other Forms of the Solution of the FGKLS Equation

**Diagonal Lindblad operator**The analysis of Equation (46) shows that the roots coincide when- $|{e}_{12}{|}^{2}=\frac{1}{54}{|{\lambda}_{1}-{\lambda}_{2}|}^{4},$and${\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}=\frac{1}{108}{|{\lambda}_{1}-{\lambda}_{2}|}^{4}$,then$${s}_{1}={s}_{2}={s}_{3}=-\frac{|{\lambda}_{1}-{\lambda}_{2}{|}^{2}}{3}$$The solution is of the form:$$\rho (t)={\rho}_{p}+{e}^{-\frac{|{\lambda}_{1}-{\lambda}_{2}{|}^{2}}{3}{c}^{2}t}{v}_{1}+t{e}^{-\frac{|{\lambda}_{1}-{\lambda}_{2}{|}^{2}}{3}{c}^{2}t}{v}_{2}+{t}^{2}{e}^{-\frac{|{\lambda}_{1}-{\lambda}_{2}{|}^{2}}{3}{c}^{2}t}{v}_{3},$$
- Provided$|{\lambda}_{1}-{\lambda}_{2}{|}^{4}>48{|{e}_{12}|}^{2}+12{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}$and$\begin{array}{l}{\left[\frac{1}{4}|{\lambda}_{1}-{\lambda}_{2}{|}^{4}-12{|{e}_{12}|}^{2}-3{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}\right]}^{3}\\ =|{\lambda}_{1}-{\lambda}_{2}{|}^{4}{\left[-\frac{1}{8}|{\lambda}_{1}-{\lambda}_{2}{|}^{4}+9{|{e}_{12}|}^{2}-\frac{9}{2}{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}\right]}^{2},\end{array}$
- (a)
- If$\begin{array}{l}|{\lambda}_{1}-{\lambda}_{2}{|}^{2}[-\frac{1}{8}|{\lambda}_{1}-{\lambda}_{2}{|}^{4}+9{|{e}_{12}|}^{2}\\ -\frac{9}{2}{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}]>0\end{array}$then$$\begin{array}{ccc}& & {s}_{1}={s}_{2}=-\frac{|{\lambda}_{1}-{\lambda}_{2}{|}^{2}}{3}\hfill \\ & & +\frac{1}{3}\sqrt{\frac{1}{4}|{\lambda}_{1}-{\lambda}_{2}{|}^{4}-12{|{e}_{12}|}^{2}-3{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}}\hfill \end{array}$$$$\begin{array}{ccc}& & {s}_{3}=-\frac{|{\lambda}_{1}-{\lambda}_{2}{|}^{2}}{3}\hfill \\ & & -\frac{2}{3}\sqrt{\frac{1}{4}|{\lambda}_{1}-{\lambda}_{2}{|}^{4}-12{|{e}_{12}|}^{2}-3{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}}\hfill \end{array}$$
- (b)
- If$\begin{array}{l}|{\lambda}_{1}-{\lambda}_{2}{|}^{2}[-\frac{1}{8}|{\lambda}_{1}-{\lambda}_{2}{|}^{4}+9{|{e}_{12}|}^{2}\\ -\frac{9}{2}{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}]<0,\end{array}$then$$\begin{array}{l}{s}_{1}={s}_{2}=-\frac{|{\lambda}_{1}-{\lambda}_{2}{|}^{2}}{3}\\ -\frac{1}{3}\sqrt{\frac{1}{4}|{\lambda}_{1}-{\lambda}_{2}{|}^{4}-12{|{e}_{12}|}^{2}-3{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}}\end{array}$$$$\begin{array}{l}{s}_{3}=-\frac{|{\lambda}_{1}-{\lambda}_{2}{|}^{2}}{3}\\ +\frac{2}{3}\sqrt{\frac{1}{4}|{\lambda}_{1}-{\lambda}_{2}{|}^{4}-12{|{e}_{12}|}^{2}-3{\left|\Delta e+\frac{1}{2}i({\lambda}_{1}{\lambda}_{2}^{*}-{\lambda}_{1}^{*}{\lambda}_{2})\right|}^{2}}\end{array}$$

The solution is of the form:$$\rho (t)={\rho}_{p}+{e}^{{s}_{1}{c}^{2}t}{\tilde{v}}_{1}+t{e}^{{s}_{1}{c}^{2}t}{\tilde{v}}_{2}+{e}^{{s}_{3}{c}^{2}t}{\tilde{v}}_{3}$$

**Lindblad operator of a Jordan block form**In the following, we will use the notation:$E\equiv {({e}_{11}-{e}_{22})}^{2}$, $K\equiv \left(\frac{1}{2}\lambda +i{e}_{21}\right)\left(\frac{1}{2}{\lambda}^{*}-i{e}_{12}\right)$.- The analysis of Equation (55) shows that all three roots coincide$${s}_{1}={s}_{2}={s}_{3}=-\frac{2}{3}$$$$\rho (t)={\rho}_{p}+{e}^{-\frac{2}{3}{c}^{2}t}{v}_{1}+t{e}^{-\frac{2}{3}{c}^{2}t}{v}_{2}+{t}^{2}{e}^{-\frac{2}{3}{c}^{2}t}{v}_{3}$$
- Two of three roots coincide if $E+4K<\frac{1}{12}$ and ${\left(\frac{1}{4}-3E-12K\right)}^{3}={\left(\frac{1}{8}+\frac{9}{2}E-9K\right)}^{2}$.Then, two situations are possible:
- (a)
- $\frac{1}{36}+E-2K>0$, then$${s}_{1}={s}_{2}=-\frac{2}{3}+\frac{1}{3}\sqrt{\frac{1}{4}-3E-12K}$$$${s}_{3}=-\frac{2}{3}-\frac{2}{3}\sqrt{\frac{1}{4}-3E-12K}$$
- (b)
- $\frac{1}{36}+E-2K<0$, then$${s}_{1}={s}_{2}=-\frac{2}{3}-\frac{1}{3}\sqrt{\frac{1}{4}-3E-12K}$$$${s}_{3}=-\frac{2}{3}+\frac{2}{3}\sqrt{\frac{1}{4}-3E-12K}$$

In both situations, the solution has the form:$$\rho (t)={\rho}_{p}+{e}^{{s}_{1}{c}^{2}t}{\tilde{v}}_{1}+t{e}^{{s}_{1}{c}^{2}t}{\tilde{v}}_{2}+{e}^{{s}_{3}{c}^{2}t}{\tilde{v}}_{3}$$

## Appendix C. Lindblad Operator of a Jordan Block Form: An Example

- Born approximation: we assume that, due to a weak coupling between system and environment, the total density matrix is close to a factorized one:$$\rho \simeq {\rho}_{S}\otimes {\rho}_{env}.$$The environmental density matrix is usually assumed to be stationary.
- Born–Markov approximation: we assume that the correlation functions in the environment,$${\langle {B}_{i}^{\u2020}(t){B}_{j}(\tau )\rangle}_{env}={Tr}_{env}\left({\rho}_{env}{B}_{i}^{\u2020}(t){B}_{j}(\tau )\right),$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& {\int}_{0}^{t}d\tau {\langle {B}_{i}^{\u2020}(t){B}_{j}(\tau )\rangle}_{env}{A}_{S,l}(t){A}_{S,k}(\tau ){\rho}_{S}(\tau )\simeq \hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \left[{\int}_{0}^{+\infty}d\tau {\langle {B}_{i}^{\u2020}(\tau ){B}_{j}(0)\rangle}_{env}{A}_{S,l}(t){A}_{S,k}(t-\tau )\right]{\rho}_{S}(t)\hfill \end{array}$$
- Rotating wave approximation: we assume that the relaxation time scale of the system to the equilibrium with the environment is much smaller than the time scale of the system. We introduce the basis of the system Hamiltonian eigenoperators defined as$$[{H}_{S},A(\omega )]=-\omega A(\omega ),$$

- ${\mathsf{\Gamma}}_{++}={\mathsf{\Gamma}}_{--}=0$, and matrix ${\mathsf{\Gamma}}_{ab}$ is arbitrary. This is equivalent to the Lindbladian with$$L=\left(\begin{array}{cc}0& c\\ \overline{c}& 0\end{array}\right),\phantom{\rule{1.em}{0ex}}L={L}^{\u2020},$$
- ${\mathsf{\Gamma}}_{++}\ne 0,{\mathsf{\Gamma}}_{--}=0$, the contribution of the matrix ${\mathsf{\Gamma}}_{ab}$ vanishes. This is equivalent to the Lindbladian with$$L=\left(\begin{array}{cc}0& 2c\\ 0& 0\end{array}\right),\phantom{\rule{1.em}{0ex}}{|c|}^{2}=\frac{{\mathsf{\Gamma}}_{++}}{2}$$
- ${\mathsf{\Gamma}}_{--}\ne 0,{\mathsf{\Gamma}}_{++}=0$, the contribution of the matrix ${\mathsf{\Gamma}}_{ab}$ vanishes. This is equivalent to the Lindbladian with$$L=\left(\begin{array}{cc}0& 0\\ 2c& 0\end{array}\right),\phantom{\rule{1.em}{0ex}}{|c|}^{2}=\frac{{\mathsf{\Gamma}}_{--}}{2}.$$

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Andrianov, A.A.; Ioffe, M.V.; Izotova, E.A.; Novikov, O.O.
The Franke–Gorini–Kossakowski–Lindblad–Sudarshan (FGKLS) Equation for Two-Dimensional Systems. *Symmetry* **2022**, *14*, 754.
https://doi.org/10.3390/sym14040754

**AMA Style**

Andrianov AA, Ioffe MV, Izotova EA, Novikov OO.
The Franke–Gorini–Kossakowski–Lindblad–Sudarshan (FGKLS) Equation for Two-Dimensional Systems. *Symmetry*. 2022; 14(4):754.
https://doi.org/10.3390/sym14040754

**Chicago/Turabian Style**

Andrianov, Alexander A., Mikhail V. Ioffe, Ekaterina A. Izotova, and Oleg O. Novikov.
2022. "The Franke–Gorini–Kossakowski–Lindblad–Sudarshan (FGKLS) Equation for Two-Dimensional Systems" *Symmetry* 14, no. 4: 754.
https://doi.org/10.3390/sym14040754