# Control Landscape of Measurement-Assisted Transition Probability for a Three-Level Quantum System with Dynamical Symmetry

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## Abstract

**:**

## 1. Introduction

## 2. Measurement-Assisted Quantum Control

## 3. Formulation of the Problem

## 4. Spin-1 Representation Theory

## 5. Transition Probabilities Driven by Coherent Control Assisted by One Non-Selective Measurement

## 6. Kinematic Quantum Control Landscape of the Transition Probability

**Lemma 1.**

**Proof.**

- (1)
- ${\beta}_{2}=\frac{\pi}{2}$. In this case,$$\left\{\begin{array}{c}{\mathfrak{L}}_{1}{(\Omega ,{\beta}_{1},\frac{\pi}{2})}_{{\beta}_{1}}^{\prime}=-\frac{1}{4}sin2{\beta}_{1}\hfill \\ {\mathfrak{L}}_{1}{(\Omega ,{\beta}_{1},\frac{\pi}{2})}_{{\beta}_{2}}^{\prime}=cos\Omega sin{\beta}_{1}{sin}^{2}\frac{{\beta}_{1}}{2}\hfill \end{array}\right..$$$$\begin{array}{cc}\hfill {H}_{{\mathfrak{L}}_{1}}\left(-\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2}\right)=& \left(\begin{array}{ccc}0& 0& \frac{1}{2}\\ 0& \frac{1}{2}& 0\\ \frac{1}{2}& 0& \frac{1}{2}\end{array}\right),\hfill \\ \hfill {H}_{{\mathfrak{L}}_{1}}\left(\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2}\right)=& \left(\begin{array}{ccc}0& 0& -\frac{1}{2}\\ 0& \frac{1}{2}& 0\\ -\frac{1}{2}& 0& \frac{1}{2}\end{array}\right).\hfill \end{array}$$
- (2)
- $\Omega =\pi $.In this case,$$\left\{\begin{array}{c}{\mathfrak{L}}_{1}{(\pi ,{\beta}_{1},{\beta}_{2})}_{{\beta}_{1}}^{\prime}=\frac{1}{8}\left(1+3cos2{\beta}_{2}\right)sin2{\beta}_{1}+\frac{1}{4}\left(cos{\beta}_{1}-cos2{\beta}_{1}\right)sin2{\beta}_{2}=0\hfill \\ {\mathfrak{L}}_{1}{(\pi ,{\beta}_{1},{\beta}_{2})}_{{\beta}_{2}}^{\prime}=\frac{1}{8}\left(1+3cos2{\beta}_{1}\right)sin2{\beta}_{2}+\frac{1}{2}sin{\beta}_{1}cos2{\beta}_{2}\left(1-cos{\beta}_{1}\right)=0\hfill \end{array}\right..$$$$\left\{\begin{array}{c}\left(1+3cos2{\beta}_{2}\right)sin2{\beta}_{1}=2\left(cos2{\beta}_{1}-cos{\beta}_{1}\right)sin2{\beta}_{2}\hfill \\ 4sin{\beta}_{1}cos2{\beta}_{2}\left(cos{\beta}_{1}-1\right)=\left(1+3cos2{\beta}_{1}\right)sin2{\beta}_{2}\hfill \end{array}\right..$$$$\frac{cos{\beta}_{1}(1+3cos2{\beta}_{2})}{2cos2{\beta}_{2}(cos{\beta}_{1}-1)}=\frac{2(cos2{\beta}_{1}-cos{\beta}_{1})}{1+3cos2{\beta}_{1}},$$$$\frac{1}{cos2{\beta}_{2}}=\frac{4(cos2{\beta}_{1}-cos{\beta}_{1})(cos{\beta}_{1}-1)}{cos{\beta}_{1}(1+3cos2{\beta}_{1})}-3=\frac{2(2{x}^{2}-1-x)(x-1)}{x(3{x}^{2}-1)}-3.$$Let us express $tan2{\beta}_{2}$ from the second equation in (19):$$tan2{\beta}_{2}=4sin{\beta}_{1}\frac{cos{\beta}_{1}-1}{1+3cos2{\beta}_{1}}=2\sqrt{1-{x}^{2}}\frac{x-1}{3{x}^{2}-1}.$$$$\frac{1}{{cos}^{2}2{\beta}_{2}}=1+{tan}^{2}2{\beta}_{2},$$$${\left(\frac{2(2{x}^{2}-1-x)(x-1)}{x(3{x}^{2}-1)}-3\right)}^{2}=1+4\frac{{(x-1)}^{2}(1-{x}^{2})}{{(3{x}^{2}-1)}^{2}}.$$Equation (20) can be solved analytically. Finally, we have$${(x+1)}^{2}\left(x-\frac{-1-\sqrt{5}}{2}\right)\left(x-\frac{-1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{6}}{5}\right)\left(x-\frac{1+\sqrt{6}}{5}\right)=0.$$$$\begin{array}{c}\hfill \left[\begin{array}{c}\left({\beta}_{1}^{I},{\beta}_{2}^{I}\right)=\left(arccos\left(\frac{1+\sqrt{6}}{5}\right),\frac{1}{2}arccos\left(\frac{1}{1-\sqrt{6}}\right)\right)\approx (0.810,1.166)\hfill \\ \left({\beta}_{1}^{II},{\beta}_{2}^{II}\right)=\left(arccos\left(\frac{1-\sqrt{6}}{5}\right),\frac{1}{2}arccos\left(\frac{1}{1+\sqrt{6}}\right)\right)\approx (1.865,0.638)\hfill \\ \left({\beta}_{1}^{III},{\beta}_{2}^{III}\right)=\left(arccos\left(\frac{-1+\sqrt{5}}{2}\right),\pi -\frac{1}{2}arctan\left(2\sqrt{2+\sqrt{5}}\right)\right)\approx (0.905,2.475)\hfill \end{array}.\right.\end{array}$$$$\begin{array}{c}\hfill {\mathfrak{L}}_{1}(\Omega ,{\beta}_{1},{\beta}_{2})=\left[\begin{array}{c}\frac{1}{4},\phantom{\rule{1.em}{0ex}}\left(\pi ,{\beta}_{1}^{III},{\beta}_{2}^{III}\right)\hfill \\ \frac{3}{50}(9-\sqrt{6}),\phantom{\rule{1.em}{0ex}}\left(\pi ,{\beta}_{1}^{I},{\beta}_{2}^{I}\right)\hfill \\ \frac{3}{50}(9+\sqrt{6}),\phantom{\rule{1.em}{0ex}}\left(\pi ,{\beta}_{1}^{II},{\beta}_{2}^{II}\right)\hfill \end{array}.\right.\end{array}$$$$\left\{\begin{array}{c}{\mathfrak{L}}_{1}{(\pi ,{\beta}_{1},{\beta}_{2})}_{\Omega \Omega}^{\u2033}=-\frac{1}{2}{sin}^{2}\frac{{\beta}_{1}}{2}sin{\beta}_{1}sin2{\beta}_{2}\hfill \\ {\mathfrak{L}}_{1}{(\pi ,{\beta}_{1},{\beta}_{2})}_{{\beta}_{1}{\beta}_{1}}^{\u2033}=\frac{1}{4}\left[\left(2sin2{\beta}_{1}-sin{\beta}_{1}\right)sin2{\beta}_{2}+cos2{\beta}_{1}\left(3cos2{\beta}_{2}+1\right)\right]\hfill \\ {\mathfrak{L}}_{1}{(\pi ,{\beta}_{1},{\beta}_{2})}_{{\beta}_{1}{\beta}_{2}}^{\u2033}={sin}^{2}\frac{{\beta}_{1}}{2}\left(2cos{\beta}_{1}+1\right)cos2{\beta}_{2}-\frac{3}{4}sin2{\beta}_{1}sin2{\beta}_{2}\hfill \\ {\mathfrak{L}}_{1}{(\pi ,{\beta}_{2},{\beta}_{2})}_{{\beta}_{2}{\beta}_{2}}^{\u2033}=\frac{1}{4}\left(3cos2{\beta}_{1}+1\right)cos2{\beta}_{2}-2{sin}^{2}\frac{{\beta}_{1}}{2}sin{\beta}_{1}sin2{\beta}_{2}\hfill \end{array}\right..$$$$\begin{array}{cc}\hfill {H}_{{\mathfrak{L}}_{1}}(\pi ,{\beta}_{1}^{I},{\beta}_{2}^{I})& =\left(\begin{array}{ccc}\frac{1}{250}(13\sqrt{6}-42)& 0& 0\\ 0& \frac{1}{100}(23\sqrt{6}-32)& \frac{1}{50}(-8-13\sqrt{6})\\ 0& \frac{1}{50}(-8-13\sqrt{6})& \frac{1}{5}(\sqrt{6}-4)\end{array}\right)\hfill \\ \hfill {H}_{{\mathfrak{L}}_{1}}(\pi ,{\beta}_{1}^{II},{\beta}_{2}^{II})& =\left(\begin{array}{ccc}\frac{1}{250}(-42-13\sqrt{6})& 0& 0\\ 0& \frac{1}{100}(-32-23\sqrt{6})& \frac{1}{50}(13\sqrt{6}-8)\\ 0& \frac{1}{50}(13\sqrt{6}-8)& \frac{1}{5}(-4-\sqrt{6})\end{array}\right)\hfill \\ \hfill {H}_{{\mathfrak{L}}_{1}}(\pi ,{\beta}_{1}^{III},{\beta}_{2}^{III})& =\left(\begin{array}{ccc}\frac{1}{4}(7-3\sqrt{5})& 0& 0\\ 0& \frac{1}{2}(\sqrt{5}-3)& \frac{1}{4}(1+\sqrt{5})\\ 0& \frac{1}{4}(1+\sqrt{5})& \frac{1}{4}(\sqrt{5}-1)\end{array}\right)\hfill \end{array}$$$$\begin{array}{c}\hfill \left[\begin{array}{c}{\mathfrak{L}}_{1}(\pi ,{\beta}_{1}^{III},{\beta}_{2}^{III})=\frac{1}{4}-\mathrm{saddle}\mathrm{point}\hfill \\ {\mathfrak{L}}_{1}(\pi ,{\beta}_{1}^{I},{\beta}_{2}^{I})=\frac{3}{50}(9-\sqrt{6})\approx 0.393-\mathrm{saddle}\phantom{\rule{4.pt}{0ex}}\mathrm{point}\hfill \\ {\mathfrak{L}}_{1}(\pi ,{\beta}_{1}^{II},{\beta}_{2}^{II})=\frac{3}{50}(9+\sqrt{6})\approx 0.687-\mathrm{global}\mathrm{maximum}\hfill \end{array}\right.\end{array}$$
- (3)
- $\Omega =0$.All the critical points are easily found from (21):$$\begin{array}{c}\hfill \left[\begin{array}{c}\left({\beta}_{1}^{I},{\beta}_{2}^{I}\right)=\left(arccos\left(\frac{1+\sqrt{6}}{5}\right),\pi -\frac{1}{2}arccos\left(\frac{1}{1-\sqrt{6}}\right)\right)\approx (0.810,1.975)\hfill \\ \left({\beta}_{1}^{II},{\beta}_{2}^{II}\right)=\left(arccos\left(\frac{1-\sqrt{6}}{5}\right),\pi -\frac{1}{2}arccos\left(\frac{1}{1+\sqrt{6}}\right)\right)\approx (1.865,2.503)\hfill \\ \left({\beta}_{1}^{III},{\beta}_{2}^{III}\right)=\left(arccos\left(\frac{-1+\sqrt{5}}{2}\right),\frac{1}{2}arctan\left(2\sqrt{2+\sqrt{5}}\right)\right)\approx (0.905,2.475)\hfill \end{array}.\right.\end{array}$$$$\left\{\begin{array}{c}{\mathfrak{L}}_{1}{(0,{\beta}_{1},{\beta}_{2})}_{\Omega \Omega}^{\u2033}=\frac{1}{2}{sin}^{2}\frac{{\beta}_{1}}{2}sin{\beta}_{1}sin2{\beta}_{2}\hfill \\ {\mathfrak{L}}_{1}{(0,{\beta}_{1},{\beta}_{2})}_{{\beta}_{1}{\beta}_{1}}^{\u2033}=\frac{1}{4}\left[\left(sin{\beta}_{1}-2sin2{\beta}_{1}\right)sin2{\beta}_{2}+cos2{\beta}_{1}\left(3cos2{\beta}_{2}+1\right)\right]\hfill \\ {\mathfrak{L}}_{1}{(0,{\beta}_{1},{\beta}_{2})}_{{\beta}_{1}{\beta}_{2}}^{\u2033}=-\frac{3}{4}sin2{\beta}_{1}sin2{\beta}_{2}-{sin}^{2}\frac{{\beta}_{1}}{2}\left(2cos{\beta}_{1}+1\right)cos2{\beta}_{2}\hfill \\ {\mathfrak{L}}_{1}{(0,{\beta}_{2},{\beta}_{2})}_{{\beta}_{2}{\beta}_{2}}^{\u2033}=2sin{\beta}_{1}sin2{\beta}_{2}{sin}^{2}\frac{{\beta}_{1}}{2}+\frac{1}{4}\left(3cos2{\beta}_{1}+1\right)cos2{\beta}_{2}\hfill \end{array}\right..$$$$\begin{array}{cc}\hfill {H}_{{\mathfrak{L}}_{1}}(0,{\beta}_{1}^{I},{\beta}_{2}^{I})& =\left(\begin{array}{ccc}\frac{1}{250}(13\sqrt{6}-42)& 0& 0\\ 0& \frac{1}{100}(23\sqrt{6}-32)& \frac{1}{50}(8+13\sqrt{6})\\ 0& \frac{1}{50}(8+13\sqrt{6})& \frac{1}{5}(\sqrt{6}-4)\end{array}\right)\hfill \\ \hfill {H}_{{\mathfrak{L}}_{1}}(0,{\beta}_{1}^{II},{\beta}_{2}^{II})& =\left(\begin{array}{ccc}\frac{1}{250}(-42-13\sqrt{6})& 0& 0\\ 0& \frac{1}{100}(-32-23\sqrt{6})& \frac{1}{50}(8-13\sqrt{6})\\ 0& \frac{1}{50}(8-13\sqrt{6})& \frac{1}{5}(-4-\sqrt{6})\end{array}\right)\hfill \\ \hfill {H}_{{\mathfrak{L}}_{1}}(0,{\beta}_{1}^{III},{\beta}_{2}^{III})& =\left(\begin{array}{ccc}\frac{1}{4}(7-3\sqrt{5})& 0& 0\\ 0& \frac{1}{2}(\sqrt{5}-3)& \frac{1}{4}(-1-\sqrt{5})\\ 0& \frac{1}{4}(-1-\sqrt{5})& \frac{1}{4}(\sqrt{5}-1)\end{array}\right).\hfill \end{array}$$$$\begin{array}{c}\hfill \left[\begin{array}{c}{\mathfrak{L}}_{1}(0,{\beta}_{1}^{III},{\beta}_{2}^{III})=\frac{1}{4}-\mathrm{saddle}\mathrm{point}\hfill \\ {\mathfrak{L}}_{1}(0,{\beta}_{1}^{I},{\beta}_{2}^{I})=\frac{3}{50}(9-\sqrt{6})\approx 0.393-\mathrm{saddle}\mathrm{point}\hfill \\ {\mathfrak{L}}_{1}(0,{\beta}_{1}^{II},{\beta}_{2}^{II})=\frac{3}{50}(9+\sqrt{6})\approx 0.687-\mathrm{global}\mathrm{maximum}\hfill \end{array}.\right.\end{array}$$

**Lemma 2.**

**Proof.**

**Lemma 3.**

**Proof.**

**Lemma 4.**

**Proof.**

**Lemma 5.**

**Lemma 6.**

**Proof.**

**Remark 1.**

**Theorem 1.**

**Corollary 1.**

## 7. Comparison with other Cases of Measurement-Assisted Transition Probabilities

**Theorem 2.**

**Proof.**

**Theorem 3.**

**Proof.**

## 8. Discussion

## Author Contributions

## Funding

## Data Availability Statement

## Acknowledgments

## Conflicts of Interest

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**Figure 1.**Energy level structure and allowed transitions between the energy levels of a three-level quantum system with dynamical symmetry. The maximum transition probability $|1\rangle \to |2\rangle $ using only coherent control is $1/2$.

**Figure 2.**Kinematic control landscapes of the transition probability ${P}_{1\to 2}(U\left(1\right),{\mathcal{M}}_{|i\rangle \langle i|},U\left(2\right))$. (

**Left**): $i=1$ and $\Omega =0$ (i.e., plot of the function ${\mathfrak{L}}_{1}(0,{\beta}_{1},{\beta}_{2})$; see below). (

**Right**): $i=2$ (i.e., plot of the function $\mathfrak{M}({\beta}_{1},{\beta}_{2})$; see below). The left landscape has second-order traps (see Theorem 1), whereas the right landscape is completely trap-free (see Theorem 3).

**Table 1.**Critical points, their types and values of the objective function ${\mathfrak{L}}_{1}(\Omega ,{\beta}_{1},{\beta}_{2})$.

$(\mathbf{\Omega},{\mathit{\beta}}_{1},{\mathit{\beta}}_{2})$ | ${\mathit{P}}_{1\to 2}$ | Type of Point |
---|---|---|

$\left(\pm \frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2}\right)$ | $1/4$ | Saddle point |

$\left(\pi ,arccos\left(\frac{-1+\sqrt{5}}{2}\right),\pi -\frac{1}{2}arctan\left(2\sqrt{2+\sqrt{5}}\right)\right)$, $\left(0,arccos\left(\frac{-1+\sqrt{5}}{2}\right),\frac{1}{2}arctan\left(2\sqrt{2+\sqrt{5}}\right)\right)$ | $1/4$ | Saddle point |

$\left(\pi ,arccos\left(\frac{1+\sqrt{6}}{5}\right),\frac{1}{2}arccos\left(\frac{1}{1-\sqrt{6}}\right)\right)$, $\left(0,arccos\left(\frac{1+\sqrt{6}}{5}\right),\pi -\frac{1}{2}arccos\left(\frac{1}{1-\sqrt{6}}\right)\right)$ | $\frac{3}{50}(9-\sqrt{6})\approx 0.393$ | Saddle point |

$\left(\pi ,arccos\left(\frac{1-\sqrt{6}}{5}\right),\frac{1}{2}arccos\left(\frac{1}{1+\sqrt{6}}\right)\right)$, $\left(0,arccos\left(\frac{1-\sqrt{6}}{5}\right),\pi -\frac{1}{2}arccos\left(\frac{1}{1+\sqrt{6}}\right)\right)$ | $\frac{3}{50}(9+\sqrt{6})\approx 0.687$ | Global maximum |

**Table 2.**Critical points, their types and values of the objective function ${\mathfrak{L}}_{\mathfrak{2}}({\alpha}_{1},{\beta}_{1},{\gamma}_{1},{\beta}_{2},{\gamma}_{2})$.

$({\mathit{\alpha}}_{1},{\mathit{\beta}}_{1},{\mathit{\gamma}}_{1},{\mathit{\beta}}_{2},{\mathit{\gamma}}_{2})$ | ${\mathit{P}}_{1\to 2}$ | Type of Point |
---|---|---|

$(0,{\beta}_{1},0,\frac{\pi}{2},{\gamma}_{2})$, ${\beta}_{1}\in (0,\pi )$, ${\gamma}_{2}\in (-\pi ,\pi ]$ | $1/2$ | Second-order trap |

**Table 3.**Critical points, their types and values of the objective function ${\mathfrak{L}}_{4}({\alpha}_{1},{\beta}_{1},{\gamma}_{1},{\alpha}_{2},{\beta}_{2},{\gamma}_{2})$.

$({\mathit{\alpha}}_{1},{\mathit{\beta}}_{1},{\mathit{\gamma}}_{1},{\mathit{\alpha}}_{2},{\mathit{\beta}}_{2},{\mathit{\gamma}}_{2})$ | ${\mathit{P}}_{1\to 2}$ | Type of Point |
---|---|---|

$(0,{\beta}_{1},0,0,{\beta}_{2},0)$, ${\beta}_{1},{\beta}_{2}\in (0,\pi )$ | 0 | Global minima |

**Table 4.**Critical points, their types and values of the objective function ${\mathfrak{L}}_{5}\left(\beta \right)$.

$\mathit{\beta}$ | ${\mathit{P}}_{1\to 2}$ | Type of Point |
---|---|---|

$\pi /2$ | $1/2$ | Second-order trap |

**Table 5.**Critical points, their types and values of the objective function ${\mathfrak{L}}_{6}(\alpha ,\beta ,\gamma )$.

$(\mathit{\alpha},\mathit{\beta},\mathit{\gamma})$ | ${\mathit{P}}_{1\to 2}$ | Type of Point |
---|---|---|

$(0,\beta ,0)$, $\beta \in (0,\pi )$ | $1/2$ | Second-order trap |

**Table 6.**Critical points, their types and values of the objective function. Asterisk indicates that the corresponding variable can take any value in its range.

$(\mathit{U}\left(1\right),\mathit{U}\left(2\right)):({\mathit{\alpha}}_{1},{\mathit{\beta}}_{1},{\mathit{\gamma}}_{1},{\mathit{\alpha}}_{2},{\mathit{\beta}}_{2},{\mathit{\gamma}}_{2})$ | ${\mathit{P}}_{1\to 2}$ | Type of Point |
---|---|---|

$(YZY,YZY):(0,{\beta}_{1}^{*},0,0,{\beta}_{2}^{*},0)$ | 0 | Global minima |

$(\mathbb{I},\mathbb{I}):(0,0,0,0,0,0)$ | 0 | Global minima |

$(ZYZ,ZYZ):({\alpha}_{1}^{*},\frac{\pi}{2},{\gamma}_{1}^{*},{\alpha}_{2}^{*},\frac{\pi}{2},\pm \frac{\pi}{2}-{\alpha}_{1}^{*})$ | $1/4$ | Saddle point |

$(ZYZ,ZYZ):({\alpha}_{1}^{*},arccos\left(\frac{-1+\sqrt{5}}{2}\right),{\gamma}_{1}^{*},{\alpha}_{2}^{*},\pi -\frac{1}{2}arctan\left(2\sqrt{2+\sqrt{5}}\right),\pi -{\alpha}_{1}^{*})$ | $1/4$ | Saddle point |

$(ZYZ,ZYZ):({\alpha}_{1}^{*},arccos\left(\frac{-1+\sqrt{5}}{2}\right),{\gamma}_{1}^{*},{\alpha}_{2}^{*},\frac{1}{2}arctan\left(2\sqrt{2+\sqrt{5}}\right),-{\alpha}_{1}^{*})$ | $1/4$ | Saddle point |

$(ZYZ,ZYZ):({\alpha}_{1}^{*},arccos\left(\frac{1+\sqrt{6}}{5}\right),{\gamma}_{1}^{*},{\alpha}_{2}^{*},\frac{1}{2}arccos\left(\frac{1}{1-\sqrt{6}}\right),\pi -{\alpha}_{1}^{*})$ | $\frac{3}{50}(9-\sqrt{6})$ | Saddle point |

$(ZYZ,ZYZ):({\alpha}_{1}^{*},arccos\left(\frac{1+\sqrt{6}}{5}\right),{\gamma}_{1}^{*},{\alpha}_{2}^{*},\pi -\frac{1}{2}arccos\left(\frac{1}{1-\sqrt{6}}\right),-{\alpha}_{1}^{*})$ | $\frac{3}{50}(9-\sqrt{6})$ | Saddle point |

$(YZY,ZYZ):(0,{\beta}_{1}^{*},0,{\alpha}_{2}^{*},\frac{\pi}{2},{\gamma}_{2}^{*})$ | $1/2$ | Second-order trap |

$(\mathbb{I},ZYZ):(0,0,0,0,\frac{\pi}{2},0)$ | $1/2$ | Second-order trap |

$(\mathbb{I},YZY):(0,0,0,0,{\beta}_{2}^{*},0)$ | $1/2$ | Second-order trap |

$(ZYZ,\mathbb{I}):(0,\frac{\pi}{2},0,0,0,0)$ | $1/2$ | Second-order trap |

$(YZY,\mathbb{I}):(0,{\beta}_{1}^{*},0,0,0,0)$ | $1/2$ | Second-order trap |

$(ZYZ,ZYZ):({\alpha}_{1}^{*},arccos\left(\frac{1-\sqrt{6}}{5}\right),{\gamma}_{1}^{*},{\alpha}_{2}^{*},\frac{1}{2}arccos\left(\frac{1}{1+\sqrt{6}}\right),\pi -{\alpha}_{1}^{*})$ | $\frac{3}{50}(9+\sqrt{6})$ | Global maxima |

$(ZYZ,ZYZ):({\alpha}_{1}^{*},arccos\left(\frac{1-\sqrt{6}}{5}\right),{\gamma}_{1}^{*},{\alpha}_{2}^{*},\pi -\frac{1}{2}arccos\left(\frac{1}{1+\sqrt{6}}\right),-{\alpha}_{1}^{*})$ | $\frac{3}{50}(9+\sqrt{6})$ | Global maxima |

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## Share and Cite

**MDPI and ACS Style**

Elovenkova, M.; Pechen, A.
Control Landscape of Measurement-Assisted Transition Probability for a Three-Level Quantum System with Dynamical Symmetry. *Quantum Rep.* **2023**, *5*, 526-545.
https://doi.org/10.3390/quantum5030035

**AMA Style**

Elovenkova M, Pechen A.
Control Landscape of Measurement-Assisted Transition Probability for a Three-Level Quantum System with Dynamical Symmetry. *Quantum Reports*. 2023; 5(3):526-545.
https://doi.org/10.3390/quantum5030035

**Chicago/Turabian Style**

Elovenkova, Maria, and Alexander Pechen.
2023. "Control Landscape of Measurement-Assisted Transition Probability for a Three-Level Quantum System with Dynamical Symmetry" *Quantum Reports* 5, no. 3: 526-545.
https://doi.org/10.3390/quantum5030035