# Entangled States Are Harder to Transfer than Product States

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## Abstract

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## 1. Introduction

## 2. n-QST Fidelity for Independent Amplitude-Damping Channels

## 3. n-QST Fidelity as a Function of Entanglement

#### 3.1. Two Qubits

#### 3.2. Three Qubits

#### 3.2.1. Three-Qubit Pure-State Entanglement

- Product states. All ${J}_{i}=0$, resulting into $A-B-C$ (class 1). All entanglement measures vanish.
- Biseparable states. All ${J}_{i}=0$ except (i) ${J}_{1}$ for $A-BC$, (ii) ${J}_{2}$ for $B-AC$, and (iii) ${J}_{3}$ for $C-AB$ (class 2a). Only the concurrence for one single pair of qubits is different from zero.
- W-states. ${C}_{\mathrm{GME}}>0$ and ${\tau}_{3}=0$
- ${J}_{4}=0$ and ${J}_{1}{J}_{2}+{J}_{1}{J}_{3}+{J}_{2}{J}_{3}=\sqrt{{J}_{1}{J}_{2}{J}_{3}}=\frac{{J}_{5}}{2}$ (class 3a).
- ${J}_{4}=0$ and $\sqrt{{J}_{1}{J}_{2}{J}_{3}}=\frac{{J}_{5}}{2}$ (class 4a).

- GHZ-states. ${C}_{\mathrm{GME}}>0$ and ${\tau}_{3}>0$, with 5 possible cases:
- All ${J}_{i}=0$ except ${J}_{4}$ (class 2b).
- ${J}_{1}={J}_{2}={J}_{5}=0$, or ${J}_{1}={J}_{3}={J}_{5}=0$ or ${J}_{2}={J}_{3}={J}_{5}=0$ (class 3b).
- ${J}_{2}={J}_{5}=0$ or ${J}_{3}={J}_{5}=0$ (class 4b).
- ${J}_{1}{J}_{4}+{J}_{1}{J}_{2}+{J}_{1}{J}_{3}=\sqrt{{J}_{1}{J}_{2}{J}_{3}}=\frac{{J}_{5}}{2}$ (class 4c).
- $\sqrt{{J}_{1}{J}_{2}{J}_{3}}=\frac{\left|{J}_{5}\right|}{2}$ and ${\left({J}_{4}+{J}_{5}\right)}^{2}-4\left({J}_{1}+{J}_{4}\right)\left({J}_{2}+{J}_{4}\right)\left({J}_{3}+{J}_{4}\right)=0$ (class 4d),

#### 3.2.2. Fidelity of 3-QST

- class 1 (product state)$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& {\u2329{F}_{\mathrm{c}1}\u232a}_{{U}^{\otimes 3}}={\u2329{F}_{1}\u232a}^{3}\hfill \end{array}$$$$\begin{array}{cc}\hfill \phantom{\rule{1.em}{0ex}}& \u2329{F}_{\mathrm{c}1}\u232a={\u2329{F}_{1}\u232a}^{3}\phantom{\rule{4pt}{0ex}}\hfill \end{array}$$
- class 2a (biseparable states)$$\begin{array}{cc}\hfill {\u2329{F}_{\mathrm{c}2\mathrm{a}}\u232a}_{{U}^{\otimes 3}}& ={\u2329{F}_{1}\u232a}^{3}-2\u2329{F}_{1}\u232a\left(\u2329{F}_{1}\u232a-\frac{1}{2}\right)\left(1-\u2329{F}_{1}\u232a\right){C}_{jk}^{2}\hfill \end{array}$$$$\begin{array}{cc}\hfill \u2329{F}_{\mathrm{c}2\mathrm{a}}\u232a& ={\u2329{F}_{1}\u232a}^{3}-\frac{{R}_{3}}{3}\hfill \end{array}$$
- class 2b (GHZ-states)$$\begin{array}{cc}\hfill {\u2329{F}_{\mathrm{c}2\mathrm{b}}\u232a}_{{U}^{\otimes 3}}& ={\u2329{F}_{1}\u232a}^{3}-3\u2329{F}_{1}\u232a\left(\u2329{F}_{1}\u232a-\frac{1}{2}\right)\left(1-\u2329{F}_{1}\u232a\right){\tau}_{3}^{2}\hfill \end{array}$$$$\begin{array}{cc}\hfill \u2329{F}_{\mathrm{c}2\mathrm{b}}\u232a& ={\u2329{F}_{1}\u232a}^{3}-{R}_{3}\hfill \end{array}$$
- class 3a (${J}_{4}=0$ and ${J}_{1}{J}_{2}+{J}_{1}{J}_{3}+{J}_{2}{J}_{3}=\sqrt{{J}_{1}{J}_{2}{J}_{3}}=\frac{{J}_{5}}{2}$)$$\begin{array}{ccc}\hfill {\u2329{F}_{\mathrm{c}3\mathrm{a}}\u232a}_{{U}^{\otimes 3}}& =& {\u2329{F}_{1}\u232a}^{3}-2\u2329{F}_{1}\u232a\left(\u2329{F}_{1}\u232a-\frac{1}{2}\right)\left(1-\u2329{F}_{1}\u232a\right)\left({C}_{BC}^{2}+{C}_{AC}^{2}+{C}_{AB}^{2}\right)\hfill \\ \hfill \u2329{F}_{\mathrm{c}3\mathrm{a}}\u232a& =& {\u2329{F}_{1}\u232a}^{3}-{R}_{3}\hfill \end{array}$$
- Class 3b (${J}_{1}={J}_{2}={J}_{5}=0$ or ${J}_{1}={J}_{3}={J}_{5}=0$ or ${J}_{2}={J}_{3}={J}_{5}=0$)$$\begin{array}{ccc}\hfill {\u2329{F}_{\mathrm{c}3\mathrm{b}}\u232a}_{{U}^{\otimes 3}}& =& {\u2329{F}_{1}\u232a}^{3}-\u2329{F}_{1}\u232a\left(\u2329{F}_{1}\u232a-\frac{1}{2}\right)\left(1-\u2329{F}_{1}\u232a\right)\left(2{C}_{BC}^{2}+3{\tau}_{3}^{2}\right)\hfill \\ \hfill \u2329{F}_{\mathrm{c}3\mathrm{b}}\u232a& =& {\u2329{F}_{1}\u232a}^{3}-\frac{4}{3}{R}_{3}\hfill \end{array}$$
- Class 4a ${J}_{4}=0$ and $\sqrt{{J}_{1}{J}_{2}{J}_{3}}=\frac{{J}_{5}}{2}$$$\begin{array}{ccc}\hfill {\u2329{F}_{\mathrm{c}4\mathrm{a}}\u232a}_{{U}^{\otimes 3}}& =& {\u2329{F}_{1}\u232a}^{3}-2\u2329{F}_{1}\u232a\left(\u2329{F}_{1}\u232a-\frac{1}{2}\right)\left(1-\u2329{F}_{1}\u232a\right)\left({C}_{BC}^{2}+{C}_{AC}^{2}+{C}_{AB}^{2}\right)\hfill \\ \hfill \u2329{F}_{\mathrm{c}4\mathrm{a}}\u232a& =& {\u2329{F}_{1}\u232a}^{3}-{R}_{3}\hfill \end{array}$$
- Class 4b (${J}_{2}={J}_{5}=0$ or ${J}_{3}={J}_{5}=0$)$$\begin{array}{ccc}\hfill {\u2329{F}_{\mathrm{c}4\mathrm{b}}\u232a}_{{U}^{\otimes 3}}& =& {\u2329{F}_{1}\u232a}^{3}-\u2329{F}_{1}\u232a\left(\u2329{F}_{1}\u232a-\frac{1}{2}\right)\left(1-\u2329{F}_{1}\u232a\right)\left(2\left({C}_{BC}^{2}+{C}_{AC}^{2}\right)+3{\tau}_{3}^{2}\right)\hfill \\ \hfill \u2329{F}_{\mathrm{c}4\mathrm{b}}\u232a& =& {\u2329{F}_{1}\u232a}^{3}-\frac{5}{3}{R}_{3}\hfill \end{array}$$
- Class 4c (${J}_{1}{J}_{4}+{J}_{1}{J}_{2}+{J}_{1}{J}_{3}+{J}_{2}{J}_{3}=\sqrt{{J}_{1}{J}_{2}{J}_{3}}=\frac{{J}_{5}}{2}$)$$\begin{array}{ccc}\hfill {\u2329{F}_{\mathrm{c}4\mathrm{c}}\u232a}_{{U}^{\otimes 3}}& =& {\u2329{F}_{1}\u232a}^{3}-\u2329{F}_{1}\u232a\left(\u2329{F}_{1}\u232a-\frac{1}{2}\right)\left(1-\u2329{F}_{1}\u232a\right)\left(2\left({C}_{BC}^{2}+{C}_{AC}^{2}+{C}_{AB}^{2}\right)+3{\tau}_{3}^{2}\right)\hfill \\ \hfill \u2329{F}_{\mathrm{c}4\mathrm{c}}\u232a& =& {\u2329{F}_{1}\u232a}^{3}-2{R}_{3}\hfill \end{array}$$
- Class 4d ($\sqrt{{J}_{1}{J}_{2}{J}_{3}}=\frac{\left|{J}_{5}\right|}{2}$ and ${\left({J}_{4}+{J}_{5}\right)}^{2}-4\left({J}_{1}+{J}_{4}\right)\left({J}_{2}+{J}_{4}\right)\left({J}_{3}+{J}_{4}\right)=0$)$$\begin{array}{ccc}\hfill {\u2329{F}_{\mathrm{c}4\mathrm{c}}\u232a}_{{U}^{\otimes 3}}& =& {\u2329{F}_{1}\u232a}^{3}-\u2329{F}_{1}\u232a\left(\u2329{F}_{1}\u232a-\frac{1}{2}\right)\left(1-\u2329{F}_{1}\u232a\right)\left(2\left({C}_{BC}^{2}+{C}_{AC}^{2}+{C}_{AB}^{2}\right)+3{\tau}_{3}^{2}\right)\hfill \\ \hfill \u2329{F}_{\mathrm{c}4\mathrm{d}}\u232a& =& {\u2329{F}_{1}\u232a}^{3}-2{R}_{3}\hfill \end{array}$$

#### 3.3. Four Qubits

## 4. Discussion

## Author Contributions

## Funding

## Data Availability Statement

## Conflicts of Interest

## Abbreviations

GHZ | Greenberger–Horne–Zeilinger |

GME | Genuine multipartite entanglement |

QST | Quantum-state transfer |

n-QST | n-qubit quantum-state transfer |

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**Figure 1.**A quantum router. A dispatch center, encircled in red, creates an n-qubit entangled state (red spheres) with the aim to send each party to a different receiver (green spheres) along independent quantum channels (blue spheres).

**Figure 3.**(

**Left**) Reduction factor for the average fidelity in the presence of entanglement as in Equation (25). (

**Right**) Average fidelity for the three-qubit classes. The dotted, vertical line is at ${\u2329F\u232a}_{1}=\frac{1}{2}\left(1+\frac{1}{\sqrt{3}}\right)\simeq 0.789$.

**Figure 4.**(

**Left**) Reduction factor for the average fidelity in the presence of entanglement as in Equation (43). (

**Right**) Average fidelity for the entangled classes as reported in Equation (42). The blue and red dotted, vertical lines are, respectively, at ${\u2329F\u232a}_{1}=0.82$ and ${\u2329F\u232a}_{1}=0.85$.

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## Share and Cite

**MDPI and ACS Style**

Apollaro, T.J.G.; Lorenzo, S.; Plastina, F.; Consiglio, M.; Życzkowski, K. Entangled States Are Harder to Transfer than Product States. *Entropy* **2023**, *25*, 46.
https://doi.org/10.3390/e25010046

**AMA Style**

Apollaro TJG, Lorenzo S, Plastina F, Consiglio M, Życzkowski K. Entangled States Are Harder to Transfer than Product States. *Entropy*. 2023; 25(1):46.
https://doi.org/10.3390/e25010046

**Chicago/Turabian Style**

Apollaro, Tony J. G., Salvatore Lorenzo, Francesco Plastina, Mirko Consiglio, and Karol Życzkowski. 2023. "Entangled States Are Harder to Transfer than Product States" *Entropy* 25, no. 1: 46.
https://doi.org/10.3390/e25010046