Thermally stable multipartite entanglements in the frustrated Heisenberg hexagon

TL;DR

This paper investigates the stability of multipartite entanglement in a frustrated antiferromagnetic Heisenberg hexagon, revealing that frustration enhances and sustains entanglement at all temperatures.

Contribution

It provides an exact solution for the model and analyzes how frustration influences the robustness of multipartite entanglement against thermal effects.

Findings

Multipartite entanglement is thermally stable due to frustration.

Frustration opposes bipartite but promotes multipartite entanglement.

Entangled states survive at any temperature in the model.

Abstract

Thermally stable quantum states with multipartite entanglements led by frustration are found in the antiferromagnetic spin-1/2 Heisenberg hexagon. The model has been solved exactly to obtain all analytic expressions of eigenvalues and eigenfunctions. Detection and characterizations for various types of entanglements have been carried out in terms of concurrence and entanglement witnesses based on several thermodynamic observables. Variations of entanglement properties with respect to temperature and frustration are discussed. Even though the frustration opposes the bipartite entanglement, it favors the multipartite entanglement. Entangled states exhibit robustness against the thermal effects in the presence of frustration and they are found to survive at any temperature.

Tables

Table 1. Table 1: Energy eigenvalues of the spin-1/2 J 1 subscript 𝐽 1 J_{1} - J 2 subscript 𝐽 2 J_{2} - J 3 subscript 𝐽 3 J_{3} Heisenberg hexagon

${𝑺}_{𝐓}$	${𝑺}_{𝐓}^{𝒛}$	Energy eigenvalues	${𝝀}_{𝒓}$	$𝒑$	${𝑺}_{𝐓}$	${𝑺}_{𝐓}^{𝒛}$	Energy eigenvalues	${𝝀}_{𝒓}$	$𝒑$
3	3	$E_1=$ $\frac{3}{2}\left(J_1+J_2+\frac{1}{2}{J}_3\right)$	1	1	1	0	$E_{33}=$ $\frac{1}{2}\left(J_1-3J_2-\frac{1}{2}{J}_3\right)$	-1	1
3	2	$E_2=$ $\frac{3}{2}\left(J_1+J_2+\frac{1}{2}{J}_3\right)$	1	1	1	0	$E_{34}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3-d_2\right)$	1	3
2	2	$E_3=$ $J_1-\frac{1}{4}{J}_3$	-1	3	1	0	$E_{35}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3+d_2\right)$	1	3
2	2	$E_4=$ $J_1-\frac{1}{4}{J}_3$	-1	3	1	0	$E_{36}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3-d_2\right)$	1	3
2	2	$E_5=$ $\frac{1}{2}\left(-J_1+3J_2-\frac{1}{2}{J}_3\right)$	-1	1	1	0	$E_{37}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3+d_2\right)$	1	3
2	2	$E_6=$ $\frac{3}{4}{J}_3$	1	3	0	0	$E_{38}=$ $\frac{3}{2}\left(-J_1-J_2+\frac{1}{2}{J}_3\right)$	1	1
2	2	$E_7=$ $\frac{3}{4}{J}_3$	1	3	0	0	$E_{39}=$ $-\frac{1}{2}\left(J_1+3J_2+\frac{1}{2}{J}_3\right)$	-1	3
3	1	$E_8=$ $\frac{3}{2}\left(J_1+J_2+\frac{1}{2}{J}_3\right)$	1	1	0	0	$E_{40}=$ $-\frac{1}{2}\left(J_1+3J_2+\frac{1}{2}{J}_3\right)$	-1	3
2	1	$E_9=$ $J_1-\frac{1}{4}{J}_3$	-1	3	0	0	$E_{41}=$ $-J_1-\frac{5}{4}{J}_3+\frac{1}{2}{d}_3$	-1	1
2	1	$E_{10}=$ $J_1-\frac{1}{4}{J}_3$	-1	3	0	0	$E_{42}=$ $-J_1-\frac{5}{4}{J}_3-\frac{1}{2}{d}_3$	-1	1
2	1	$E_{11}=$ $\frac{3}{4}{J}_3$	1	3	3	-1	$E_{43}=$ $\frac{3}{2}\left(J_1+J_2+\frac{1}{2}{J}_3\right)$	1	1
2	1	$E_{12}=$ $\frac{3}{4}{J}_3$	1	3	2	-1	$E_{44}=$ $J_1-\frac{1}{4}{J}_3$	-1	3
2	1	$E_{13}=$ $\frac{1}{2}\left(-J_1+3J_2-\frac{1}{2}{J}_3\right)$	-1	1	2	-1	$E_{45}=$ $J_1-\frac{1}{4}{J}_3$	-1	3
1	1	$E_{14}=$ $-J_1-\frac{1}{4}{J}_3+\frac{1}{2}{d}_1$	1	1	2	-1	$E_{46}=$ $\frac{3}{4}{J}_3$	1	3
1	1	$E_{15}=$ $-J_1-\frac{1}{4}{J}_3-\frac{1}{2}{d}_1$	1	1	2	-1	$E_{47}=$ $\frac{3}{4}{J}_3$	1	3
1	1	$E_{16}=$ $-J_1-\frac{1}{4}{J}_3$	-1	3	2	-1	$E_{48}=$ $\frac{1}{2}\left(-J_1+3J_2-\frac{1}{2}{J}_3\right)$	-1	1
1	1	$E_{17}=$ $-J_1-\frac{1}{4}{J}_3$	-1	3	1	-1	$E_{49}=$ $-J_1-\frac{1}{4}{J}_3+\frac{1}{2}{d}_1$	1	1
1	1	$E_{18}=$ $\frac{1}{2}\left(J_1-3J_2-\frac{1}{2}{J}_3\right)$	-1	1	1	-1	$E_{50}=$ $-J_1-\frac{1}{4}{J}_3-\frac{1}{2}{d}_1$	1	1
1	1	$E_{19}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3-d_2\right)$	1	3	1	-1	$E_{51}=$ $-J_1-\frac{1}{4}{J}_3$	-1	3
1	1	$E_{20}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3+d_2\right)$	1	3	1	-1	$E_{52}=$ $-J_1-\frac{1}{4}{J}_3$	-1	3
1	1	$E_{21}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3-d_2\right)$	1	3	1	-1	$E_{53}=$ $\frac{1}{2}\left(J_1-3J_2-\frac{1}{2}{J}_3\right)$	-1	1
1	1	$E_{22}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3+d_2\right)$	1	3	1	-1	$E_{54}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3-d_2\right)$	1	3
3	0	$E_{23}=$ $\frac{3}{2}\left(J_1+J_2+\frac{1}{2}{J}_3\right)$	1	1	1	-1	$E_{55}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3+d_2\right)$	1	3
2	0	$E_{24}=$ $J_1-\frac{1}{4}{J}_3$	-1	3	1	-1	$E_{56}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3-d_2\right)$	1	3
2	0	$E_{25}=$ $J_1-\frac{1}{4}{J}_3$	-1	3	1	-1	$E_{57}=$ $-\frac{1}{4}\left(J_1+3J_2+J_3+d_2\right)$	1	3
2	0	$E_{26}=$ $\frac{3}{4}{J}_3$	1	3	3	-2	$E_{58}=$ $\frac{3}{2}\left(J_1+J_2+\frac{1}{2}{J}_3\right)$	1	1
2	0	$E_{27}=$ $\frac{3}{4}{J}_3$	1	3	2	-2	$E_{59}=$ $J_1-\frac{1}{4}{J}_3$	-1	3
2	0	$E_{28}=$ $\frac{1}{2}\left(-J_1+3J_2-\frac{1}{2}{J}_3\right)$	-1	1	2	-2	$E_{60}=$ $J_1-\frac{1}{4}{J}_3$	-1	3
1	0	$E_{29}=$ $-J_1-\frac{1}{4}{J}_3+\frac{1}{2}{d}_1$	1	1	2	-2	$E_{61}=$ $\frac{1}{2}\left(-J_1+3J_2-\frac{1}{2}{J}_3\right)$	-1	1
1	0	$E_{30}=$ $-J_1-\frac{1}{4}{J}_3-\frac{1}{2}{d}_1$	1	1	2	-2	$E_{62}=$ $\frac{3}{4}{J}_3$	1	3
1	0	$E_{31}=$ $-J_1-\frac{1}{4}{J}_3$	-1	3	2	-2	$E_{63}=$ $\frac{3}{4}{J}_3$	1	3
1	0	$E_{32}=$ $-J_1-\frac{1}{4}{J}_3$	-1	3	3	-3	$E_{64}=$ $\frac{3}{2}\left(J_1+J_2+\frac{1}{2}{J}_3\right)$	1	1

$d_1=\sqrt{5J_1^2+9J_2^2+4J_3^2-10J_1{J}_2-8J_2{J}_3}$,	$d_2=\sqrt{17J_1^2+9J_2^2+16J_3^2-10J_1{J}_2-24J_1{J}_3-8J_2{J}_3}$,
$d_3=\sqrt{13J_1^2+9J_2^2+4J_3^2-18J_1{J}_2-8J_1{J}_3}$.

Table 3. Table 2: Eigenstates of the spin-1/2 J 1 subscript 𝐽 1 J_{1} - J 2 subscript 𝐽 2 J_{2} - J 3 subscript 𝐽 3 J_{3} Heisenberg hexagon

${𝑺}_{𝐓}^{𝒛}$	Eigenstates
3	${\Psi }_1=$ $|{\psi }_n^3⟩$
2	${\Psi }_2=$ $\frac{1}{\sqrt{6}}$ $\sum _{n=1}^6|{\psi }_n^2⟩$
2	${\Psi }_3=$ $\frac{1}{\sqrt{12}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^{n-1}|{\psi }_n^2⟩+3\sum _{n=3,6}{\left(-1\right)}^n|{\psi }_n^2⟩\right)$
2	${\Psi }_4=$ $\frac{1}{2}$ $\left(\sum _{n=1}^2|{\psi }_n^2⟩-\sum _{n=4}^5|{\psi }_n^2⟩\right)$
2	${\Psi }_5=$ $\frac{1}{\sqrt{6}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^{n-1}|{\psi }_n^2⟩\right)$
2	${\Psi }_6=$ $\frac{1}{2}$ $\left(\sum _{n=3,6}|{\psi }_n^2⟩-\sum _{n=2,5}|{\psi }_n^2⟩\right)$
2	${\Psi }_7=$ $\frac{1}{\sqrt{12}}$ $\left(3\sum _{n=1,4}|{\psi }_n^2⟩-\sum _{n=1}^6|{\psi }_n^2⟩\right)$
1	${\Psi }_8=$ $\frac{1}{\sqrt{15}}$ $\left(\sum _{n=1}^6\left({|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)+\sum _{n=1}^3{|{\psi }_n^1⟩}_2\right)$
1	${\Psi }_9=$ $\frac{1}{4}$ $\left(\sum _{n=1}^6{\left(-1\right)}^{n-1}{|{\psi }_n^1⟩}_0+3\sum _{n=1,4}{\left(-1\right)}^n{|{\psi }_n^1⟩}_0+\sum _{n=2}^3{|{\psi }_n^1⟩}_1-\sum _{n=5}^6{|{\psi }_n^1⟩}_1\right)$
1	${\Psi }_{10}=$ $\frac{1}{\sqrt{48}}$ $\left(3\sum _{n=2}^3{|{\psi }_n^1⟩}_0-3\sum _{n=5}^6{|{\psi }_n^1⟩}_0+\sum _{n=1}^6{\left(-1\right)}^n{|{\psi }_n^1⟩}_1+3\sum _{n=1,4}{\left(-1\right)}^{n-1}{|{\psi }_n^1⟩}_1\right)$
1	${\Psi }_{11}=$ $\frac{1}{\sqrt{48}}$ $\left(3\sum _{n=2,5}\left({|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)-\sum _{n=1}^6\left({|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)+2\sum _{n=2}^3{|{\psi }_n^1⟩}_2-4{|1⟩}_2\right)$
1	${\Psi }_{12}=$ $\frac{1}{4}$ $\left(\sum _{n=1,4}\left({|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)-\sum _{n=3,6}\left({|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)+2\sum _{n=2}^3{\left(-1\right)}^n{|{\psi }_n^1⟩}_2\right)$
1	${\Psi }_{13}=$ $\frac{1}{\sqrt{6}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^{n-1}{|{\psi }_n^1⟩}_1\right)$
1	${\Psi }_{14}=$ $\frac{1}{{\alpha }_1\sqrt{6}}$ $\left(\sum _{n=1}^6\left(C_{11}{|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)+\sqrt{2}{C}_{12}\sum _{n=1}^3{|{\psi }_n^1⟩}_2\right)$
1	${\Psi }_{15}=$ $\frac{1}{{\alpha }_1^{\prime }\sqrt{6}}$ $\left(\sum _{n=1}^6\left(C_{11}^{\prime }{|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)+\sqrt{2}{C}_{12}^{\prime }\sum _{n=1}^3{|{\psi }_n^1⟩}_2\right)$
1	${\Psi }_{16}=$ $\frac{1}{\sqrt{48}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^n{|{\psi }_n^1⟩}_0+3\sum _{n=1,4}{\left(-1\right)}^{n-1}{|{\psi }_n^1⟩}_0+3\sum _{n=2}^3{|{\psi }_n^1⟩}_1-3\sum _{n=5}^6{|{\psi }_n^1⟩}_1\right)$
1	${\Psi }_{17}=$ $\frac{1}{4}$ $\left(\sum _{n=2}^3{|{\psi }_n^1⟩}_0-\sum _{n=5}^6{|{\psi }_n^1⟩}_0+\sum _{n=1}^6{\left(-1\right)}^{n-1}{|{\psi }_n^1⟩}_1+3\sum _{n=1,4}{\left(-1\right)}^n{|{\psi }_n^1⟩}_1\right)$
1	${\Psi }_{18}=$ $\frac{1}{\sqrt{6}}$ $\sum _{n=1}^6{\left(-1\right)}^n{|{\psi }_n^1⟩}_0$
1	${\Psi }_{19}=$ $\frac{1}{{\alpha }_2\sqrt{12}}$ $\left(3\sum _{n=2,5}\left(C_{21}{|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)-\sum _{n=1}^6\left(C_{21}{|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)\right)$
	$+\frac{1}{{\alpha }_2\sqrt{6}}{C}_{22}\left(2{|1⟩}_2-\sum _{n=2}^3{|{\psi }_n^1⟩}_2\right)$

${𝑺}_{𝐓}^{𝒛}$	Eigenstates
1	${\Psi }_{20}=$ $\frac{1}{{\alpha }_2^{\prime }\sqrt{12}}$ $\left(3\sum _{n=2,5}\left(C_{21}^{\prime }{|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)-\sum _{n=1}^6\left(C_{21}^{\prime }{|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)\right)$
	$+\frac{1}{{\alpha }_2^{\prime }\sqrt{6}}{C}_{22}^{\prime }\left(2{|1⟩}_2-\sum _{n=2}^3{|{\psi }_n^1⟩}_2\right)$
1	${\Psi }_{21}=$ $\frac{1}{2{\alpha }_2}$ $\left(\sum _{n=1,4}\left(C_{21}{|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)-\sum _{n=3,6}\left(C_{21}{|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)\right)$
	$+\frac{1}{{\alpha }_2\sqrt{2}}{C}_{22}\sum _{n=2}^3{\left(-1\right)}^{n-1}{|{\psi }_n^1⟩}_2$
1	${\Psi }_{22}=$ $\frac{1}{2{\alpha }_2^{\prime }}$ $\left(\sum _{n=1,4}\left(C_{21}^{\prime }{|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)-\sum _{n=3,6}\left(C_{21}^{\prime }{|{\psi }_n^1⟩}_0+{|{\psi }_n^1⟩}_1\right)\right)$
	$\frac{1}{{\alpha }_2^{\prime }\sqrt{2}}{C}_{22}^{\prime }\sum _{n=2}^3{\left(-1\right)}^{n-1}{|{\psi }_n^1⟩}_2$
0	${\Psi }_{23}=$ $\frac{1}{\sqrt{20}}$ $\left(\sum _{n=1}^6\left({|{\psi }_n^0⟩}_0+{|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)+\sum _{n=1}^2{|{\psi }_n^0⟩}_3\right)$
0	${\Psi }_{24}=$ $\frac{1}{\sqrt{24}}$ $\left(2\sum _{n=1}^2{|{\psi }_n^0⟩}_0-2\sum _{n=4}^5{|{\psi }_n^0⟩}_0+\sum _{n=2}^3\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)-\sum _{n=5}^6\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)\right)$
0	${\Psi }_{25}=$ $\frac{1}{\sqrt{72}}$ $\left(6\sum _{n=3,6}{\left(-1\right)}^n{|{\psi }_n^0⟩}_0+3\sum _{n=1,4}{\left(-1\right)}^{n-1}\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)\right)$
	$+\frac{1}{\sqrt{72}}\sum _{n=1}^6{\left(-1\right)}^n\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2-2{|{\psi }_n^0⟩}_0\right)$
0	${\Psi }_{26}=$ $\frac{1}{\sqrt{8}}$ $\left(\sum _{n=2,5}\left({|{\psi }_n^0⟩}_2-{|{\psi }_n^0⟩}_1\right)+\sum _{n=3,6}\left({|{\psi }_n^0⟩}_1-{|{\psi }_n^0⟩}_2\right)\right)$
0	${\Psi }_{27}=$ $\frac{1}{\sqrt{24}}$ $\left(3\sum _{n=1,4}\left({|{\psi }_n^0⟩}_1-{|{\psi }_n^0⟩}_2\right)+\sum _{n=1}^6\left({|{\psi }_n^0⟩}_2-{|{\psi }_n^0⟩}_1\right)\right)$
0	${\Psi }_{28}=$ $\frac{1}{6}$ $\left(\sum _{n=1}^6{\left(-1\right)}^{n-1}\left({|{\psi }_n^0⟩}_0+{|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)+3\sum _{n=1}^2{\left(-1\right)}^{n-1}{|{\psi }_n^0⟩}_3\right)$
0	${\Psi }_{29}=$ $\frac{1}{{\alpha }_3\sqrt{12}}$ $\left(\sum _{n=1}^6\left(\sqrt{2}{C}_{31}{|{\psi }_n^0⟩}_0+C_{32}\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)\right)+\sqrt{6}\sum _{n=1}^2{|{\psi }_n^0⟩}_3\right)$
0	${\Psi }_{30}=$ $\frac{1}{{\alpha }_3^{\prime }\sqrt{12}}$ $\left(\sum _{n=1}^6\left(\sqrt{2}{C}_{31}^{\prime }{|{\psi }_n^0⟩}_0+C_{32}^{\prime }\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)\right)+\sqrt{6}\sum _{n=1}^2{|{\psi }_n^0⟩}_3\right)$
0	${\Psi }_{31}=$ $\frac{1}{\sqrt{24}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^n\left({|{\psi }_n^0⟩}_1-{|{\psi }_n^0⟩}_2\right)+3\sum _{n=1,4}{\left(-1\right)}^{n-1}\left({|{\psi }_n^0⟩}_1-{|{\psi }_n^0⟩}_2\right)\right)$
0	${\Psi }_{32}=$ $\frac{1}{\sqrt{8}}$ $\left(\sum _{n=5}^6\left({|{\psi }_n^0⟩}_2-{|{\psi }_n^0⟩}_1\right)+\sum _{n=2}^3\left({|{\psi }_n^0⟩}_1-{|{\psi }_n^0⟩}_2\right)\right)$
0	${\Psi }_{33}=$ $\frac{1}{\sqrt{12}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^{n-1}\left({|{\psi }_n^0⟩}_1-{|{\psi }_n^0⟩}_2\right)\right)$
0	${\Psi }_{34}=$ $\frac{C_4}{{\alpha }_4\sqrt{8}}$ $\left(\sum _{n=3,6}\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)-\sum _{n=2,5}\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)\right)$
	$+\frac{1}{2{\alpha }_4}\left(\sum _{n=2,5}{|{\psi }_n^0⟩}_0-\sum _{n=1,4}{|{\psi }_n^0⟩}_0\right)$

${𝑺}_{𝐓}^{𝒛}$	Eigenstates
0	${\Psi }_{35}=$ $\frac{C_4^{\prime }}{{\alpha }_4^{\prime }\sqrt{8}}$ $\left(\sum _{n=3,6}\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)-\sum _{n=2,5}\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)\right)$
	$+\frac{1}{2{\alpha }_4^{\prime }}\left(\sum _{n=2,5}{|{\psi }_n^0⟩}_0-\sum _{n=1,4}{|{\psi }_n^0⟩}_0\right)$
0	${\Psi }_{36}=$ $\frac{C_4}{{\alpha }_4\sqrt{24}}$ $\left(3\sum _{n=1,4}\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)-\sum _{n=1}^6\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)\right)$
	$+\frac{1}{{\alpha }_4\sqrt{12}}\left(3\sum _{n=3,6}{|{\psi }_n^0⟩}_0-\sum _{n=1}^6{|{\psi }_n^0⟩}_0\right)$
0	${\Psi }_{37}=$ $\frac{C_4^{\prime }}{{\alpha }_4^{\prime }\sqrt{24}}$ $\left(3\sum _{n=1,4}\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)-\sum _{n=1}^6\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)\right)$
	$+\frac{1}{{\alpha }_4^{\prime }\sqrt{12}}\left(3\sum _{n=3,6}{|{\psi }_n^0⟩}_0-\sum _{n=1}^6{|{\psi }_n^0⟩}_0\right)$
0	${\Psi }_{38}=$ $\frac{1}{\sqrt{12}}$ $\left(\sum _{n=1}^6\left({|{\psi }_n^0⟩}_2-{|{\psi }_n^0⟩}_1\right)\right)$
0	${\Psi }_{39}=$ $\frac{1}{2}$ $\left(\sum _{n=3,6}{\left(-1\right)}^n{|{\psi }_n^0⟩}_0+\sum _{n=1,4}{\left(-1\right)}^n\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)\right)$
	$+\frac{1}{6}\sum _{n=1}^6{\left(-1\right)}^{n-1}\left({|{\psi }_n^0⟩}_0+{|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)$
0	${\Psi }_{40}=$ $\frac{1}{\sqrt{12}}$ $\left(\sum _{n=1}^2{|{\psi }_n^0⟩}_0-\sum _{n=4}^5{|{\psi }_n^0⟩}_0+\sum _{n=5}^6\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)-\sum _{n=2}^3\left({|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)\right)$
0	${\Psi }_{41}=$ $\frac{1}{{\alpha }_5\sqrt{12}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^{n-1}\left(\sqrt{2}{C}_{51}{|{\psi }_n^0⟩}_0+{|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)+\sqrt{6}{C}_{52}\sum _{n=1}^2{\left(-1\right)}^{n-1}{|{\psi }_n^0⟩}_3\right)$
0	${\Psi }_{42}=$ $\frac{1}{{\alpha }_5^{\prime }\sqrt{12}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^{n-1}\left(\sqrt{2}{C}_{51}^{\prime }{|{\psi }_n^0⟩}_0+{|{\psi }_n^0⟩}_1+{|{\psi }_n^0⟩}_2\right)+\sqrt{6}{C}_{52}^{\prime }\sum _{n=1}^2{\left(-1\right)}^{n-1}{|{\psi }_n^0⟩}_3\right)$
-1	${\Psi }_{43}=$ $\frac{1}{\sqrt{15}}$ $\left(\sum _{n=1}^6\left({|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)+\sum _{n=1}^3{|{\psi }_n^{-1}⟩}_2\right)$
-1	${\Psi }_{44}=$ $\frac{1}{4}$ $\left(\sum _{n=1}^6{\left(-1\right)}^{n-1}{|{\psi }_n^{-1}⟩}_0+3\sum _{n=1,4}{\left(-1\right)}^n{|{\psi }_n^{-1}⟩}_0+\sum _{n=2}^3{|{\psi }_n^{-1}⟩}_1-\sum _{n=5}^6{|{\psi }_n^{-1}⟩}_1\right)$
-1	${\Psi }_{45}=$ $\frac{1}{\sqrt{48}}$ $\left(3\sum _{n=2}^3{|{\psi }_n^{-1}⟩}_0-3\sum _{n=5}^6{|{\psi }_n^{-1}⟩}_0+\sum _{n=1}^6{\left(-1\right)}^n{|{\psi }_n^{-1}⟩}_1+3\sum _{n=1,4}{\left(-1\right)}^{n-1}{|{\psi }_n^{-1}⟩}_1\right)$
-1	${\Psi }_{46}=$ $\frac{1}{\sqrt{48}}$ $\left(3\sum _{n=2,5}\left({|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)-\sum _{n=1}^6\left({|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)+2\sum _{n=2}^3{|{\psi }_n^{-1}⟩}_2-4{|-1⟩}_2\right)$
-1	${\Psi }_{47}=$ $\frac{1}{4}$ $\left(\sum _{n=1,4}\left({|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)-\sum _{n=3,6}\left({|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)+2\sum _{n=2}^3{\left(-1\right)}^n{|{\psi }_n^{-1}⟩}_2\right)$
-1	${\Psi }_{48}=$ $\frac{1}{\sqrt{6}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^{n-1}{|{\psi }_n^{-1}⟩}_1\right)$
-1	${\Psi }_{49}=$ $\frac{1}{{\alpha }_1\sqrt{6}}$ $\left(\sum _{n=1}^6\left(C_{11}{|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)+\sqrt{2}{C}_{12}\sum _{n=1}^3{|{\psi }_n^{-1}⟩}_2\right)$
-1	${\Psi }_{50}=$ $\frac{1}{{\alpha }_1^{\prime }\sqrt{6}}$ $\left(\sum _{n=1}^6\left(C_{11}^{\prime }{|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)+\sqrt{2}{C}_{12}^{\prime }\sum _{n=1}^3{|{\psi }_n^{-1}⟩}_2\right)$

${𝑺}_{𝐓}^{𝒛}$	Eigenstates
-1	${\Psi }_{51}=$ $\frac{1}{\sqrt{48}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^n{|{\psi }_n^{-1}⟩}_0+3\sum _{n=1,4}{\left(-1\right)}^{n-1}{|{\psi }_n^{-1}⟩}_0+3\sum _{n=2}^3{|{\psi }_n^{-1}⟩}_1-3\sum _{n=5}^6{|{\psi }_n^{-1}⟩}_1\right)$
-1	${\Psi }_{52}=$ $\frac{1}{4}$ $\left(\sum _{n=2}^3{|{\psi }_n^{-1}⟩}_0-\sum _{n=5}^6{|{\psi }_n^{-1}⟩}_0+\sum _{n=1}^6{\left(-1\right)}^{n-1}{|{\psi }_n^{-1}⟩}_1+3\sum _{n=1,4}{\left(-1\right)}^n{|{\psi }_n^{-1}⟩}_1\right)$
-1	${\Psi }_{53}=$ $\frac{1}{\sqrt{6}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^n{|{\psi }_n^{-1}⟩}_0\right)$
-1	${\Psi }_{54}=$ $\frac{1}{{\alpha }_2\sqrt{12}}$ $\left(3\sum _{n=2,5}\left(C_{21}{|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)-\sum _{n=1}^6\left(C_{21}{|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)\right)$
	$+\frac{1}{{\alpha }_2\sqrt{6}}{C}_{22}\left(2{|-1⟩}_2-\sum _{n=2}^3{|{\psi }_n^{-1}⟩}_2\right)$
-1	${\Psi }_{55}=$ $\frac{1}{{\alpha }_2^{\prime }\sqrt{12}}$ $\left(3\sum _{n=2,5}\left(C_{21}^{\prime }{|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)-\sum _{n=1}^6\left(C_{21}^{\prime }{|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)\right)$
	$+\frac{1}{{\alpha }_2^{\prime }\sqrt{6}}{C}_{22}^{\prime }\left(2{|-1⟩}_2-\sum _{n=2}^3{|{\psi }_n^{-1}⟩}_2\right)$
-1	${\Psi }_{56}=$ $\frac{1}{2{\alpha }_2}$ $\left(\sum _{n=1,4}\left(C_{21}{|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)-\sum _{n=3,6}\left(C_{21}{|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)\right)$
	$+\frac{1}{{\alpha }_2\sqrt{2}}{C}_{22}\sum _{n=2}^3{\left(-1\right)}^{n-1}{|{\psi }_n^{-1}⟩}_2$
-1	${\Psi }_{57}=$ $\frac{1}{2{\alpha }_2^{\prime }}$ $\left(\sum _{n=1,4}\left(C_{21}^{\prime }{|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)-\sum _{n=3,6}\left(C_{21}^{\prime }{|{\psi }_n^{-1}⟩}_0+{|{\psi }_n^{-1}⟩}_1\right)\right)$
	$+\frac{1}{{\alpha }_2^{\prime }\sqrt{2}}{C}_{22}^{\prime }\sum _{n=2}^3{\left(-1\right)}^{n-1}{|{\psi }_n^{-1}⟩}_2$
-2	${\Psi }_{58}=$ $\frac{1}{\sqrt{6}}$ $\sum _{n=1}^6|{\psi }_n^{-2}⟩$
-2	${\Psi }_{59}=$ $\frac{1}{\sqrt{12}}$ $\left(\sum _{n=1}^6{\left(-1\right)}^{n-1}|{\psi }_n^{-2}⟩+3\sum _{n=3,6}{\left(-1\right)}^n|{\psi }_n^{-2}⟩\right)$
-2	${\Psi }_{60}=$ $\frac{1}{2}$ $\left(\sum _{n=1}^2|{\psi }_n^{-2}⟩-\sum _{n=4}^5|{\psi }_n^{-2}⟩\right)$
-2	${\Psi }_{61}=$ $\frac{1}{\sqrt{6}}$ $\sum _{n=1}^6{\left(-1\right)}^{n-1}|{\psi }_n^{-2}⟩$
-2	${\Psi }_{62}=$ $\frac{1}{2}$ $\left(\sum _{n=3,6}|{\psi }_n^{-2}⟩-\sum _{n=2,5}|{\psi }_n^{-2}⟩\right)$
-2	${\Psi }_{63}=$ $\frac{1}{\sqrt{12}}$ $\left(3\sum _{n=1,4}|{\psi }_n^{-2}⟩-\sum _{n=1}^6|{\psi }_n^{-2}⟩\right)$
-3	${\Psi }_{64}=$ $|{\psi }_n^{-3}⟩$

$C_{11}=$ $\frac{\left(2J_1+2J_3\right)\left(J_1-J_2+2J_3-d_1\right)-8J_1{J}_2}{8J_2^2-\left(3J_1+J_2-d_1\right)\left(J_1-J_2+2J_3-d_1\right)}$,	$C_{12}=$ $\frac{2\sqrt{2}\left(J_2\left(2J_1+2J_3\right)-J_1\left(3J_1+J_2-d_1\right)\right)}{\left(3J_1+J_2-d_1\right)\left(J_1-J_2+2J_3-d_1\right)-8J_2^2}$,
$C_{11}^{\prime }=$ $\frac{\left(2J_1+2J_3\right)\left(J_1-J_2+2J_3+d_1\right)-8J_1{J}_2}{8J_2^2-\left(3J_1+J_2+d_1\right)\left(J_1-J_2+2J_3+d_1\right)}$,	$C_{12}^{\prime }=$ $\frac{2\sqrt{2}\left(J_2\left(2J_1+2J_3\right)-J_1\left(3J_1+J_2+d_1\right)\right)}{\left(3J_1+J_2+d_1\right)\left(J_1-J_2+2J_3+d_1\right)-8J_2^2}$,
$C_{21}=$ $\frac{8J_1{J}_2-2\left(2J_3-J_1\right)\left(-J_1+J_2+4J_3-d_2\right)}{\left(3J_1-J_2-d_2\right)\left(-J_1+J_2+4J_3-d_2\right)-8J_2^2}$,	$C_{22}=$ $\frac{2\sqrt{2}\left(J_1\left(3J_1-J_2-d_2\right)-2J_2\left(2J_3-J_1\right)\right)}{\left(3J_1-J_2-d_2\right)\left(-J_1+J_2+4J_3-d_2\right)-8J_2^2}$,
$C_{21}^{\prime }=$ $\frac{8J_1{J}_2-2\left(2J_3-J_1\right)\left(-J_1+J_2+4J_3+d_2\right)}{\left(3J_1-J_2+d_2\right)\left(-J_1+J_2+4J_3+d_2\right)-8J_2^2}$,	$C_{22}^{\prime }=$ $\frac{2\sqrt{2}\left(J_1\left(3J_1-J_2+d_2\right)-2J_2\left(2J_3-J_1\right)\right)}{\left(3J_1-J_2+d_2\right)\left(-J_1+J_2+4J_3+d_2\right)-8J_2^2}$,
$C_{31}=$ $\frac{\sqrt{3}\left(J_3\left(3J_1+J_2+2J_3-d_1\right)-2J_1\left(2J_2+J_1\right)\right)}{2{\left(2J_2+J_1\right)}^2-\left(3J_1+J_2+2J_3-d_1\right)\left(3J_1-J_2+J_3-d_1\right)}$,	$C_{32}=$ $\frac{\sqrt{6}\left(J_1\left(3J_1-J_2+J_3-d_1\right)-J_3\left(2J_2+J_1\right)\right)}{2{\left(2J_2+J_1\right)}^2-\left(3J_1+J_2+2J_3-d_1\right)\left(3J_1-J_2+J_3-d_1\right)}$,
$C_{31}^{\prime }=$ $\frac{\sqrt{3}\left(J_3\left(3J_1+J_2+2J_3+d_1\right)-2J_1\left(2J_2+J_1\right)\right)}{2{\left(2J_2+J_1\right)}^2-\left(3J_1+J_2+2J_3+d_1\right)\left(3J_1-J_2+J_3+d_1\right)}$,	$C_{32}^{\prime }=$ $\frac{\sqrt{6}\left(J_1\left(3J_1-J_2+J_3+d_1\right)-J_3\left(2J_2+J_1\right)\right)}{2{\left(2J_2+J_1\right)}^2-\left(3J_1+J_2+2J_3+d_1\right)\left(3J_1-J_2+J_3+d_1\right)}$,
$C_4=$ $\frac{2\sqrt{2}\left(J_1-J_2\right)}{3J_1+J_2-4J_3+d_2}$,	$C_4^{\prime }=$ $\frac{2\sqrt{2}\left(J_1-J_2\right)}{3J_1+J_2-4J_3-d_2}$,
$C_{51}=$ $\frac{\sqrt{2}\left(3J_1{J}_3-\left(2J_2-J_1\right)\left(-J_1+3J_2+J_3-d_3\right)\right)}{\left(3J_1-J_2-J_3-d_3\right)\left(-J_1+3J_2+J_3-d_3\right)-3J_3^2}$,	$C_{52}=$ $\frac{\sqrt{6}\left(J_1\left(3J_1-J_2-J_3-d_3\right)-J_3\left(2J_2-J_1\right)\right)}{3J_3^2-\left(3J_1-J_2-J_3-d_3\right)\left(-J_1+3J_2+J_3-d_3\right)}$,
$C_{51}^{\prime }=$ $\frac{\sqrt{2}\left(3J_1{J}_3-\left(2J_2-J_1\right)\left(-J_1+3J_2+J_3+d_3\right)\right)}{\left(3J_1-J_2-J_3+d_3\right)\left(-J_1+3J_2+J_3+d_3\right)-3J_3^2}$,	$C_{52}^{\prime }=$ $\frac{\sqrt{6}\left(J_1\left(3J_1-J_2-J_3+d_3\right)-J_3\left(2J_2-J_1\right)\right)}{3J_3^2-\left(3J_1-J_2-J_3+d_3\right)\left(-J_1+3J_2+J_3+d_3\right)}$,
${\alpha }_1=$ $\sqrt{1+C_{11}^2+C_{12}^2}$,	${\alpha }_1^{\prime }=$ $\sqrt{1+C_{11}^{\prime 2}+C_{12}^{\prime 2}}$,
${\alpha }_2=$ $\sqrt{1+C_{21}^2+C_{22}^2}$,	${\alpha }_2^{\prime }=$ $\sqrt{1+C_{21}^{\prime 2}+C_{22}^{\prime 2}}$,
${\alpha }_3=$ $\sqrt{1+C_{31}^2+C_{32}^2}$,	${\alpha }_3^{\prime }=$ $\sqrt{1+C_{31}^{\prime 2}+C_{32}^{\prime 2}}$,
${\alpha }_4=$ $\sqrt{1+C_4^2}$,	${\alpha }_4^{\prime }=$ $\sqrt{1+C_4^{\prime 2}}$,
${\alpha }_5=$ $\sqrt{1+C_{51}^2+C_{52}^2}$,	${\alpha }_5^{\prime }=$ $\sqrt{1+C_{51}^{\prime 2}+C_{52}^{\prime 2}}$.

Equations

$$H$$

$$H_{\mathrm{NF}}$$

$$H_{\mathrm{F}}$$

$$\mathcal{F}=\mathrm{avg}\frac{⟨\mathcal{G}|H_{\mathrm{F}}|\mathcal{G}⟩}{|⟨\mathcal{G}|H_{\mathrm{NF}}|\mathcal{G}⟩|},$$

$$\rho (T)=\frac{1}{Z}\sum _{n=1}^{64}{e}^{-\beta E_n}{\rho }^n;{\rho }^n=|{\Psi }_n⟩⟨{\Psi }_n|,$$

$${\rho }_{ij}(T)=\frac{1}{Z}\sum _{n=1}^{64}{e}^{-\beta E_n}{\rho }_{ij}^n,$$

$$\displaystyle\rho^{\rm G}_{ij}=\left[{\begin{array}[]{cccc}a&0&0&f\\ 0&b_{1}&z&0\\ 0&z^{*}&b_{2}&0\\ f^{*}&0&0&d\end{array}}\right].$$

$${\text{CN}}_{ij}=2\text{max}\left(0,|z|-\sqrt{ad}\right).$$

$${\chi }_{\alpha }=\frac{{\left(g{\mu }_{\mathrm{B}}\right)}^2}{k_{\mathrm{B}}T}\left(⟨M_{\alpha }^2⟩-⟨M_{\alpha }{⟩}^2\right),$$

$${\chi }_{\alpha }=\frac{{\left(g{\mu }_{\mathrm{B}}\right)}^2}{k_{\mathrm{B}}T}\left(\sum _{i,j=1}^N⟨S_{\alpha }^i{S}_{\alpha }^j⟩-⟨\sum _{i=1}^N{S}_{\alpha }^i{⟩}^2\right).$$

$$\chi =\frac{{\left(g{\mu }_{\mathrm{B}}\right)}^2}{k_{\mathrm{B}}T}\left(\frac{N}{4}+\frac{2}{3}\sum _{i<j}⟨{\vec{S}}_i.{\vec{S}}_j⟩\right).$$

$$\chi \ge \frac{(g{\mu }_B{)}^2}{k_{B}T}.$$

$$|{\psi }_n^3⟩=T^{n-1}|3⟩\left(n=1\right),|3⟩=|\uparrow \uparrow \uparrow \uparrow \uparrow \uparrow ⟩,$$

$$|{\psi }_n^2⟩=T^{n-1}|2⟩\left(n=1,2,3,4,5,6\right),|2⟩=|\uparrow \uparrow \uparrow \uparrow \uparrow \downarrow ⟩,$$

$${|{\psi }_n^1⟩}_0=T^{n-1}{|1⟩}_0\left(n=1,2,3,4,5,6\right),{|1⟩}_0=|\uparrow \uparrow \uparrow \downarrow \downarrow \uparrow ⟩,$$

$${|{\psi }_n^1⟩}_1=T^{n-1}{|1⟩}_1\left(n=1,2,3,4,5,6\right),{|1⟩}_1=|\downarrow \uparrow \uparrow \uparrow \downarrow \uparrow ⟩,$$

$${|{\psi }_n^1⟩}_2=T^{n-1}{|1⟩}_2\left(n=1,2,3\right),{|1⟩}_2=|\downarrow \uparrow \uparrow \downarrow \uparrow \uparrow ⟩,$$

$${|{\psi }_n^0⟩}_0=T^{n-1}{|0⟩}_0\left(n=1,2,3,4,5,6\right),{|0⟩}_0=|\uparrow \uparrow \uparrow \downarrow \downarrow \downarrow ⟩,$$

$${|{\psi }_n^0⟩}_1=T^{n-1}{|0⟩}_1\left(n=1,2,3,4,5,6\right),{|0⟩}_1=|\uparrow \uparrow \downarrow \downarrow \uparrow \downarrow ⟩,$$

$${|{\psi }_n^0⟩}_2=T^{n-1}{|0⟩}_2\left(n=1,2,3,4,5,6\right),{|0⟩}_2=|\uparrow \downarrow \uparrow \downarrow \downarrow \uparrow ⟩,$$

$${|{\psi }_n^0⟩}_3=T^{n-1}{|0⟩}_3\left(n=1,2\right),{|0⟩}_3=|\uparrow \downarrow \uparrow \downarrow \uparrow \downarrow ⟩,$$

$${|{\psi }_n^{-1}⟩}_0=T^{n-1}{|-1⟩}_0\left(n=1,2,3,4,5,6\right),{|-1⟩}_0=|\downarrow \downarrow \downarrow \uparrow \uparrow \downarrow ⟩,$$

$${|{\psi }_n^{-1}⟩}_1=T^{n-1}{|-1⟩}_1\left(n=1,2,3,4,5,6\right),{|-1⟩}_1=|\uparrow \downarrow \downarrow \downarrow \uparrow \downarrow ⟩,$$

$${|{\psi }_n^{-1}⟩}_2=T^{n-1}{|-1⟩}_2\left(n=1,2,3\right),{|-1⟩}_2=|\uparrow \downarrow \downarrow \uparrow \downarrow \downarrow ⟩,$$

$$|{\psi }_n^{-2}⟩=T^{n-1}|-2⟩\left(n=1,2,3,4,5,6\right),|-2⟩=|\downarrow \downarrow \downarrow \downarrow \downarrow \uparrow ⟩,$$

$$|{\psi }_n^{-3}⟩=T^{n-1}|-3⟩\left(n=1\right),|-3⟩=|\downarrow \downarrow \downarrow \downarrow \downarrow \downarrow ⟩.$$

Full text

I INTRODUCTION

The field of quantum information processing experiences a mammoth growth in the last two decades Amico ; HorodeckisRMP ; Guuhne . Entanglement emerges as the most useful quantity among the quantum correlations through an extensive investigations in this field. Vast amounts of works involve in detection, characterization, distillation and quantification of entanglements associated in various quantum systems. Nowadays, there are plethora of ideas which pave the way to realize the more secure and faster information processing tools as well as the more stable and efficient quantum communication networks. These technological innovations include cryptography Ekert , dense coding Bennett1 , teleportation Bennett2 and many more. Interacting spin models consist of both small clusters and long chains where the spins are interacting through the exchange interactions can serve as the platforms to verify the outcome of those proposals QSTNE . Thermal stability of the entangled state, on the other hand, is the main concern to make those protocols operational at room temperature.

Besides those achievements, quantum correlations exhibit dramatic changes in their values when the system undergoes a quantum phase transition (QPT) Sachdev ; Somma ; JVidal ; Wu . Again, value of those correlations can be obtained exactly for the spin models as well as the locations of QPTs can be identified more precisely. In addition, real materials are also available those could serve as the macroscopic realizations of any specific spin models. One of the example of such material is polyoxovanadate compound, (NHEt)3 [V${}_{8}^{\rm IV}$V${}_{4}^{\rm V}$As8O40(H2O)]$\cdot$H2O Procissi . The magnetic properties of this compound are faithfully explained by considering a four-spin cluster, in which four spin-1/2 degrees are arranged on the vertices of a square and they are interacting with the nearest one with isotropic antiferromagnetic (AFM) Heisenberg exchange couplings. QPT occurs at a definite point for this model in the presence of diagonal exchange interaction Bose . Experimental evidence suggests that entanglement can affect the macroscopic properties of solids. The observed values of specific heat and magnetic susceptibility for the compound LiHOxY1-xF4 predict that those can be explained if entanglement of the relevant quantum states are considered explicitly Ghosh . Thermodynamic observables of macroscopic system, like internal energy Dowling , susceptibility Wiesniak and structure factor Krammer serve as the entanglement witness (EW), since the measurement of those quantities eventually leads to the detection of entanglement. For example, the magnetic structure of deuterated copper nitrate Cu(NO3)22.5D2O has been considered as composed of uncoupled spin-1/2 bond alternating AFM Heisenberg chains and for this material susceptibility acts as EW Brukner . A rigorous study of entanglement properties for Heisenberg spin chains in the thermodynamic limit is a theoretical challenge since the eigenvalues and eigenfunctions are not known exactly in every case. In many cases, exact results are obtainable whenever the spin chain is mapped onto a spinless fermionic models Korepin ; Mezzadri ; Osborne ; GVidal ; Damski . Bipartite and multipartite entanglement properties of various spin models have been investigated at finite temperatures Wooters_prl80 ; Wooters_pra63 ; Bruss ; Glaser ; Osterloh ; Bose . But the analytic derivations of all entanglement properties for an isolated cluster containing few spins is possible so far as it is exactly diagonalizable. Moreover, spin cluster with higher values of exchange strength can enhance the stability of entangled state at room temperature which is facing a real challenge nowadays. A spin-cluster material, copper carboxylate {Cu2(O2CH)4}{Cu(O2CH)2(2-methylpyridine)2} is found recently which supports entanglement above room temperature Souza .

In this article, a cluster of six spins with all-to-all two-spin AFM exchange interactions is considered which gives rise to quantum states with multipartite entanglement those can withstand thermal agitations. The model has been solved exactly to obtain all analytic expressions of eigenvalues and eigenfunctions. Symmetry of each eigenstate is studied by exploiting the six-fold rotational invariance of the Hamiltonian. Bipartite entanglements have been characterized with the help of concurrence (CN) while the multipartite entanglements are studied by introducing several EWs. The model Hamiltonian is introduced in the Sec. II along with the characterization of frustration embedded in it. The properties of thermal CN have been discussed in the section III. Detection of bipartite and multipartite entanglements in terms of EWs based on susceptibility, fidelity and internal energy is presented in the Sec. IV while Sec. V holds a comprehensive discussion on the results.

II The SPIN-$\frac{1}{2}$ AFM $J_{1}$-$J_{2}$-$J_{3}$ HEISENBERG HEXAGON

Spin-$\frac{1}{2}$ AFM Heisenberg Hamiltonian on the hexagonal cluster is defined by

[TABLE]

$\vec{S}_{i}$ is the spin-1/2 operator at the position $i$. In this model, every spin is interacting with all other spins via the AFM exchange interactions. As a result, three topologically different exchange couplings, say, nearest neighbor (NN), next nearest neighbor (NNN) and further neighbor (FN) or diagonal exchanges appear whose strengths are $J_{1}$, $J_{2}$ and $J_{3}$, respectively. Geometrical view of this spin model is shown in the Fig. 1(a). Frustration appears in a magnetic system when all the AFM bonds are not energetically minimized in the classical ground state simultaneously. In this model, $J_{2}$ is frustrating, while $J_{1}$ and $J_{3}$ are non-frustrating. With this view, the total Hamiltonian, $H$, (Eq. 1) is decomposed into two parts, non-frustrated ($H_{\rm NF}$) and frustrated ($H_{\rm F}$). Frustration does not appear in this system if $J_{2}$ is assumed negative (ferromagnetic). For $J_{2}<(J_{1}+J_{3}/2)$, the classical ground state of this model is a doublet, where each state is connected to other by flipping the spins in every site. One of such state, $|\mathcal{G}\rangle$, is shown in the Fig. 1(c), in which the adjacent spins are antiparallel. In this case, energy minimization for both $H_{\rm NF}$ and $H_{\rm F}$ cannot be taken place simultaneously in the ground state. Energy minimization of an AFM bond occurs when the spin alignments around this bond is antiparallel. As a result, energy corresponding to $H_{\rm NF}$ with respect to the ground state $|\mathcal{G}\rangle$ is minimized but that of $H_{\rm F}$ is maximized with respect to the same $|\mathcal{G}\rangle$. The frustration of this model can be characterized by using the quantity, frustration degree ($\mathcal{F}$) which is defined as Sen ,

[TABLE]

where “avg” denotes the averaging over all possible ground states. In this model, $\mathcal{F}=J_{2}/(J_{1}+J_{3}/2)$. For the frustrated system $\mathcal{F}>0$, while it is non-frustrated when $\mathcal{F}\leq 0$. The higher value of $\mathcal{F}$ corresponds to the stronger frustration. The variation of $\mathcal{F}$ in the $J_{2}$-$J_{3}$ parameter space is shown in the Fig. 2 (a). $\mathcal{F}$ is found to increase (decrease) with the increase of $J_{2}$ ($J_{3}$). The maximum value of $\mathcal{F}$ is unity which appears at the point $J_{3}=0$ over the line $J_{2}=J_{1}$ in the parameter space. This particular point is labeled by the letter M in the parameter space (Fig. 1 (d)). Therefore, the system is maximally frustrated at the point M. On the other hand, the minimum value of $\mathcal{F}$ is zero for this AFM model which is observed over the line $J_{2}=0$, where the system is said to be non-frustrated. Thus, effects of magnetic frustration on the entanglement properties can be studied with this model.

The Hamiltonian, Eq. 1, commutes with total spin operator, $S_{\rm T}$, as well as the $z$-component of the total spin, $S^{z}_{\rm T}$. As a result, the Hamiltonian may be spanned in the different subspaces of $S^{z}_{\rm T}$ to obtain analytic expressions of eigenvalues and eigenfunctions. The exact analytic expressions of all 64 eigenstates ($\Psi_{n},\;n=1,2,3,\cdots,64$) and corresponding energy eigenvalues ($E_{n}$) are available in the Appendix A. Those states essentially comprise to five singlets ($S_{\rm T}=0$), nine triplets ($S_{\rm T}=1$), five quintets ($S_{\rm T}=2$) and one septet ($S_{\rm T}=3$). Five distinct singlets are denoted by the eigenstates $\Psi_{38}$, $\Psi_{39}$, $\Psi_{40}$, $\Psi_{41}$ and $\Psi_{42}$ in the Appendix A. Among the five, two singlets, $\Psi_{42}$ and $\Psi_{38}$ can be expressed by two distinct combinations of dimer states which are known as resonating valence bond (RVB) states. Those two particular singlets are defined by $\Psi_{\rm RVB}$ and $\Psi_{\rm RVB}^{\prime}$. The arrangements of dimer states in $\Psi_{\rm RVB}$ ($\Psi_{42}$) and $\Psi_{\rm RVB}^{\prime}$ ($\Psi_{38}$) are shown in the Fig. 1 (e). Ground state is always a total spin singlet. All the five singlets participate in four different manners to constitute the ground state in the whole parameter space. Thus, depending on the combinations of singlets in the ground states, $J_{2}$-$J_{3}$ parameter space is decomposed into four segments. $\Psi_{\rm RVB}$ and $\Psi_{\rm RVB}^{\prime}$ are the ground states (non-degenerate) in the regions, R1 ($J_{1}+J_{3}>2J_{2}$) and R2 ($J_{1}+J_{3}<2J_{2}$), respectively. $\Psi_{\rm RVB}$ and $\Psi_{\rm RVB}^{\prime}$ form the doubly degenerate ground state over the line, L ($J_{1}+J_{3}=2J_{2}$), junction of the two regions, R1 and R2. And all the five singlets constitute the ground state (five-fold degenerate) at the point P ($J_{1}=J_{3}=J_{2}$). Positions of R1, R2, L and P on the parameter space are shown in the Fig. 1 (d). The area of R1 is three times larger than that of R2. A first order QPT occurs across the line L as well as at the point P, where the ground state cross over takes place.

In addition, the Hamiltonian possesses another useful symmetry, where it is invariant under the rotation by $60^{\circ}$, (Fig. 1(b)). For the counter clockwise rotation by $60^{\circ}$, a rotational operator, $\hat{R}$, can be defined as $\hat{R}\left|S_{1}S_{2}S_{3}S_{4}S_{5}S_{6}\right\rangle=\left|S_{2}S_{3}S_{4}S_{5}S_{6}S_{1}\right\rangle$, where $\left|S_{1}S_{2}S_{3}S_{4}S_{5}S_{6}\right\rangle=\left|S_{1}^{z}\right\rangle\otimes\left|S_{2}^{z}\right\rangle\otimes\left|S_{3}^{z}\right\rangle\otimes\left|S_{4}^{z}\right\rangle\otimes\left|S_{5}^{z}\right\rangle\otimes\left|S_{6}^{z}\right\rangle$, in which $\left|S_{i}^{z}\right\rangle$ is the spin state at site $i$. So, $\hat{R}^{(n)}$ be the successive $\hat{R}$ operation by $n$ times, such that $\hat{R}^{(6)}$ is the identity operation which leaves any state unchanged. Each eigenstate ($\Psi$) of the Hamiltonian has some definite rotational property, which can be characterized in terms of an eigenvalue equation, like $\hat{R}^{(n)}|\Psi\rangle=\lambda_{r}|\Psi\rangle$, where $\lambda_{r}$’s are the eigenvalues of the rotational operator $\hat{R}^{(n)}$. $\lambda_{r}$ can assume the value either $+1$ or $-1$ for the minimum number ($p$) of $\hat{R}$ operations on a definite state. Obviously, for the same state $\lambda_{r}$ is always $+1$ for $2p$ number of $\hat{R}$ operations. The states with $\lambda_{r}=+1$ for $p$ number of $\hat{R}$ operations have even parity (symmetric) while those with $\lambda_{r}=-1$ have odd parity (antisymmetric). It is found that, every eigenstate has definite values of both $p$ and $\lambda_{r}$, and thus has definite parity. 36 states have even parity while the remaining 28 states have odd parity. Values of $p$ and $\lambda_{r}$ for all eigenstates are shown in the Tab. I. It is observed that $p$ takes up either 1 or 3 and never takes up 2, 4 and 5. For $\Psi_{\rm RVB}$, $\lambda_{r}=-1$ and $p=1$, while, for $\Psi_{\rm RVB}^{\prime}$, $\lambda_{r}=1$ and $p=1$. Thus, $\Psi_{\rm RVB}^{\prime}$ does not change sign under any number of $\hat{R}$ operations, while $\Psi_{\rm RVB}$ changes sign for odd numbers of $\hat{R}$ operations. So, $\Psi_{\rm RVB}$ is antisymmetric, whereas, $\Psi_{\rm RVB}^{\prime}$ is symmetric under the rotation by $60^{\circ}$.

III Thermal Concurrence

For the Heisenberg hexagon, thermal state density matrix has been written down as

[TABLE]

where $Z$ is the partition function of the system. $\beta^{-1}=k_{\rm B}T$, where $k_{\rm B}$ and $T$ are the Boltzmann constant and temperature, respectively. Eigenvalues, $E_{n}$ and the corresponding eigenstates, $\Psi_{n}$ are shown in the Appendix A. Similarly, the reduced thermal state density matrix $\rho_{ij}(T)$ can be written as,

[TABLE]

where the reduced density matrix, $\rho_{ij}^{n}$ is obtained from $\rho^{n}$ by tracing out the remaining four spin degrees of freedom, those are not located at the sites $i$ and $j$. CN is one of the simplest measure to quantify the entanglement between two qubits when they sit at two different sites in the surrounding of other interacting spins and that can be derived from the expression of $\rho_{ij}(T)$. At $T=0$, $\rho_{ij}(T)$ reduces to $\rho^{\rm G}_{ij}$, where $\rho^{\rm G}=\left|\Psi_{\rm G}\right\rangle\left\langle\Psi_{\rm G}\right|$ and $\Psi_{\rm G}$ is the ground state. $\Psi_{\rm G}$ becomes equal to $\Psi_{42}$ and $\Psi_{38}$ for the regions R1 and R2, respectively. On the line L, $\rho^{\rm G}=(\left|\Psi_{38}\right\rangle\left\langle\Psi_{38}\right|+\left|\Psi_{42}\right\rangle\left\langle\Psi_{42}\right|)/2$. Similarly, at P $\rho^{\rm G}=(\left|\Psi_{38}\right\rangle\left\langle\Psi_{38}\right|+\left|\Psi_{39}\right\rangle\left\langle\Psi_{39}\right|+\left|\Psi_{40}\right\rangle\left\langle\Psi_{40}\right|+\left|\Psi_{41}\right\rangle\left\langle\Psi_{41}\right|+\left|\Psi_{42}\right\rangle\left\langle\Psi_{42}\right|)/5$. Depending on the positions of the sites $i$ and $j$, only three different types of $\rho^{\rm G}_{ij}$ can be constructed. They are $\rho^{\rm G}_{\rm NN}$, $\rho^{\rm G}_{\rm NNN}$ and $\rho^{\rm G}_{\rm FN}$, when the sites $i$ and $j$ are NN, NNN and FN, respectively. For example, there is six distinct pairs of NN sites for different values of $i$ and $j$ ({$ij$}), say, {12}, {23}, {34}, {45}, {56} and {61}. $\rho^{\rm G}_{ij}$ is same for all these six NN pairs by virtue of the rotational symmetry of hexagon. So, they are abbreviated as $\rho^{\rm G}_{\rm NN}$. The similar argument holds true for other combinations, NNN and FN. NNN corresponds to six distinct pairs while FN corresponds to only three pairs. The general form of two-qubit $\rho^{\rm G}_{ij}$ in the space of $S^{z}$ diagonal basis states, $\{\left|\uparrow\uparrow\right\rangle,\,\left|\uparrow\downarrow\right\rangle,\,\left|\downarrow\uparrow\right\rangle,\,\left|\downarrow\downarrow\right\rangle\}$, looks like,

[TABLE]

By expressing $\rho^{\rm G}_{ij}$ in this form one can define the spin reversed reduced density matrix as, $\overline{\rho^{\rm G}_{ij}}=(\sigma_{y}\otimes\sigma_{y})(\rho^{\rm G})^{{}^{*}}_{ij}(\sigma_{y}\otimes\sigma_{y})$, where $\sigma_{y}$ is the Pauli matrix. Then concurrence between the sites $i$ and $j$ (CNij) is given by CN${}_{ij}={\rm max}\{\lambda_{1}-\lambda_{2}-\lambda_{3}-\lambda_{4},0\}$, where $\lambda_{i}$s are the square roots of the eigenvalues of the non-Hermitian matrix $\rho^{\rm G}_{ij}\,\overline{\rho^{\rm G}_{ij}}$, in descending order Wooters_prl80 . Since $S^{z}_{\rm T}$ is the good quantum number, the element $f$ in $\rho^{\rm G}_{ij}$ (Eq. 9) vanishes. As a result, the expression of concurrence looks simpler, which is given by Wooters_pra63

[TABLE]

CNij measures the pairwise entanglement between two spins at sites $i$ and $j$, which varies from CN${}_{ij}=0$ for a separable state to CN${}_{ij}=1$ for a maximally entangled state. Variations of CNNN and CNFN for four different locations in the parameter space are shown in Fig. 2 (b) and (c), respectively. CNNNN is zero everywhere which means that concurrence between NNN sites does not survive. CNNN is found to obey the relation CN${}_{\rm NN}=-\frac{1}{2}[4E_{\rm G}/N+1]$, for $J_{2}=J_{3}=0$ where $E_{\rm G}$ is the ground state energy of $S=\frac{1}{2}$ AFM Heisenberg chain with $N$ sites and periodic boundary condition Wooters_pra63 . Similar types of relations for CNNNN and CNFN are not found. CN${}_{\rm NN}=0.434$ over the line $J_{2}=J_{3}$ which is also maximum. This particular line is marked by the dashed line OP in the parameter space (Fig. 1 (d)). CNNN vanishes in the region R2. CNNN suffers a jump over the line L, which is the signature of a first-order QPT. In R1, for fixed value of the frustrating bond ($J_{2}$), CNNN increases with $J_{3}$ up to the line OP, where it acquires the maximum value. With further increase of $J_{3}$, it begins to decrease. On the other hand, CNFN is zero throughout the region R2 in addition to the portion of R1 where $J_{3}\leq(0.87J_{2}+0.14)$. In the region R1, for fixed $J_{2}$, CNFN increases with the increase of $J_{3}$ but for fixed $J_{3}$, it decreases with increasing $J_{2}$. The maximum value of CNFN is observed over the line $J_{3}=J_{1}$ barring the point P. There is no effect of frustration on CNFN in the locations R2 and L. On the other hand, they tend to decrease with the increase of $J_{2}$ in R1.

The thermal state concurrence (TCN) has been derived from $\rho_{ij}(T)$ by using Eqs. (10). The variations of TCNNN with respect to $k_{\rm B}T/J_{1}$ for the line L including the point P has been displayed in Fig. 2 (d). TCN decreases with temperature and exactly vanishes at the critical temperature $T^{ij}_{\rm c}$. Non-zero values for $T^{\rm NN}_{\rm c}$ and $T^{\rm FN}_{\rm c}$ have been observed while $T^{\rm NNN}_{\rm c}$ is always zero. Variations of $T^{\rm NN}_{\rm c}$ and $T^{\rm FN}_{\rm c}$ have been shown in Fig. 3 (a) and (b), respectively. For a fixed $J_{3}$, both $T^{\rm NN}_{\rm c}$ and $T^{\rm FN}_{\rm c}$ decrease very fast with $J_{2}$ whereas for fixed $J_{2}$, they both increase slowly with $J_{3}$. The variations of both $T^{\rm NN}_{\rm c}$ and $T^{\rm FN}_{\rm c}$ with respect to $J_{2}$ indicate that frustration opposes the bipartite entanglement in this system.

IV entanglement witnesses: susceptibility, fidelity and internal energy

In 1996, Horodecki et. al. formulate the necessary and sufficient conditions for separability of a bipartite system Horodeckis . This formulation leads to the existence of a particular EW which is essentially a measure of violation of Bell inequality Terhal . For a magnetic system, it has been shown that magnetic susceptibility can serve as an EW which can be applied without complete knowledge of the Hamiltonian Wiesniak . For an isolated $N$-spin cluster, which is SU(2) invariant and translationally symmetric, the condition of untangled states has been put forward in term of an inequality Brukner . For the Heisenberg Hamiltonian, which is isotropic in the spin space, the magnetic susceptibility along a particular direction, $\alpha$, ($\alpha=x,y,z$) is given by

[TABLE]

where $M_{\alpha}=\sum\limits_{i=1}^{N}S^{i}_{\alpha}$ is the magnetization along the direction $\alpha$, $g$ is the g-factor and $\mu_{B}$ is the Bohr magneton. Thus,

[TABLE]

Since the Hamiltonian is isotropic in the spin space, $\chi=\chi_{x}=\chi_{y}=\chi_{z}$, or, $\chi=\frac{1}{3}\left(\chi_{x}+\chi_{y}+\chi_{z}\right)$, and $\langle\sum\limits_{i=1}^{N}S^{i}_{\alpha}\rangle=0$, $\chi$ can be expressed as

[TABLE]

The second term in the expression of $\chi$, i. e., the sum within the expectation value in Eq. 11, acts as the all-to-all spin interaction term. Alternately, in this particular case, this sum is equivalent to the Hamiltonian (Eq. 1), at the point P when $J_{1}=1$, say, $H_{\rm P}$. As a result, $\sum\limits_{i<j}\langle\vec{S}_{i}.\vec{S}_{j}\rangle=\langle H_{\rm P}\rangle$ corresponds to the ground state energy at the point P for $J_{1}=1$. Due to AFM spin interaction the ground state expectation value of $H_{\rm P}$ is always negative. So, $\langle H_{\rm P}\rangle$ makes a negative contribution to $\chi$. And the maximum negative value of $\langle H_{\rm P}\rangle$ is equal to the ground state energy of $H_{\rm P}$ itself. It has been discussed in the next section that minimum energy of the separable states is negative and equivalent to the ground-state energy of the corresponding classical Hamiltonian. For $N=6$, $\langle H_{\rm P}\rangle=-3/4$. For any general separable states, $\langle H_{\rm P}\rangle$ always make lesser contribution to $\chi$ in comparison to the separable state of minimum energy. Therefore, for a single cluster of $N=6$ spin the condition of untangled states has been given by the inequality

[TABLE]

Curves describing the variation of $\chi/(g^{2}\mu^{2}_{\rm B}J_{1})$ against $k_{\rm B}T/J_{1}$ arising from the above equality, Eq. 12 and the same variation resulting from Eq. 11 intersect at a critical temperature, $T_{\rm c}$, below which the system is entangled. Thus, $\chi$, (Eq. 11) acts as an EW. The variations of $\chi/(g^{2}\mu^{2}_{\rm B}J_{1})$ against $k_{\rm B}T/J_{1}$ representing Eqs. 11 and 12 have been shown in Fig. 4 (a). Eq. 11 has been evaluated for $N=6$ where only NN interactions are considered. Two curves intersects at $T_{\rm c}\approx 1.43J_{1}/K_{B}$. The variation of TCNNN with respect to $k_{\rm B}T/J_{1}$ has been shown in Fig. 4 (b), where only NN interactions are considered. This variation indicates that $T_{\rm c}^{\rm NN}\approx 0.802J_{1}/k_{\rm B}$, where $T_{\rm c}^{\rm NN}$ is that critical temperature beyond which the bipartite entanglement does not exist. By comparing the values of $T_{\rm c}$ and $T_{\rm c}^{\rm NN}$, it is obvious that only multipartite entanglement is present in the system in the intermediate temperature range $T_{\rm c}^{\rm NN}<T<T_{\rm c}$. Thus, below $T_{\rm c}^{\rm NN}$, both bipartite and multipartite entanglements are present while they vanish above $T_{\rm c}$. The variation of $k_{\rm B}T_{\rm c}/J_{1}$ for the AFM Heisenberg hexagon has been shown in Fig. 5 (a). $T_{\rm c}$ has the maximum value at the point P when $J_{1}=J_{2}=J_{3}$, i. e., where all-to-all interactions of equal strength are present. The minimum value of $T_{\rm c}$ appears when $J_{2}=J_{3}=0$, i. e., where only NN interactions are present. With the increase of both $J_{2}$ and $J_{3}$, $T_{\rm c}$ increases steadily. But the rate of increase of $T_{\rm c}$ with respect to $J_{2}$ is more than that of $J_{3}$, which means that frustration enhances the multipartite entanglement in the system.

In order to investigate the presence of six-qubit entanglement in the AFM Heisenberg hexagon, the state preparation fidelity, $F$ is defined as, $F(\rho)=\left\langle\Psi_{GHZ}\right|\rho(T)\left|\Psi_{GHZ}\right\rangle$, where $\left|\Psi_{GHZ}\right\rangle=\frac{1}{\sqrt{2}}\left(\left|\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\right\rangle-\left|\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\right\rangle\right)$ is the six-spin Greenberger-Horne-Zeilinger (GHZ) state Wang . The sufficient condition for the presence of six-particle entanglement in this six-qubit system is given by the inequality, $F(\rho)>\frac{1}{2}$ Sackett ; Bennett . For the hexagonal system with $J_{2}=J_{3}=0$, variation of $F(\rho)$ against $k_{\rm B}T/J_{1}$ has been shown in Fig. 4 (b). The variation of ground state fidelity $F$ in the parameter space is shown in Fig. 5 (b). The value of $F$ is fixed over the line OP and that value of $F$ is $0.458$. The maximum value of $F$ at zero temperature is $1/2$ which is observed over another line $J_{3}=J_{1}$ except the point P. $F$, however, vanishes over the entire region R2. The value of $F$ just over the line L is fixed, and it suffers a sudden jump, which is the manifestation of QPT. $F$ decreases with the increase of $T$ throughout the parameter space. Since $F(\rho)\leq\frac{1}{2}$, the six-spin entanglement is absent in the ground as well as the thermal states at all temperatures. On the other hand, for $S=\frac{1}{2}$ AFM Heisenberg tetramer with NN interaction, $F=\frac{2}{3}>\frac{1}{2}$, which indicates the presence of four-particle entanglement in ground state Wang ; Bose . In general, $F$ increases with $J_{3}$ for fixed $J_{2}$ and decrease with $J_{2}$ for fixed $J_{3}$. Therefore, frustration opposes the six-spin entanglement in this case.

Another kind of detection for EW has been introduced by Dowling and others based on a comparison between the internal energy ($U(T)$) at finite temperature, $T$, and the minimum separable energy ($E_{\rm sep}$) Dowling . The entanglement gap energy, $G_{\rm E}$ is defined by $G_{\rm E}(T)=E_{\rm sep}-U(T)$, at non-zero temperature while that at zero temperature is given by $G_{\rm E}(0)=E_{\rm sep}-E_{\rm G}$, where $E_{\rm G}$ is the ground state energy. $U(T)$ is given by $U(T)=-\frac{1}{Z}\frac{\partial Z}{\partial\beta}$. The multipartite entanglement would be present in the system at non-zero temperature, whenever $G_{\rm E}(T)>0$. With the increase of $T$, $G_{\rm E}(T)$ decreases since $U(T)$ increases with $T$. Obviously, there would be a limiting value of $T$ above which $G_{\rm E}(T)<0$. This critical value of temperature, known as the entanglement gap temperature ($T_{\rm E}$) is define by, $U(T_{\rm E})=E_{\rm sep}$. Therefore, below $T_{\rm E}$ multipartite entanglement is non-zero. Thus a thermal state is entangled if $T<T_{\rm E}$. To formalize this analysis, an EW, $Z_{\rm EW}$, a Hermitian operator is introduced such that Tr[$Z_{\rm EW}\rho_{\rm ent}$]$<0$, when there exists an entangled state, $\rho_{\rm ent}$. It is noted that $Z_{\rm EW}$ witnesses multipartite entanglement in $\rho_{\rm ent}$. Therefore, positive entanglement gap, $G_{\rm E}(T)>0$, defines the EW by the equation $Z_{\rm EW}=H-IE_{\rm sep}$, where $I$ is the identity matrix on the Hilbert space. Hence, Tr[$Z_{\rm EW}\rho_{\rm sep}$]$=$Tr[$H\rho_{\rm sep}$]$-E_{\rm sep}\geq 0$, when $\rho_{\rm sep}$ is any separable state while $E_{sep}$ is the lowest possible energy for a separable state. On the other hand, for the ground state, $\rho_{\rm G}$, Tr[$Z_{\rm EW}\rho_{\rm G}$]$=E_{\rm G}-E_{\rm sep}<0$ at $T=0$. Thus, $Z_{\rm EW}$ serves as an EW.

Generally variational methods are being employed to find the lowest possible energy for a separable state of spin chains. Otherwise, it has been noted that for AFM Heisenberg spin cluster with all-to-all couplings of same strengths, a minimum energy separable state is given by that classical spin configuration where the total spin vector is zero Dowling . In order to find the separable state with minimum energy in this case, we introduce the most general form of separable state, like, $|\psi_{\rm sep}\rangle=\prod_{j}|S_{j}\rangle$, $j=1,2,3,\cdots,6$, where $|S_{j}\rangle=\cos{\theta_{j}}|\uparrow\rangle+e^{i\phi_{j}}\sin{\theta_{j}}|\downarrow\rangle$, $0\leq\theta_{j}\leq\pi$, and $0\leq\phi_{j}\leq 2\pi$. $E_{\rm sep}$ is obtained by minimizing $\langle\psi_{\rm sep}|H|\psi_{\rm sep}\rangle$ with respect to both $\theta_{j}$ and $\phi_{j}$. By using simplex minimizing procedure Nelder_Mead , $E_{\rm sep}$ is found to equal to $-\frac{3}{2}(J_{1}-J_{2})-\frac{3}{4}J_{3}$, which essentially corresponds to $\theta_{j}=\frac{2\pi j}{6}$ and $\phi_{j}=0$. The symmetry in the Hamiltonian is responsible for the symmetric solutions. The solutions always correspond to the classical spin configuration with the total spin vector is zero, although the condition of all-to-all couplings of same strength is mostly violated except the point P. The variations of $E_{\rm sep}/J_{1}$ and $k_{\rm B}T_{\rm E}/J_{1}$ are shown in Figs. 5 (c) and (d), respectively. Usually $E_{\rm sep}$ is negative everywhere except the extreme point, M, over the line $J_{1}=J_{2}$, where $E_{\rm sep}$ becomes zero. At the point M, the value of frustration degree, $\mathcal{F}$ is the maximum and $T_{\rm E}$ tends to $\infty$, which is shown in the Fig. 5 (d). The value of $k_{\rm B}T_{\rm c}/J_{1}$ at this point is 2.395. The bipartite entanglement vanishes over the same line including this point. Therefore, at this point multipartite entanglement survives at all temperatures in the absence of bipartite entanglement. Besides this particular point, M, entangled states are found to exist at high temperatures in the vicinity of the point. Thus, the entanglement in quantum states in this particular region exhibits a robustness to the thermal noise. It appears from the expression of $E_{\rm sep}$ that positive contribution to $E_{\rm sep}$ only comes from the NNN frustrating bond, $J_{2}$. So, in the absence of NNN bond, $E_{\rm sep}$ is always negative which gives rise to very low $T_{\rm E}$. Therefore, the presence of frustration leads to the high values of $T_{\rm E}$. This observation shows that the frustration induces the multipartite entanglement in this spin cluster in such a manner that it does sustain against the thermal agitation. On the other extreme point P over the same line $J_{1}=J_{2}$, it is found that $T_{\rm E}=T_{\rm c}=2.862J_{1}/K_{\rm B}$. Since the bipartite entanglement vanishes over this line only multipartite entanglement survives in the system at P for $0<T<T_{\rm E}$. The equality between $T_{\rm E}$ and $T_{\rm c}$ results from the fact that at this point the effective spin interactions are defined on a non-bipartite graph or lattice for which EW based on thermal energy detects only the multipartite entanglement. Now consider another point O ($J_{2}=J_{3}=0$) in the parameter space, where $E_{\rm sep}=-1.5J_{1}$, and the value of $T_{\rm E}$ is 0.802$J_{1}/K_{\rm B}$ which is identical to that of $T_{\rm c}^{\rm NN}$. This is due to the fact that at this point the resulting spin interactions are defined on a bipartite graph or lattice and EW based on thermal energy in this case detects only the bipartite entanglement.

V Discussion

The spin-$\frac{1}{2}$ AFM Heisenberg hexagon with all-to-all exchange couplings is considered to investigate the variety of entanglement properties. Four different locations, R1, R2, L and P have been identified where the nature of ground states are different while QPT occurs over the line L including the point P. By exploiting its six-fold rotational symmetry three different kinds of CNs, CNNN, CNNNN and CNFN are introduced and those give totally different results. Both $T_{\rm c}^{\rm NN}$ and $T_{\rm c}^{\rm FN}$ decrease with the increase of $J_{2}$ (Fig. 3) and ultimately vanish over the lines $J_{2}/J_{1}=1$ and $J_{3}/J_{1}=0.87J_{2}/J_{1}+0.14$, respectively. Those observations reveal the fact that the frustration opposes the bipartite entanglement in this system. Multipartite entanglements of this model have been studied where susceptibility, fidelity and internal energy serve as the EWs. Multipartite entanglements survive up to the temperature, $T_{\rm c}$ which is always higher than the $T_{\rm c}^{\rm NN}$. Thus, the bipartite entanglement diminishes due to thermal agitation more quickly than the multipartite entanglement. Frustration leads to the higher values of $T_{\rm c}$, so it favors the multipartite entanglements. Fidelity measurement indicates that this model exhibits no six-spin entanglement even in zero temperature. Survival of multipartite entanglement at finite temperature has been studied in terms of internal energy as EW. Entanglement is found to persist at high temperatures in this system in the vicinity to the point M, where value of $\mathcal{F}$ is the maximum. It appears that frustration is responsible for the robustness of quantum entanglement against the thermal effect around this point. Existence of the multipartite entanglement at finite temperatures is found in this system where the bipartite entanglement vanishes at non-zero temperatures. EW in terms of susceptibility can detect the existence of both bipartite and multipartite entanglements collectively at finite temperatures. On the other hand, EW in terms of internal energy can detect bipartite and multipartite entanglements separately for the cases when the spin interactions are defined on bipartite and non-bipartite graphs or lattices, respectively. For this model, EW based on internal energy detects only the bipartite entanglement at the point O in the parameter space, and that measures only the multipartite entanglement at the other point P. Therefore, at the point O, $T_{\rm c}^{\rm NN}=T_{\rm E}$. Similarly, $T_{\rm c}$ becomes equal to $T_{\rm E}$ only at the point P, where only multipartite entanglement survives and measured separately by the EWs based on susceptibility and internal energy. It further appears that EW based on internal energy detects the collective existence of both bipartite and multipartite entanglements everywhere in the $J_{2}$-$J_{3}$ parameter space except the points O and P. Therefore, development of more effective EWs is necessary for precise detection of different types of entanglements separately. Engineering of entangled quantum state at room temperature is a new challenge. So, the frustrated AFM spin models could shed light in this direction.

The inelastic neutron scattering study on Cu3WO6 reveals that spin-1/2 Cu2+ ions are arranged on the vertices of hexagons in its crystalline state Hase . Dynamic structure factor predicts the magnitudes of $J_{1}$, $J_{2}$ and $J_{3}$ are 78.5K, 50.4K and 40.0K, respectively. As they satisfy the relation $J_{1}+J_{3}>2J_{2}$, this system belongs to the region R1 having the RVB ground state, $\Psi_{42}$. Position of this compound in the $J_{2}$-$J_{3}$ parameter space is identified by the point C. Estimations of various quantities for Cu3WO6 yield following values. $\mathcal{F}=0.51$, CN${}_{\rm NN}=0.43$, $k_{\rm B}T^{\rm NN}_{\rm c}/J_{1}=0.50$, $F=0.44$, $k_{\rm B}T_{\rm c}/J_{1}=2.20$, $E_{\rm sep}/J_{1}=-0.92$ and $k_{\rm B}T_{\rm E}/J_{1}=1.33$. Hence, this material is no more suitable to yield thermally stable multipartite entanglement. Therefore, in our opinion synthesis of new AFM compounds whose compositions as well as structures are very close to Cu3WO6 or other one such that $J_{2}\geq J_{1}+J_{3}/2$ becomes necessary for the production of thermally stable multipartite entanglement.

VI ACKNOWLEDGMENTS

MD acknowledges the UGC fellowship, no. 524067 (2014), India. AKG acknowledges a BRNS-sanctioned research project, no. 37(3)/14/16/2015, India.

VII Author contribution statement

MD did the analytical work and AKG did the numerical work. The manuscript was prepared jointly by both the authors.

Appendix A ENERGY EIGENVALUES AND EIGENSTATES

In this section, we provide the analytic expressions of all eigenvectors and corresponding eigenvalues of the Hamiltonian, (Eq. 1). All energy eigenvalues with definite values of $S_{\rm T}$, $S^{z}_{\rm T}$, $\lambda_{r}$ and $p$ have been enlisted in the Tab. I. To express the eigenstates following notations have been used.

[TABLE]

Here $T$ is a unitary cyclic right shift operator such that $T\left|abcdef\right\rangle=\left|fabcde\right\rangle$, where $\left|abcdef\right\rangle=\left|a\right\rangle\otimes\left|b\right\rangle\otimes\left|c\right\rangle\otimes\left|d\right\rangle\otimes\left|e\right\rangle\otimes\left|f\right\rangle$. All the energy eigenstates are enlisted in the Tab. II.