Effects of external magnetic field on thermal entanglement in a spin-one chain with three particles

Abstract

In this paper we have investigated the effect of temperature, magnetic field, and exchange coupling on the thermal entanglement in a spin chain with three particles in terms of the measure of entanglement called negativity. The results show that the entanglement decreases with increase in temperature. Also, we have found that under a uniform magnetic field in constant temperature, the entanglement decreases while passing through some maxima. We have found that increasing exchange coupling of any two particles decreases the entanglement of the other two particles. Finally, we have compared our system with a two-particle system and found that in presence of a magnetic field the increase in number of particles leads to the increase in the entanglement.

Appendix

By using Eq. (5) we find the Hamiltonian matrix as the values of parameters for case $J_{1} = 0.5,\,\,\,J_{2} = 1$. Since this matrix is very large we present only the non-zero elements as

$$ H_{1\;2} = H_{1\;4} = H_{1\;10} = H_{2\;1} = H_{2\;3} = H_{2\;5} = H_{2\;11} = H_{3\; 2} = H_{3\; 6} = H_{3 \;11} = H_{3\;12} = H_{4\;1} = H_{4\;5} = H_{4\;7} = H_{4 \;13} = H_{5\; 2} = H_{5 \;4} = H_{5\; 6} = H_{5\; 8} = H_{5 \;14} = H_{6\; 3} = H_{6\; 5} = H_{6\; 9} = H_{6\; 15} = H_{7 \;4} = H_{7 \;8} = H_{7 \;16} = H_{8\; 5} = H_{8 \;7} = H_{8\; 9} = H_{8 \;17} = H_{9 \;6} = H_{9\; 8} = H_{9\; 18} = H_{10 \;1} = H_{10\; 11} = H_{10\; 13} = H_{10\; 19} = H_{11\; 2} = H_{11\; 10} = H_{11\; 12} = H_{11\; 14} = H_{11\; 20} = H_{12 \;3} = H_{12 \;11} = H_{12\; 15} = H_{12\; 21} = H_{13\; 4} = H_{13 \;10} = H_{13 \;14} = H_{13\; 16} = H_{13\; 22} = H_{14\; 5} = H_{14 \;11} = H_{14\; 13} = H_{14\;15} = H_{14\; 17} = H_{14 \;23} = H_{15 \; 6} = H_{15 \; 12} = H_{15 \; 14} = H_{15 \; 18} = H_{15 \;24} = H_{16 \; 7} = H_{16\; 13} = H_{16\; 17} = H_{16\; 25} = H_{17 \; 8} = H_{17\; 14} = H_{17\; 16} = H_{17\; 18} = H_{17\; 26} = H_{18\; 9} = H_{18\; 15} = H_{18\; 17} = H_{18\; 27} = H_{19\; 10} = H_{19\; 20} = H_{19\; 22} = H_{20\; 11} = H_{20\; 19} = H_{20\; 21} = H_{20\; 23} = H_{21\; 12} = H_{21\; 20} = H_{21\; 24} = H_{22\; 13} = H_{22\; 19} = H_{22\; 23} = H_{22\; 25} = H_{23\; 14} = H_{23\; 20} = H_{23\; 22} = H_{23\; 24} = H_{23\; 26} = H_{24 \; 15} = H_{24\; 21} = H_{24\; 23} = H_{24\; 27} = H_{25\; 16} = H_{25\; 22} = H_{25 \; 26} = H_{26\; 17} = H_{26 \; 23} = H_{26\; 25} = H_{26 \; 27} = H_{27\; 18} = H_{27\; 24} = H_{27\; 26} = {\user2 r} $$

$$ H_{2 \;10} = H_{6 \;14} = H_{8 \;16} = H_{9 \;17} = H_{10\; 2} = H_{11\; 3} = H_{11\; 19} = H_{12\; 20} = H_{13\; 5} = H_{14\; 6} = H_{14\; 22} = H_{15\; 23} = H_{16\; 8} = H_{17\; 9} = H_{17\; 25} = H_{18\; 26} = H_{19\; 11} = H_{20\; 12} = H_{22\; 14} = H_{23\; 15} = H_{25\; 17} = H_{26\; 18} = {\mathbf 1} $$

$$ H_{2 \;4 } = H_{3 \;5} = H_{4 \;2} = H_{4 \;10} = H_{5\; 3} = H_{5\; 7} = H_{5\; 11} = H_{6 \;8} = H_{6\; 12} = H_{7\; 5} = H_{7\; 13} = H_{8\; 6} = H_{8 \; 14} = H_{8 \; 16} = H_{9\; 15} = H_{10\; 4} = H_{11\; 5} = H_{11\; 13} = H_{12\; 6} = H_{12\; 14} = H_{13\; 7} = H_{13\; 11} = H_{13\; 19} = H_{14\; 8} = H_{14\; 12} = H_{14\; 16} = H_{14\; 20} = H_{15\; 9} = H_{15\; 17} = H_{15 \; 21} = H_{16 \; 14} = H_{16\; 22} = H_{17\; 15} = H_{17\; 23} = H_{18\; 24} = H_{19\; 13} = H_{20 \; 14} = H_{20\; 22} = H_{21\; 15} = H_{21\; 23} = H_{22 \; 16} = H_{22 \; 20} = H_{23\; 17} = H_{23 \; 21} = H_{23\; 25} = H_{24\; 18} = H_{24\; 26} = H_{25\; 23} = H_{26\;24} = \frac{{\mathbf 1} }{{\mathbf 2} } $$

where $r = \frac{1}{\sqrt 2 }b$ and b is the magnetic field. Now, we find the eigenvalues and the related eigenvector as follow. The coefficients m_i's and h_i's are functions of the magnetic field.

Table 1