Timeline for Why are there gapless excitations in the anti-ferromagnetic Heisenberg model while the true ground state is a singlet?
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Aug 12, 2016 at 20:30 | comment | added | Ruben Verresen | I think this answer is misleading or at least missing the point: it is a correct analysis of the 2D case, but it's not analogous to the 1D case which the question asks about. In particular in 1D there is no ground state degeneracy for the Heisenberg AFM, even in the Td limit. I.e. the AFM Heisenberg chain does not have spontaneous symmetry breaking in any sense. | |
| May 15, 2014 at 10:22 | comment | added | Kai Li | And I think the local moments $\left \langle S_i \right \rangle$ can also be used as an order parameter to describe the ground states, where $\left \langle \cdot \right \rangle$ is again the ensemble average. | |
| May 15, 2014 at 10:12 | comment | added | Kai Li | @ Abhimanyu I tend to think the reason for saying a Neel order may be due to, e.g., the long range correlations $\left \langle S_iS_j \right \rangle$ in the thermodynamic limit, where $\left \langle \cdot \right \rangle$ represents the ensemble average at zero temperature, which contains the expectation values with respect to both singlet ground state and $SU(2)$ symmetry breaking ground states. | |
| May 15, 2014 at 6:35 | comment | added | Abhimanyu | I like your answer. I would also like to better understand the point you brought up- Why is the broken symmetry state considered the ground state rather than a singlet ground state in the thermodynamic limit | |
| May 13, 2014 at 15:20 | history | edited | Kai Li | CC BY-SA 3.0 |
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| May 13, 2014 at 15:14 | history | edited | Kai Li | CC BY-SA 3.0 |
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| May 13, 2014 at 13:03 | history | edited | Kai Li | CC BY-SA 3.0 |
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| May 13, 2014 at 10:12 | history | edited | Kai Li | CC BY-SA 3.0 |
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| May 13, 2014 at 10:07 | history | answered | Kai Li | CC BY-SA 3.0 |