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Haldane's conjecture states that the integer spin antiferromagnetic Heisenberg chains have a gap in the excitation spectrum. However, the dispersion relation of the antiferromagnetic spin wave is $\omega_k\sim k$ in the long wave length limit, meaning that the excitation energy could be zero. What is the matter?

Qmechanic
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2 Answers2

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Spin wave theory simply does not apply for 1D spin system. The starting point of the spin wave theory is a magnetically ordered ground state. But Mermin-Wagner theorem states that 1D spin system can not order even at zero temperature, due to the strong quantum fluctuation. So 1D Heisenberg model does not lead to an antiferromagnetically ordered ground state, and hence the spin wave is not well defined, and the spin fluctuation does not follow the dispersion relation $\omega\sim k$. It is known[1] that 1D spin chain is gapped, as conjectured by Haldane.

[1] Z.-C. Gu and X.-G. Wen, Phys. Rev. B 80, 155131 (2009).

Everett You
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    This is mostly a good answer, except that you need to clarify that integer 1D spin chains are known to be gapped (and the $S=1/2$ chain is known to be gapless). – wsc Apr 04 '13 at 22:29
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    @Everett Mermin-Wigner or Mermin-Wagner? –  Apr 05 '13 at 06:27
  • @ChrisWhite Sorry. Should be Mermin-Wagner. Thanks for pointing out. – Everett You Apr 05 '13 at 06:37
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    @wsc Thanks very much for your comments. I have heard of some numerical proved, but not aware of if there is any analytic prove. Could you point out more references to help improve the answer? – Everett You Apr 05 '13 at 06:42
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    There is no fully rigorous proof of the Haldane gap for the integer spin HAF. (But if you are happy to accept some continuum limits etc., Haldane's original work already contains the argument.) – Norbert Schuch Apr 05 '13 at 07:38
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    I agree that the problem is that spin-wave theory does not work if the ground state is not AFM ordered (which it isn't for the S=1 HAF), but does this follow from the Mermin-Wagner theorem? According to http://www.scholarpedia.org/article/Mermin-Wagner_Theorem, "It says: At finite temperatures, the quantum spin-S Heisenberg model with isotropic and finite-range exchange interactions on one- or two-dimensional lattices can be neither ferro- nor antiferromagnetic.", but I don't see why it also applies to the ground state? – Norbert Schuch Apr 05 '13 at 07:47
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    @NorbertSchuch Because there is a quantum-statistics correspondence that 1D quantum ground state = 2D statistical ensemble, if the spin system can not order at finite temperature in 2D, it also can not order at zero temperature in 1D. – Everett You Apr 05 '13 at 12:43
  • @hlew Yes, spin wave theory follows from the spontaneous symmetry broken, according to the Goldstone theorem. – Everett You Apr 05 '13 at 12:46
  • @EverettYou But Mermin-Wagner holds for quantum systems in 1D and 2D? What would be the 2D classical ensemble corresponding to the 1D Heisenberg AFM? – Norbert Schuch Apr 05 '13 at 13:22
  • @NorbertSchuch You may take a look at the Wikipedia about imaginary time (http://en.wikipedia.org/wiki/Imaginary_time). It is well known that "using a Wick rotation one can show that the Euclidean quantum field theory in (D + 1)-dimensional spacetime is nothing but quantum statistical mechanics in D-dimensional space." – Everett You Apr 06 '13 at 07:54
  • @EverettYou To start with, we're not talking about a quantum field theory here. In any case, this duality wouldn't resolve my question what Mermin-Wagner tells us about ground states of 1D quantum systems (since I don't see what it tells us about 2D classical systems). On an unrelated note, I think you should edit the last sentence in the answer since the Haldane conjecture is not rigorously proven (beyond what has been shown by Haldane + numerics). – Norbert Schuch Apr 06 '13 at 09:04
  • @EverettYou, but there are even experimental data in books showing $\omega\sim k$ ! – richard Dec 09 '13 at 16:17
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    Norbert is correct and Mermin-Wagner cannot be applied here. For example, ferromagnetism can be realized as a ground-state order in 1-d at zero temperature due to its dispersion is k square rather than linearity. – Yuan Yao Oct 07 '19 at 15:54
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To clarify and correct some of the above points ...

There is no long range order in 1D systems at zero temperature, as explicitly proved by Pitaevskii and Stringari in 1991; see 'Uncertainty Principle, Quantum Fluctuations and Broken Symmetries', J. of Low Temp. Phys. 85, 377. But I agree with @NorbertSchuch in that I doubt if Mermin-Wagner says this.

The correspondence between classical and quantum systems, which @EverettYou explains slightly mistakenly, holds mostly at zero-temperature criticality i.e. the effective $D+1$-th dimension in the classical model is infinite in extent only at $T=0$ of the quantum model.

To answer the original question, spin-wave theory always builds on a mean-field solution, which is almost always based on an educated guess. If the starting guess is incorrect, so will most of what results follow.

MaviPranav
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  • As mentioned above, how to understand the 1D ferromagnetic order according to Mermin-Wagner theroem? – ZJX Nov 01 '19 at 16:16