Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Where do you find that the dispersion relation for spin waves on the integer spin antiferromagetic Heisenberg chain is $\omega_k\sim k$? – Norbert Schuch Apr 4 '13 at 7:59
It can be found in any text book on solid state theory. The dispersion relation in the long wavelength limit is $\omega_k\sim k$ for the Heisenberg antiferromagnet. For the Heisenberg ferromagnet, it is $\omega_k\sim k^2$ – hlew Apr 4 '13 at 11:04
You are sure it says that for the integer spin HAF? – Norbert Schuch Apr 4 '13 at 12:05
Yes, in spin wave theory, the dispersion relation for HAF is $\omega_k\sim k$ for any spin quantum number. – hlew Apr 4 '13 at 13:01
Well, then you just showed that the Haldane conjecture is wrong ;-) – Norbert Schuch Apr 4 '13 at 13:02
up vote 9 down vote accepted

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).

share|cite|improve this answer
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 4 '13 at 22:29
@Everett Mermin-Wigner or Mermin-Wagner? – Chris White Apr 5 '13 at 6:27
@ChrisWhite Sorry. Should be Mermin-Wagner. Thanks for pointing out. – Everett You Apr 5 '13 at 6:37
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, "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 5 '13 at 7:47
@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 5 '13 at 12:43

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.