The one dimensional SU(2) Heisenberg quantum spin chain is known to be described by the 1+1d O(3) nonlinear $\sigma$ model with a $\Theta$ term, following the action $$S=\int\mathrm{d}^2x\frac{1}{g}(\partial_\mu n)^2+\frac{i\Theta}{8\pi}\epsilon_{\mu\nu}\epsilon_{abc}n_a\partial_\mu n_b\partial_\nu n_c,$$ where $\Theta=2\pi s$ is related to the spin $s$ on each site. On the other hand, Haldane conjectured that the integer spin chain is gapped, while the half-integer spin chain is gapless (or degenerated?)

My question is how to understand the conjecture based on the topological $\Theta$ term? Or explicitly, how to calculate the ground state degeneracy and the low energy spectrum from the action given above? Is it even possible to write down the ground state wave functions for topological field theory? Is there a systematic way to analyze the low energy spectrum of a topological field theory in general?

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    $\begingroup$ More on Haldane's conjecture: physics.stackexchange.com/q/59986/2451 $\endgroup$
    – Qmechanic
    Apr 10, 2013 at 9:03
  • $\begingroup$ @Qmechanic Thanks for reminding. I am aware of the post which I have answered. But in my answer, I did not provide an argument for Haldane's conjecture, which is what I am asking for here. $\endgroup$ Apr 10, 2013 at 12:55
  • $\begingroup$ If the spin chain is gapless , is the ground state degeneracy still well defined ? Thanks. $\endgroup$
    – Kai Li
    Apr 10, 2013 at 17:27
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    $\begingroup$ @K-boy For gapless spin chain, the ground state degeneracy is not well defined. I was told that when $\Theta=\pi$ the ground state is either gapless, or gapped but 2-fold degenerated. I have not figured out how to show this explicitly. $\endgroup$ Apr 10, 2013 at 19:07
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    $\begingroup$ @K-boy Following Lieb-Schultz-Mattis (LSM), SU(2) invariant tinv. chains with half-integer spin have at least two states with low energy, i.e., they are either gapless or have a degenerate ground state. An example for the former is the spin-1/2 Heisenberg chain, and for the latter the Majumdar-Ghosh model (which breaks tinv.). I would say the degeneracy is always related to local symmetry breaking, since the degeneracy appears also for PBC. Note that LSM has been generalized to 2D (cond-mat/0305505), in which case the degeneracy can be related to topological order (e.g. in topo. spin liquids). $\endgroup$ Apr 11, 2013 at 10:59

1 Answer 1


It's a good question and I don't think there is a simple way of seeing it from the action. I just checked Fradkin (section 7.7 and 7.8) and through an RG analysis he shows that any half-integer spin behaves the same as the spin-$\frac{1}{2}$ case. But for the latter he then refers to the exact Bethe ansatz solution to show it is gapless!

However, I wanted to bring to your attention a very nice result that seems quite under-appreciated:

The ``projective is gapless''-theorem:

If you have a spin chain that

  • is translation invariant for a unit cell, and
  • is symmetric for a projective representation on the unit cell,

then it is gapless or breaks one of the above symmetries!

Clearly the half-integer Heisenberg spin chain satisfies the above conditions, so we can conclude is either gapless or breaks one of its symmetries. Furthermore, since $SU(2)$ is continuous, breaking it would imply gaplessness (or you could argue that it can't be broken at all due to Coleman-Mermin-Wagner, but such a strong statement is not needed here), so we can conclude: either the half-integer Heisenberg spin chain is gapless or spontaneously breaks translation symmetry. So that doesn't show gaplessness, but it gets pretty damn close.

The first instance of the above theorem was indeed due to Lieb, Schultz and Mattis (as noted by Schuch), however the above statement is quite more general, but the really cool part is that the proof is incredibly simple, once you get friendly with Matrix Product States! I'll give a short outline how to arrive at it, starting from scratch (note that by collapsing the unit cell we can limit to the case where the unit cell is one site):

  1. The ground state of any gapped spin chain can be written as a certain tensor, called a Matrix Product State (MPS)
  2. To prove the theorem, let us assume our ground state is gapped, translation invariant and has this projective symmetry. From this we derive a contradiction. The assumption means our ground state is described by a translation invariant MPS.
  3. Fact for any translation invariant MPS with an on-site symmetry group $G$: if you apply the symmetry operation on just one site, then it is equivalent to doing a certain operation $U$ on the entanglement that site has with everything to its left and an operation $U^\dagger$ on the entanglement with everything on its right. This $U$ is a (possibly projective) representation of $G$
  4. So acting $G$ is the same as acting $U$ and $U^\dagger$ at the same time, but whereas $U$ can be projective, performing both at the same time must be a linear representation (think of how $\frac{1}{2} \otimes \frac{1}{2} = 0 \oplus 1$)
  5. Conclusion: this means the on-site symmetry must be linear, but by assumption it is projective. Contradiction. Hence one of the assumptions in step 2 must be wrong :)

Reference: I think Schuch, Perez-Garcia, Cirac and Chen, Gu, Wen were the first to observe this

EDIT: I recently learned that this ``projective is gapless''-theorem was already proven in 2001 by Matsui !

  • $\begingroup$ Really nice argument! Can you remind me why all gapped states in 1D must have MPS representations? $\endgroup$ Feb 10, 2016 at 7:38
  • $\begingroup$ Sure. First note any state whose entanglement obeys the area law (i.e. in 1D this means the entanglement between a subblock and the rest of the system has to be independent of subsystem size) is equivalent to there being a Matrix Product State representation (or in higher dimension PEPS). This is by construction: MPS is basically rewriting a wavefunction based on the assumption that the entanglement between a site and the rest of the system is bounded by some number. Moreover, we intuitively expect any system with finite correlation length (hence a gapped phase) to obey that area law. [cont.] $\endgroup$ Feb 10, 2016 at 14:11
  • $\begingroup$ [cont.] There are actually counter-examples to that reasoning, called quantum data hiding states. However, Hastings ( arxiv.org/abs/0705.2024 ) rigorously proved that the entanglement law is obeyed in gapped one dimensional systems. (So this proves that data hiding states cannot be the ground state of a gapped local Hamiltonian.) So this proves there is an MPS representation. $\endgroup$ Feb 10, 2016 at 14:15

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