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The true ground state of the anti ferromagnetic quantum Heisenberg Model (nearest neighbor only)is known to be a singlet (I think this is Liebs theorem.)

Since a singlet is invariant under rotations, the ground state doesn't break the rotational symmetry of the Hamiltonian (which the classical Neel state does break)

So Goldstone's theorem should not apply as we have not spontaneously broken any continuous symmetry in the ground state. Yet we do have anti-ferromagnetic magnons which are gapless excitations.

Why do we have the gapless excitations without breaking any symmetry?

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  • $\begingroup$ to "we do have anti-ferromagnetic magnons which are gapless excitations" : not always true if Haldanes conjecture is true : physics.stackexchange.com/q/59986 $\endgroup$
    – jjcale
    Commented May 13, 2014 at 21:35
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    $\begingroup$ OK. I clarify - consider my question only for the case of half integer spins. $\endgroup$
    – Abhimanyu
    Commented May 13, 2014 at 22:44
  • $\begingroup$ A modern viewpoint on why it is gapless (although its origin traces back to Lieb-Schultz-Mattis) is presented here: arxiv.org/abs/1503.07292 $\endgroup$
    – Ruben Verresen
    Commented Sep 30, 2017 at 15:37

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You are correct that Goldstone's theorem does not apply to the 1-D Heisenberg chain, or indeed to any local 1-D system, because there is no continuous symmetry breaking in 1-D for local Hamiltonians. But Goldstone modes are not the only possible kinds of gapless excitations - you don't need continuous SSB for a system to be gapless. There are plenty of examples (other than the Heisenberg chain) of gapless systems that don't break any continuous symmetries, like critical systems, algebraic spin liquids, or systems with Fermi surfaces.

Put another way, Goldstone's theorem says that continuous SSB implies gapless excitations, but the converse of Goldstone's theorem is not true: gapless excitations do not imply continuous SSB.

The low-energy excitations of the antiferromagnetic Heisenberg chain are "spinons," which can be roughly thought of as the free spins that arise from breaking spin singlets in the ground state. Their gaplessness can be proven using the Bethe ansatz, or by using the fact that the emergent low-energy QFT description is a Wess-Zumino-Witten model, which is a CFT and therefore gapless, as an energy gap would set a scale and break conformal invariance.

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  • $\begingroup$ hi, when people say "Heisenberg chain", do we always mean the "antiferromagnetic Heisenberg model" rather than the "ferromagnetic Heisenberg model" ? $\endgroup$
    – Kai Li
    Commented May 12, 2016 at 16:14
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    $\begingroup$ @KaiLi No, just saying "Heisenberg chain" could mean either ferromagnetic or antiferromagnetic. I've edited my answer to clarify that the last paragraph only applies to antiferromagnets. (Ferromagnetic chains also spontaneously break the SU(2) rotation symmetry and have corresponding Goldstone modes, but these are spin-1 "magnons" which are very different and simply correspond to a single spin flip.) $\endgroup$
    – tparker
    Commented May 12, 2016 at 16:51
  • $\begingroup$ To add an even more simple example of a gapless system without SSB: sound waves in classical fluids. $\endgroup$
    – KF Gauss
    Commented Oct 22, 2019 at 5:25
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    $\begingroup$ @KFGauss The notion of a "gapless excitation" doesn't really apply for a classical system with continuous degrees of freedom like a classical fluid, because everything is continuous, so no such system can ever possibly have an excitation gap. $\endgroup$
    – tparker
    Commented Oct 22, 2019 at 13:18
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I think this is a very good and deep question that maybe related to Prof.Wen's question, and I would try to answer you in my understanding.

Let's take the nearest-neighbor spin-1/2 antiferromagnetic Heisenberg Model on the square lattice as an example, where the symmetry breaking Neel state only emerges in the thermodynamic limit.

As you have mentioned, note that only when the system has finite size, (i.e., the square lattice constitutes of two sublattices A and B with equal sizes $N_A=N_B$, and hence the total number of spins $N=N_A+N_B$ is even), the ground state is unique and is exact a singlet state with $SU(2)$ spin-rotation symmetry (Marshall,1955; Lieb and Mattis, 1962). However, as the system size becomes large, there are many low-lying excited states with very small energy gap $\Delta$ above the singlet ground state, and these low-lying states break the $SU(2)$ spin-rotation symmetry(i.e., they may be triplet states). More subtlely, as $N$ approaches $\infty $, those nearly degenerate ground states would 'collapse' into the ground state in the thermodynamic limit ($\Delta\rightarrow 0$), indicating that the Neel state is in fact a superposition of many nearly degenerate ground states in the thermodynamic limit. Thus, in the strict thermodynamic limit, there exists an $SU(2)$ symmetry breaking state of the 'highly' degenerate ground states.

Indeed, this is a nontrivial example of spontaneous symmetry breaking since the exact ground state of the finite system does not break $SU(2)$ spin-symmetry while there emerges spontaneous symmetry breaking (due to the nearly degenerate ground states) in the thermodynamic limit. The above argument is just a very rough picture, and I also feel it is somewhat difficult to understand how the ground state degeneracy happens for a gapless system in the thermodynamic limit? Moreover, I also get another question: Theoretically, as there are 'highly' degenerate ground states containing both $SU(2)$ symmetric singlet state and symmtery-breaking Neel states in the thermodynamic limit, why we are used to saying the ground state of the antiferromagnetic Heisenberg Model on the square lattice is a Neel state rather than a singlet state?

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  • $\begingroup$ 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 $\endgroup$
    – Abhimanyu
    Commented May 15, 2014 at 6:35
  • $\begingroup$ @ 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. $\endgroup$
    – Kai Li
    Commented May 15, 2014 at 10:12
  • $\begingroup$ 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. $\endgroup$
    – Kai Li
    Commented May 15, 2014 at 10:22
  • $\begingroup$ 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. $\endgroup$
    – Ruben Verresen
    Commented Aug 12, 2016 at 20:30

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