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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?

Kai Li
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2 Answers2

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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.

tparker
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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?

Kai Li
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