Question

I came across the question of SSB of a discrete symmetry (say, Z₂) symmetry in a QFT in d=1+1 dimensions, at finite temperature, and I have trouble making sense of two different viewpoints which seem to lead to contradicting results. Can you point me to the fallacy in my reasoning?

Let us assume that at zero temperature, at infinite volume, (that is, on R₂, in an Euclidean setting) the Z₂ is spontaneously broken (for instance, as a consequence of an 't Hooft anomaly). Let us also assume that the first excitations on top of each vacuum are gapped with mass M
. Finite temperature means periodic Euclidean time, τ∼τ+β
. This turns our (Euclidean) spacetime manifold into R×S1
. One observable that is sensitive to SSB is the Euclidean partition function ZE(β)
: the leading order in the large β expansion will give us the ground state degeneracy.

Viewpoint 1: Let us quantize the theory on the line. The full Hilbert space on the line splits in 2 superselection sectors, H=H0⊕H1, and states on each sector are obtained by adding excitations on top of the respective vacuum. The two sectors are exactly identical. At leading order, then, we would find ZE(β)=2+O(e−βM), where M is the gap.

Viewpoint 2: Let us quantize the theory on the circle. This is equivalent to studying a QFT in finite volume and zero temperature. Being the volume finite, there cannot be SSB. There is a unique invariant ground state, let us call it |+⟩. (There is also another state, |−⟩, almost degenerate to it. The gap between the two can be obtained à la Coleman: ΔE=Ke−S₀, where S₀ is the instanton action, and S₀=βm, since I expect the instanton (of mass m) to be invariant under rotation in the compact direction. K is a dimensionful constant that depends on the details of the theory.) Let us now try to argue for the behavior of ZE(β). Indeed, we now should compute Tre−LH, in the limit L→[infinity] (here L is interpreted as inverse temperature, which is taken to infinity since we want to end up on the infinite cylinder). No matter what, however, only the ground state will contribute in this limit, and ZE(β)∼1.

My intuition for the resolution to the contradiction would be the following: Viewpoint 1 is right and Viewpoint 2 is wrong, because the limit L→[infinity]
is discontinuous, and if I kept L
finite and then sent it to infinity also in Viewpoint 1 I would have obtained the same (wrong) answer. However, I do not see a way to fix Viewpoint 2 in such a way to give the correct result, if such a way exists.

Is this it? Or am I being wrong in both cases?

Answer 1

Viewpoint 1 is correct, and Viewpoint 2 contains a fallacy. The correct behavior of the Euclidean partition function ZE(β) in the context of a Z₂ symmetry in a 1+1-dimensional QFT at finite temperature on a circle is ZE(β) = 2 + O(e^(-βM)), where M is the gap.

Step-by-step explanation:

Viewpoint 1 correctly considers the theory's quantization on the line, acknowledging the superselection sectors H=H0⊕H1. This leads to ZE(β) = 2 + O(e^(-βM)), which accurately accounts for the ground state degeneracy and the effect of excitations over each vacuum state. However, Viewpoint 2's reasoning regarding quantization on the circle fails due to the incorrect assumption that the theory on a finite circle corresponds to a QFT at zero temperature in finite volume, disregarding the impact of finite temperature on a compactified dimension. This approach overlooks the subtleties introduced by the thermal circle and erroneously concludes ZE(β) ∼ 1, considering only the ground state contribution in the limit of infinite circumference. In reality, the limit process applied in Viewpoint 2 fails to capture the essential aspects of finite temperature, leading to an incorrect evaluation of the partition function.

In essence, the fallacy lies in assuming that the finite volume of the circle suppresses spontaneous symmetry breaking at finite temperature, disregarding the influence of temperature on a compact dimension. The correct understanding emerges from acknowledging the distinct implications of quantization on the line versus the circle in the context of finite temperature effects on symmetry breaking. This crucial distinction accounts for the observed discrepancy in the behavior of ZE(β).