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  Why does the Coleman-Mermin–Wagner–Hohenberg theorem exclude discrete symmetry?

+ 5 like - 0 dislike

The Mermin–Wagner theorem (also known as Mermin–Wagner–Hohenberg theorem or Coleman theorem) states that continuous symmetries cannot be spontaneously broken at finite temperature in systems with sufficiently short-range interactions in low dimensions. In particular at zero temperature $T=0$, if we have a quantum system, the theorem states:

The continuous symmetries cannot be spontaneously broken in 1+1d quantum systems.

Apparently, discrete finite symmetries can still be spontaneously broken in 1+1d quantum systems.

Why does the reasoning of the theorem work for the continuous symmetry but not for the discrete finite symmetries? 

If we take a  $\mathbb{Z}_N$ discrete symmetry with a large $N$, it is still not a U(1) continuous symmetry. Why does the argument not work in $\mathbb{Z}_N$?

asked Nov 4, 2016 in Theoretical Physics by RKKY (325 points) [ revision history ]

2 Answers

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If a continuous symmetry is spontaneously broken, then the system contains a Goldstone boson, which in not possible in 2d because massless scalar fields have IR divergent behaviour in 2d.

If a discrete symmetry is spontaneously broken, there is no Goldstone boson so the previous argument does not apply.

The most famous system with a spontaneously broken discrete symmetry in 2d is the Ising model ($\phi^4$ field theory)(the discrete symmetry being $\mathbb{Z}/2$).

answered Nov 4, 2016 by 40227 (5,140 points) [ revision history ]

sorry $\phi^4$ looks like to be compatible with a larger U(1) symmetry? It can be continuous.

$\phi$ is  a real scalar field so there is no $U(1)$ symmetry (even if $\phi$ were complex, $\phi^4$ is not invariant under a general $U(1)$ rotation $\phi \mapsto e^{i \theta} \phi$).

+ 3 like - 0 dislike

To add to 40227's answer, there's also a domain-wall proof of the theorem, of which the basic intuition is to rotate a finite region of spins with arbitrarily small energy cost, achieved by creating a domain wall of finite thickness, and in there carefully massaging a smooth interpolation between the rotated spin region and the unrotated ones.

If the symmetry group were discrete, there's no way to massage a smooth interpolation, and hence always at least a finite cost for creating domain walls.

answered Nov 5, 2016 by Jia Yiyang (2,640 points) [ no revision ]

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