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Several papers set up a model of fermions in 2D and calculate superconducting $T_c$, e.g.: https://arxiv.org/pdf/1808.08635.pdf
https://arxiv.org/pdf/1406.1193.pdf
https://www.pnas.org/content/pnas/114/19/4905.full.pdf

Superconductivity implies spontaneous symmetry breaking and as far as I can tell, the interactions in these papers are not long range. Why is not $T_c=0$ per the Mermin Wanger theorem?

These works actually try to model layered superconductors in 3D where MW doesn't apply, though the models they use are in 2D. Is 3Dness somehow implicitly incorporated into their calculatons in a way that allows for spontaneous symmetry breaking at finite $T$?

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    $\begingroup$ Minor comment to the post (v2): In the future please link to abstract pages rather than pdf files. $\endgroup$ – Qmechanic Nov 18 '19 at 6:39
  • $\begingroup$ Perhaps you were right after all. I'm used to thinking of Mermin-Wagner at zero temperature, where it does not hold at 2+1D, but if you turn on finite temperature I'm not so sure. $\endgroup$ – octonion Nov 18 '19 at 15:13
  • $\begingroup$ Ok I see. Indeed what I'm confused about is why the superconducting transition happens at a finite temperature. If $T_c=0$ then I agree it would be allowed by MW but in these works $T_c>0$. $\endgroup$ – Petter Nov 18 '19 at 17:06
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Short answer: They are not truly 2D dimensional systems.

Needless to say that experimentally, cuprates and other high-$T_c$ materials are all 3D.

In their mathematical models, they put bosons and fermions in 2D lattices... but there are no bosons and fermions in 2D! The statistics in 2D in fractional, and involves anyons.
There might also be other specific comments about the use of mean field theory and some statistical integrals that are technically defined in 3D.

So you see that they are implicitly assuming 3D elements, and just casting the problem on a 2D background for simplicity.

By the way: Mermin-Wagner just states that there cannot be any long-range order at $T\neq 0$ for $d\leq 2$. This is not forbidding a phase transition: some phases are not distinguished by a symmetry and hence do not need to break one. The most famous example is the Berezinskii-Kosterlitz-Thouless transition, which involves vortex binding/unbinding but which does allow for superfluidity.

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  • $\begingroup$ I do not quite get what you mean the answer is. Are you saying that the model they use is in some sense three dimensional? How do I see that? Or are you saying that the superconducting phase they find only has quasi-long range order? $\endgroup$ – Petter Nov 18 '19 at 15:27
  • $\begingroup$ The models they use are intrinsically three dimensional, they are just casting it on a 2D lattice. You can see it by them having bosons and fermions (only defined in 3D). $\endgroup$ – SuperCiocia Nov 18 '19 at 16:07
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    $\begingroup$ What do you mean bosons and fermions are only defined in 3D? You can have bosons and fermions in more or less than three dimensions, and additionally in 2D you can have anyons. $\endgroup$ – Petter Nov 18 '19 at 16:57
  • $\begingroup$ I meant the fact that they are not considering anyons and only considering bosons and fermions as the only two statistical distributions. $\endgroup$ – SuperCiocia Nov 19 '19 at 2:51
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    $\begingroup$ Just a comment. Although in a strict 2d world, there can be fractional statistics, fermions and bosons can exist. $\endgroup$ – Zhiqiang Wang Dec 30 '19 at 22:36

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