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Several papers set up a model of fermions in 2D and calculate superconducting $T_c$, e.g.: https://arxiv.org/abs/1808.08635
https://arxiv.org/abs/1406.1193
https://doi.org/10.1073/pnas.1620651114

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 $T$?

Qmechanic
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Petter
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1 Answers1

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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.

SuperCiocia
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