Why can superconductivity exist in 2D since Mermin-Wagner should forbit it? This question was asked here before, but I don't think anyone gave a satisfactory answer, so let me revisit it.

I have read both the rigorous proof of BCS/Hartree-Fock (1) and that of the quantum Mermin-Wagner (2). Of course, the two rigorous statements don't exactly exclude each other. However, on a conceptual level, it seems hard to reconcile. Indeed, superconductivity breaks $U(1)\cong SO(2)$ gauge symmetry at some finite temperature $T>0$ for all spatial dimension $d\ge 1$, but this is in principle forbidden by Mermin-Wagner since it doesn't allow any continuous symmetry breaking at finite temperatures for $d\le 2$.

There's also this explanation, but I don't quite understand its answer. Mathematically speaking, a 3D system that is translationally invariant in 1 dimension is equivalent to a 2D system so I don't quite understand why it matters. Although anyons are the correct way to describe statistics in a 2D system, the mathematical formalism of fermions/bosons do exist (not to mention that fermions/bosons are also types of anyons). Finally, the only way for the $U(1)$-symmetry to not have long range order (the phase of the complex order parameter $\psi$) seems to be that the amplitude $|\psi|=0$, so superconductivity indeed seems to be a long-range order phase transition (correct me if I'm wrong).

  1. Bach, V., Lieb, E.H. & Solovej, J.P. Generalized Hartree-Fock theory and the Hubbard model. J Stat Phys 76, 3–89 (1994). https://doi.org/10.1007/BF02188656

  2. http://www.scholarpedia.org/article/Mermin-Wagner_Theorem


3 Answers 3


All the answers here and in the other question do not address the important difference between superconductivity and superfluidity: namely that the Nambu-Goldstone modes in superconductors are not gapless. The latter is an assumption for the validity of the Mermin-Wagner-Hohenberg-Coleman theorem, and therefore it does not apply.

The question of whether superconductivity can exist in a strictly 2D system turns out to be very interesting. Let's go in steps:

  1. The Mermin-Wagner-Hohenberg-Coleman theorem precludes true long-range order in 2 dimensions (at finite temperature) or 1 dimension (at zero temperature). The reason, as is clear from Coleman's proof, is that that the fluctuations of linearly dispersing scalar modes, c.q. Nambu-Goldstone modes, in 2+0 or 1+1D are so violent as to preclude their existence altogether. This is due to an infrared divergence, so this applies to long wavelengths/large systems.

  2. However, the Nambu-Goldstone modes in a superconductor are gapped by the Anderson-Higgs mechanism (due to the coupling to gauge fields, the electromagnetic field). Therefore, there is no infrared divergence (the $k^2$ term in the denominator is replaced by $k^2 + (\hbar\omega_{\rm{p}}/c)^2$, with $\Delta$ the superconducting gap with $\omega_{\rm{p}}$ the plasma frequency). The theorem does not apply.

  3. So there seems to be no obstruction to superconductivity in any low dimension. But this does not take into account the topological defects (vortices). In 2D superfluids, there is the BKT phase transition between a low-temperature, quasi-long-range ordered phase, where vortex pairs are bound, and a high-temperature, disordered phase, where vortex pairs are unbound. The transition temperature is set by the balance between the energy cost of a vortex pair (which grows logarithmically with system size) and the entropy gain of having thermally excited pairs (which also grows logarithmically with system size). But in superconductors, the size of a vortex is capped by the inverse of the superconducting energy gap $\Delta$. Conversely, the entropy gain is unaffected. Therefore the argument leading to the BKT phase transition does not apply, and the transition temperature is pushed to zero as the system size grows. In an infinite volume, vortices are unbound at any temperature.

So the conclusion seems to be that, despite the non-applicability of the MWHC-theorem, superconductivity cannot exist in an infinite, strictly 2D system.

In real life, however, there are many examples of quasi-2D systems (even monolayers) that exhibit all the signs of superconductivity, including dissipationless current and a form of Meissner effect. The reason is that the electromagnetic field is not restricted to 2D. The field lines permeate out of the 2D layer. This causes the in-plane penetration depth $\lambda_{\rm{2D}}$ to become very large. To lowest order, one gets: $$ \lambda_{\rm{2D}} = \frac{\lambda_{\rm{L}}}{d} $$ where $\lambda_{\rm{L}}$ is the usual London penetration depth depending on the superconducting order parameter, and $d$ is the thickness of the system. For a very thin sample, the penetration depth diverges. In other words, it is a very strong type-II superconductor.

If the penetration depth becomes larger than the linear size of the system, it is effectively a neutral superfluid as far as transverse electromagnetic effects are concerned. This also implies that the vortex energy again depends logarithmically on the system size, and the BKT criterion applies. In fact, the BKT transition has been observed in many quasi-2D superconductors.

To my knowledge, none of this has been investigated really well, or at all. It would make for an interesting research project.

TL;DR The Mermin-Wagner-Hohenberg-Coleman theorem does not apply. However, due to vortex unbinding, strictly 2D superconductors would not exist. In reality, the electromagnetic field is always 3D and turns a 2D superconductor into a neutral superfluid with BKT transition.

  • $\begingroup$ I haven't quite done the algebra, but since the energy spectrum only depends on the magnitude $|\Delta|^2$, shouldn't there also be a goldstone mode related to the $U(1)$ symmetry of $\Delta$? Is it because the energy is gapped? $\endgroup$ Commented Aug 9, 2021 at 21:55
  • $\begingroup$ And in the context of Landau-Ginzburg, if we assume that the amplitudes $|\psi|$ are uniform so that the $U(1)$-gauge contributes to the energy via $|\nabla \theta|^2$, then it seems that the standard "proof" for Mermin-Wagner applies. $\endgroup$ Commented Aug 9, 2021 at 22:04
  • $\begingroup$ Okay, my language was a bit sloppy. The gap of the NG modes is not the single-particle gap $\Delta$, but the plasma frequency $\omega_p$. This depends not on the condensate amplitude $| \Delta |$ but on the number of paired or unpaired electrons. It is the $U(1)$ of $\Delta$ that would be the NG mode, but it is coupled to the electron plasma that gives it a gap, which is actually very large, of the order of 1eV (compared with the single-particle gap which is typically of the order of 50 meV). $\endgroup$ Commented Aug 10, 2021 at 12:45
  • $\begingroup$ That sounds interesting, but would you be able to elaborate in your answer? I'm not quite familiar with the plasma frequency and how it's related to the problem. $\endgroup$ Commented Aug 10, 2021 at 18:05
  • $\begingroup$ I corrected my answer slightly. I feel elaborating on the plasma frequency is beyond the scope of the question; it's a basic concept of the free electron model. How it's related to superconductivity was elucidated by P.W. Anderson's 1958 paper. This can be found in any good textbook. Here are some lecture notes with a calculation of the fluctuations in 3D and 2D (page 21) $\endgroup$ Commented Aug 12, 2021 at 1:19

The heuristic argument for Mermin-Wagner is that, if the continuous symmetry were spontaneously broken, then there would be Goldstone bosons, which are gapless excitations. But in $d \leq 2$, these Goldstone bosons would lead to an IR divergence, which is inconsistent with the initial assumption of symmetry breaking.

However, in practice Mermin-Wagner may not prohibit seeing what looks like a spontaneously broken continuous symmetry in conditions that are accessible in experiments. In a finite system of linear size $L$, the maximum wavelength of an excitation will be $O(L)$, so to get the IR divergence in Mermin-Wagner you have to have a sufficiently large system. More precisely, you can estimate a temperature-dependent lengthscale $L_{\mathrm{IR}}(T)$, where for systems of size $L > L_{\mathrm{IR}}(T)$, Mermin-Wagner physics will destroy the spontaneous symmetry breaking at temperatures above $T$.

The question is then how $L_{\mathrm{IR}}(T)$ scales with $T$. It turns out this is related to the type of IR divergence due to the Goldstone bosons. In $d=2$, there is only a very weak logarithmic divergence, so $L_{\mathrm{IR}}(T)$ increases exponentially with the ratio $T_{0}/T$, where $T_{0}$ is a characteristic temperature of the system. This characteristic temperature is not necessarily associated with scales associated with the symmetry breaking, e.g. the critical temperature for superconductivity, but is often much larger, given by say the Debye temperature in a solid. That means it is not too hard to go to temperatures in the lab where the ratio $T_{0}/T$ is large, which in turn means the cutoff scale $L_{\mathrm{IR}}(T)$ is astronomically large, and so Mermin-Wagner physics has no bearing on what happens in the lab.

Anthony Leggett's lecture notes are a good reference on everything I have said here. As an example, he estimates that to spontaneously break continuous spatial translation symmetry, i.e. to form crystalline order, in $d=2$ and at a lab temperature $T\sim 1\mathrm{K}$, if we use a typical Debye temperature of the order of a few hundred Kelvin, then the cutoff lengthscale $L_{\mathrm{IR}}(T)$ is of order the distance from the Earth to the Moon!

  • $\begingroup$ You might also find this answer helpful. $\endgroup$
    – anon1802
    Commented Aug 4, 2021 at 9:28

Leggett does not mention superconductivity, where the situation is even worse than for crystal order. The sample would have to be larger than the observable universe for long-wavelength fluctuations to have an effect on the SC Tc in 2D. See our paper Physical limitations of the Hohenberg–Mermin–Wagner theorem, also available on https://arxiv.org/abs/2107.09714.


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