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In class Professor claimed that 1D superconductivity does not exist even at zero temperature. I did a preliminary search and found papers on 1D superconductors. Did I or the Prof make a mistake or could somebody point to me a proof to this claim published somewhere?

Many thanks!

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    $\begingroup$ There is no long-range superconducting order in 1D even $T=0$ (the Mermin-Wagner theorem for spontaneous breaking of continuous symmetries), the closest one can get is the quasi-long-range order, which means that the correlation function of the order parameter decays slowly. You may see 1D superconductors in recent literatures by coupling a 1D system to a 3D superconductor and the superconducting order is induced by proximity effect, which does not contradict the previous statements. $\endgroup$
    – Meng Cheng
    Commented Dec 20, 2015 at 0:20
  • $\begingroup$ Thanks a lot for the quick reply, Dr Cheng! But does the Mermin-Wagner theorem also apply to zero temperature? The proof of the Mermin-Wagner theorem that I found (via Bogoliubov's inequality) requires finite temperature. Is there a more general proof of the theorem that covers zero temperature? $\endgroup$
    – PhysicsMath
    Commented Dec 20, 2015 at 0:31
  • $\begingroup$ I actually do not know a proof directly for 1D system at T=0, although one can imagine a 1D quantum system is mapped to a 2D classical system and Mermin-Wigner theorem should apply. There are field theory argument, like the famous paper by Sidney Coleman. $\endgroup$
    – Meng Cheng
    Commented Dec 20, 2015 at 2:11
  • $\begingroup$ I see. Thanks for the information Dr Cheng! Actually to me the question of whether we can remove the ``at finite temperature'' restriction to the statement of the Mermin-Wagner theorem is absolutely more important than the current question of 1D superconductivity. Because that would implies the Mermin-Wagner theorem covers quantum fluctuation as well. $\endgroup$
    – PhysicsMath
    Commented Dec 21, 2015 at 0:11
  • $\begingroup$ At zero temperature you can have quantum phase transitions, and in principle nothing restrict these ones to exist in low dimensions. See the book by Sachdev for instance. Note your question is completely misleading since no-one made a mistake ... $\endgroup$
    – FraSchelle
    Commented Dec 21, 2015 at 13:04

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In 1-d you can find BCS and charge density wave ground states at the mean field level, but these states are destroyed by quantum fluctuations and the correct ground state is a disordered phase called the Luttinger liquid, see, for example, http://arxiv.org/abs/cond-mat/9510014 .

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  • $\begingroup$ Thanks for pointing to the right direction, Thomas! Do you happen to know any proof of 1D long-range order destroyed by quantum fluctuation without using the Mermin-Wagner theorem + quantum-classical mapping? $\endgroup$
    – PhysicsMath
    Commented Dec 21, 2015 at 0:42
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Q: Why no 1D superconductivity?

A: By quantum fluctuation long-range order (LRO) does not exist. But power law decay of correlation functions is allowed as can be calculated by Luttinger liquid model.

Q: Why quasi-1D superconductivity?

A: LRO isn't required for superconductivity. There is still a critical temperature below which one can tell the difference between an exponential decay and a power law decay for the correlation function.

Sorry for the confusion (1D vs. quasi-1D)!

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