Superconductivity and zero resistance

[Jimmy87]

Hi, I am fairly new to superconductivity (introductory college level). I have covered and grasped the basics but was wondering if someone could shed some light on what specifically causes materials to have zero resistance when they become superconducting. I know that cooper pairs form when the material is in a low energy state (10-100 kelvin). These cooper pairs have integer spins meaning they form composite bosons. The lattice vibrational energy (phonons) are also bosons. But what is it about these quantum mechanical effects that specifically causes zero resistance to arise? Does it have something to do with Pauli's Exclusion Principle as i have seen that some websites mention this but do not explain its significance in superconductivity?

 

[Drakkith]

From wiki: http://en.wikipedia.org/wiki/Cooper_pair

Cooper originally considered only the case of an isolated pair's formation in a metal. When one considers the more realistic state of many electronic pair formations, as is elucidated in the full BCS Theory, one finds that the pairing opens a gap in the continuous spectrum of allowed energy states of the electrons, meaning that all excitations of the system must possesses some minimum amount of energy. This gap to excitations leads to superconductivity, since small excitations such as scattering of electrons are forbidden.[6] The gap appears due to many-body effects between electrons feeling the attraction.

Does this help?

 

[DrDu]

The gap is not crucial for zero resistivity as there are also zero gap superconductors.
The point is that the Cooper pairs form a Bose-Einstein condensate. In a current carrying superconductor, all cooper pairs move with the same velocity. This state is metastable as breaking ( and decelerating ) only some Cooper pairs corresponds to excited states. Only if there are sufficiently many excited states carrying zero momentum so that these can form an alternative Bose-Einstein condensate of lower energy, superconductivity will break down. The probability for such a number of pairs being excited by scattering at the same time by chance is arbitrary low. If there are too few excited states to form a Bose-Einstein condensate on their own they have no other possibility to recombine into the original current carrying condensate.

 

[vatlychatran]

The gap is not crucial for zero resistivity as there are also zero gap superconductors.
The point is that the Cooper pairs form a Bose-Einstein condensate. In a current carrying superconductor, all cooper pairs move with the same velocity. This state is metastable as breaking ( and decelerating ) only some Cooper pairs corresponds to excited states. Only if there are sufficiently many excited states carrying zero momentum so that these can form an alternative Bose-Einstein condensate of lower energy, superconductivity will break down. The probability for such a number of pairs being excited by scattering at the same time by chance is arbitrary low. If there are too few excited states to form a Bose-Einstein condensate on their own they have no other possibility to recombine into the original current carrying condensate.

What are zero gap superconductors? Could you give some examples? Do you mean zero gap in some directions but not in others as d-wave cuprates?

 

[DrDu]

I think the following article is quite illuminating:
Brun Hansen, Infinite conductivity of ordinary and gapless superconductors. Physica Vol. 39, pp. 271-292, Year 1968

 

[zhanhai]

I have just read through the BCS paper, which gives basically the same explanation (possibly except the Bose-Einstein condensate). I do not think it is a microscopic explanation, since it is too general. So I searched "zero resistance" in PF and as many as 135 posts (including this one) turned out, which should be an indication that the question of "why zero resistance" has not been properly explained.

Why "all cooper pairs move with the same velocity"?

That a pair has non-zero velocity should mean that it is not at the lowest energy? Then some of its energy should be allowed to be transferred to another object (such as a phonon mode).

"Scattering of individual electrons... can only produce fluctuations about the current". Why does not scattering lead to resistance ? ( as it would do in normal state ?)

Sure, the paired state with a current has the lowest free energy with that current, so any deviation from this paired state is not favored. But this is a general statement. "The opening of the gap leads to that the free energy is the lowest " is an explanation as it is. Then why scattering does not lead to lose of electrons' energy (conversion to phonon energy or so on)?

 

[DrDu]

So I searched "zero resistance" in PF and as many as 135 posts (including this one) turned out, which should be an indication that the question of "why zero resistance" has not been properly explained.

Did you reach to this conclusion before or after reading at least part of these 135 posts?

 

[ZapperZ]

I have just read through the BCS paper, which gives basically the same explanation (possibly except the Bose-Einstein condensate). I do not think it is a microscopic explanation, since it is too general. So I searched "zero resistance" in PF and as many as 135 posts (including this one) turned out, which should be an indication that the question of "why zero resistance" has not been properly explained.

This is a faulty conclusion. We get asked repeatedly the SAME question on many different topics on here. We get new members all the time who either did not browse old threads, or did not know these questions have been asked ad nauseum in this forum because they did not do a search. Heck, even the stuff we put on our FAQ get asked again.

None of these have any indication that these things have not been properly explained!

Zz.

 

[Henryk]

Jimmy87
This thread reminds me of the time when I was a graduate student taking a theoretical solid state course. We discussed BCS theory in detail, Cooper pairs, condensation, energy gap, etc. And then I posed the same question to the professor: how do we get zero resistance out of all of this?
The professor answer was (if I remember correctly) " You have to start with a gauge-invariant Hamiltonian".
Apparently, your question has an answer but the answer is beyond the scope of a graduate course on this subject.

 

[zhanhai]

Did you reach to this conclusion before or after reading at least part of these 135 posts?

I had read through each of the posts on the first page of the search result list before I wrote my previous replying post.

 

[DrDu]

I can only recommend again the article by Bruns i cited in #5.

 

[DrDu]

Your post made me think about this question again and I think it might be helpful to consider analogous phenomena in other systems with broken symmetry: Consider a magnet being brought into a magnetic field at some angle. If the system is in vacuum, it will oscillate almost forever. This is the analogon of the persistent current in a superconductor. You may ask why it isn't destroyed by single spins flipping into an orientation aligned with the external field. The reason is the same as with a superconductor. While those flips are possible and do occur, they don't lead to an energetically more favourable situation as the flipped spins lack stabilization by interaction with spins of similar orientation. You would have to flip all spins at the same time (or at least a considerable fraction of them) to get an energetically more stable situation. However, this is extremely improbable to happen in a macroscopic system.

 

[cgk]

I do not think that the question of "why zero resistivity?" is in any sense trivial or generally well explained in solid state textbooks. After searching myself and asking the same question (how do you get from BCS to zero resistivity?) to a number of high level active researchers in this domain of theoretical solid state physics, the most frequent answer I came up with was "this is complicated."[1]

One researcher pointed out the following: According to Yang (http://dx.doi.org/10.1103/RevModPhys.34.694, 1962), long-range charge order (a measurable property of the 2nd order reduced density matrix) implies the existence of a Meisner effect, which in turn is a core feature of super-conductivity. So, for this reason, articles dealing with super-conductivity typically investigate these charge correlations. The relationship to zero-resistivity is still non-obvious to me, but it is at least a start...

I have not yet read the Hansen article DrDu mentioned, but will do so right away.

[1] (of course, after the usual Bose condensation of Cooper Pairs, etc, stuff, which--however--does not easily translate to an ab initio theory. These things deal in quasi-particles, after all, and should be visible as emergent phenomena in real first principles wave functions which deal with actual interacting electrons. My point was that if I calculate such wave functions on models which by conventional wisdom should be super-conducting, I either do not get anything resembling an BCS state as ground state solution, and even if I did, I would not see how this would make for zero resistivity).

 

[DrDu]

Cjk, do you know the article:
Weinberg, Steven. "Superconductivity for particular theorists." Progress of Theoretical Physics Supplement 86 (1986): 43-53.