I Basic questions about superconductivity

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fluidistic

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From what I'm reading about superconductivity, Cooper pairs, which are responsible for superconductivity (SC) arise from the interaction between electrons and phonons. Also, from what I've read, it is generally said that SC occurs below a certain critical temperature T_c (let's stick to type I superconductors for the sake of simplicity).

Let's say that we cool down a SC material. We cool it down so much that its temperature is now a millionth of a degree Kelvin above the absolute 0. At such low temperature, I am guessing there are very few phonons, and that they have large wavelength (compared to the size of the material).
First question: Are these few and long wavelength phonons enough to ensure that the material stays superconducting? In other words, is it possible to find a positive temperature below T_c such that SC would stop?

Second question: I have been told that theoretically, at absolute 0, the crystal would be perfectly static. I assume this means no phonon. Thus I assume that theoretically (if the absolute 0 was reachable), SC cannot exist at absolute 0. Is this true?

Third question: Is it the electrons that are responsible for the phonons that sustain SC? If that's the case, then indeed SC would hold for any temperature below T_c. It also means it's not possible to cool down the material at 0 K, and possibly even at extremely low temperature. I.e. it would not be possible to cool down the material below a certain positive temperature because the electrons would form some Cooper pairs which would disturb the lattice no matter what we do to cool down the material. Or is it the temperature that is the mere responsible for the phonons that can sustain SC?
 
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DrDu

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The phonons responsible for superconductivity are virtual phonons, not real ones. If you put a charge somewhere into the lattice, the lattice will be deformed due to the attraction or repulsion of the atomic nuclei. This deformation can be described in terms of virtual phonons. The other electrons in the lattice will also respond to this charge, however, their response will be subtle different from that of the nuclei because the electrons due to the Pauli principle have some resistance against any force which tries to localize them. This leads to the phenomenon that at some distances around the perturbing charge, a charge of the same type will feel attraction. The phenomenon is called overscreening and is the principle behind the attractive force leading to superconductivity.
 

fluidistic

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The phonons responsible for superconductivity are virtual phonons, not real ones. If you put a charge somewhere into the lattice, the lattice will be deformed due to the attraction or repulsion of the atomic nuclei. This deformation can be described in terms of virtual phonons. The other electrons in the lattice will also respond to this charge, however, their response will be subtle different from that of the nuclei because the electrons due to the Pauli principle have some resistance against any force which tries to localize them. This leads to the phenomenon that at some distances around the perturbing charge, a charge of the same type will feel attraction. The phenomenon is called overscreening and is the principle behind the attractive force leading to superconductivity.
Wow, so the Internet is clogged with misinformation about basics description of what SC is about.
But hmm, then there is still a deformation of the lattice due to the electrons. It would mean that at 0 K, SC cannot exist.... or not? At 0 K the crystal should be in its ground state. I wonder if the ground state of a SC material would have a non periodic lattice structure, discarding all defects. In other words I wonder whether at 0 K an idealized (defect-free) SC material would have a periodic lattice structure or not, due to the Cooper pairs distorting the structure dynamically.
 

DrDu

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The total wavefunction of the crystal is still periodic at T=0, but it makes no sense to assume that the positions of the nuclei form a regular lattice as the nuclei are quantum objects like the electrons and their position is unsharp due to the uncertainty principle.
 

ZapperZ

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Wow, so the Internet is clogged with misinformation about basics description of what SC is about.
But hmm, then there is still a deformation of the lattice due to the electrons. It would mean that at 0 K, SC cannot exist.... or not? At 0 K the crystal should be in its ground state. I wonder if the ground state of a SC material would have a non periodic lattice structure, discarding all defects. In other words I wonder whether at 0 K an idealized (defect-free) SC material would have a periodic lattice structure or not, due to the Cooper pairs distorting the structure dynamically.
The superconducting state is easiest to be solved when one considers T = 0 K! Beginning students work through the BCS theory at T=0K. In fact, it gets to be more complicated theoretically when one deals with T>0K. So your claim that it can't exist at T=0K is incorrect.

Zz.
 
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Deformations around an electron are part of the lowest energy state.
 

fluidistic

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Deformations around an electron are part of the lowest energy state.
I see. But then wouldn't the entropy not vanish at 0 K? It's supposed to vanish with a perfectly regular crystalline shape at 0 K. Now, if the ground state contains the deformations near electrons (and it's dynamics, i.e. varies in time), I see no way for S to vanish at 0 K for SC materials. Am I missing something?
 

DrDu

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In the ground state of e.g. a hydrogen atom, neither the proton nor the electron are also not at rest at the center of mass but fluctuate quantum mechanically around it. The situation is not different in a superconductor or any solid.
 
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There is nothing wrong with degenerate ground states (with non-zero entropy), but I don't think a superconductor needs that. The electrons don't have a fixed position in the crystal, they are spread out. Only the correlation of the deformation is local.
 

Henryk

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The entropy of the superconducting state is zero and superconducting states doesn't conduct heat (It has been verified experimentally).
Now, back to the first question, what happens to superconductivity when temperature is so low that phonons are not excited (that's how I understand).
Frankly, I wish I asked that question when was taking course in superconductivity. Well, I didn't and all I got was some sort of "phonon-mediated attraction" between Cooper pairs. There was also a hand waving arguments of "overscreening" of electric field but not much more details.
So, what do I think about the original question?
Consider this: two oxygen atoms, each with its own electron states, once they are close together, they form an oxygen molecule O2. How come?
Because you can create an electron states with lower energy than that of two single oxygen atoms.
Similarity, there is an quantum state involving two electrons(the Cooper pair) and lattice displacements (phonons) that has lower energy than two individual atoms and non-interacting phonons.
Does it explain anything? Absolutely not. The only thing is that it leads to calculations that agree with experimental results.
 

fluidistic

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The entropy of the superconducting state is zero and superconducting states doesn't conduct heat (It has been verified experimentally).
I guess you mean the Cooper paired electrons have 0 entropy and that they're not involved at all in the heat transfer. I already knew this, but how does this help at all?

Henryk said:
Now, back to the first question, what happens to superconductivity when temperature is so low that phonons are not excited (that's how I understand).
Frankly, I wish I asked that question when was taking course in superconductivity. Well, I didn't and all I got was some sort of "phonon-mediated attraction" between Cooper pairs. There was also a hand waving arguments of "overscreening" of electric field but not much more details.
So, what do I think about the original question?
Consider this: two oxygen atoms, each with its own electron states, once they are close together, they form an oxygen molecule O2. How come?
Because you can create an electron states with lower energy than that of two single oxygen atoms.
Similarity, there is an quantum state involving two electrons(the Cooper pair) and lattice displacements (phonons) that has lower energy than two individual atoms and non-interacting phonons.
Does it explain anything? Absolutely not. The only thing is that it leads to calculations that agree with experimental results.
Uh... there's nothing new there either. But you do mention that there are lattice displacements (phonons) and you do not mention virtual phonons at all, which apparently are the ones responsible for superconductivity.
 

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