BCS and the formation of Copper Pairs


by paulzhen
Tags: copper, formation, pairs
paulzhen
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#1
Dec13-12, 03:13 AM
P: 21
Hi,

I am trying to understand the formation of Copper pairs in a superconductor. However, 2 major explainations I found in the internet:

1, Cooper pairs are formed by emitting and receiving phonons between 2 electrons.

2, another explaination is, when an electron drifting between the lattice, the vicinity lattice will be attracted slightly toward to that electron, and increase the charge density in that area, hence another electron will be attracted to the area and become a Copper pair with the 1st electron.

which one is correct??

And, why the momentums of the electrons must be "equal and oppsite" to form a Cooper pair?

Thanks a lot for your help!!
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DrDu
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#2
Dec13-12, 03:44 AM
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The two explanations are equivalent. Phonons are quanta of lattice deformation. So emission and absorption of phonons corresponds to the generation and deletion of lattice deformations. The coupling strength of the electrons to the phonons depends on the change in background charge due to the deformation.

The momenta have to be equal and opposite to generate a pair without net momentum.
If not, you get a condensate which carries a definite current which is also possible but does not correspond to the energetical minimum.
paulzhen
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#3
Dec13-12, 05:58 AM
P: 21
Thanks DrDu!

By the way, why people use "phonon absorbing&emitting" to explain the attraction between electrons? or electron and lattice?

DrDu
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#4
Dec13-12, 06:19 AM
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BCS and the formation of Copper Pairs


Quote Quote by paulzhen View Post
Thanks DrDu!

By the way, why people use "phonon absorbing&emitting" to explain the attraction between electrons? or electron and lattice?
Because that's the way you calculate it.
Darwin123
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#5
Dec14-12, 11:44 AM
P: 741
Quote Quote by paulzhen View Post
Thanks DrDu!

By the way, why people use "phonon absorbing&emitting" to explain the attraction between electrons? or electron and lattice?
The Feynmann diagrams describing the Cooper pair show these processes. However, one has to be cautious in interpreting these diagrams. You are probably picturing an exchange of real phonons. In that case, all the conservation laws would be strictly valid at all times.

The phonons that are being exchanged between the two electrons are virtual phonons. This is a little different than an exchange of real phonons. For instance, conservation of momentum is momentarily violated when a virtual phonon is absorbed or emitted.

Both visual pictures are "accurate" in terms of quantum mechanics. The force binding the Cooper pair can be pictured both as an exchange of virtual phonons and as a deformation of the lattice. However, the interaction deviates from classical mechanics in either case. Pictures that rely on classical concepts can only go so far.
zhanhai
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#6
Dec18-12, 12:49 PM
P: 25
Quote Quote by Darwin123 View Post
The Feynmann diagrams describing the Cooper pair show these processes. However, one has to be cautious in interpreting these diagrams. You are probably picturing an exchange of real phonons. In that case, all the conservation laws would be strictly valid at all times.

The phonons that are being exchanged between the two electrons are virtual phonons. This is a little different than an exchange of real phonons. For instance, conservation of momentum is momentarily violated when a virtual phonon is absorbed or emitted.

Both visual pictures are "accurate" in terms of quantum mechanics. The force binding the Cooper pair can be pictured both as an exchange of virtual phonons and as a deformation of the lattice. However, the interaction deviates from classical mechanics in either case. Pictures that rely on classical concepts can only go so far.
In my understanding, "deformation of lattice" is a classical description and may not be used in quantum mechanical explanation.

Momentum needs not be conserved at the mediating state(s) of a many phonon process, but should be conserved between the initial/final states of all process.

So a question is whether the phonon process is single phonon process or many phonon process, no matter virtual or real. For the two electrons at at k and -k, it seems that one of the electrons can only exchange state with the other electron by a many phonon process (?). But the question is how binding energy is generated by the phonon process(es). I cannot understand it.

In the original paper, it is said that a negative term occurs in H' so a binding state exists. And where the negative term comes from, the paper says it is from another paper.
DrDu
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#7
Dec18-12, 02:38 PM
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Quote Quote by zhanhai View Post
In my understanding, "deformation of lattice" is a classical description and may not be used in quantum mechanical explanation.
I don't see why not. In QM the deformations become operators but we also carry on calling "position" "position" and "momentum" "momentum" in QM although they are operators.

Btw: To which original paper are you referring?
Darwin123
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#8
Dec18-12, 05:59 PM
P: 741
Quote Quote by zhanhai View Post
In my understanding, "deformation of lattice" is a classical description and may not be used in quantum mechanical explanation.
The phrase “lattice deformation” does not imply a system that is fully classical. The nuclei part of the lattice is much more massive then the valence electrons in the chemical bonds that hold the lattice together. Therefore, one can use an approximation of quantum mechanics where the nuclei and core electrons of the lattice move according to classical dynamics, but the electrons are behaving as a wave in quantum mechanics.

So an "explanation" can have "quantum" electrons and "classical" phonons. However, the composite particle called a Cooper pair is governed mostly by quantum mechanics. The Cooper pair is a quantum mechanical object because it is so large compared to the spacing between the Cooper pairs. Basically, the size of the Cooper pair is an indeterminacy of its position. So if you want to treat the Cooper pair as a quasiparticle, you have to take the quantum mechanics into account. So we have a "quantum" Cooper pair.

What you are possibly saying is that perturbation techniques aren’t entirely valid under the conditions of “lattice deformation”. However, there are other approimxations. The adiabatic approximation and the WKB approximation are valid under the conditions of a “slowly moving” lattice.

The electrons are moving far faster than the nuclei under these conditions. The indeterminacy in the position of the nuclei are much smaller than the undeterminacy of the valence electrons. Therefore, one can use “hybrid” mechanics where nuclei with core electrons are “classical particles” and both conduction electrons and valence electrons are “quantum waves”.

The distorted lattice, consisting of displaced nuclei, can be pictured as generate and electric field. A higher concentration of positive charges (the nuclei) are the source of an electric field that is moving outward from the points of greatest concentration of nuclei. The conduction electrons are in a potential that is caused by this electric field. So the electrons can “move” coherently with the deformed lattice.

Two electrons can move together toward the region where the positive charge density is greatest. However, the electrons are also pulling at the nuclei. They are making the regions of high density nuclei less dense. Although this is a self consistent picture classically, one can get more accuracy by assuming that the electrons at least are quantum mechanical.

So basically two electrons are interacting through the lattice deformation, otherwise called a phonon.

Here is a link that claims that variational and WKB approximation are valid with Cooper pairs. You can assume that the electrons are quantum mechanical.
http://en.wikipedia.org/wiki/Perturb...m_mechanics%29
“An example of this phenomenon may be found in conventional superconductivity, in which the phonon-mediated attraction between conduction electrons leads to the formation of correlated electron pairs known as Cooper pairs. When faced with such systems, one usually turns to other approximation schemes, such as the variational method and the WKB approximation. This is because there is no analogue of a bound particle in the unperturbed model and the energy of a soliton typically goes as the inverse of the expansion parameter. However, if we "integrate" over the solitonic phenomena, the nonperturbative corrections in this case will be tiny; of the order of or in the perturbation parameter g. Perturbation theory can only detect solutions "close" to the unperturbed solution, even if there are other solutions (which typically blow up as the expansion parameter goes to zero).”



http://arxiv.org/ftp/arxiv/papers/1012/1012.0879.pdf
“ The binding energy of superconducting electrons dominates the superconducting transition temperature in the corresponding material. Under an electric field, superconducting electrons move coherently with lattice distortion wave and periodically exchange their excitation energy with chain lattice, that is, the superconducting electrons transfer periodically between their dynamic bound state and conducting state, so the superconducting electrons cannot be scattered by the chain lattice, and supercurrent persists in time. Thus, the intrinsic feature of superconductivity is to generate an oscillating current under a dc voltage.”

You mustn’t think that the Cooper pairs are individual particles. Actually, they are squashed together. Therefore, position of each electron is highly uncertain. So even if you think of the lattice deformation as classical, the electrons are not classical.
http://www.desy.de/f/students/lectur...schmueser1.pdf
“The binding energy of a Cooper pair turns out to be small, 10��4��10��3 eV, so low temperatures are needed to preserve the binding in spite of the thermal motion. According to Heisenberg’s Uncertainty Principle a weak binding is equivalent to a large extension of the composite system, in this case the above-mentioned d = 100 �� 1000 nm. As a consequence, the Cooper pairs in a superconductor overlap each other. In the space occupied by a Cooper pair there are about a million other Cooper pairs. Figure 22 gives an illustration. The situation is totally different from other composite systems like atomic nuclei or atoms which are tightly bound objects and well-separated from another. The strong overlap is an important prerequisite of the BCS theory because the Cooper pairs must change their partners frequently in order to provide a continuous binding.”
DrDu
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#9
Dec19-12, 01:53 AM
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Quote Quote by Darwin123 View Post
The adiabatic approximation and the WKB approximation are valid under the conditions of a “slowly moving” lattice.

The electrons are moving far faster than the nuclei under these conditions. The indeterminacy in the position of the nuclei are much smaller than the undeterminacy of the valence electrons. Therefore, one can use “hybrid” mechanics where nuclei with core electrons are “classical particles” and both conduction electrons and valence electrons are “quantum waves”.
That's not quite right. Superconductivity is carried entirely by electronic excitations in a very thin shell around the Fermi surface which is thinner than the maximal phonon frequency, the Debye frequency. For these electrons, the picture is rather that of an inverted adiabatic approximation, where the phonons follow adiabatically the electronic motion.

The adiabatic approximation is formalized in solid state theory in the form of Migdals theorem. The russians believed too long in Migdals theorem which is the reason why they didn't find the mechanism behind superconductivity before BCS.
Darwin123
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#10
Dec19-12, 09:09 AM
P: 741
Quote Quote by DrDu View Post
That's not quite right. Superconductivity is carried entirely by electronic excitations in a very thin shell around the Fermi surface which is thinner than the maximal phonon frequency, the Debye frequency. For these electrons, the picture is rather that of an inverted adiabatic approximation, where the phonons follow adiabatically the electronic motion.

The adiabatic approximation is formalized in solid state theory in the form of Migdals theorem. The russians believed too long in Migdals theorem which is the reason why they didn't find the mechanism behind superconductivity before BCS.
Thank you! So the phonons are quantum and the electrons are classical. I didn't understand that point before.
zhanhai
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#11
Dec19-12, 11:36 AM
P: 25
Quote Quote by DrDu View Post
I don't see why not. In QM the deformations become operators but we also carry on calling "position" "position" and "momentum" "momentum" in QM although they are operators.
In so far that an electron in a crystal system has a defined k value, the "size" of the electron is infinite (or the same as the size of the crystal, to be practical).


Quote Quote by DrDu View Post
Btw: To which original paper are you referring?
http://prola.aps.org/abstract/PR/v104/i4/p1189_1
DrDu
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#12
Dec19-12, 11:47 AM
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1. I thought you were talking about lattice deformations, not electrons.
2. Electrons in a lattice do not have a definite k value. Rather the k-states form a generalized basis into which every electronic state can be resolved.
zhanhai
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#13
Dec19-12, 12:12 PM
P: 25
Phonon is the unit by which lattice exchanges energy/momentum with other object (electron, neutron, or so on). On the other hand, the number of phonons a lattice mode has is proportional to the square of magnitude of the lattice mode (in the form of 1/2+n or so). In that sense, if a real phonon is taken by an electron, then the lattice mode of the phonon loses one phonon, and its magnitude/intensity/energy decreases accordingly. If this change in the magnitude of lattice mode is called "deformation", then I would not disagree with "lattice deformation"; but as far as the phonon exchange is virtual, there is still no actual "lattice deformation". And I tend to believe that the phonon exchange associated with electron pairing (and binding energy) is virtual.
DrDu
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#14
Dec19-12, 01:28 PM
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"Virtual" does not mean that there is no deformation. Compare with the electromagnetic field.
The Coulomb field is a very real electric field, however it is due to the exchange of virtual photons.
M@2
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#15
Feb13-13, 07:28 PM
P: 67
Quote Quote by DrDu View Post
Superconductivity is carried entirely by electronic excitations in a very thin shell around the Fermi surface which is thinner than the maximal phonon frequency, the Debye frequency. For these electrons, the picture is rather that of an inverted adiabatic approximation, where the phonons follow adiabatically the electronic motion.

The adiabatic approximation is formalized in solid state theory in the form of Migdals theorem. The russians believed too long in Migdals theorem which is the reason why they didn't find the mechanism behind superconductivity before BCS.
There are superconductors where very thick shell around the Fermi surface.

The russians believed too long in Migdals theorem which is the reason why they didn't find the mechanism behind superconductivity before BCS.
I think the russians believed too long to fermi liquid theory which is the reason why they didn't find the mechanism behind superconductivity!!!!
:)
DrDu
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#16
Feb14-13, 01:49 AM
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Quote Quote by M@2 View Post
There are superconductors where very thick shell around the Fermi surface.
In which ones are you thinking and are they well described by BCS theory?
M@2
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#17
Feb16-13, 01:03 AM
P: 67
Quote Quote by DrDu View Post
In which ones are you thinking and are they well described by BCS theory?
Excuse, notwithstanding to anybody opinion, even to Nobel Prize Commetee, i think (as Hirsch, as Vasiliev, as Bernd Matthias, a
well-respected solid state experimentalist who had been
making superconducting materials in his lab for many
years, as love.Minich.Ru) BCS theory is completely wrong.

As this idea is completely forbidden in PF, i become silent.


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