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Can real-space-eigenstates of conduction electrons in crystal cause formation of electronic singlet pairs?


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Posted

Crystals may contain electronic real-space-eigenstates as ground states, which are spatially much larger than one unit cell, such as impurity states, standing waves at Brillouin zone edges, Anderson localization, etc. Every eigenstate is usually occupied by two conduction electrons with opposite spins, forming a singlet pair. Notably: if the eigenstate is limited in real space, then the excitation energy of each singlet electron is not necessarily negligible, so below a certain temperature the singlet pair can be lasting. Isn't this a long-debated pairing mechanism in superconductors ?

Posted

I noticed that this is a cut and paste from physics stack exchange.  I also see that you have posed the same question in other sites but you have no follow up, just the single post and no further replies.  Do plan on replying to comments on your post?

Posted
5 hours ago, StanislavDolgopolov said:

Crystals may contain electronic real-space-eigenstates as ground states, which are spatially much larger than one unit cell, such as impurity states, standing waves at Brillouin zone edges, Anderson localization, etc. Every eigenstate is usually occupied by two conduction electrons with opposite spins, forming a singlet pair. Notably: if the eigenstate is limited in real space, then the excitation energy of each singlet electron is not necessarily negligible, so below a certain temperature the singlet pair can be lasting. Isn't this a long-debated pairing mechanism in superconductors ?

For these to be “conduction electrons”, wouldn’t the state have to extend throughout the crystal?  But would they then be treated as paired?

Posted
2 hours ago, Bufofrog said:

I noticed that this is a cut and paste from physics stack exchange.  I also see that you have posed the same question in other sites but you have no follow up, just the single post and no further replies.  Do plan on replying to comments on your post?

Yes, I would discuss the topic with community, since it seems to be important for general understanding of superconductivity.

50 minutes ago, exchemist said:

For these to be “conduction electrons”, wouldn’t the state have to extend throughout the crystal?  But would they then be treated as paired?

The conduction electrons are extended throughout the crystal before they fall into a local state. Inside the local state two electrons must be a singlet pair.

Posted
11 minutes ago, StanislavDolgopolov said:

Yes, I would discuss the topic with community, since it seems to be important for general understanding of superconductivity.

The conduction electrons are extended throughout the crystal before they fall into a local state. Inside the local state two electrons must be a singlet pair.

But if they fall into a localised state (presumably of lower energy if they "fall" into it) they cease to be conduction electrons, surely? In effect they go into the valence band, don't they?   

Posted
2 minutes ago, exchemist said:

But if they fall into a localised state (presumably of lower energy if they "fall" into it) they cease to be conduction electrons, surely? In effect they go into the valence band, don't they?   

Yes, the electrons in a local well are no longer usual conduction electrons. They form a new band - band of local states, which is usually much narrower than the conduction band.

Posted
15 minutes ago, StanislavDolgopolov said:

Yes, the electrons in a local well are no longer usual conduction electrons. They form a new band - band of local states, which is usually much narrower than the conduction band.

Well then, if they are localised in the valence band they won't participate in superconducting behaviour, will they?

I'm not a solid state physicist but my understanding is conduction requires a continuum of delocalised states, so that there is no energy gap to be jumped when an electron is given a bit of extra energy.

Posted
46 minutes ago, exchemist said:

Well then, if they are localised in the valence band they won't participate in superconducting behaviour, will they?

I'm not a solid state physicist but my understanding is conduction requires a continuum of delocalised states, so that there is no energy gap to be jumped when an electron is given a bit of extra energy.

Yes. The electrons don't jump the energy gap. Every electron pair is a combined particle with zero spin, that is a boson. The boson population at low temperature can condense into one ground state (Bose-Einstein-Condensate), then all states of bosons are energetically identical, so wave functions of bosons may be overlapped in real space and every boson can tunnel in the common space without energy loss (since it is in ground state). Nothing new. Superfluidity works this way.

Posted
6 minutes ago, StanislavDolgopolov said:

Yes. The electrons don't jump the energy gap. Every electron pair is a combined particle with zero spin, that is a boson. The boson population at low temperature can condense into one ground state (Bose-Einstein-Condensate), then all states of bosons are energetically identical, so wave functions of bosons may be overlapped in real space and every boson can tunnel in the common space without energy loss (since it is in ground state). Nothing new. Superfluidity works this way.

Well no, in general valence band electrons can't tunnel. The barriers are too high and too thick. You would need very special circumstances for tunnelling to be possible, I think. 

Posted
16 minutes ago, exchemist said:

Well no, in general valence band electrons can't tunnel. The barriers are too high and too thick. You would need very special circumstances for tunnelling to be possible, I think. 

The special circumstance is the degenerative wave function of bosonic pairs. 

The second circumstance is the fact that the local state band is not identical to a normal valence band. In a valence band the barriers are pinned to each atom, whereas boundaries of every local state can fluctuate, so in an external potential the local state can be flexible in space.  

Posted
26 minutes ago, StanislavDolgopolov said:

The special circumstance is the degenerative wave function of bosonic pairs. 

The second circumstance is the fact that the local state band is not identical to a normal valence band. In a valence band the barriers are pinned to each atom, whereas boundaries of every local state can fluctuate, so in an external potential the local state can be flexible in space.  

I find difficulty making sense of this. 

Posted
10 hours ago, exchemist said:

I find difficulty making sense of this. 

OK, consider a sample. An impurity state covers up to 1000 sites, and all these atoms vibrate at zero temperature, so impurity state boundaries are not fixed in space. Two conduction electrons are bound with the impurity state and form a coherent BEC with electron pairs of other impurity states. Coherent wave functions of BEC bosons are strongly overlapped in space, physical boundaries of electronic states are flexible in space; there are no obstacles for a paired charge transport between neighboring impurity states.

If we consider Anderson localization, there are no fixed impurities at all, so every local state can drift due to real-space-vibrations of atom potentials. Electron pairs drift together with their states, the BEC provides that the pair drift is dissipationless.

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