Dear Ijaz ,

many thanks for the references. In all books the start point of the theory is the indistinguishability of particles. However, the electrons may be distinguishable. Example : some Helium atoms are put in an electronic Fermi sea; obviously, the local electrons of Helium are not identical to the free electrons of the Fermi sea. Identical situation is in a crystal, where some electrons are delocalized, and some electrons are local on-site.

Thus, the main principle of the second quantization is violated.

Yes, systems are well described by BE or FD statistics if all particles are identical and exchangeable. But in a crystal the electrons may form 2 distinguishable particle groups : (i) free electrons, (ii) local electrons, which are permanently pinned to atoms. The particle exchange between 2 groups needs an energy for delocalization of local electrons. Thus we cannot say that the particles are identical and exchangeable. Probably, we should use the second quantisation for every group separately, although both groups are fermionic. Anyway thank you for the helpful discussion. Stanislav

Dear Prof. Stanislav Dolgopolov ,

Interesting question. Please check the following external pdf notes:

http://www.lassp.cornell.edu/clh/Book-sample/1.1.pdf

They base their argument about realistic hamiltonians nicely, by saying that the particles usually move on lattices, not in the continuum real space.

They continue to elaborate arguing that there is a duality of direct space and Fourier space, quoting: the density operator is local in a direct space but not in momentum space (it mixes different wavevectors); on the other hand, the Bloch occupation operator is local in Fourier space, but nonlocal in real space.

Probably that duality makes the second quantization formalism work well for valence electrons, these electrons are not part of the sea of conduction electrons.

Dear Prof. Contreras,

It seems your conclusion "valence electrons, these electrons are not part of the sea of conduction electrons" is applicable for conduction and superconduction electrons. Only one sample:

We consider a system containing a non-dissipative pair (e1,e2) and two normal (dissipative) electrons e3, e4 . We investigate two cases, A and B:

A. The non-dissipative pair (e1,e2) is permanent in time. Then an initial non-zero momentum of the pair is also permanent in time. Obviously, this permanent momentum is a supercurrent;

B. The non-dissipative pair (e1,e2) is not permanent in time, i.e. a recombination is possible: e1, e2 become normal, e3, e4 become non-dissipative and back. But at every time moment there are one non-dissipative pair and two normal electrons:

(e1,e2)non-diss + e3 + e4 e1 + e2 + (e3,e4)non-diss

In the case B the initial non-zero momentum of the pair (e1,e2) dissipates, because the electrons e1,e2 become periodically dissipative and there is no external force to give to the new created pair (e3,e4) exactly the same momentum, which the pair (e1,e2) had. The same is valid for the pair (e3,e4) since the pairs are identical. If there is no external force to push the pair along the current direction, then the total momentum of the system dissipates and the current vanishes. Thus non-permanent pairs cannot keep a supercurrent. Notable is the fact that the Fock states in both cases A and B are identical due to the indistinguishability of particles, whereas the case A is superconducting and the case B is dissipative.

The conclusion from the sample: normal and SC electrons are distinguishable and, thus, should belong to different Fock spaces.

A bit more details about the issue in the note Preprint Limitations of Second Quantization Notation for Description ...

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Thank you for the post, Prof. Stanislav Dolgopolov, and I apologize for my late reply, somehow I missed your answer for 2 months.

I will be back as soon as I read and undestant the reference. Thank you.

Dr. Stanislav Dolgopolov, I read two of your publications, the physical picture of the pairing in those is a mean-field approach.

What I cannot figure out is from what kind of ground state, the electrons can be classified into two categories?

Dissipative and not dissipative, both pairs come from nearby the Fermi level?

Thank you.

Dear Prof. Contreras,

thank you for the message and questions.

The ground states of SC electrons are permanent singlets, condensed as bosons into a BEC. The bosons are non-dissipative below the condensation temperature TBEC. Above TBEC bosons can occupy higher states in the bosonic spectrum, so an energy dissipation is possible. Dissipative and not dissipative, both pairs emerge from the Fermi surface of normal (single) electrons. However, the paired electron have a higher kinetic energy than the kinetic energy of normal electrons on the Fermi surface, whereas the total energy of every paired electron is lower than the total energy of normal electrons on the Fermi surface. It is possible because the paired state have a lower potential than the single state. The same case we see in the model "helium atoms in the free electron sea"; indeed, every electron of helium is in a deep potential well with a large kinetic energy, so the singlet state in helium is favorable and stable. In superconductors the potential well is not very deep, but the principle is the same.

Would be happy to answer further questions. Stanislav

Thank you for the clear and neat answer, Prof. Stanislav Dolgopolov.

Now I understand, yes, the paired electrons are stable, therefore, have lower energy for the bounded pair to form. Now is the TBEC the same transition superconducting Tc?

Some time ago answering a question of non-equilibrium superconductors, I kept from a Russian Thesis in 1D nonequilibrium superconductors (I cannot remember the reference), the following figure which I guess reflects somehow what you explained there can be uncompensated e-h pairs.

I agree with your reasoning for the helium model.

Dear Prof. Contreras,

thank you for the answer.

TBEC is indeed related with Tc . We should also take into account the pairing temperature T*. In Fermi metals BEC occurs simultaneously with the pairing, at T*, because the pair density is sufficient for BEC; so Tc=T*=TBEC.

In strongly correlated systems (for instance, in underdoped cuprates) the density of singlet electrons on Fermi surface may be insufficient for BEC at T*, so BEC occurs at a lower T, that is TBEC=Tc

Thank you, Prof. Stanislav Dolgopolov, yes in BCS superconductors, they both are the same, Tc and TBEC, somehow if I remember my course in superconductivity (we used D Gennes and LifPit IX Vol books) the resistivity is zero, so we do have 2e charged particles (a supercurrent), and somehow this is important. Superconductivity is a different phenomenon from BEC. Their ground states are totally different.

You see, these classical books did separate well the two phenomena (also Abrikosov book on metals). Something that I guess, became forgotten with High-Tc compounds. And I guess it has been due to the persistence to fit with one theory, the phase diagram of High Tc.

Thank you for the nice discussion.

Dear Prof. Contreras,

many thanks for your helpful discussion.

Now is better understandable, what is going on (in the physics science).

Stanislav

My pleasure, Dear Prof. Stanislav Dolgopolov.

Thank you also for raising several important issues regarding models to describe a quantum electron description in metals following the second quantization:

First, implications of local and no local states in crystals and its possible implications for the second quantization picture.

Second, certain ambiguity between the two spaces, the Fourier k and the real r spaces.

Thank you also for the example with superconductors.

Dear Prof. Contreras,

Unfortunately, very few people know, that they know very few.

For further discussions you can use my mail [email protected]

Best, Stanislav

Thank you so much, Prof. Stanislav Dolgopolov. I will.