# What is BCS approximation?

1. Jan 27, 2006

### Neitrino

Dear PF,

Would you please be so kind and help me with one question?

Ive put my question in attached word file since my LATEX does no show me the formulas I type.

Would you pls have a glance on my question and give any feedback?

Thanks a lot
Georeg

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2. Jan 29, 2006

### Neitrino

One question pls

Since there is no any reply Im very shame, seems I posted silly question here, but could anyone advise me when it is posssible to substitute product of fields with their propagator: (psi_alpha)(psi_beta) ->
<0|T(psi_alpha)(psi_beta)|0>. I ask this question because there was one equation of motion (in PDF attachment) where the product of fermion fileds was substituted with propagator of these fileds?

Thanks and sorry if one thinks what a stupid questions these are.

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3. Jan 30, 2006

### nbo10

This isn't a silly questions it's a difficult question, probaly only a few people on here could answer. I know BCS but I don't know QFT and Greens functions that well.

THe BCS approximation is that there are many copper pairs in the groundstate and the pair operator has a finite expectation value, and it's fluctuations are small.

4. Feb 11, 2006

### Physics Monkey

In the BCS approximation and similar schemes one replaces operators by their average values. In the case of the BCS theory the goal is to approximate the four fermion term that appears in the Hamiltonian with the best two fermion term you can come up with. You end up with a well known self consistency condition for the so called superconducting gap which is really just the average value of a sum of pairs of fermion operators. Schematically, you have something like $$c^+_1 c^+_2 c_3 c_4 \rightarrow c^+_1 \langle c^+_2 c_3 \rangle c_4$$ going on with similar terms for other possible pairings. Until the advent of BCS theory people had always paired a creation operator with a destruction operator when approximating the four fermion term with average values. If you know something about band structure then it isn't hard to convince yourself that the pairing of a creation operator and a destruction operator really represents some kind of Hartree-Fock scheme for calculating the band structure. Part of the great insight of BCS theory was realizing that terms like $$\langle c^+_1 c^+_2 \rangle c_3 c_4$$, which don't conserve particle number, can also be nonzero. Diagonalizing the new approximate Hamiltonian leads immediately to all the wonderful results of the BCS theory. So, if I may try to answer your question at last, clearly what you're doing is replacing an operator by its average value. The motivations for this replacement are varied. As was noted above, you might expect the average value to be large compared to the fluctuations. In the case of BCS theory, you have an intractable Hamiltonian which you would like to approximate in a self consistent way. By replacing pairs of operators with their average value your turn a four fermion interaction term into a two fermion term which can be diagonalized.

If this doesn't answer your question at all, please let me know and I can try to be more specific.

Last edited: Feb 11, 2006
5. Feb 15, 2006

### Neitrino

Thanks for your reply I'll go thro it & hope it will help me.

George

6. Feb 23, 2006

### akhmeteli

As far as quantum statistical physics is concerned, the situation is as follows.
The Bardeen - Cooper - Schrieffer (BCS) theory of superconductivity uses a Hamilonian with a four-fermion interaction term. To replace this Hamiltonian with an approximating one, which only includes products of two fermion operators (and is thus tractable), some products of two fermions (q-numbers) are replaced by c-numbers. Is this a correct procedure? N.N. Bogoliubov rigorously proved that the ground state energies of the BCS Hamiltonian and the approximating Hamiltonian coincide in the thermodynamical limit (i.e. when the number of particles N tends to infinity). Later N.N. Bogolubov Jr. rigorously proved that the free energies of the BCS Hamiltonian and the (appropriately chosen) approximating Hamiltonian coincide in the thermosynamical limit for arbitrary temperatures.
If you have more specific questions, I might be able to answer them.
As for QFT specifics, you should ask people more knowledgeable in QFT than I am.

7. Feb 23, 2006

### reilly

Allow me, please, to say much the same thing as did Physics Monkey, but in a slightly different way.

The hardest part of the BCS theory, at least in my opinion, was showing that electrons near the Fermi Surface attracted rather than repelled each other. They reasonably assumed that the electron-electron potential near the Fermi Surface was constant in momentum space. This means that the potential is given by a huge matrix of 1s, multiplied by -V, the potential (V>0).

Constant throughout momentum space means concentrated in configuration space. -- its like a very strong negative potential centered at X=0.

So, roughly speaking, this hole can suck all the electrons into itself, and creates a bound state, which involves all the electrons. The other solutions, give particle-like states; Cooper Pairs in particular, with positive excitation energies. Further, the ground state has every possible electron state filled, so the only excitations possible are hole-electron pairs, which due to the positive excitation energy (energy gap) behave like 'free particles'" .
Regards,
Reilly Atkinson

8. Feb 24, 2006

### nbo10

Some of the statements in post #7 are confusing and seem to be incorrect. The wording could be the reason that they seem to be incorrect.

9. Feb 24, 2006

### reilly

/

So, lay it on me please. I'm far from being a BCS expert; that I might have misstated things is indeed a possibility,
Thanks and regards,
Reilly Atkinson

10. Feb 24, 2006

### nbo10

I think your general thinking is correct, but details are odd.

(*Start a unimportant detail*).
A constant pairing potential didn't start as a reasonable assumption. It was an approximation that could be solved. And in the end it turned out that is was a good approximation. (*End a unimportant detail*). What is a huge matrix of 1s? I've never come across this terminology.

I have no idea what this means. Pairing is carried out in momentum space, but I'm not sure of your defintion of configuration space and where it comes into play.
What hole? Are you talking about the effective potential? Even so, only a small fraction of the fermi sea is involved in pairing.

(*Other details that aren't that important for the over all disscussion*)

11. Feb 24, 2006

### reilly

Real quick:

A matrix of ones? Just an unnormalized Random Walk transition matrix, in which steps of any length are allowed.

As in:

..11111111111111111111111111111111111111.....
11111111111111111111111111111111111111
11111111111111111111111111111111111111
11111111111111111111111111111111111111

11111111111111111111111111111111111111
11111111111111111111111111111111111111

Configuration space, as usual, is the space of positions, of where stuff is. Heisenberg and Fourier tell us that a function in a very limited domain, say in configuration space, has an extremely large domain in conjugate space, mometrum space in our example.

So in configuration space, the system sees a potential highly localized at X=0. Now, going back to potential wells in beginning QM, it is visually evident that a localized potential will look like a hole. So, the stuff of interest falls into that hole -- I'm providing a metaphorical description of the system's dynamics -- I do apologize for an overly loose use of "hole"..
Regards,
Reilly Atkinson