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I'm studying Richardson Model in second quantization. There are many initial points that I don't understand:

  • We supposed that an attractive force between 2 electrons exists, due to electron-phonon interaction: an electron generates a phonon that is absorbed by the other electron, and we prove that the resulting potential is attractive. But is this really stronger than the Coulomb repulsive potential? How can one estimate this?

  • Why does it neglect the attractive force between electrons with opposite spin? ($\pm$) They have an opposite magnetic momentum too...

  • Why do we only consider the attractions between electrons with opposite spin?

  • Why do we suppose $k_f ~R ~\ll ~1$ , where $k_f$ is (Fermi momentum)/$\hbar$, and $R$ the interaction distance?

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Do you have a reference for this Richardson model? I don't think I'm familiar with it... Also, this needs some serious clarification; by my reading, your second and third questions are contradictory. –  wsc Feb 15 '11 at 3:44
    
I meant that we considered only actraction between opposite spin electrons, but not for magnetic force dipole-dipole, because the model doesn`t consider them, but only actraction due to phonons...I know a reference online in italian language, I`ll tray to find another in english language... –  Boy Simone Feb 15 '11 at 9:19
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The Richardson model is an exact analytic solution of the BCS system. You can find a nice summary of Richardson's solution on pg. 2 of this paper, which also contains references to Richardson's original papers. –  user346 Feb 15 '11 at 14:23
    
Neat, but very strange... He defines local pair operators (actually people do this all the time in high-Tc, but there we know the pairs are smallish) but real BCS Cooper pairs are extremely large! In fact highly local pairing, while fine in a model context, feels very unphysical in the BCS regime since Coulomb won't be well screened and the attraction is instantaneous... –  wsc Feb 15 '11 at 15:10
    
@space_cadet: thank you –  Boy Simone Feb 15 '11 at 15:15
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1 Answer

up vote 4 down vote accepted

I will just answer the first part of question: is phonon attraction stronger than Coulomb?

Short answer: No.

Longer answer: Nothing (in condensed matter) is ever stronger than the Coulomb force.

Longest answer: There are two aspects to consider. First is the self-screening of the electrons, which will add a mass term to the photons, giving a Yukawa-esque potential ($e^{-mr}/r$). Second, although we always work in momentum space, it's actually unhelpfully abstract in this case. You need to imagine what the force looks like in real space and time. The Coulomb force is a 1/r^2 instaneous repulsion, but it decays very fast (instaneous) in time. The screened Yukawa/Thomas-Fermi force decays less slowly (exponentially with scale $m$) in time. The phonon attraction, however, is a retarded force, meaning it only acts some finite time after the electron has left. Thus if you imagine a plot of the potential as as function of spacetime, there is a singularity at the origin, but there is a small dip after it! In terms of dynamics this is even easier to understand: electrons cause small lattice deformations as they move around, but the lattice deformations need time to be set up and relax again, and this time-scale is larger than the time it takes an electron to get across the distortion; thus another electron sees a small, attractive hole into which it could fall.

Because the Fermi liquid is so unstable (it has too many low energy degrees of freedom) any attractive interaction in any form will push it towards a broken symmetry state, such as superconductivity (or charge density waves, etc.).

So although the phonon attraction isn't "stronger", it can beat Coulomb repulsion. It just has to be clever about it.

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Thanks for the clarificaation. You said "CAN beat Coulomb repulsion". When phonon attraction beats it and when not? We can with a simple model estimate the value of this attraction? –  Boy Simone Feb 15 '11 at 13:09
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@Boy Simone: first principles prediction is hard, if not impossible. The Eliashberg theory of conventional BCS theory will get many things almost right, but essentially depend on knowing what the electron-phonon coupling actually is --- which is next to impossible to measure. As a rule in condensed matter, all parameters are effective: they are all renormalised by whatever you haven't considered. Theories of superconductivity give predictions between parameters, but then relating them to measurements is another layer of complexity. –  genneth Feb 16 '11 at 9:41
    
@Boy Simone: A more-or-less systematic way of attacking the problem is by using renormalization group theory, where starting from a generic Hamiltonian, high-energy modes are integrated out and an effective theory for electrons around the fermi-surface (low-energy regime) is derived. The integration can of course only be done perturbatively, so for weakly interacting fermions. During this process the coupling constants of the theory renormalize. If I remember correctly, the coulomb interaction is an irrelevant perturbation and the constant goes to zero (this is essentially the statement of –  Heidar Mar 1 '11 at 11:25
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fermi liquid theory). The effective attractive electron-electron interaction that originates from electron-phonon interactions, is a relevant perturbation and the coupling constant grows as high energy modes are integrated out. Thus even it is much smaller than the coulomb interactions in the microscopic theory, it is much bigger in the effective low-energy theory. A nice treatment of this is given by Shankar: rmp.aps.org/abstract/RMP/v66/i1/p129_1 –  Heidar Mar 1 '11 at 11:28
    
Should a thread be effectively closed like this by an acceptence of an answer, if the posts says he's only answered some part of the question? –  NiftyKitty95 Feb 4 '12 at 20:52
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