Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cooper pairs - 2nd quantization

  1. Oct 15, 2008 #1
    Hi all,

    I am looking at (elementary) theory of superconductivity. In particular, I am looking at the calculation showing that a (however small) attractive interaction makes the Fermi sea unstable.

    Kittel's "Introduction to solid state physics" (7 ed) sketches this calculation in Appendix H. I'm more or less happy of Kittel's version, which seems to follow quite closely the original derivation by Cooper.

    The same subject is treated in chapter 9 of Plischke and Bergersen's "Equilibrium Statistical Physics", although in 2nd quantization.
    I found a couple of errors in their derivation, which however happen to cancel out to produce the same result as in Kittel (except perhaps for a qualitatively irrelevant 1/2 factor).

    I can follow most of their calculations, but there's a point I am not really getting.
    The eigenvalue equation on their trial state produces the following equation

    0 = [E- 2 \epsilon(\mathbf{k})] \alpha_{\mathbf{k}} + v \sum_{\mathbf{q}}\alpha_{\mathbf{k}+\mathbf{q}},\qquad \epsilon_F \leq \epsilon(\mathbf{q}) \leq \epsilon_F + \hbar \omega_D

    where [tex]\epsilon(\mathbf{q})[/tex] is the free particle energy and the k's are outside the Fermi sphere (actually the book has a minus in front of v and the 2 in front of [tex]\epsilon[/tex] is missing).

    Now the key point is to introduce a "constant" that allows to solve for the energy E. The book takes the continuum limit and sets

    \sum_{\mathbf{q}}\alpha_{\mathbf{k}+\mathbf{q}} = \int_0^{\hbar \omega_D} d \epsilon \rho(\epsilon) \alpha(\epsilon) = \Lambda

    and here comes my problem. Is this exact or there's an underlying, undiscussed assumption?
    Actually I can see two assumptions, the second of which I find rather disturbing

    1) the coefficients [tex]\alpha_{\mathbf{k}}[/tex] actually depend only on the energy (scalar) and not on the momentum (vector).
    2) [tex]\Lambda[/tex] does not depend on [tex]\mathbf{k}[/tex].

    Accepting the above equation one gets

    \alpha(\epsilon) = \frac{v \Lambda}{2 \epsilon(\mathbf{k})-E}

    which gives

    1 = v \int_0^{\hbar \omega_D} d \epsilon \frac{\rho(\epsilon)}{[2 \epsilon(\mathbf{k})-E}

    Solving this for E gives basically the final result as in Kittel.

    But, as I say, the point 2 above puzzles me a lot. Is this a further assumption? What is its justification?
    Is this some kind of truncated self-consistence?

    The derivation by Kittel (Cooper) seems to be immune from this point, but I suspect it might be hidden in the assumption made about the matrix element.

    Can someone give me an hint?

    PS I have 2nd edition of Plischke's book. I'm trying to get hold of the 3rd edition, which could contain a more detailed discussion or, at least, a smaller number of errors...
  2. jcsd
  3. Oct 15, 2008 #2
    apologies, my browser seems to have some problem with this forum...
    Ignore this reply (but not the problem above!).
  4. Oct 15, 2008 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    I had a bit of a tough time trying to follow the notation and where exactly in the variational method you are at. The only think that I can guess is that your [itex]\Lambda[/itex] is actually the energy gap.

    If that is the case, then to address #2, this is due to the fact that the s-wave gap in BCS superconductors are isotropic, i.e. no dependence on k, unlike the d-wave gap in high-Tc superconductors. So such an assumption is perfectly valid for BCS superconductors.

    If you want a good reference text, I strongly recommend Tinkham's classic "Intro to Superconductivity". He covers both the variational method derivation AND the field-theoretic method to get the BCS theory.

  5. Oct 15, 2008 #4


    User Avatar
    Homework Helper

    If the sum is over all \vec q, then Lambda does *not* depend on \vec k because I can simply change dummy variables from q to p=q+k.
    \sum_{\mathbf{q}}f(\mathbf{q}+\mathbf{k})=\sum_{\mathbf{p}}f(\mathbf{p})\equiv \Lambda\;.

    So, then it is easy to see that only depends on the energy.
  6. Oct 15, 2008 #5


    User Avatar
    Homework Helper

    Sigh... the forums are just not letting me preview posts. The last line above is supposed to read:

    "...that alpha only depends on energy".
  7. Oct 16, 2008 #6
    Hi ZapperZ,

    and thanks for replying!

    Sorry about the notation, I was following Plischke's, but I had not enough time to
    copy the whole derivation, so I just mentioned the main points.
    I'm attaching the relevant book section, which fits in a single page.

    No, that's not the energy gap. It is a constant that appears in most of the derivations I've seen. It is --- or it bears strict relation to --- what Kittel calls C in appendix H.

    Thanks, I'll look up into that.


    Attached Files:

  8. Oct 16, 2008 #7
    Hi olgranpappy,

    and thanks for your suggestion.

    Unfortunately the point is exactly that the sum is not on all [tex]\vec q[/tex], otherwise
    my "problem" would have been trivial, as you suggest.
    I've tried to specify the allowed values of [tex]\vec q[/tex] besides the first equation:
    the integration domain is a thin shell outside the Fermi sphere.
    Again, apologies for the "condensed" notation, but I thought that this particular constraint was quite standard, since I've seen it in every derivation I've found.

  9. Oct 16, 2008 #8
    On second thoughts, I think that olgranpappy's reply is correct, although the "dumminess" of
    k does not trivially ensue from the fact that the sum includes all q's.

    As far as I understand, the point is that the sum in Eq. (9.6) of the attached excerpt includes all q's such that q+k lies in a shell surrounding the Fermi sphere. In this case, I see that [tex]\alpha_{k+q}[/tex] can be safely changed into [tex]\alpha_{q}[/tex].

    I was mislead by the way Plischke put the condition on v(q). He says

    and I got that as a constraint on q. If that was the case, k would not be a dummy index.

    However now I'm almost sure he means that the constraint applies to all of the matrix elements involved
    in the derivation of (9.6). This is obtained by applying (E-H) to

    |\psi\rangle = \sum_{\mathbf{k}'} \alpha_{\mathbf{k}'} |\mathbf{k}'\rangle

    and subsequently projecting onto [tex]\langle \mathbf{k}|[/tex], where I introduced the

    |\mathbf{k}\rangle = c^\dag_{\mathbf{k} \uparrow} c^\dag_{-\mathbf{k} \downarrow} |F\rangle

    Working out all of the anticommutation one finds that if the interacting potential is not vanishing
    only when the above kets meet the constraint, then the sum in 9.6 is actually on all q's such that q+k lies in a shell surrounding the Fermi sphere. Then the constant [tex]Lambda[/tex]
    actually does not depend on k and the following calculations make sense, except for the errors which,
    somewhat luckily, basically cancel out. It almost seems like Plischke vaguely recalled the calculation and
    handwavingly managed to obtain the correct result. I think that this section is in need of some rewriting.

    Thanks a lot for your help, and apologies for having been a bit sloppy.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Cooper pairs - 2nd quantization
  1. Cooper Pairing (Replies: 24)

  2. Cooper pairs (Replies: 2)