7
$\begingroup$

I understand that Cooper pair in low-temperature superconductivity are formed by electron-phonon interaction. Normally one then assumes that electrons of opposite momentum and spin are paired. This is what i dont understand. In my understanding two electrons with opposite momentum should seperate and thus not feel an attractive force due to electron-phonon interaction.

$\endgroup$
  • $\begingroup$ I can't recall where, but I believe that this has to do with an electron following the trail left behind by the other in the pair (and somehow, the optimal momentum pairing for this interaction is oppositely directed but equal in magnitude). I'm curious to see this formally proved by others, and how this idea of one electron following the trail of the other, for lack of better words, follows, as an intuitive explanation. $\endgroup$ – eqb May 7 '14 at 12:44
  • $\begingroup$ Electron pairs do separate. Cooper pairs have short lifetime (I can't really recall values but as far as I remember it's on the order of tens of microseconds), but they are continuously created. $\endgroup$ – Jarosław Komar May 8 '14 at 12:25
  • 1
    $\begingroup$ Try not to think about it as a dynamical process (even though that's what we humans tend to do all the time). The Cooper instability does not couple opposite momentum states dynamically, but only when one try to do a pertubation calculation starting from the Fermi sea. The real SC state has nothing to do with the initial naive state, compare the Fer,i gas wavefunction to that of the BCS ansatz ! The dynamical process of the instability is of course an interesting problem, which is a somewhat active domain of research nowadays. $\endgroup$ – Adam May 8 '14 at 13:27
7
$\begingroup$

It is indeed a counterintuitive fact. Let us go slowly through the argument.

You start with a degenerate gaz of fermions. They are piled-up in the momentum space, up to the so-called Fermi level. It means that the last electrons entering the game have the Fermi energy, which is pretty huge ($\sim 10\;000$ Kelvin). Also, these electrons are the only accessible ones, since the ones deep inside the Fermi sea are frozen up by the other ones.

On top of this stable situation (this is called a metal after all), you allow coupling of two-electrons through phonon exchange. This happens locally in space, since this is the simpler hypothesis. This is the usual naive picture: an electron creates a vibration of the lattice (i.e. the electron releases a phonon) which perturbs the next electron flowing (i.e. the next electron catches up the phonon), but this picture is confusing and you should not bother yourself with I believe.

Let us come back to our two-electrons interaction mediated locally by a phonon. It turns out that the interaction is slightly attractive for some parameters. You can verify that this small attraction leads to a huge effect : two electrons which were previously at the Fermi level collapse (so to say) to the zero-energy state (picturesquely said, they lost roughly $10\;000$ Kelvin in the process, but this also is a naive picture). This is called the Cooper instability of the Fermi sea. In fact, the electrons do not collapse, they bound to each other, like in a covalent bound if you wish. And so the kinetic energy of the bound state is zero, although the two electrons forming the bound state were previously at the Fermi energy... so how to reconcile these two ideas ? Well, you say that the former electrons were in states $k_{F}$ and $-k_{F}$, the Fermi momentum, such that the bound state has momentum $k_{F}+(-k_{F})=0$ is at rest ! So you indeed couple two electrons with opposite momenta into a bound state with no momentum.

Since the interaction is local in space, the bounding can only happens for singlet pairing: the former independent electrons were in opposite spin states, thanks to Pauli's principle.

Now, it is customary to call this state a Cooper pair, and to note it as $\left|k_{F},\uparrow;-k_{F},\downarrow\right\rangle $.

So far so good for two electrons and one phonon, what about the condensate ? You can verify that the condensate of paired fermions minimised the energy of the system of interacting electrons. Bardeen, Cooper and Schrieffer first demonstrated it, and so is the BCS explanation for superconductivity.

So to conclude: yes the two electrons forming the Cooper pair have opposite momenta (and spin, but I think your question was mostly about the momentum) but the resulting momentum of the bound state is still zero. The trick is that a Cooper pair is a novel object, which has not so much to do with the properties of the two electrons it's made with, and that's why the condensate is at rest of course ! It's even better to not see a Cooper pair as the pairing of 2 electrons at all, since it is a new state. Say differently, the Cooper instability absorbs two particles (picturesquely called electrons) at momentum $\pm k_{F}$ and with opposite spins and create a new particle (picturesquely coined Cooper pair) with zero momentum and zero spin. Then spin and momentum are conserved (hourra !). To conserve the charge, the new particle (the Cooper pair) must have twice the charge of the former ones (the electrons).

$\endgroup$
1
$\begingroup$

As best I recall (what, me look in Wikipedia? :-) ), The point is that superconductivity and superfluids only can occur for boson-ic particles (thanks, Pauli!). By pairing off electrons, or even $He^3$ atoms, the paired items behave as though they are bosons, thus allowing state degeneracy and all that.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.