How do we know that superconductivity is due to Cooper pairs as opposed to pairs of more than 2 electrons?

Question

I am wondering how do we know that superconductivity is due to Cooper pairs? I.e. due to pairing between 2 electrons, as opposed to say, pairs of 6 electrons.

If a pair of electrons consists of 6 electrons, with momentum -p, p, p and -p, -p, p, we fall back to the case of 2 fermions with momentum p and -p. Which is similar to the case of Cooper pairs, which are also made of fermions with opposite momentum. In both cases the pair is a boson with null momentum. However it is well accepted that only Cooper pairs made of 2 electrons are responsible for superconductivity. How can we rule out the pair of 6 electrons (and possible more)?

I understand that the probability of such pairs to form would be much smaller than a pairing of 2 electrons, but is it really 0? If it isn't 0, why aren't such pairs contributing to superconductivity?

Answer 1

This is an important issue as applied to nuclear matter, because 2-nucleon pairing can also be viewed as 6-quark condensation, and some light nuclei can be modeled as aggregates of alpha particles. I’ll try to outline the theoretical reason why 2-body pairing is favored over 4 or 6-body condensation, but I’ll leave discussion of experimental evidence to others.

Standard 2-body pairing with overall momentum $\mathbf{p}-\mathbf{p}=0$ is favored by the roughly constant density of states near the Fermi surface, which allows arbitrarily weak attractions to induce pairing. In 6-body aggregation, the momenta do not divide up rigidly as $(\mathbf{p}+\mathbf{p}-\mathbf{p})+(-\mathbf{p}-\mathbf{p}+\mathbf{p})=0$, which represents a tiny corner of momentum space. They could divide up just anyhow, subject to $\sum{{{\mathbf{p}}_{i}}}=0$, and the density of states would vanish as some power of excess energy, i.e., total kinetic energy minus $6{{E}_{F}}$. The states closest to the Fermi surface are particularly important because of the energy denominator of the gap equation.

The basic difference between Cooper pairing and perturbative correlations is evident from the joint occupancy function of two momenta, i.e., the probability that both are occupied. Absent interactions, all states up to the Fermi level would be filled, with the trivial consequence of perfect correlation between opposite momenta. Given interactions, perturbative correlations would fuzz-up the Fermi surface, whereas Cooper pairing would manifest as a delta-function spike along $\mathbf{p}+\mathbf{q}=0$. Perturbative effects would still add fuzz, but not broaden the spike. Perturbative 6-body correlations would contribute to the fuzz, not to the spike.

(I have used the term delta-function loosely, to describe narrowness, not height, since probabilities cannot exceed one. This is a peculiarity of a finite system with discrete states. If you were to fix the momentum of the spin-up fermion, you would get a distribution of spin-down fermions, which would tend to a delta function ${{\delta }^{3}}(\mathbf{p}+\mathbf{q})$ in addition to a continuous function in the continuum limit of an infinite system.)

Answer 2

Ultimately, this is obviously an empirical question. In weak coupling only two particle states condense, but in strong coupling a variety of condensates are possible. In some cases (involving composite fermions or bosons), it is also a question of nomenclature. Do we view the condensation of neutron pairs in dilute nuclear matter as pair condensation (nn), or as six-quark condensation ($(qqq)(qqq)$). We tend to think of it as pair condensation, because the scale of neutron pairing (1-2 MeV) is well separated from the QCD confinement scale, but at higher density this separation may not be so clear.

A nice formal explanation was provided using renormalization group methods by Polchinski and Shankar. These authors write down the most general effective action for fermions in the vicinity of a Fermi surface, and then study the scaling of operators as the momenta are scaled towards $k_F$. They find that six, eight, etc fermion operators are formally irrelevant, which is why there is no six fermion pairing in weak coupling. The basic reason is the one mentioned by Bert: Near a Fermi surface there is less and less phase space for $n$-body operators.

Most four-fermion operators are irrelevant, with the exception of those that describe the BCS (back-to-back) and zero sound (forward scattering) channels. These are marginal. The marginal zero sound operators describe a possible Fermi liquid theory fixed point, parametrized by Landau Fermi liquid parameters. At one-loop the BCS channel becomes marginally relevant (logarithmically growing) for attractive interactions, and marginally irrelevant for repulsive interactions. The logarithmic growth of the BCS interaction leads to a Landau pole, signalling the breakdown of Landau theory, at the BCS scale $\sim \exp(-1/g^2)$. There are some subtlelties if the interaction is not short range (in the presence of gauge fields) that were studied by Polchinski and others. These do not change the conclusion that in weak coupling the BCS channel dominates (but they modify Fermi liquid theory).

Answer 3

I understand that the 2 electron "pairing" is just more probable but groups of larger number of electrons are theoretically allowed. Multiple-electron (four) grouping has been considered for cuprate superconductors in the 90s but that idea lost momentum since.

Most of all, we should bear in mind that in physics we (try to) explain natural phenomena through abstract concepts. This means "when in doubt measure it!!!" The data has the final word. There are many papers showing that the boson responsible for condensation in most (all?) regular superconductors has a charge of 2e and the mass of two electrons.

I like one particular paper that shows that Cooper pairs are truly "pairs" (i.e. two) [Lafarge et al. Nature 365, 422 (1993)]. This paper shows that when an electron comes into a metal-superconductor junction it has to "wait" for a second electron to arrive before getting into the superconductor. In other words, carriers can only exist in a superconductor in multiples of 2 electrons.

Answer 4

On the experimental side, one experiment that confirms pairs of electrons is flux quantization. The magnetic flux, $\Phi$ through a superconducting ring is quantized in units of: $$\frac{2\pi\hbar c}{q}$$ Experimentally, q is measured to be 2e. See Kittel's discussion of flux quantization.