Why is high temperature superconductivity so hard to solve?

Question

The phenomenon of high temperature superconductivity has been known for decades, particularly layered cuprate superconductors. We know the precise lattice structure of the materials. We know the band theory of electrons and how electronic orbitals mix. But yet, theoreticians still haven't solved high Tc superconductivity yet. What is the obstacle to solving it? What are we missing?

Answer 1

One problem is that band theory isn't everything! Crucially, band theory completely neglects the interactions between electrons. The fact that often one can do this and obtain near correct results is actually amazing, and worth several lecture courses to flesh out the reasons. However, it cannot always be correct. In many materials the electron-electron interaction dominates --- a good example is the so-called Mott insulator, where by band structure calculations you would think you get a half-filled band and so a conductor, but because the electrons repel each other so strongly you actually get a grid-locked lattice of electrons which cannot move, because moving any of them would put two electrons on top of each other! The cuprates are known to be Mott insulating when they are undoped; this is good evidence that interactions are very important. Unfortunately, without the massively simplifying assumption that electrons are independent (i.e. non-interacting) it is an almost intractable problem to describe their behaviour; indeed, we know from other strongly interacting systems such as fractional quantum hall systems that it's possible to end up with no electrons at all, but fractions of them --- the possibilities for novel electronic structure are really unimaginable.

The 2nd problem which has plagued the field is more technical, which simply that the materials don't behave in universal ways! Although we can point to many superficially similar aspects to many cuprates, it's actually not the case that quantitatively they are the same. For instance, the fabled "linear scaling" is actually incredibly hard to really get --- it's very sensitive on impurities, precise doping levels, etc. The flip-side to this is that if we just look at the qualitative features and ask "what theories predict these?" we actually have quite a few --- marginal Fermi liquid theories, quantum critical theories, strongly coupled gauge theories, Gutzwiller projection theory, etc. All of these will give a superconducting dome, with conducting behaviour at high doping, insulating at low, and some form of anomolous transport. However, experimental signatures are actually very hard to pin down without controversy about what really has been measured, so the debate continues.

In addition, the historically long argument has created some unpleasant sociology; some (many?) would claim that actually things are pretty settled, and that their favourite theory is clearly superior. This hasn't helped a consensus to form.

Answer 2

In perturbation theory, we split the Hamiltonian into a "free" part, and an "interacting" part, with the magnitude of the interacting part being much smaller than that of the free part. If there are no degeneracies in the free Hamiltonian — at least for the ground state which we're interested in — the analysis is relatively straightforward. If we have a mild degeneracy, we can first project the interacting Hamiltonian to the free ground eigenspace, diagonalize it, and work with the diagonalized basis. But what if the ground state is highly degenerate?

An example would be the fractional quantum Hall effect. In the presence of a uniform background magnetic field, the kinetic part of the Hamiltonian — i.e the free part — decomposes into Landau levels, with an infinite degeneracy for the lowest Landau level. With a partially filled lowest Landau level and the Pauli exclusion principle for electrons, we have an extremely large ground eigenspace for the free Hamiltonian. The interaction Hamiltonian is due to the Coulomb repulsion between electrons, which is much weaker. However, diagonalizing the projection of the interaction Hamiltonian onto the lowest Landau level is extremely difficult.

Now, let's get on to the more complicated case of a doped Mott insulator. Even with the fermion hopping terms neglected, the dynamics are more complicated. For a slight doping of a Mott insulator, we still get a huge degeneracy.

Answer 3

For strongly correlated systems — where perturbation theory and mean field theory breaks down — even if you know the exact form of the Hamiltonian, and regulate it over a lattice, it can still be extremely difficult to compute the ground state, or even figure out its qualitative behavior.

Being a quantum system, we can't simulate it using a classical computer. You need a quantum computer for that.