It shows how as one increases the number of deuterium atoms (replacing H) on the BEDT-TTF molecules (shown below) one gradually moves from the superconducting (S.C.) phase into the antiferromagnetic Mott insulator (A.F.I) phase.
This transition is analogous to what happens if one decreases the pressure or replaces the Br with Cl, i.e. it is generally associated with a change in the intermolecular spacings. Theory suggests this corresponds to increasing U/W [ratio of the Hubbard U to the bandwidth W]. It may also be due to a decrease in electronic frustration.
First, this must be a quantum nuclear effect (e.g. due to zero point motion) because the chemical forces [potential energy surfaces] for H and D are identical. This is the Born-Oppenheimer approximation.
Second, the relevant molecular orbitals are centred [i.e. have the dominant electronic density] on the S and C atoms and not the H atoms.
Third, one can expect some sort of geometric isotope effect where there are small changes in bond lengths and crystal lattice parameters with H to D substitution. However, generally these effects are very small, particularly for systems in which hydrogen bonding plays a small role. Here the H bonding is with the anion. Thus it is hard to see how these small geometric changes might produce large enough changes in the band structure parameters t and t' in the relevant Hubbard model.
Further insight might be gained by doing DFT-based calculations of the parameters t and t' for the geometries of the H and D crystals determined by x-ray diffraction.
One also needs to understand how these structures change as a function of temperature. It is quite possible that the effects of H/D substitution are amplified at low temperatures. See: http://arxiv.org/abs/1111.4999
ReplyDeleteIf one believes that the parameters of the Hamiltonian can fluctuate by 5-10% due to subtle structural changes on cooling, then it is not too hard to imagine H/D substitution could have a large effect.
A systematic study would be very interesting!
Could this not be explained by the changing symmetry of the vibronic (phonon) modes, due to changing the masses in the normal modes? By breaking the symmetry in the mass distribution of the molecule I think you would be changing the symmetry of the vibronic modes vs. the electronic wavefunction, and causing degenerate states vibronic states to separate off as the symmetry is lowered.
ReplyDeleteThe fully deuterated molecule should possess the same symmetries of vibrations, but with some frequencies rescaled by up to half.
The 'isotope effect' is enormously magnified, for normal modes with small reduced masses (i.e. lots of hydrogen motion).
I would imagine that both these effects could quite significantly change the magnitude of the electron-phonon coupling.
Do you know of anyone actually trying the systematic study? Were the H and D crystal structures actually solved from XRD?