ICTP2012: Innovations in Strongly Correlated Electronic Systems

Theory of magnetic structures in layered iridates; band or Mott
by Hae-Young Kee (Toronto Univ.)
blogged by Ryotaro Arita

Hae-Young starts with the periodic table, where she notes that the spin-orbit coupling of 5d elements is stronger  than that of 3d or 4d elements. In this talk, she will focus on the so-called Ruddelsden-Propper series of Ir oxides.

Then she moves to the crystal structure of Sr_{n+1}Ir_{n}O_{3n+1}. Here, she mentions that Sr2IrO4
is iso-structural to the mother compound of high T_c cuprate, La2CuO4 or Sr2RuO4. She shows the transport (resistivity as a function of temperature) data which indicates that the system is insulating for n=1 and 2, but bad metallic for n=∞. As for the magnetic property, she mentions that the ground state of Sr214 is canted AF.

Now her main subject is the "origin of magnetism and metal-insulator transition".

To address this question, she start with the level diagram. Here we see that the combination of crystal field splitting and strong spin-orbit coupling forms J_eff=1/2 and 3/2 band. Since the number of d electrons in each Ir atom is five, the J_eff=1/2 band is half-filling. If we follow the band theory, the system is a metal. On the other hand, if the electron correlation is strong enough, then the system should become a Mott insulator, like La2CuO4.

From the X-ray absorption spectra for Sr2IrO4, (by looking at the ratio between L3 and L2), we can say that the spin-orbit coupling is indeed strong, and the low-energy states around the Fermi level are formed by J_eff=1/2 states. The optical conductivity tells us that the system is an insulator, so that experimentally, it is suggested that Sr2IrO4 is a spin-orbit driven Mott insulator.

Given this situation, she raises the following two questions:
(1) For Sr214 and Sr327: Are they spin-orbit band or Mott insulator ?
(2) For SrIrO3: Is it semimetal and topological insulator?
She is going to answer these questions by means of a microscopic calculation.

First, she explains the detail of the effective model for Sr214. The perovskite octahedra in the unit cell have distortion and rotation, so that the unit cell contains 4 Ir layers, each layer has 4 Ir atoms. Considering this complexity, the one-body part of the Hamiltonian was derived by the Slater-Koster approach. The band dispersion of Sr2IrO4 is shown, which captures the essential feature of the ab initio calculations. Next she discusses what happens if the Hubbard U is introduced. She shows a phase diagram, where Uc (the critical U for metal-insulator transition) is about 1, but U for Sr2IrO4 is about 1.8. She comments that the magnetic pattern of the antiferromagnetic phase is canted AF (in the plane).

On the other hand, for n=2, she emphasizes that bilayer hopping is as strong as that of in-plane, and the band dispersion of the one-body part of the effective Hamiltonian already has a gap around the Fermi level. This is in high contrast with Sr214, and Sr327 has a possibility of being a band insulator. Indeed, according to her phase diagram,  Uc sits on the border between spin-orbit band insulator and AF insulator, so that she concludes that Sr3IrO7 is not a Mott insulator. This is the answer to the first question.

She also demonstrates that the magnetic pattern of Sr214 (canted AF) and Sr327 (collinear AF) can be understood in terms of the effective spin model (derived by the 1/U expansion for the multi-orbital Hubbard model). The spin model contains the Heisenberg isotropic exchange, DM, anisotropic exchange terms.

Q: How large is the spin dependent hopping ?
A: It depends on the distortion in the crystal structure.

Then she moves to the second question on SrIrO3. If we look at the crystal structure, we find that it has Pbnm lattice symmetry, i.e., there are rotation and tilting of the perovskite octahedra in the unit cell. To describe this complexity, she needs to introduce 3 Pauli matrices in the effective model.

She shows the result of LDA+SOC+U. Around U point in the Brillouin zone, the four J_eff=1/2 bands form two interpenetrated pairs of cones. These cones form two Dirac-like points and a circular line node at the Fermi energy. The density of states around the Fermi level is very small, and we need large Hubbard U to make the system a magnetic insulator. She shows a phase diagram against SOC and U, for which she emphasizes that SOC stabilizes the semimetal phase.

Lastly she proposes that if we make a superlattice of Sr2IrRhO6, then we will have phase transitions between nodal semimetal -> strong topological insulator -> band insulator.

Q: Z2 topological invariant calculated ?
A: Yes, by looking at the parity of wave functions

Q: Why Sr214 is not SC but Sr2RuO4 is SC?
A: SO is large for Sr214, while Sr2RuO4 has weaker SOC