1
Finding Unusually Strong Spin-Orbit Coupling in Post-Perovskite CaIrO 3 via X-ray Magnetic Circular Dichroism Luke G. Marshall 1 Jinguang Cheng 1,2 , Jianshi Zhou 1 , John B. Goodenough 1 , Daniel Haskel 3 , Michel van Veenendaal 3,4 Financial Support From: Special Thanks To: Advanced Photon Source Beam Line 4-ID-D Supported by the US Department of Energy, Office of Science, Office of Basic Energy Science, under contract No. DE- Ac02-06cH11357. 1 Texas Materials Institute, The University of Texas at Austin 2 The Institute of Physics, The Chinese Academy of Sciences 3 Advanced Photon Source, Argonne National Laboratory 4 Department of Physics, Northern Illinois University Motivation: Post-Perovskite (pPv) Perovskite (Pv) pPv has edge sharing octahedra along a, and corner sharing along c: For 5d 5 Ir, we would expect metallic behavior along M-M exchanges in edge-shared direction, but pPv CaIrO 3 shown to be Mott insulator. So where is the energy gap coming from? O 1 O 2 Ir z Octahedral Bonds: Two Short, Four Long 2 X Ir—O 1 : 1.9786(13) Å 4 X Ir—O 2 : 2.0543(13) Å Ref: J.-G. Cheng et al., PRB 83, 064401 (2011) (site axes) xy yz±izx " L•S The octahedral distortion shows that the orbital angular momentum is not quenched. Large spin-orbit coupling (SOC) can win over Ir-O-Ir " bonds. 1.978Ǻ 2.054Å C 134.5° Ir O1 O2 O2 Ref: N. A. Bogdanov et al., PRB 85, 235147 (2012) Ref: B.J. Kim, et al., PRL 101, 076402 (2008) Proposed Model: As SOC is taken into account, t 2g states effectively correspond to L=1 states CaIr 4+ O 3 : LS t 2g 5 e g 0 Initial Conclusions: 1. pPv CaIrO 3 has edge-sharing octahedra along a, and corner-sharing along c. 2. Octahedral site distortion forms 2 short and 4 long Ir-O bonds, which puts the hole in the degenerate yz and zx orbitals. Therefore, the orbital angular momentum is not quenched by the crystal field and SOC splits yz/zx the states. 3. The observation of dominant coupling to the orbital moment in the magnetized state is consistent with an energy gap generated by strong spin-orbit coupling. Additional Ref: http://ssrl.slac.stanford.edu/stohr/xmcd.htm Experimental Techniques – XANES & XMCD: # F 2p 3/2 [4] L + S 2p 1/2 [2] L - S X-ray Absorption Near Edge Spectroscopy (XANES) Fe 3d "* band M=0 Energy Intensity M=0 L 3 L 2 L 3 L 2 1. To explain how XANES and XMCD can be useful tools to probe SOC, we first start with a simple example of Fe-metal (no SOC). 2. Unpolarized photons excite photoelectrons from the 2p states to empty part of 3d band while scanning x-ray energy. Resulting data shows white peaks at lower-energy L 3 edge and higher-energy L 2 . L 3 counts are twice L 2 counts because of degeneracy of 2p 3/2 state compared to 2p 1/2 . There is no magnetic field. X-ray Magnetic Circular Dichroism (XMCD) M≠0 Energy Intensity LH M=0 L 3 L 2 3. A magnetic field applied parallel to the beam. The field splits the 3d band as half of the electronic states are stabilized and the other half destabilized. 4. Left-Handed (LH) circularly polarized x-rays are used. Because of the polarization, the x-rays will only “see” half of the electrons – those that have the same direction as the handedness of the beam. 5. Since there are more states to enter, absorption intensities are enhanced. Energy Intensity L 3 L 2 LH M=0 RH M≠0 6. Then the same is repeated for Right-Handed (RH) circularly polarized light. The absorption intensities show a reduction commiserate with the smaller available bandwidth. Energy Intensity L 3 L 2 LH M=0 RH 2p 3/2 [4] L + S 2p 1/2 [2] L - S 7. It is important now to correct the sign of the L 2 edge. Because of SOC in the 2p orbitals, the 2p 1/2 electrons produce the opposite picture of the 3d band structure as the 2p 3/2 . Energy Intensity L 2 L 3 8. The dichroism signal is the difference between the LH and RH signals. Above is typical XMCD for no SOC. Results and Analysis: Compared to the results for Fe that has no SOC, these results are quite striking: L 3 dichroism is slightly smaller than L 2 , and the sign of the dichroism is the same. Is this evidence of SOC? M≠0 If we draw out our Mott-insulator density of states similar to above we get the picture to the left. In this case, we expect a larger contribution from L from the 2p photoelectrons rather than S, because of the energy gap. XMCD Sum Rules: Exchange Hamiltonian: H exch = αL z + βS z I L3 c − 2I L2 c I L3 c + I L2 c = 4 S z +14 T z 3 L z Branching Ratio: BR = I L3 I L2 = 2 + r ( ) 1 − r ( ) r = L • S n h Modeling XMCD:
