Partial Excitation NiO Example

asked by Myron Huzan (2021/02/14 16:55)

I am interested in the replicability of partial excitations, specifically the NiO example (XAS L2,3 partial excitations) which includes a ligand-field description.

From the spectra printed in the examples the spectral features aren't fully replicable to that of the total, i.e eg + t2g != total. Is this a limitation of the transition matrices used for the partial excitations or is something else occurring? This extends to the ligand field description, is there a scaling required for the d8, d9 spectra since the satellite feature of this spectra (~1eV, Fig 3) the d8 satellite is drastically greater than that of the total.

How reliably can this partial excitation be extended to other symmetries, not simply Oh.

Thank you in advance!

Answers

, 2021/02/15 10:13, 2021/02/18 17:02

Dear Myron,

For a given Hamiltonian $H$ and transition operator $T$ the spectrum exciting from state $\psi$ is given as $G(\omega) = \left \langle \psi \left| T^{\dagger} \frac{1}{\omega - H - \mathrm{i} \Gamma/2} T \right| \psi \right \rangle$. The partial excitations are implemented as a separation of the operator $T$. Instead of allowing all excitations we allow either excitations in the $e_g$ or $t_{2g}$ orbitals, or any other separation. With $T=T_1+T_2$ we find $G(\omega) = \left \langle \psi \left| (T_1 + T_2)^{\dagger} \frac{1}{\omega - H - \mathrm{i} \Gamma/2} (T_1 + T_2) \right| \psi \right\rangle$. Writing this out yields 4 functions

$G_{11}(\omega) = \left \langle \psi \left| T_1^{\dagger} \frac{1}{\omega - H - \mathrm{i} \Gamma/2} T_1 \right| \psi \right\rangle$

$G_{22}(\omega) = \left \langle \psi \left| T_2^{\dagger} \frac{1}{\omega - H - \mathrm{i} \Gamma/2} T_2 \right| \psi \right\rangle$

$G_{12}(\omega) = \left \langle \psi \left| T_1^{\dagger} \frac{1}{\omega - H - \mathrm{i} \Gamma/2} T_2 \right| \psi \right\rangle$

$G_{21}(\omega) = \left \langle \psi \left| T_2^{\dagger} \frac{1}{\omega - H - \mathrm{i} \Gamma/2} T_1 \right| \psi \right\rangle$

And the total spectrum is the sum of all four. In other words, the reason that the sum of the diagonal spectra is not equal to the total spectrum is due to the interference terms. As you noted from above the fact that I just talk about operator 1 and 2 allows you to do this for any symmetry.

Best wishes, Maurits

, 2021/02/15 14:48

Dear Maurits,

Thank you for the prompt response and that does make sense, is this currently implementable in Quanty since from what I can tell CreateSpectra uses individual operators and doesn't calculate G_12, G21.

Kind Regards, Myron

, 2021/02/15 15:05, 2021/02/18 17:03

Yes it is implemented. i.e. you can calculate all 4 components in one shot. If you use for the example on NiO in crystal-field and partial spectra (http://www.quanty.org/documentation/tutorials/nio_crystal_field/xas_l23_partial_excitations) use the option {“Tensor”,true}

XASSpectra = CreateSpectra(XASHamiltonian, {TXASxeg,TXASxt2g}, psiList[1], {{"Emin",-10}, {"Emax",20}, {"NE",3000}, {"Gamma",0.1}, {"Tensor",true}})

you should get 4 spectra, representing the 2 by 2 matrix defined above. For an example see the calculation of XAS as a conductivity tensor http://www.quanty.org/documentation/tutorials/nio_crystal_field/xas_l23_as_conductivity_tensor and the documentation of createspextra http://www.quanty.org/documentation/language_reference/functions/createspectra

Maurits

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