How to define a spin system of lanthanide ions in easyspin?
Posted: Thu Jun 07, 2018 9:05 am
Hello, I am a new user of easyspin. I met problems in treating lanthanide ions using easypin.
As the document "EPR Reference> Orbital angular momentum and spin-orbit interaction" said to treat lanthanide ions, the soc Hamiltonian usually include higher orders terms (lambda n*(sigma*L*S)^n).
And in the user guide "The spin system", in the soc part an example was given as below:
---
% three electron spins, with linear soc parameter of 170, 250 cm-1 and 0:
Sys.soc = [170; 250; 0]*clight*1e-4;
% as above, plus the first spin has a cubic soc parameter of 10 cm-1:
Sys.soc = [170 0, 10; 250, 0, 0; 0, 0, 0]*clight*1e-4;
------
Is there a mistake that a comma between 170 and 0 is lost? Hence the right format is Sys.soc = [1st order, 2nd order, 3rd order, 4th order, ....], corresponding to lambda n?
If I use this definition with lambda n, the soc of excited states will be also considered and the system will be very large. Is there any method/approximation to accelerate the calculations?
As the document "EPR Reference> Orbital angular momentum and spin-orbit interaction" said to treat lanthanide ions, the soc Hamiltonian usually include higher orders terms (lambda n*(sigma*L*S)^n).
And in the user guide "The spin system", in the soc part an example was given as below:
---
% three electron spins, with linear soc parameter of 170, 250 cm-1 and 0:
Sys.soc = [170; 250; 0]*clight*1e-4;
% as above, plus the first spin has a cubic soc parameter of 10 cm-1:
Sys.soc = [170 0, 10; 250, 0, 0; 0, 0, 0]*clight*1e-4;
------
Is there a mistake that a comma between 170 and 0 is lost? Hence the right format is Sys.soc = [1st order, 2nd order, 3rd order, 4th order, ....], corresponding to lambda n?
If I use this definition with lambda n, the soc of excited states will be also considered and the system will be very large. Is there any method/approximation to accelerate the calculations?