Spin system definition, total angular momentum J

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usrmb1
Newbie
Posts: 4
Joined: Tue Feb 23, 2021 2:51 am

Spin system definition, total angular momentum J

Post by usrmb1 »

Hallo,
I have a question about spin system definition. We declare S and L, for example:
Sys = struct('S', 3/2, 'L', 6, 'g', [gr gw]); % one spin 1 associated S = 3/2 (3 electrons), L = 6.
S and L there are the same, however the total angular momentum J may be different. Two different rare-earth elements:
Nd3+, 4f3, ground state: 4I9/2 (S = 3/2, L = 6, J = L-S = 9/2)
Er3+, 4f11 ground state 4I15/2 (S = 3/2, L = 6, J = L+S = 15/2)
How to distinguish between the two elements in the calculation, for example in procedure resfields(…)?
How do I include the difference in the total angular momentum J in calculations? Does it matter?

Best regards and thanks in advance for your answer and help

Stefan Stoll
EasySpin Creator
Posts: 1106
Joined: Mon Jul 21, 2014 10:11 pm
Location: University of Washington

Re: Spin system definition, total angular momentum J

Post by Stefan Stoll »

If you provide both S and L, then all states for all possible J, from |S-L| to S+L, are included in the simulation. The energy splittings are determined by all the spin Hamiltonian parameters that you provide in the spin system. With a temperature in Exp.Temperature, the intensity of the transitions will depend on the Boltzmann populations. Often, only the lowest-energy J manifold is thermally populated, so that you will get a spectrum from that manifold only.

However, you might also consider simulating systems using effective spin. For example, Er(III) probably gives X-band EPR only between the two lowest states of the 4I_15/2 term. So instead of using (S=3/2,L=6) or using (J=15/2), you should be able to use (S_eff=1/2), with effective spin Hamiltonian parameters taken from literature. Note that this S_eff approach might not work in every case though.

usrmb1
Newbie
Posts: 4
Joined: Tue Feb 23, 2021 2:51 am

Re: Spin system definition, total angular momentum J

Post by usrmb1 »

Thank you for your complete and comprehensive answer. That clarifies the problem.

Kind regards

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