EasySpin forum

Dear developer

I tried to use EasySpin 6.0.0-dev.31 for a dinuclear system in which one center has spin and angular momentum and the second center has only spin.

cm = 100*clight/1e6; % Conversion constant from cm-1 to MHz
Sys.S = [3/2 1/2];
Sys.L = [1 0];
Sys.soc = [392; 0]*cm; 
Sys.g = [2.0; 2.0];
J12=10; %cm-1
Sys.J = J12*cm; 

stev2 = cm*[0 0 100 0 0];
stev4 = cm*[0 0 0 0 1 0 0 0 0];
stev6 = cm*[0 0 0 0 0 0 1 0 0 0 0 0 0];

Sys.CF2 = [stev2];
Sys.CF4 = [stev4];
Sys.CF6 = [stev6];

however, there is error:

Error using sham
Field Sys.CF6 has to have 2 rows since there are 2 orbital angular momenta.

If I change CFx definitions to contain also second row like this:

Sys.CF2 = [stev2; 0 0 0 0 0];
Sys.CF4 = [stev4; 0 0 0 0 0 0 0 0 0];
Sys.CF6 = [stev6; 0 0 0 0 0 0 0 0 0 0 0 0 0];

then there is another error:

Error using sop
First input must contain valid spin quantum numbers (1/2, 1, 3/2, etc).

Error in eeint

Error in sham

So the question is how to properly define a polynuclear system in which only some centers have angular momentum?

Thank you for your kind help in advance

R.

This is currently not working, but I agree this should be possible. We'll fix this.

It's fixed now in 6.0.0-dev.32

Thank you very much for your quick response.
It is working now.

Hello

I have found a discrepancy in meff for Yb3+ ion.

cm = 100*clight/1e6; % Conversion constant from cm-1 to MHz
%Yb3+ LS
Sys.S = [1/2];
Sys.L = [3];
Sys.orf = [1.0];
Sys.soc = [2910]*cm; 
Sys.g = [2.0023];

Sys.CF2 = cm*[0 0 0 0 0];

Opt.Output = 'mueff'; Opt.Units = 'SI';
Exp.Temperature = 300;
Exp.Field = 200;
Exp.CrystalOrientation = []; 
Opt.GridSize = 19;  

mueff =  curry(Sys,Exp,Opt)

%Yb3+ JM
Sys2.S = [7/2];
Sys2.g = [8/7];
mueff2 =  curry(Sys2,Exp,Opt)

mueff calculated for L-S using orbital and spin quantum numbers results in the value of 4.2516.
mueff calculated for JM results in the value of 4.5356

I guess that these two numbers should be the same.
Can you please point me to the mistake in my input?

R.

The reason for this discrepancy was that EasySpin had an unconventional sign choice in the orbital Zeeman interaction term.

It is now fixed in 6.0.0-dev.33. The example below now gives essentially identical mueff for both methods. Note that you need to use a negative Sys.soc to get the J=7/2 multiplet as the ground multiplet.

clear, clc

cm = 100*clight/1e6; % Conversion constant from cm-1 to MHz

%Yb3+ LS
SysLS.S = 1/2;
SysLS.L = 3;
SysLS.orf = 1;
SysLS.g = 2;
SysLS.soc = -2910*cm; 

Opt.Output = 'mueff';
Opt.Units = 'SI';

Exp.Temperature = 300;
Exp.Field = 200;

mueff = curry(SysLS,Exp,Opt)

%Yb3+ JM
SysJ.S = 7/2;
SysJ.g = 8/7;
mueff2 =  curry(SysJ,Exp,Opt)

Dear Stefan

thank you very much for the updates. However, I guess I have found some bug regarding the calculation of magnetic moment - muBM for dimer with L.

Here is input:

clear, clc, clf; close all;
cm = 100*clight/1e6; % Conversion constant from cm-1 to MHz
%Co-V
Sys.S = [3/2 1/2];
Sys.L = [1 0];
Sys.soc = [-180 0]*cm; 
Sys.g = [2.0; 1.95];
J12=10; %cm-1
Sys.J = J12*cm; 

stev2 = cm*[0 0 10 0 0];
stev4 = cm*[0 0 0 0 0 0 0 0 0];
stev6 = cm*[0 0 0 0 0 0 0 0 0 0 0 0 0];

Sys.CF2 = [stev2; 0 0 0 0 0];
Sys.CF4 = [stev4; 0 0 0 0 0 0 0 0 0];
Sys.CF6 = [stev6; 0 0 0 0 0 0 0 0 0 0 0 0 0];


Exp.Temperature = linspace(2,300,100);
Exp.Field = 200;
muBM =  curry(Sys,Exp);
Opt.Output = 'muBM'; Opt.Units = 'SI';
muBM2 =  curry(Sys,Exp,Opt);

figure(1)
plot(Exp.Temperature,muBM,Exp.Temperature,muBM2);


Opt.Output = 'chimolT'; Opt.Units = 'CGS';
chiT =  curry(Sys,Exp,Opt);
figure(2)
plot(Exp.Temperature,chiT);

If you use "chiT" form for curry, the calculations seem working o.k. (Figure 2)
However, when you want to calculate muBM, you get negative values of muBM, which is non-sense. (Figure 1)

I need muBM values to calculate mean susceptibility.

Thank you in advance for your kind help

R.

Can you write down the Hamiltonian for this system that you want EasySpin to use? (You can skip all the Stevens operator terms.) Then we can look which part EasySpin implements differently or incorrectly.

The Hamiltonian should be e.g. for
V(IV)-Co(II) system, where V(IV) has S=1/2 and L=0, octahedral Co(II) is described by Griffith-Figgis Hamiltonian, S=3/2 and L=1.
However, when one L is zero, muMB is negative, as posted before.

If I change system to Co(II)-Co(II) like this

clear, clc, clf; close all;
cm = 100*clight/1e6; % Conversion constant from cm-1 to MHz
%Co-Co
Sys.S = [3/2 3/2];
Sys.L = [1 1];
Sys.soc = [-180; -180]*cm; 
Sys.g = [2.0; 2.0];
J12=-10; %cm-1
Sys.J = J12*cm; 

Exp.Temperature = linspace(2,300,100);
Exp.Field = 200;
muBM =  curry(Sys,Exp);
Opt.Output = 'muBM'; Opt.Units = 'SI';
muBM2 =  curry(Sys,Exp,Opt);

figure(1)
plot(Exp.Temperature,muBM,Exp.Temperature,muBM2);


Opt.Output = 'chimolT'; Opt.Units = 'CGS';
chiT =  curry(Sys,Exp,Opt);
figure(2)
plot(Exp.Temperature,chiT);

then muMB is positive.

So problem is when center has L=0.

R.

Thanks. It would be helpful if you spell out the Hamiltonian that you are expecting, including the Zeeman and SOC term. This way, we can check where EasySpin deviates from this.