EasySpin forum

Posted: **Thu May 25, 2017 9:29 am**

Dear all,
I am having some troubles in understanding the way ES handle the case of a pair of exchange coupled spin centers with hyperfine coupling. I used the following input to simulate a pair of S=1/2, I=7/2 with isotropic interaction. The hyperfine interaction is internded to be localized on each center, i.e I1A1S1+I2A2S2, with no cross terms (i guess the problem is here). This is the relevant part of the input file:

Code: Select all

clear all
convfact=(clight*100)/1e6;     %
V_MA119.S = [1/2, 1/2];
gpv = [1.98 1.98 1.94];       % g principal values
V_MA119.g = [gpv;gpv];
V_MA119.Nucs = '51V,51V';
V_MA119.A = [0.006 0.006 0.017 0 0 0;0 0 0 0.006 0.006 0.017]*convfact;
V_MA119.J=0;

What I cannot understand is why, by setting J=0, I do not get the same energy pattern of a single S=1/2, I=7/2 with the same hyperfine parameters. This is evident by using levelsplot for two cases, which I would have considered as identical:

Code: Select all

V_single.S = 1/2;
gpv = [1.98 1.98 1.94];       % principal values
V_single.g = gpv;
V_single.Nucs = '51V';
V_single.A = [0.006 0.006 0.017 ]*convfact
levelsplot(V_single,'z',[0 450])
figure
levelsplot(V_MA119,'z',[0 450])

Thanks in advance for any suggestion.

Posted: **Thu May 25, 2017 9:39 am**

This is not related to the hyperfine coupling. Setting the exchange coupling in the two-electron-spin system to zero does not make it a one-electron system. You still have both spins, and each of them has a Zeeman interaction with the external field. With J=0, there is just no zero-field splitting. Here is a minimal example:

Code: Select all

clear

Sys2.S = [1/2, 1/2];
Sys2.g = [2, 2];
Sys2.J = 0;

Sys1.S = 1/2;
Sys1.g = 2;

subplot(2,1,1)
levelsplot(Sys2,'z',[0 450])
subplot(2,1,2)
levelsplot(Sys1,'z',[0 450])

Posted: **Mon May 29, 2017 9:35 am**

Dear Stefan,
many thanks for your indications. I agree with you that two spins S=1/2 with zero interaction are providing a different energy pattern compared to a single spin S=1/2, and the same is true for two S=1/2, I=7/2 compared to a single one. However, the EPR spectrum the two S=1/2 give in your example is the same as for the single S=1/2 (except for an obvious intensity factor). This is following expectations, since the two hamiltonians depict the same physical situation, i.e. independent spins with the same SH parameters. I tested this with a somehow simpler system (S=1/2, I=1/2, isotropic g) and it apparently works.

Code: Select all

clear
Exp.mwFreq=9.45
Exp.Range=[300 400]
Sys2.S = [1/2, 1/2];
Sys2.A=[400 100 25 0 0 0;0 0 0 400 100 25]
Sys2.g = [2, 2];
Sys2.J = 0;
Sys2.Nucs='1H,1H'
Sys2.lw=1
Sys1.S = 1/2;
Sys1.Nucs='1H'
Sys1.A=[400 100 25]
Sys1.g = 2;
Sys1.lw=1
[B2,sim2]=pepper(Sys2,Exp);
[B1,sim1]=pepper(Sys1,Exp);
plot(B2,sim2,'g',B1,sim1*4,'r')

So, what is puzzling me and led me thinking there was something wrong in my input, is the non-coincidence of the EPR spectra in the case I presented before; for the two VO2+centers with J=0, some weak but definite resonances which are not there in the case of a single center appear. Is this simply a matter of low precision? I tried to increase Threshold/nKnots but with no apparent effect (see below).

Code: Select all

clear all
tic
convfact=(clight*100)/1e6;     %da cm-1 a Mhz
V_MA119t11.S = [1/2, 1/2];
gpv = [1.98 1.98 1.94];       % principal values
V_MA119t11.g = [gpv;gpv];
V_MA119t11.Nucs = '51V,51V';
V_MA119t11.A = [0.0062 0.0062 0.0173 0 0 0; 0 0 0 0.0062 0.0062 0.0173]*convfact
V_MA119t11.lw=[0.4 0.4];
V_MA119t11.HStrain=[5. 5. 5.];  
V_MA119t11.ee=0
Exp=struct('mwFreq',9.45,'nPoints',2048,'Range',[260 420],'Temperature',50,'Harmonic',1); %campo in mT

Options.nKnots = 91
Options.Threshold = 0.1



Vsingolo.S=1/2
Vsingolo.g=gpv
Vsingolo.Nucs = '51V';
Vsingolo.A = [0.0062 0.0062 0.0173]*convfact
Vsingolo.lw=[0.4 0.4];
Vsingolo.HStrain=[5. 5. 5.];  

[B,spec1] = pepper(V_MA119t11,Exp,Options);
[B,spec2] = pepper(Vsingolo,Exp,Options);

figure('Name','Spectrum','NumberTitle','off');
plot(B,spec1,'g', B,spec2*2,'r','LineWidth', 1);  
xlim(Exp.Range)
xlabel('magnetic field (mT)');
ylabel('Signal');
toc

Thanks again for your time and patience.

Posted: **Mon May 29, 2017 10:15 pm**

EasySpin has a hard time dealing with the many degeneracies and level crossings in this case. The built-in transition preselection, which is used to narrow down the set of transitions (level pairs) to just those that will give appreciable EPR intensity, fails. The only way around in this case is to use Options.Threshold = 0, which turns the preselection off. This will search all level pairs for possible EPR resonances.

Here is a simplified test example. If you run it, you will see that the spectra are visually identical for the 1-nucleus and 2-nuclei systems only if you set Opt.Threshold to zero. The remaining discrepancy can be reduced by increasing the number of orientations.

Code: Select all

clear, clc

gpv = [1.96];
Apv = [200 500];
Nuc = '14N'; % run with 14N, 63Cu, 55Mn, 51V

Sys1.S = 1/2;
Sys1.g = gpv;
Sys1.Nucs = Nuc;
Sys1.A = Apv;
Sys1.lw = 1;

Sys2.S = [1/2 1/2];
Sys2.g = [gpv; gpv];
Sys2.Nucs = [Nuc ',' Nuc];
Sys2.A = [Apv, 0 0; 0 0, Apv];
Sys2.lw = Sys1.lw;
Sys2.ee = 0;

Exp.mwFreq = 9.45;
Exp.Range = [280 410];
Exp.nPoints = 5000;
Exp.Temperature = 50;
Exp.Harmonic = 0;

Opt.nKnots = 91;
Opt.Threshold = 0;

[B,spec2] = pepper(Sys2,Exp,Opt);
[B,spec1] = pepper(Sys1,Exp,Opt);
spec1 = spec1*2;

subplot(3,1,[1 2]);
plot(B,spec1,B,spec2);
xlim(Exp.Range);
legend('1 Nuc','2 Nuc'); legend boxoff

subplot(3,1,3)
plot(B,spec2-spec1);
xlim(Exp.Range);
xlabel('magnetic field (mT)');

Posted: **Tue May 30, 2017 12:28 am**

Thanks a lot, that is completely clear now.

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