Passage to the limit: biradical → triplet

[trukhan]

Hello everyone!
As far as I understand, the EPR spectrum of the biradical in the limiting case of a large exchange interaction constant should transform into the spectrum of the triplet state. (It is only necessary to redefine some constants of the Spin Hamiltonian.)
However, in this example, no matter how large J is, there are still significant differences in the spectra of the triplet and biradical.

Code: Select all

%EasySpin 6.0.0-dev.30
clear, clf, clc
Exp.mwFreq = 9.5;
Exp.Range = [100 500];
Exp.nPoints = 8192;
Opt.GridSize = [99 3];
g = [2 2 2];
Axy = 70; 
Az = 200;
D = 2000;
J = 1e10;
lw = 3;
Nucs = '63Cu,63Cu';

SysT.S = 1;
SysT.g = g;
SysT.D = D;
SysT.Nucs = Nucs;
SysT.A = [Axy Axy Az; Axy Axy Az];
SysT.lw = lw;
[B,specT] = pepper(SysT,Exp,Opt);

SysB.S = [1/2 1/2];
SysB.g = [g; g];
SysB.A = 2*[Axy Axy Az 0 0 0; 0 0 0 Axy Axy Az];
SysB.Nucs = Nucs;
SysB.J = J;
SysB.dip = 2/3*D; 
SysB.lw = lw;
[B,specB] = pepper(SysB,Exp,Opt);

plot(B,specT,B,specB);
legend('triplet','biradical');
axis tight; 

BiradicalTriplet2Cu.png (25.4 KiB) Viewed 2662 times

Is there anything I can do to make the spectra match?
In this regard, the question also arises: In what cases and to what extent one can trust the calculated spectra of biradicals?

It is interesting to note that if we make the nuclei not quite equivalent (introduce a small difference in HFC), the coincidence of the spectra will be much better!

---

I also noticed that each new calculation of the biradical spectrum in the example above gives noticeably different spectra. How is this possible?

Code: Select all

%EasySpin 6.0.0-dev.30
clear, clf, clc
Exp.mwFreq = 9.5;
Exp.Range = [130 450];
Exp.nPoints = 8192;
Opt.GridSize = [99 3];
g = [2 2 2];
Axy = 70; 
Az = 200;
D = 2000;
J = 1e10;
lw = 3;
Nucs = '63Cu,63Cu';

SysB.S = [1/2 1/2];
SysB.g = [g; g];
SysB.A = 2*[Axy Axy Az 0 0 0; 0 0 0 Axy Axy Az];
SysB.Nucs = Nucs;
SysB.J = J;
SysB.dip = 2/3*D; 
SysB.lw = lw;
[B,specB1] = pepper(SysB,Exp,Opt);
[B,specB2] = pepper(SysB,Exp,Opt);
[B,specB3] = pepper(SysB,Exp,Opt);

plot(B,specB1,B,specB2,B,specB3);
legend('biradical 1st run','biradical 2nd run','biradical 3rd run');
axis tight; 

BiradicalTriplet2Cu_BBB.png (31.55 KiB) Viewed 2662 times

It seems to me that in this case it will be problematic to determine the Spin Hamiltonian parameters of the experimental spectrum...

[Matt Krzyaniak]

On the surface everything seems to be working correct, if you turn off the nuclear spins, the spectra are effectively identical; the same goes for shifting to Opt.Method = 'Hybrid', which treats the nuclear spin states as perturbations.

I would guess this comes about from how the nuclear spins states factor into spectra, you've defined all of the tensors as colinear, but I'm not sure how that plays out with mixing in the full matrix diagonalization, there may be some small off diagonal elements which are shifting things around. You may not have enough points to adequately smooth the spectra out.

With regard to the spectra being different each time, ES adds a little bit of noise to the Hyperfine values in order to break potential degeneracies when doing the full matrix diagonalization.

[Stefan Stoll]

I think the main issue here is that the two spin-3/2 with identical hyperfine tensors lead to a large number of degeneracies and energy level anticrossing that throw off EasySpin's automatic transition selection in resfields (which is called by pepper).

Turn automatic transition selection off withOpt.Threshold = 0, and you should get essentially identical spectra for the biradical and the triplet.