EasySpin forum

Posted: **Mon Mar 23, 2020 9:15 am**

Hello,

I am trying to simulate to similar species with highly varying anisotropies. The experimental intensities of the g|| component are of similar intensities, however, due to the different g||, g_|_ values the calculated intensities are very different.

The calculated transition probabilities are identical, which means that the calculated spectrum intensity should be defined by some other parameter.

Other than playing around with Sys.weight to get roughly equal intensities, is there a way to make the g|| components strictly equal?

An indicative code is:

Code: Select all

clear all; close all;
Sys1.S = 1/2;
Sys1.g = [2 2.2]; % Isotropic system
Sys1.lwpp = [2 0];
Sys2.S = 1/2;
Sys2.g = [1.5 2.2]; % Highly anisotropic system
Sys2.lwpp = [2 0];

% Calculate probability amplitudes
H1 = sham(Sys1,[0,0,0.001]); % Generate the Hamiltonian matrix of the system
[Vt1,Ener1]=eig(H1); % Generate the right-hand vectors (Vt = |>) and the energy eigenvalues (Ener) of the Hamiltonian (in MHz)
H2 = sham(Sys2,[0,0,0.001]); % Generate the Hamiltonian matrix of the system
[Vt2,Ener2]=eig(H2); % Generate the right-hand vectors (Vt = |>) and the energy eigenvalues (Ener) of the Hamiltonian (in MHz)

% Transitions
u1 = Vt1(:,1); % State 1 is the ground state within the ground doublet
v1 = Vt1(:,2); % State 2 is the excited state within the ground doublet
u2 = Vt2(:,1); % State 1 is the ground state within the ground doublet
v2 = Vt2(:,2); % State 2 is the excited state within the ground doublet

[S1x,S1y] = sop(Sys1,'x','y');
[S2x,S2y] = sop(Sys2,'x','y');

% Transition 1->2 (u->v)
trans_mom_1x = u1'*S1x*v1; % <u|Sx|v> Transition moment
trans_mom_1y = u1'*S1y*v1; % <u|Sy|v> Transition moment
prob1 = (abs(trans_mom_1x)^2 + abs(trans_mom_1y)^2) / 2; % Transition amplitude for x, |<u|Sx|v>|
trans_mom_2x = u2'*S2x*v2; % <u|Sx|v> Transition moment
trans_mom_2y = u2'*S2y*v2; % <u|Sy|v> Transition moment
prob2 = (abs(trans_mom_2x)^2 + abs(trans_mom_2y)^2) / 2; % Transition amplitude for x, |<u|Sx|v>|

Exp.Temperature = 4; Exp.mwFreq = 9.4; Exp.Range_Xpowder = [0 5000]; Exp.nPoints = 2000;

[B1, spc1] = pepper(Sys1,Exp);
[B2, spc2] = pepper(Sys2,Exp);

plot(B1,spc1,B2,spc2);

Posted: **Tue Mar 24, 2020 12:25 pm**

EasySpin's EPR spectra are scaled by the number of spins. So if you have two species, A with a narrow powder spectrum and B with a wide powder spectrum, then the intensity of B will be lower for the same number of spins.

Sys.weight is indeed the correct way to adjust the relative amount of the two species.

Posted: **Wed Mar 25, 2020 7:57 am**

Thanks Stefan,

Yes, now that you explain it, it is clear. That should indeed be the case for a powder sample, resonating with at all possible orientations.

However, when I add experimental conditions to simulate a single crystal...

Code: Select all

Exp.CrystalSymmetry = 146;        % space group R3
Exp.MolFrame = [0 0 0]*pi/180;  % molecular frame tilted towards the bc plane
cori0 = [0 0 0]*pi/180;              % initial crystal orientation in lab frame
nRot_L = [1;0;0];               % rotation axis = x axis of lab frame i.e. EPR tube axis based on frame definitions
rho = rot*pi/180;               % rotation angle
cori = rotatecrystal(cori0,nRot_L,rho);   % rotate crystal by rho around nRot
Exp.CrystalOrientation = cori;

...the intensity of the high-anisotropy system is still smaller. Since we are talking about an oriented sample, the total number of spins should be the same (by default Sys1.weight = Sys1.weight = 1), and they all resonate at the same field.

However, even at rot = 0, the high-anisotropy system gives a weaker signal. Is that expected?

Posted: **Wed Mar 25, 2020 8:53 am**

The calculated intensity contains three factors: (a) the quantum-mechanical transition probability, (b) a temperature-dependent polarization factor (Boltzmann), (c) the 1/g factor for field sweeps. The differences you observe are probably a combination of (a) and (c).

Posted: **Wed Mar 25, 2020 9:04 am**

Thanks for the rapid reply.

-My code also calculates the transition probabilities to check for (a). These are identical, so it shouldn't be that.
-The experimental conditions (hence the temperature) are the same, so it shouldn't be that either.
-So I guess the 1/g factor considers an average g between gpar and gperp which probably explains the observed difference.

Posted: **Wed Mar 25, 2020 9:55 am**

Try to simulate a frequency-swept spectrum. In that case, there is no 1/g factor.

Posted: **Wed Mar 25, 2020 11:08 am**

Weird...

I did a frequency-swept simulation with Exp.Field = 100; Exp.mwRange = [0 10]; and with lwpp = [400 0] MHz line width. The same thing.

Even the Sys2.weight required to give approximately equal intensities is the same (Sys2.weight = 26).

Posted: **Wed Mar 25, 2020 11:58 am**

Please post complete (minimal) script.

Posted: **Wed Mar 25, 2020 12:00 pm**

This is a bare-bones script based on the previous one:

Code: Select all

clear all; close all;

Sys1.S = 1/2;
Sys1.g = [1.5 2.2];
Sys1.lwpp = [400 0];
Sys1.weight = 1;
Sys2.S = 1/2;
Sys2.g = [0.3 2.2]
Sys2.lwpp = [400 0];
Sys2.weight = 26;
rot = 0;

Exp.CrystalSymmetry = 146;        % space group R3
Exp.MolFrame = [0 0 0]*pi/180;  % molecular frame tilted towards the bc plane
cori0 = [0 0 0]*pi/180;              % initial crystal orientation in lab frame
nRot_L = [1;0;0];               % rotation axis = x axis of lab frame i.e. EPR tube axis based on frame definitions
rho = rot*pi/180;               % rotation angle
cori = rotatecrystal(cori0,nRot_L,rho);   % rotate crystal by rho around nRot
Exp.CrystalOrientation = cori;

Exp.Temperature = 300; Exp.nPoints = 5000;
% Frequency scan
Exp.Field = 100; Exp.mwRange = [0 10];

[B1, spc1] = pepper(Sys1,Exp);
[B2, spc2] = pepper(Sys2,Exp);

plot(B1,spc1,B2,spc2);

Posted: **Wed Mar 25, 2020 1:33 pm**

The quantum mechanical transition probability is proportional to the square of the g factor perpendicular to the static field. If you put the field along z in your two systems, the g values perpendicular to that are 1.5 and 0.3. (1.5/0.3)^2 = 25, so that explains the difference in intensities.

Qualitatively, the microwave field can only act on the spin through the g value along the mw field direction. The smaller that g value, the lower the effect of the mw on the spins.

EasySpin forum

Absolute intensity of calculated spectra as a function of g-anisotropy

Absolute intensity of calculated spectra as a function of g-anisotropy

Re: Absolute intensity of calculated spectra as a function of g-anisotropy

Re: Absolute intensity of calculated spectra as a function of g-anisotropy

Re: Absolute intensity of calculated spectra as a function of g-anisotropy

Re: Absolute intensity of calculated spectra as a function of g-anisotropy

Re: Absolute intensity of calculated spectra as a function of g-anisotropy

Re: Absolute intensity of calculated spectra as a function of g-anisotropy

Re: Absolute intensity of calculated spectra as a function of g-anisotropy

Re: Absolute intensity of calculated spectra as a function of g-anisotropy

Re: Absolute intensity of calculated spectra as a function of g-anisotropy