levelsplot for a powder spectrum

[sc262]

I want to look at the energy level diagram and allowed transitions using levelsplot for a powder spectrum. Can anyone explain me how can I do that? There are various options for orientation using Ori. What is the default value for it if we do not provide any orientation?

[joscha_nehrkorn]

You can't! What exactly should a energy level diagram for a powder be? Apart from isotropic cases this does not exists, so it is impossible to calculate.

Here is the sample code to figure out what happens, if you don't provide any orientation:

Code: Select all

Sys = struct();
levelsplot(Sys,[],100);

[Stefan Stoll]

For a powder sample, there is not a single energy level diagram. If the spin system is anisotropic, then every orientation gives a different energy level diagram. What you can do is to choose a few directions and overlay the plots. Here is an example:

Code: Select all

clear, clf

Sys.S = 1;
Sys.D = [1000 200]; % MHz

B = 0:5:100; % mT
Ex = levels(Sys,'x',B);
Ey = levels(Sys,'y',B);
Ez = levels(Sys,'z',B);

hold on
hx = plot(B,Ex/1e3);
hy = plot(B,Ey/1e3);
hz = plot(B,Ez/1e3);
set(hx,'Color','r');
set(hy,'Color',[0 0.5 0]);
set(hz,'Color','b');
xlabel('magnetic field (mT)');
ylabel('energy (GHz)');
box on
legend([hx(1) hy(1) hz(1)],'B||x','B||y','B||z','location','sw');
legend boxoff

[sc262]

You can't! What exactly should a energy level diagram for a powder be? Apart from isotropic cases this does not exists, so it is impossible to calculate.

Here is the sample code to figure out what happens, if you don't provide any orientation:

Code: Select all

Sys = struct();
levelsplot(Sys,[],100);

I have tested the script already without any orientation but I could not understand it.

%test example for EPR transitions
clear all
Param = struct('mwFreq',9.5,'Range',[0 700],'nPoints',2048);
Param.Temperature= 10;
Sys = struct('S',3/2,'g',[2],'lw',[1]);
Sys.D = [0.413 0.011]*29979; %MHz

%without giving orientation
levelsplot(Sys,Param.Range,Param.mwFreq);

[sc262]

For a powder sample, there is not a single energy level diagram. If the spin system is anisotropic, then every orientation gives a different energy level diagram. What you can do is to choose a few directions and overlay the plots. Here is an example:

Code: Select all

clear, clf

Sys.S = 1;
Sys.D = [1000 200]; % MHz

B = 0:5:100; % mT
Ex = levels(Sys,'x',B);
Ey = levels(Sys,'y',B);
Ez = levels(Sys,'z',B);

hold on
hx = plot(B,Ex/1e3);
hy = plot(B,Ey/1e3);
hz = plot(B,Ez/1e3);
set(hx,'Color','r');
set(hy,'Color',[0 0.5 0]);
set(hz,'Color','b');
xlabel('magnetic field (mT)');
ylabel('energy (GHz)');
box on
legend([hx(1) hy(1) hz(1)],'B||x','B||y','B||z','location','sw');
legend boxoff

Thanks Stefan. Actually I was trying to understand the powder spectrum by considering all possible orientations. I am not sure if it is right. I think I would need some scaling factor. The figure is attached herewith.

%test example for EPR transitions
clear all
Param = struct('mwFreq',9.5,'Range',[0 700],'nPoints',2048);
Param.Temperature= 10;
Sys = struct('S',3/2,'g',[2],'lw',[1]);
Sys.D = [0.413 0.011]*29979; %MHz

for i = 0:0.1:pi/2
for j = 0:0.1:pi/2
Ori = ;
levelsplot(Sys,Ori,Param.Range,Param.mwFreq);
hold all;
end
end