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Direction of magnetic field
Posted: Thu Feb 05, 2015 4:03 am
by Smirnova38
Hello, experts. I have a question again
I want to simulate the spectrum for crystall CaF2 with Zn-impurity. Ion Zn(+) is located in the center of F-cube, so I have 8 equivalent nuclei of F and one electron (S=1/2) interacting with them. When crystal with z axis aligned with B0, it means we work with <100> direction, the spectrum should being consisted of 9 lines. And I was getting it, when I used garlic. But! I don't understand why the result with pepper-function is not the same. Maybe it causes perturb1-method???
Also I have a serious problems with orientations! I need to simulate the spectrum for <110> direction of magnetic field. But changing code line with Exp.Orientations, my spectrum doesn't feel it. It is equivalent to the first one (see the figure for pepper-function). I don't know where is the main mistake.
Code: Select all
clear
S=1/2;
g=2.0;
Sys.Nucs = '19F';
Sys.n = [8];
Sys.A = 20;
Sys.lwpp = 0.06;
Exp.mwFreq = 9.5;
Exp.nPoints = 2^13;
garlic(Sys,Exp);
Code: Select all
clear all;
Sys.S = 1/2;
Sys.g = 2;
A = [20];
Sys = nucspinadd(Sys,'19F',A);
Sys = nucspinadd(Sys,'19F',A);
Sys = nucspinadd(Sys,'19F',A);
Sys = nucspinadd(Sys,'19F',A);
Sys = nucspinadd(Sys,'19F',A);
Sys = nucspinadd(Sys,'19F',A);
Sys = nucspinadd(Sys,'19F',A);
Sys = nucspinadd(Sys,'19F',A);
Sys.lwpp = 0.6;
Exp.mwFreq = 9.5;
Exp.Range = [330 350];
Exp.Orientations = [0;0;0];
Exp.nPoints = 2^13;
Opt.Method = 'perturb1';
pepper (Sys,Exp,Opt);
Re: Direction of magnetic field
Posted: Thu Feb 05, 2015 11:40 am
by Matt Krzyaniak
You should get the same result from both pepper and garlic in this particular circumstance, Though the code you posted for pepper has a different linewidth, fix that and they are basically equivalent.
In terms of the field direction, unless you include anisotropy in the hyperfine interaction the field direction and exp.orientations will have to effect on the spectrum.
Re: Direction of magnetic field
Posted: Thu Feb 05, 2015 10:32 pm
by Smirnova38
Matt Krzyaniak wrote:You should get the same result from both pepper and garlic in this particular circumstance, Though the code you posted for pepper has a different linewidth, fix that and they are basically equivalent.
In terms of the field direction, unless you include anisotropy in the hyperfine interaction the field direction and exp.orientations will have to effect on the spectrum.
Thank you, Matt! I will try to remake my code!
Re: Direction of magnetic field
Posted: Fri Feb 06, 2015 8:46 am
by Matt Krzyaniak
I had tried my hand at it a little bit yesterday and it all ultimately depends upon the principal axis system you feel like working in but something like:
Code: Select all
clear
Sys.S = 1/2;
Sys.g = 2;
% axial hyperfine interaction with some isotropic component
A = 20*[1 1 -2] + 30;
% theta and phi for the vertices of a cube(I think this is correct) since
% the hyperfine is axial the first euler angle doesn't matter also the
% sign got lost on some of them but realistically it doesn't matter
theta = [54.7 125.3 54.7 125.3 54.7 125.3 54.7 125.3 ]*pi/180 ;
phi = [45 45 -45 -45 -45 -45 45 45]*pi/180;
% for loop to make the script a bit cleaner
for i=1:8
Sys = nucspinadd(Sys,'19F',A,[0 theta(i) phi(i)]);
end
Sys.lwpp = 0.1;
Exp.mwFreq = 9.5;
Exp.Range = [330 350];
%Exp.Orientations = [0;0;0];
Exp.nPoints = 2^13;
Opt.Method = 'perturb1';
pepper (Sys,Exp,Opt);
Then if you play with Orientations, namely the second value you'll observe significant spectral changes.
Re: Direction of magnetic field
Posted: Wed Feb 11, 2015 9:44 pm
by Smirnova38
Matt Krzyaniak wrote: Then if you play with Orientations, namely the second value you'll observe significant spectral changes.
Matt, hello
I succeed in orientations. Changing Exp.Orientations = [0;45;0]*pi/180 I get true spectrum for <110>-direction and Exp.Orientations = [45;54.7;0]*pi/180 for <111>-direction. Thank you for the timely advice!!!
Re: Direction of magnetic field
Posted: Fri Feb 13, 2015 11:58 pm
by Smirnova38
Can I use function "Exp.CrystalSymmetry" describing my equivalent 8 nuclei of F in vertices of cube? I have tried to add code-line this way, program doesn't work without theta and phi
Code: Select all
Sys.lwpp = 0.1;
Exp.mwFreq = 9.5;
Exp.Range = [325 350];
Exp.CrystalSymmetry = 207;
Exp.Orientations = [45;54.7;0]*pi/180;
Exp.nPoints = 2^13;
Opt.Method = 'perturb1';
pepper (Sys,Exp,Opt);
Probably, I don't understand comletely the meaning of function
In many crystals, there are several symmetry-related sites with identical paramagnetic centers differing only in their orientation in the crystal. You can tell pepper about this by providing the crystal symmetry in the field Exp.CrystalSymmetry, e.g.
Exp.CrystalSymmetry = 'P21/m'; % space group symbol
Exp.CrystalSymmetry = 11; % space group number (between 1 and 230)
Exp.CrystalSymmetry = 'C2h'; % point group, Schönflies notation
Exp.CrystalSymmetry = '2/m'; % point group, Hermann-Mauguin notation
Re: Direction of magnetic field
Posted: Sat Feb 14, 2015 12:01 pm
by Stefan Stoll
With Exp.Crystalsymmetry, ES will calculate the spectra of all symmetry-related centers in your crystal and add them up. However, you have to define one spin center completely. In your case, that's the center with S=1/2 coupled to its 8 neighboring 19F. Depending on the crystal space group, this same type of center might occur in different positions and orientations throughout the crystal.
Re: Direction of magnetic field
Posted: Sun Feb 15, 2015 11:54 pm
by Smirnova38
Stefan Stoll wrote:With Exp.Crystalsymmetry, ES will calculate the spectra of all symmetry-related centers in your crystal and add them up. However, you have to define one spin center completely. In your case, that's the center with S=1/2 coupled to its 8 neighboring 19F. Depending on the crystal space group, this same type of center might occur in different positions and orientations throughout the crystal.
Stefan, considering your advice I think, that the definition of spin center is same to the Matt's code
Code: Select all
clear
Sys.S = 1/2;
Sys.g = 2;
A = 20*[1 1 -2] + 30;
theta = [54.7 125.3 54.7 125.3 54.7 125.3 54.7 125.3 ]*pi/180 ;
phi = [45 45 -45 -45 -45 -45 45 45]*pi/180;
for i=1:8
Sys = nucspinadd(Sys,'19F',A,[0 theta(i) phi(i)]);
end
but instead of theta and phi I want to describe the symmetry of crystal. In my opinion, it might look like that for the <111>-direction, but the spectrum is the same <100>-direction.
Code: Select all
clear
Sys.S = 1/2;
Sys.g = 2;
A = 20*[1 1 -2] + 30;
for i=1:8
Sys = nucspinadd(Sys,'19F',A);
end
Sys.lwpp = 0.1;
Exp.mwFreq = 9.5;
Exp.Range = [325 350];
Exp.CrystalSymmetry = 207;
Exp.Orientations = [45;54.7;0]*pi/180;
Exp.nPoints = 2^13;
Opt.Method = 'perturb1';
pepper (Sys,Exp,Opt);
Why doesn't the program feel Exp.Orientations - function?
Re: Direction of magnetic field
Posted: Mon Feb 16, 2015 10:19 pm
by Stefan Stoll
There is no way around defining the coupled spin system of your local paramagnetic center completely, even in a crystal of high symmetry. This means you have to include the S=1/2 and all 8 hyperfine couplings to the 8 neighboring fluorine atoms in Sys. Only once you have that, you can bring the crystal symmetry into play via the crystal's space group in Exp.CrystalSymmetry. The symmetry of the space group will then result in several symmetry-related (S=1/2)+8xF centers. These might have different orientations and then give different EPR spectra if there is sufficient g anisotropy.
Re: Direction of magnetic field
Posted: Wed Feb 25, 2015 2:13 am
by Smirnova38
Stefan Stoll wrote: This means you have to include the S=1/2 and all 8 hyperfine couplings to the 8 neighboring fluorine atoms in Sys. Only once you have that, you can bring the crystal symmetry into play via the crystal's space group in Exp.CrystalSymmetry. .
Stefan, I couldn't realize your advice in program because I didn't understand that. I have the loop describing 8 equivalent nuclei of F. This loop has for each fluorine atom the hyperfine coupling A as well as parameters of spin-system. Whad does it stand for "you have to include the S=1/2 and all 8 hyperfine couplings"? I can't imagine this code line.