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Pepper with negative hyperfine values?

Posted: Wed Apr 26, 2017 3:42 pm
by thurberk
Hi, running pepper with negative hyperfine values for some axes gives spectra that do not match my expectations, or my own Matlab code. I will paste in an explicit example below. I expect this spectra to have a narrow central line (for 14N = 0), and the two (14N = +-1) other peaks to form a Pake doublet like pattern around the narrow central line. Instead it looks like the -20 MHz hyperfine axes are treated as +20?

Nx.g = [2.005 2.005 2.005];
Nx.lwpp = 0.1; %mT for field sweeps
Nx = nucspinadd(Nx,'14N',[-20 -20 49]); % MHZ
%------------------------------------------------------------
Exp1.mwFreq = 33.94465;
Exp1.Harmonic = 0;
Exp1.Range = [1199.0 1220.9893];
Exp1.nPoints = 2048;
Opt.Method = 'perturb2';
Opt2.Method = 'matrix';

% Simulate and plot frequency-sweep spectrum
[B,spec]=pepper(Nx,Exp1,Opt);
[B2,spec2]=pepper(Nx,Exp1,Opt2);

Re: Pepper with negative hyperfine values?

Posted: Wed Apr 26, 2017 4:53 pm
by Stefan Stoll
The effective hyperfine coupling for a given orientation is independent of the sign of the hyperfine principal values. Try changing the signs of the components of A in this:

Code: Select all

theta = pi/3;
phi = pi/5;
n = ang2vec(phi,theta); % direction of magnetic field
A = [-30 -10 49];       % hyperfine principal values
A = diag(A);            % full hyperfine tensor
Aeff = norm(n'*A)       % effective hyperfine coupling along n
The effective hyperfine coupling Aeff is independent of the signs. Bottom line: From CW EPR, we cannot generally determine the signs of hyperfine couplings.

Re: Pepper with negative hyperfine values?

Posted: Sun May 07, 2017 1:02 pm
by thurberk
Thanks. As you say, the effective hyperfine coupling is independent of the signs of the principal values, because the calculation of Aeff ends up as the square root of the sum of the components squared.