Rotating single crystal with two symmetry-related centres

[thanasis]

I guess this is a two-stage question

1. The first relates to the crystal symmetry and the number of different paramagnetic centres per unit cell.
E.g., in my case, in a triclinic P-1 crystal, there are two symmetry related molecules (x,y,z and -x-y-z). However, how do I instruct Easyspin that the inversion center is outside the molecule (which means that the molecules in the two asymmetric units are antiparallel) and not inside the molecule (which means that all molecules in the unit cell are aligned parallel to each other).
In axial or rhombic systems, would two antiparallely aligned magnetic centers give rise to different signals as a function of their orientation inside the magnetic field?

2. The second relates to describing the orientation of the molecule in the crystal.
Due to the low symmetry of the molecule, there is no crystallographic plane that is parallel to the molecular z-axis. The closest is the bc plane which intersects at an angle of 8o.
As it seems, I managed to mount the crystal with the bc plane horizontally (I could get very axial signals for certain rotations). For these rotations, I consider that there is a +/-8 degree angle between the molecular z-axis and the magnetic field (let's call it a tilt angle). Indeed, I could very nicely simulate those signals, so I guess my assumption is correct.
What I am trying to understand now is how to correctly describe the rotation of the crystal.
My conventions are:
-zL||B
-There is a +/-8o angle between zM and zL at zero rotation
-The crystal rotates around xL (the axis of the EPR tube)
-The rotation is clockwise

The code I have written to describe this is:

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cm=2.99793e4; % Conversion constant from cm-1 to MHz (~ 100*clight/1e6)
%The variables
gxy=1.9; %gperpendicular common for all Cu(II) ions
gz=2.22; %g|| common for all Cu(II) ions
D32=-0.087; % D value of the S=3/2 spin (in cm-1)
tilt=8; % molecular axis tilt from the bc (horizontal) plane (in degrees)
rot=90; % rotation of the crystal with respect to the magnetic field (in degrees)

%The spin system
%--------------------------------------------------------------------------
Sys.S=3/2;
Sys.g=[gxy gz];
Sys.D=D32*cm

%Experimental
Exp.Temperature = 4.5; Sys.lwpp=[10 10]; Exp.mwFreq=9.42; Exp.CenterSweep=[270 550.0]; Exp.nPoints=4096;
Exp.CrystalSymmetry = 2;        % space group P-1
Exp.MolFrame = [0 tilt 0]*pi/180;  % molecular frame tilted with respect to the bc plane
cori0 = [0 0 0]*pi/180;              % initial crystal orientation in lab frame
Exp.CrystalOrientation = cori0;
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;
pepper(Sys,Exp);
pepper(Sys,Exp); %For pepper calculation
%Ori=cori; FieldRange=[0 1000]; Freq=9.42; nPoints=10000; levelsplot(Sys,Ori,FieldRange,Freq,nPoints); %Uncomment for levelsplot calculation

It seems to behave correctly in preliminary simulations, however:
-have I grasped correctly the Easyspin conventions and Euler angle conventions?
-should the rot angle be negative or positive for clockwise rotations, i.e.

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rho = rot*pi/180

or

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rho = -rot*pi/180

-will the levelsplot calculation correctly take into account the defined orientation the way I have defined it as

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Ori=cori;

?

[Stefan Stoll]

1.
EPR spectra of molecules are invariant under two transformations: translation and inversion. So there is no need to worry about either. Invariance under translations means that only the point subgroup of the space group is needed. Invariance under inversion means that the EPR spectrum will aways be of a point group symmetry that includes inversion. For triclinic space groups P1 and P-1, the point subgroups are C1 and Ci, and the one relevant for EPR is Ci. EasySpin only simulates the spectrum once, since the second site will always give an identical spectrum.

2.
It is best to look at a specific example with small angles and then draw a diagram to see what happened

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cori0 = [0 0 0]; % Euler angles for crystal->lab frame transformation
nRot_L = [1;0;0]; % rotation axis, in lab frame coordinates
rho = 10*pi/180;
cori = rotatecrystal(cori0,nRot_L,rho);

% corresponding transformation matrix
R0_C2L = erot(cori0)
R_C2L = erot(cori) 

which gives

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R0_C2L =
     1     0     0
     0     1     0
     0     0     1
R_C2L =
    1.0000   -0.0000   -0.0000
   -0.0000    0.9848   -0.1736
    0.0000    0.1736    0.9848

In these matrices, the three columns are the three crystal axis vectors xC, yC, zC, represented in lab coordinates. Conversely, the three rows are the lab axis vectors xL, yL, zL, represented in crystal coordinates. [This is explained in the Easyspin documentation].

Now, yC changed from [0;1;0] to [0;0.98;0.17], so the rotation took yC from originally parallel to yL towards zC.

[thanasis]

Thanks a lot!

Yes, I get it better now.