EasySpin forum

Dear all,

I have used VASP to calculate the electric field gradients at the atomic nuclei.
The entries of the 3x3 V matrix are given in V/A² (Volts per square Angstroem).
How can I transform this result into the Q matrix needed in EasySpin to take the nuclear quadrupole interaction (https://easyspin.org/easyspin/documenta ... n.html#nqi) into account?

I need to know only the entries of the Q matrix. Which unit do they have?

Q is in units of MHz.

See here for the conversion equations.

Also, make sure that the calculate EFG is indeed in V/A². I am not familiar with VASP, but a lot of other quantum chemistry programs return the EFG in atomic units, Eh/(e*a0²).

Thank you for your reply.

Unfortunately I have no access to this source.

When using the quantum chemisty program ORCA, both the EFG tensor and the nuclear quadrupole tensor are returned. From this, you should be able to infer how to convert. Here is an example from a hydroxyl radical, with the EFG and NQ tensor for deuterium:

 -----------------------------------------------------------
 Nucleus   1H : A:ISTP=    1 I=  0.5 P=533.5514 MHz/au**3
                Q:ISTP=    2 I=  1.0 Q=  0.0029 barn     
 -----------------------------------------------------------
 Tensor is right-handed.

 Raw EFG matrix (all values in a.u.**-3):
            -0.0881       0.0083       0.0000
             0.0083      -0.1336       0.0000
             0.0000       0.0000       0.2216

 V(El)       0.8041       0.7557      -1.5597
 V(Nuc)     -0.8907      -0.8907       1.7813
         ----------   ----------   ----------
 V(Tot)     -0.0866      -0.1350       0.2216 
 Orientation: 
  X      -0.9848399    0.1734656    0.0000000
  Y      -0.1734656   -0.9848399    0.0000000
  Z       0.0000000   -0.0000000    1.0000000


 Quadrupole tensor eigenvalues (in MHz;Q= 0.0029 I=  1.0)
  e**2qQ            =     0.150 MHz 
  e**2qQ/(4I*(2I-1))=     0.037 MHz 
  eta               =     0.218
  NOTE: the diagonal representation of the SH term I*Q*I = e**2qQ/(4I(2I-1))*[-(1-eta),-(1+eta),2]

I am not completely sure, but I am trying to follow.

As far as I understand one would first have to diagonalise the V matrix (EFGs).
Then the diagonal Q matrix (nuclear quadrupole) can be calculated from the largest-magnitude eigenvalue and the asymmetry together with the nuclear spin and quadrupole moment.

Could diagonal Q then be transformed into full Q via the same similarity transformation that produces full V from diagonal V?

No need to diagonalize.

Here is a code snipped that converts the EFG (in V/Ų) to the nuclear quadrupole tensor (in MHz).

% assuming V is the EFG tensor in units of V Å^-2

V = V/angstrom^2;                        % V Å^-2 -> V m^-2

Nuc = '17O';                             % isotope
qmom = nucqmom(Nuc);                     % nuclear quadrupole moment, barn
qmom = qmom*barn;                        % barn -> m^2
I = nucspin(Nuc);                        % nuclear spin quantum number

P = V*echarge*qmom/2/I/(2*I-1);          % nuclear quadrupole tensor, J
P = P/planck/1e6;                        % J -> MHz
Sys.Q = P;

Would this mean the Q matrix is simply a constant times the V matrix?

Yes, as long as both are defined in the same coordinate frame.

Yes. It is the same coordinate frame.

I need this constant "X" for ¹¹ B and ¹⁴ N for the conversion of V[V/(A²)] into Q[MHz].

Q=X*V

X(¹¹ B) = 0.01636
X(¹⁴ N) = 0.02471

Can you confirm these values?
This would solve all my questions.

I get the same numbers using the code snippet

Nuc = '14N';
qmom = nucqmom(Nuc);
I = nucspin(Nuc);
X = echarge/angstrom^2*qmom*barn/2/I/(2*I-1)/planck/1e6