Quadrupolar interaction Hamiltonian

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Krishk
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Joined: Mon Feb 01, 2021 1:16 pm

Quadrupolar interaction Hamiltonian

Post by Krishk »

If I use Sys.Q=10 in MHz and lets for the purpose of discussion, assume eta=0.
Does the Sys.Q includes 4I(2I-1) term? Or this Sys.Q is simply equal to VeQ/h, where V is the electric-field gradient matrix generated by DFT calculations, and Q is (scalar) the electric quadrupole moment that can be picked from the standard tables available in literatures.
Thanks.

nwili
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Joined: Mon Oct 16, 2017 6:27 am

Re: Quadrupolar interaction Hamiltonian

Post by nwili »

This is what the documentation says: https://easyspin.org/easyspin/documenta ... tem.html#Q

Code: Select all

eeQqh = 1;     % MHz
eta = 0.2;     % unitless
I = 1;         % nuclear spin must be known!
Q = eeQqh/(4*I*(2*I-1)) * [-1+eta, -1-eta, 2]

so as I see it it includes 4I(2I-1) if you put it into Q.

Stefan Stoll
EasySpin Creator
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Joined: Mon Jul 21, 2014 10:11 pm
Location: University of Washington

Re: Quadrupolar interaction Hamiltonian

Post by Stefan Stoll »

Sys,Q is iunderstood by EasySpin's functions to be equal to e2qQ/h, as documented here. It does not include the mentioned spin-dependent scaling factor. Internally, EasySpin adds the denominator before building the quadrupole term in the spin Hamiltonian.

nwili
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Joined: Mon Oct 16, 2017 6:27 am

Re: Quadrupolar interaction Hamiltonian

Post by nwili »

The clarification in the documentation how to convert to the principal Q-values somehow confused me more than it clarified.

the Q in

Code: Select all

eeQqh = 1;     % MHz
eta = 0.2;     % unitless
I = 1;         % nuclear spin must be known!
Q = eeQqh/(4*I*(2*I-1)) * [-1+eta, -1-eta, 2]

is NOT what one should put into Sys.Q.

Sorry for the confusion.

Stefan Stoll
EasySpin Creator
Posts: 1059
Joined: Mon Jul 21, 2014 10:11 pm
Location: University of Washington

Re: Quadrupolar interaction Hamiltonian

Post by Stefan Stoll »

The documentation is correct.

There are several equivalent ways to input a given quadrupole tensor in Sys.Q. They all result in the same spin Hamiltonian. Two of them are as follows:

Code: Select all

eeQqh = 1;     % MHz
eta = 0.2;     % unitless

% Method 1: provide e^2Qq/h and eta
Sys.Q = [eeQqh eta]; 

% Method 2: provide principal values of quadrupole tensor (MHz)
I = 1;         % nuclear spin must be known!
Q = eeQqh/(4*I*(2*I-1)) * [-1+eta, -1-eta, 2]; % calculate principal values from e^2Qq/h, eta, and I
Sys.Q = Q;
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