Quadrupolar term

[Eufemio]

Dear Stefan,

I have a question regarding the quadrupolar term in easyspin. I simulated the Zeeman diagram from a paper of a Tb(III) ion, including the quadrupolar interaction. For some reason in order to obtain the simulation as in the paper I have to multiply the quadrupolar value given in the paper exactly by 4. (the quadrupolar term in the paper is: P{Iz^2 1/3I(I+1)})

I thought the quadrupolar term in easyspin and the paper were similar. Do you have any idea what is the reason for this?

Thanks in advance!

Best wishes,

Eufemio

[Stefan Stoll]

Please post a short script and the link to the paper, so we can have a look.

[Eufemio]

Sure, this is the link to the paper (http://onlinelibrary.wiley.com/doi/10.1 ... 8/abstract). The Zeeman diagram is the one in Figure 3. The Script is:

Sys.g = [3/2];
Sys.S = [6];
Sys.Nucs = '159Tb';
Sys.A = [A]*clight/1e4;
Sys.Q = 4*[Q]*clight/1e4; %To reproduce the Zeeman diagram I have to multiply the Quadrupolar term by 4!
Sys.B2 = alpha*[0 0 B20 0 0]*clight/1e4;
Sys.B4 = beta*[B44 0 0 0 B40 0 0 0 0]*clight/1e4;
Sys.B6 = gamma*[0 0 B64 0 0 0 B60 0 0 0 0 0 0]*clight/1e4;

Alpha, beta and gamma are constants related to the Tb ion.

Cheers

[Stefan Stoll]

Please post a self-containing script that gives all the constants and also the function call so that we can run it. Thanks!

[Eufemio]

Attached is the script file.
Best,
Eufemio

[Stefan Stoll]

This is correct. If you give a single number for Sys.Q, EasySpin understands it as e^2Qq/h.

The term given in the paper, P*(Iz^2-I(I+1)/3), corresponds to P/3*(-Ix^2-Iy^2+2*Iz^2). If you look at the EasySpin reference page on the nuclear quadrupole interaction, the prefactor P/3 is equal to e^2Qq/h/(4I*(2I-1)). The denominator in the last expression is 12, so this means P/3 = e^2Qq/h/12. From this e^2Qq/h = 4*P.

[Eufemio]

Perfect, thanks for the explanation Stefan!
Cheers,
Eufemio