@tsdev Hi, Dr. Toth
I am trying to use the 'type' option to add biquadratic exahange in my model (bcc lattice) :
LaSrCrO4.addcoupling('mat','Biq','bond', 4, 'type', 'biquadratic');
I assume this add a Q*(Si * Sj)^2 type term to the Hamiltonian. However the spinwave calculated by SpinW is a little different from my analytic calculations. And I found that if the defination of biqudratic is 0.5Q(Si * Sj)^2 , these two results fit perfectly.
I read your tutorial document "tutorial28.m". This document also compared the results of analystic calculation and SpinW on fcc lattice.
fcc.addmatrix('label','B','value',-0.5*Q/S^3*2)
fcc.addcoupling('mat','B','bond',1,'type','biquadratic')
According to the paper ''PHYSICAL REVIEW B 85, 054409 (2012)'' mentioned in "tutorial28.m", their Hamiltonian is:
In this example, we also have to add another factor 2 in SpinW to get consistent results.
So I want to confirm that when we use command
fcc.addcoupling('mat','B','bond',1,'type','biquadratic')
we actually add 0.5Q(Si * Sj)^2 instead of Q*(Si * Sj)^2 to Hamiltonian?
JingZhou
@tsdev Hi, Dr. Toth
I am trying to use the 'type' option to add biquadratic exahange in my model (bcc lattice) :
LaSrCrO4.addcoupling('mat','Biq','bond',4, 'type','biquadratic');I assume this add a Q*(Si * Sj)^2 type term to the Hamiltonian. However the spinwave calculated by SpinW is a little different from my analytic calculations. And I found that if the defination of biqudratic is 0.5Q(Si * Sj)^2 , these two results fit perfectly.
I read your tutorial document "tutorial28.m". This document also compared the results of analystic calculation and SpinW on fcc lattice.
fcc.addmatrix('label','B','value',-0.5*Q/S^3*2)fcc.addcoupling('mat','B','bond',1,'type','biquadratic')According to the paper ''PHYSICAL REVIEW B 85, 054409 (2012)'' mentioned in "tutorial28.m", their Hamiltonian is:
In this example, we also have to add another factor 2 in SpinW to get consistent results.
So I want to confirm that when we use command
fcc.addcoupling('mat','B','bond',1,'type','biquadratic')we actually add 0.5Q(Si * Sj)^2 instead of Q*(Si * Sj)^2 to Hamiltonian?
JingZhou