spin spiral dispersion energy with pair hamiltonian and multiple atoms per unit cell

[loudesplat]

Dear spirit developers,
I am writing to get some clarification on the pair hamiltonian when one has more than one atom per unit cell.
This is my basis and bravais vectors:

bravais_lattice hex2d120

basis
3 # number of spins in the basis
0.0 0.0 0.0 
 0.0 0.0 0.62
0.33 0.66 0.30

Firstly, the heisenberg_neighbours hamiltonian is completely unable to find even the first pair correctly (which should be between atoms 0 and 1 within the same uc), so one has no choice but to use heisenberg_neighbours, as shown below

n_shells_exchange 2
jij 		 109.8460   30.4400 

output:

###    Interaction neighbours:

n_neighbours_exchange 4

 i   j     da  db  dc           Jij      

 0   2      0  -1   0      109.84600000  
 1   2      0  -1   0       30.44000000  
 2   0      0   1   0      109.84600000  
 2   1      0   1   0       30.44000000

When specifying only interlayer exchange, aka pairs in different uc's, I am able to reproduce the theorical spin spiral dispersion (aka E(q)=sum_r J(r) cos(q*r)) with no issue:

n_interaction_pairs  	 18 
 i j   da db dc    Jij   Dij  Dijx Dijy Dijz 
1 	 1 	 -1  	 -1 	 0 	 -2.1170 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
1 	 1 	 0  	 -1 	 0 	 -2.1170 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
1 	 1 	 -1  	 0 	 0 	 -2.1170 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
1 	 1 	 1  	 -0 	 0 	 -2.1170 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
1 	 1 	 -0  	 1 	 0 	 -2.1170 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
1 	 1 	 1  	 1 	 0 	 -2.1170 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
0 	 0 	 -1  	 -1 	 0 	 -0.1985 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
0 	 0 	 0  	 -1 	 0 	 -0.1985 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
0 	 0 	 -1  	 0 	 0 	 -0.1985 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
0 	 0 	 1  	 -0 	 0 	 -0.1985 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
0 	 0 	 -0  	 1 	 0 	 -0.1985 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
0 	 0 	 1  	 1 	 0 	 -0.1985 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
2 	 2 	 -1  	 -1 	 0 	 -3.2540 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
2 	 2 	 0  	 -1 	 0 	 -3.2540 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
2 	 2 	 -1  	 0 	 0 	 -3.2540 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
2 	 2 	 1  	 -0 	 0 	 -3.2540 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
2 	 2 	 -0  	 1 	 0 	 -3.2540 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
2 	 2 	 1  	 1 	 0 	 -3.2540 	 0.0000 	 0.0000 	 0.0000  	 0.0000 

However, when trying to specify the intralayer pairs within the same uc, the spin spiral dispersion obtained from spirit no longer matches the theory:

n_interaction_pairs  	 8 
 i j   da db dc    Jij   Dij  Dijx Dijy Dijz 
1 	 0 	 0  	 0 	 0 	 54.9230 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
0 	 1 	 0  	 0 	 0 	 54.9230 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
0 	 2 	 0  	 0 	 0 	 15.2200 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
0 	 2 	 1  	 -0 	 0 	 15.2200 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
0 	 2 	 1  	 1 	 0 	 15.2200 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
2 	 0 	 -1  	 -1 	 0 	 15.2200 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
2 	 0 	 -1  	 0 	 0 	 15.2200 	 0.0000 	 0.0000 	 0.0000  	 0.0000 
2 	 0 	 0  	 0 	 0 	 15.2200 	 0.0000 	 0.0000 	 0.0000  	 0.0000 

Note that here the bond for J_1 is always perpendicular to q so it contributes a constant value of -J_1/uc and always between atoms of the same uc, I believe the issue may come from J_2.

I am wondering if specifying pairs within the same uc should be done a bit differently than in different ucs? Here I specify all lattice rotations which includes also the J_2 pairs between different ucs. Or there maybe be an issue elsewhere or something I am missing?

I hope this question makes sense.
Thank you in advance.

[Puerling]

Thank you for reaching out. The shell configuration not working correctly is something we are going to address.

Regarding the pair specification:
If your goal is to specify the neighbor shells by hand the 0 2 connections are not what what you intended. The connections between 0 2 are the edges to the lateral faces of a triangular pyramid and should look like this:

i j  da db dc  Jij
0 2   0  0  0  30.44
2 0   0  1  0  30.44
2 0   1  1  0  30.44

Equivalently you could use the six pairs that are closer to the ones you gave:

i j  da db dc  Jij
2 0   0  0  0  15.22
2 0   0  1  0  15.22
2 0   1  1  0  15.22
0 2  -1 -1  0  15.22
0 2   0 -1  0  15.22
0 2   0  0  0  15.22

but Spirit will internally deduplicate pairs that are inverses of each other. The inverse of a pair is given as:
$$(i,j,[d_a,d_b,d_c])^{-1} = (j,i,[-d_a,-d_b,-d_c])$$

Let me know whether using these pairs for J_2 fixes your issue.

[loudesplat]

Thank you for your answer! You are completely correct, I have a script that generates the input pairs and it used to work but apparently broke recently and I had not realized and did not think to check... Oops! This indeed fixes the dispersion. Sorry for PEBCAK!
Cheers
Louise