The magnetic moment of an atom at position τ (internal coordinates) is rotated by 2π(q⋅τ). For the given example and β=π/2 this implies that an increase of the q vector to q=(0.5,0.5,0.0) would yield a layered structure with an antiferromagnetic alignment of the magnetization directions along z.
How does it transform into antiferromagnetic in the z direction ？
If I calculate the hcp structure, how should q be set to ensure that the z direction changes from 0.1 to 0.5

This is the relevant tutorial I watched

In the figure, the q vector multiplied by B is equal to (1,0,0) ，but why does it represent the antiferromagnetic in the z direction ？

Thank you very much.I want to calculated the hcp structure, how should I set the q vector path from (0,0,0) to (0,0,0.5) and (0,0,0) to (0.5,0.5,0.0), so that I can calculate the spin stiffness A based on the result? Or can you give me a detailed tutorial for calculating spin stiffness A ？

I am going to calculate the material of a hcp structure

Dear developers,

I want to know, Is the direction of spiral B*q or q*B ？
For fcc, B*q or q*B is the same, while they are different for some lattice.
For example, a rhombohedral primitive cell.

If I input the structure as follows,
##############################################
2.4387895018949948 1.4080357754159085 4.7429872621175369
-2.4387895018949948 1.4080357754159085 4.7429872621175369
0.0000000000000000 -2.8160715508318170 4.7429872621175369
1.889726125
1 1 1
###############################################

The out file will give bravais matrices of real and reciprocal lattices,
4.608644 -4.608644 0.000000 ---- 0.681674 0.393565 0.233673
2.660802 2.660802 -5.321604 ---- -0.681674 0.393565 0.233673
8.962947 8.962947 8.962947 ---- 0.000000 -0.787129 0.233673

Then if spiral direction is along the z direction
For B*q, the q will be (0.0000000000 -0.8469595000 1.4264960700)
That is
( 0.681674 0.393565 0.233673) ---- (0.0000000000) ---- (0)
(-0.681674 0.393565 0.233673) ---- (-0.8469595000) = (0)
(0.000000 -0.787129 0.233673) ---- (1.4264960700) ---- (z)


For q*B, the q will be (1,1,1).
-------------( 0.681674 0.393565 0.233673)
----(1 1 1) (-0.681674 0.393565 0.233673)--------- =(0 0 z)
------------(0.000000 -0.787129 0.233673)

Thank you for your valuable time and giving me useful guidance.

Dear developers,

I checked the internal -> cartesian of real-space (A) and reciprocal space (B). The A in out file should be the transpose of true real space, while the B is the true reciprocal space.
I read the source code of cotra.f. The internal -> cartesian is wrote as :

The real space:
! **********************************************************
SUBROUTINE cotra0(r,x,amat)
! **********************************************************
! ..
! .. Array Arguments ..
REAL, INTENT (IN) :: r(3)
REAL, INTENT (IN) :: amat(3,3)
REAL, INTENT (OUT):: x(3)
!
DO i = 1,3
s = 0.0
DO j = 1,3
s = s + amat(i,j)*r(j)
ENDDO
x(i) = s
ENDDO

RETURN
END SUBROUTINE cotra0

! ***********************************************************


The reciprocal space :
!*********************************************************************
SUBROUTINE cotra3(f,g,bmat)
!*********************************************************************
! ..
! .. Array Arguments ..
REAL, INTENT (IN) :: f(3)
REAL, INTENT (IN) :: bmat(3,3)
REAL, INTENT (OUT):: g(3)
!
DO j = 1,3
g(j) = 0.0
DO i = 1,3
g(j) = g(j) + f(i)*bmat(i,j)
ENDDO
ENDDO

RETURN
END SUBROUTINE cotra3


Therefore, for the vector of spiral, It should be q*B.

For my case above, the internal q should be (1,1,1), but not (0.0000000000 -0.8469595000 1.4264960700) which is a strange vector.