A given real-space cubic lattice will have dimensions:

Such that the position of any particular cell within the infinite lattice is:

Where h, k, and l are indices.
The corresponding inverse-space lattice would be:


In the case where
:

Vectors
There are many equivalent ways to define/construct the Cartesian basis for the unit cell in real-space. The unit cell vectors can be written as:

According to this, the vectors can be written as:

which is mathematically equivalent.
According to:
The vectors are written as:

This is, again, an equivalent expression. The equivalence can be show by:

TBD: Reciprocal vector components

Calculate q_hkl generally
def q_hkl(self, h, k, l):
"""Determines the position in reciprocal space for the given reflection."""
# The 'unitcell' coordinate system assumes:
# a-axis lies along x-axis
# b-axis is in x-y plane
# c-axis is vertical (or at a tilt, depending on beta)
# Convert from (unitcell) Cartesian to (unitcell) fractional coordinates
reduced_volume = sqrt( 1 - (cos(self.alpha))**2 - (cos(self.beta))**2 - (cos(self.gamma))**2 + 2*cos(self.alpha)*cos(self.beta)*cos(self.gamma) )
#volume = reduced_volume*self.lattice_spacing_a*self.lattice_spacing_b*self.lattice_spacing_c
a = ( self.lattice_spacing_a , \
0.0 , \
0.0 )
b = ( self.lattice_spacing_b*cos(self.gamma) , \
self.lattice_spacing_b*sin(self.gamma) , \
0.0 )
c = ( self.lattice_spacing_c*cos(self.beta) , \
self.lattice_spacing_c*( cos(self.alpha) - cos(self.beta)*cos(self.gamma) )/( sin(self.gamma) ) , \
self.lattice_spacing_c*reduced_volume/( sin(self.gamma) ) )
# Compute (unitcell) reciprocal-space lattice vectors
volume = np.dot( a, np.cross(b,c) )
u = np.cross( b, c ) / volume # Along qx
v = np.cross( c, a ) / volume # Along qy
w = np.cross( a, b ) / volume # Along qz
qhkl_vector = 2*pi*( h*u + k*v + l*w )
qhkl = sqrt( qhkl_vector[0]**2 + qhkl_vector[1]**2 + qhkl_vector[2]**2 )
return (qhkl, qhkl_vector)
def q_hkl_length(self, h, k, l):
qhkl, qhkl_vector = self.q_hkl(h,k,l)
#qhkl = sqrt( qhkl_vector[0]**2 + qhkl_vector[1]**2 + qhkl_vector[2]**2 )
return qhkl