Understanding Reciprocal Lattices for Beginners

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SUMMARY

The discussion focuses on the calculation of the reciprocal lattice for a simple cubic lattice, specifically addressing the relationship between the original crystal lattice and its reciprocal lattice. The base vectors of the reciprocal lattice are defined as b1=2π (a2×a3/a1·(a2×a3)), b2=2π (a3×a1/a1·(a2×a3)), and b3=2π (a1×a2/a1·(a2×a3)). The volume V of the unit cell is determined by the scalar triple product a1·(a2×a3), which equals a³ for a simple cubic lattice with side length 'a'. The discussion emphasizes the importance of including the magnitude of the lattice vectors in calculations.

PREREQUISITES
  • Understanding of vector operations, specifically cross product and dot product.
  • Familiarity with crystal lattice concepts, particularly simple cubic lattices.
  • Knowledge of reciprocal lattice definitions and calculations.
  • Basic understanding of the geometric interpretation of volume in three-dimensional space.
NEXT STEPS
  • Study the derivation of reciprocal lattice vectors for different crystal structures.
  • Learn about the significance of the scalar triple product in crystallography.
  • Explore the applications of reciprocal lattices in X-ray diffraction techniques.
  • Investigate the relationship between reciprocal lattices and Brillouin zones.
USEFUL FOR

Students and researchers in materials science, solid-state physics, and crystallography who are seeking to understand the fundamentals of reciprocal lattices and their applications in analyzing crystal structures.

Michaelangelo
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here is a question on reciprocal lattices that I am stuck on

for a simple cubic lattice, the unit cell is defined by a1=a(1,0,0) a2 = a(0,1,0) a3 = a(0,0,1), demonstrate that the reciprocal lattice of its reciprocal lattice is the original crystal lattice.From what I've found, i think the reciprocal lattice base vectors b1 b2 b3 of the primitive vectors of the crystal lattice a1 a2 a3 is defined by (π is pi btw)

b1=2π (a2xa3/a1.a2xa3), b2=2π (a3xa1/a1.a2xa3) , b3=2π(a1xa2/a1.a2xa3)

the volume V is defined by a1.a2xa3 so i have to figure that out, but if i were to do a2xa3, would it be:
i j k
0 1 0
0 0 1

= (1-0)i + (0-0)j + (0-0)k = so i take it this would equal a(1,0,0) which is a1?, so is a1.a2xa3 basically a1.a1 or am i horribly confused?

it might be the latter but ill soldier on, if it is a1.a1 then does this not just give us the answer a(1,0,0)? or should i have got rid of the a at some point? or should i just be getting an integer? can someone please give me a gentle push in this question i really think i could do most of it myself I am just a bit confused and stressed.But from here I am stuck, i took 2 years off my degree and I've found that my basic vector calculation skills have left me completely. Do i do the cross product first or the dot? any help appreciated.
 
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Hmm... Towards the end you confused me with all the run on questions. The cross product of a2 and a3 is close. You forgot that each of those vectors have the magnitude "a". So the factor out front should be a^2 and not a.

Then taking the dot product with a1 gives you a^3. This makes sense since the volume of a cube with the side 'a' is just a^3.
 

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