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MathematicalPhysicist
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I'll write the passage from the text which doesn't seem to follow.
For q_1 and q_2 I get that they should be integers without the multiple of ##2\pi##.
Here's my reasoning:
If I plug (18.27) back into the scalar product in (18.25) and also (18.26) and use the identity of (18.24) connecting ##a_i## and ##b_j##, I get:
$$2\pi \times integer = q_1n_1a_1\cdot b_1+q_2n_2a_2\cdot b_2 = 2\pi(q_1n_1+q_2n_2)$$
Divide by ##2\pi## and get: ##integer = q_1n_1+q_2n_2##.
So ##q_i \in \mathbb{Q}##.
How did they get relation (18.28)?
Thanks.
Aschroft and Mermin said:Suppose, furthermore, that the crystal surface is a lattice plane perpendicular to the reciprocal lattice vector ##b_3##.
Choose a set of primitive vectors including ##b_3## for the reciprocal lattice, and primitive vectors ##a_i## for the direct lattice, satisfying:
(18.24) ##a_i\cdot b_j = 2\pi \delta_{ij}##.
If the electron beam penetrates so little that only scattering from the surface plane is significant, then the condition for constructive interference is that the change ##q## is the wave vector of the scattered electron satisfy:
(18.25) ##q\cdot d = 2\pi\times integer, \ \ \ \ \ q=k'-k##
for all vectors ##d## joining lattice points in the plane of the surface (cf. Eq. (6.5)).
Since such ##d## are perpendicular to ##b_3##, they can be written as:
(18.26) ##d = n_1a_1+n_2a_2##.
Writing ##q## in the general form:
(18.27) ##q= \sum_{i=1}^3 q_i b_i , ##
we find that conditions (18.25) and (18.26) require:
(18.28) ##q_1=2\pi \times integer##
##q_2 = 2\pi \times integer##
##q_3 = arbitrary##
For q_1 and q_2 I get that they should be integers without the multiple of ##2\pi##.
Here's my reasoning:
If I plug (18.27) back into the scalar product in (18.25) and also (18.26) and use the identity of (18.24) connecting ##a_i## and ##b_j##, I get:
$$2\pi \times integer = q_1n_1a_1\cdot b_1+q_2n_2a_2\cdot b_2 = 2\pi(q_1n_1+q_2n_2)$$
Divide by ##2\pi## and get: ##integer = q_1n_1+q_2n_2##.
So ##q_i \in \mathbb{Q}##.
How did they get relation (18.28)?
Thanks.