Reciprocal lattice and the metric tensor
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Transcript Reciprocal lattice and the metric tensor
Reciprocal lattice and the metric tensor
Concept of a metric and the dual space is known from the theory of relativity
-line element ds measuring the distance between 2 neighboring events in
space time reads
ds2 g dx dx
metric tensor
coordinate differentials
-in flat space time with coordinates ct, x, y, z
ds2 (cdt )2 dx2 dy2 dz 2
1
0
g
0
0
0
1
0
0
0
0
1
0
0
0
0
1
In 3D real space we can represent a vector X by its coordinates xi
according to
X xi ei x1e1 x2 e2 x3 e3
basis vectors
X xi ei x1e1 x2 e2 x3 e3
i
i j
i
i
j
i
i
Changing the basis to ei Aj e j , ei Bj e j changes the coordinates x Bj x , x Aj x
Matrix A and B are related according to
Aij Bkj ki
-quantities with a subscript transform like the basis vectors and are called covariant
-quantities with a superscript transform like the coordinates are called countervariant
Now we construct a new set of basis vectors, the countervariant basis, which is
identical to the basis of the reciprocal space
j
j
j
Consider the scalar product X ei x e j ei x ei e j x gij xi
gij ei e j
xi x j g ij
where
metric tensor
g ij g jk ki
-as we know from relativity
ds2 dxi ei dx j e j gij dxi dx j
The new reciprocal basis reads
Let’s show that the
ei
ei gij e j
form really a set of basis vectors
coordinates with respect to the reciprocal basis
X xi ei x j g ij ei
xi x j g ij
xj e j
ei gij e j
i
i
Note: in the lecture we introduced reciprocal basis vectors e 2 e
so that e e j 2 ij
i
Application in solid state physics
-we have basis vectors (not necessarily orthogonal) a 1 , a2 , a3
Metric tensor
a1 a1 a1 a2 a1 a3
g a2 a1 a2 a2 a2 a3
a a a a a a
3 1 3 2 3 2
Reciprocal lattice vectors
i
a 2 gij a j
As an example let’s consider the reciprocal lattice of the bcc lattice in real space
-We know from the conventional approach
bcc: a1=a(½, ½,-½), a2=a(-½, ½,½) and a3=a(½,- ½,½)
ex
e y ez
1
1
2
a2 a3 a 1/ 2 1/ 2 1/ 2 a2 e x e y
2
2
1/ 2 1/ 2 1/ 2
2
2
g a2 a3 1, 1, 0
1 V
a
1 1
2 2
1 1 a3
V a1 a 2 a3 a3
2 2 2
1/ 2 0
2
2
g a3 a1 0, 1, 1
2 V
a
and
2
2
g a1 a2 1, 0, 1
3 V
a
-Now we use the metric tensor
a1 a1 a2 a2 a3 a3 3a2 / 4
3 1 1
a
g
ij 4 1 3 1
1 1 3
2
g1 a 2 g11 a1 g12 a2 g13 a3
1
a1 a3 a3 a1 a2 / 4
a1 a2 a2 a1 a2 / 4
g g
ij
ij
1
a2 a3 a3 a2 a2 / 4
2 1 1
1
2 1 2 1
a
1 1 2
a/2
a / 2
a / 2
1
1
1
2
2
1
2 2 a / 2 2 a / 2 2 a / 2
a a / 2 a a / 2 a a / 2 a 0
etc.