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.