5.73 Lecture 39
39 - 1
One Dimensional Lattice: Weak Coupling Limit
See Baym “Lectures on Quantum Mechanics” pages 237-241.
Each atom in lattice represented as a 1-D V(x) that could bind an unspecified number of
electronic states.
Lattice could consist of two or more different types of atoms.
Periodic structure: repeated for each “unit cell”, of length l.
Consider a finite lattice (N atoms) but impose periodic (head-to-tail) boundary condition.
L = Nl
Each unit cell, eq:
A+
•
B–
×
A–
•
Vi(x) i-th unit cell
This is an infinitely repeated finite interval: Fourier Series
V( x) =
∞
∑
e iKnx Vn
n =−∞
K=
2π
l
“reciprocal lattice vector”
updated September 19,
2003
5.73 Lecture 39
39 - 2
Vn is the (possibly complex) Fourier coefficient of the part of V(x) that looks like a
free particle state with wave-vector Kn (momentum hKn). Note that Kn is larger
than the largest k (shortest λ) free particle state that can be supported by a lattice
of spacing l.
Kn = n
2π
,
l
first Brillouin Zone for k
−
π
l
≤k≤
π
l
We will see that the lattice
is able to exchange momentum in quanta of hnK with the
r
free particle. In 3-D, K is a vector.
To solve for the effect of V(x) on a free particle, we use perturbation theory.
1.
Define basis set.
2
p2
h
d2
H =
=−
2m
2m dx 2
V( 0 ) = cons tan t
( 0)
ψ (k0 ) = L−1/ 2e ikx
E
2.
( 0)
k
k2
=
2m
h
H =
(1)
2
∞
∑
e iKn Vn
n =−∞


= ∫ dx[L−1/ 2e − ik ′x ]∑ e iKnx Vn [L−1/ 2e ikx ]
0
 n

(
)
e ix k + Kn − k ′ Vn
Matrix elements: H
H
(1)
k ′k
1 L
= ∫ dx ∑
L 0
n
(1)
k ′k
L
integral = 0 if k + Kn − k ′ ≠ 0
∴ k ′ = k + Kn
H(k1′)k =
1
L ∑ Vn δ k ′,k + Kn = ∑ Vn δ k ′,k + Kn
L n
n
updated September 19,
2003
5.73 Lecture 39
39 - 3
Must be careful about H(kk1) ′ (relative to H(k1′)k )
1 L
(
)
H = ∫ dx ∑ e ix − k + Kn + k ′ Vn = ∑ Vn δ k ',k − Kn
L 0
n
n
but Hermitian H requires H(kk1) ′ = H(k1′)*k
(1)
kk ′
∴ ∑ Vn δ k ',k − Kn = ∑ Vn*δ k ',k + Kn
n
n
true if Vn = V−*n
So now that we have the matrix elements of H(0) and H(1), the problem is essentially
solved. All that remains is to plug into perturbation theory and arrange the results.
( 0)
(1)
Solve for ψ k = ψ k + ψ k
3.
ψ (k0 ) = L−1/ 2e ikx
−1/ 2
ψ =L
(1)
k
−1/ 2
ψ =L
(1)
k
ψ
ik ′x
(1) ik ′x
V
e
δ
H
e
−
1
/
2
′
,
−
n
k
k
Kn
′
( Σ ′ means k ′ ≠ k )
Σ ′ ( 0kk)
Σ′ ( 0)
( 0) = L
( 0)
n E
n
E k − E k − Kn
k − Ek′
i( k − Kn )x
V
e
n
Σ′ ( 0)
( 0)
n E
k − E k − Kn
(1)*
k
−1/ 2
=L
Vn*e − i( k − Kn )x
′
Σ ( 0)
( 0)
n E
k − E k − Kn
Vn* = V− n
ψ
(1)*
k
−1/ 2
=L
− i( k − Kn )x
− i( k + Kn )x
V
e
V
e
1
/
2
−
n
n
−
= L Σ ′ ( 0)
Σ′ ( 0)
( 0)
( 0)
n E
−n E
k − E k − Kn
k − E k + Kn
But n is just a dummy index, so replace –n by n.
updated September 19,
2003
5.73 Lecture 39
39 - 4
4. Use ψ k and ψ *k to compute E k = E (k0 ) + E (k1) + E (k2 ) .
Rather than use the usual formula for E(2), go back to the λn formulation of
perturbation theory.
E k = λ0E (k0 ) + λ1E (k1) + λ2E (k2 ) = ψ k λ0H( 0 ) + λ1H(1) ψ k
Retain terms only through λ2
− i( k + Kn )x
2
2


V
e
d
1 L  − ikx
h
iKmx 
n
′
Vm e
E k = ∫ dx e
+ λ Σ ( 0)
− 2m dx 2 + λ Σ

( 0)
n E
m
−
E
L 0 
k
k + Kn 
i( k − Kn ′ )x
 ikx

V
e
′
n
× e + λ Σ ′ ( 0 )

( 0)
n′ E
−
E
′

k
k − Kn 
E (k0 )
1  h2 ( 2 )  0 h2k 2
= λ −
−k L  = λ
2m

L  2m
E (k0 )
 h2 d 2  
1
iKmx
ikx
− ikx
= λ ∫ dx e
∑ e Vme + 2 terms involving  − 2m dx2  
L
m

0

d 2 ikx
2 ikx 
=
−
e
k
e 
recall

dx 2


1
1st term, only m = 0 term in sum gives nonzero integral.
2nd terms, need n or n′ = 0 term from sum, but these are excluded by Σ′.
1
E = λ LV0 = λ1V0
L
(1)
k
1
updated September 19,
2003
5.73 Lecture 39
39 - 5
i( k − Kn′ ) x
∞
∞

V
e
1
E (k2 ) = λ2 ∫ dx e − ikx ∑ Vme iKmx Σ′ (n0′)
( 0)
L 
E
E
−
n′ = −∞ k
m = −∞
k − Kn′
− i( k + Kn ) x 
 ikx 
V
e
iKm
n
Vme  e 
+ ∫ dx Σ ′ ( 0 )
∑

( 0)
n≠ 0 E


k − E k + Kn  m
0 = – k + Km + k − Kn′, requires m = n′
0 = – k − Kn + Km + k, requires m = n
1st term
2nd term
E
( 2)
k
2
2

V
V
1 2
m
m
′
+
dx
= λ ∫ dx Σ ′ ( 0 )
Σ

( 0)
( 0)
( 0)
∫
m E
m E
−
E
−
E
L 
k
k + Km 
k
k − Km
E
( 2)
k
Vn2
= 2λ Σ ′ ( 0 )
( 0)
m = – ∞ E k − E k + Kn
∞
2
Combine terms for n and − n and sum
∞
∑
n =1
E (k0 ) − E (k0±) Kn =
2
k
[
2m
h
2
1
1
+
E (k0 ) − E (k0+)Kn E (k0 ) − E k − Kn
E
( 2)
k
=
8m
h
2
∞
∑
n =1
2
Kn
[Kn ± 2k ]
2m
4m
1
= 2 2 2
2
h K n − 4k
− (k ± Kn) 2 ] =
h
Vn2
K 2n 2 − 4 k 2
updated September 19,
2003
5.73 Lecture 39
39 - 6
But there are many zeroes in this denominator as n goes 0 → ∞.
Must use degenerate perturbation theory for each small denominator.
1/ 2
2

V
E k + E k ′  E k − E k ′ 
2
 +V 
→ E ± =
± 



E k ′
2
2


E k
Recall 
V
Vn2
k
8m ∞
Ek =
+ V0 + 2 ∑
2 2
2
h n =1 K n − 4 k
2m
h
2 2
Kn
2π
nπ
= ± n= ±
except n = 0
2
2l
l
at k = 0, there are no nearby zeroes
zeroes at k = ±
dE k
dk
k =0
=
k
m
h
(minimum at k = 0)
d 2E k
h
(positive curvature)
=
2
dk k = 0 m
just like free particle!
K
, there are zeroes in denominator, so there is a gap in energy of
2
K
2 V1 at k = ±
2
2 V2 at k = ±K
At k = ±
M
2 Vn at k = ±
nK
2
What does this look like?
updated September 19,
2003
5.73 Lecture 39
39 - 7
 h2  2
E = V0 +   k
 2m
look at text Baym page 240.
updated September 19,
2003
5.73 Lecture 39
39 - 8
But we want to shift each of the segments by integer times K to left or right
so that they all fit within the
−K
K
≤k≤
“first Brillouin Zone”.
2
2
E
2V3
2V2
2V1
V0
-K/2
K/2
k diagram. Curvature gives meff
3 − D k - diagram — much more information. Tells where to find allowed
r
r
transitions as function of 3- D k vector in reciprocal lattice of lattice vector K.
Scattering of free particle off lattice. Conservation of momentum in the sense
r
r
r
k final − k initial = K.
updated September 19,
2003
k