Chap 7
Non-interacting electrons in a periodic potential
• Bloch theorem
• The central equation
• Brillouin zone
• Rotational symmetry
Dept of Phys
M.C. Chang
Bloch recalled,
The main problem was to explain how the electrons could sneak by
all the ions in a metal so as to avoid a mean free path of the order of
atomic distances. Such a distance was much too short to explain the
observed resistances, which even demanded that the mean free path
become longer and longer with decreasing temperature.
By straight Fourier analysis I found to my delight that the wave differed
from the plane wave of free electrons only by a periodic modulation.
This was so simple that I didn't think it could be much of a discovery,
but when I showed it to Heisenberg he said right away: "That's it!"
Hoddeson L – Out of the Crystal Maze, p.107
⎡ =2 2
G ⎤
(
) ⎥ψ = εψ
U
r
−
∇
+
⎢ 2m
⎣
⎦
G
G
G
U (r + R) = U (r )
Lattice Hamiltonian
U
a
• Translation operator
⎛ i G G⎞
T = exp ⎜ p ⋅ R ⎟ ,
⎝=
⎠
G
R
(T
G
a1
, TaG2 , TaG3 mutually commute
)
• Translation symmetry of the lattice → ⎣⎡ H , TRG ⎦⎤ = 0
• Simultaneous eigen-states
Hψ αβ = ε αψ αβ
|Ψ(x)|2 is the same
in each unit cell
Re-label
TRGψ αβ = CRG , βψ αβ
Hψ nkG = ε nkGψ nkG
CRG = 1; CRG CRG ' = CRG + RG '
TRGψ nkG = e ψ nkG
G G
iβ ⋅R
⇒ CRG , βG = e
G G
G G
G
iβ ⋅R
∴ψ αβ (r + R) = e ψ αβ (r )
G G
ik ⋅ R
Bloch energy,
Bloch state
orthogonality
ψ nkG ψ n ' kG ' = δ nn 'δ kGGk '
[× V ]
Bloch theorem (1928)
The electron states in a periodic potential can be written as
GG
G
G
ik ⋅r
ψ nkG (r ) = e unkG (r )
Marder’s has
a factor
1
N
where uk(r)= uk(r+R) is a cell-periodic function
Pf:
GG
G
G
− ik ⋅r
define
unkG (r ) = e ψ nkG (r )
G G
G
G
G
ik ⋅ R
then from ψ nkG (r + R ) = e ψ nkG (r )
G
G
G
⇒
unkG (r + R ) = unkG (r ).
The cell-periodic part unk(x) depends on the form of the potential.
• Effective Hamiltonian for u(r)
G
H (k )unkG = ε nkG unkG within one unit cell
2
2
GG
GG
G
G
∇
=
G
⎛
⎞
where H (k ) ≡ e − ik ⋅r Heik ⋅r =
+
+
(
)
k
U
r
⎜
⎟
2m ⎝ i
⎠
(1023 times less effort than the original Schrodinger eq.)
important
Allowed values of k are determined by the B.C.
Periodic B.C.
(3-dim case)
G
G
G
Ψ nkG (r + N i ai ) = Ψ nkG (r ), i =1,2,3
→e
G G
iNi k ⋅ai
= 1, ∀i
G G
→ N i k ⋅ ai = 2π mi , mi ∈ Z , ∀i
G m G m G m3 G
→ k = 1 b1 + 2 b2 +
b3
N1
N2
N3
G
G
G
G
⎛
⎞ 1 G G G
b
b
b
3
3
1
2
Δk=
⋅⎜
×
⎟ = b1 ⋅ (b2 × b3 )
N1 ⎝ N 2 N 3 ⎠ N
1 (2π )3 G G G
V
=
, a1 ⋅ (a2 × a3 ) = v =
N v
N
(2π )3
=
, as in the free-electron case.
V
N =N1 N 2 N 3
G G G
b1 ⋅ (b2 × b3 )
G
=N
3
Δk
Therefore, there are N k-points in a unit cell (of reciprocal lattice),
where N = total number of primitive cells in the crystal.
important
Fourier decomposition and reciprocal lattice vectors
If f(r) has lattice translation symmetry, f(r)=f(r+R), for any lattice vector R
[eg. f(r) can be the lattice potential], then it can be expanded as,
G
f (r ) =
∑e
G
all G
G G
iG ⋅r
f GG
where G is the reciprocal lattice vector.
Pf:
Fourier expanson,
GG
G
G
ik ⋅r
f (r ) = ∑
e f (k )
G
k
G G GG
G
G G
G
ik ⋅ R ik ⋅r
(
)
=
(
).
f ( r + R) = ∑
e
e
f
k
f
r
G
k
G G
ik ⋅ R
Therefore, e
G
= 1 for ∀ R
G
G
G G
k = hb1 + kb2 + lb3
G
= Ghkl
∀ h, k , l
The expansion above is very general, it applies to
• all types of Bravais lattice (e.g. bcc, fcc, tetragonal, orthorombic...)
• in every dimension (1, 2, and 3)
All you need to do is find out the reciprocal lattice vectors G
A simple example:
electron density (or potential, or cell-periodic function)
of a 1-dim lattice
ρ(x)
G
a1 = ax ,
G
→ b1 = (2π / a ) x
x
a
G
G
Gn = nb1 ,
G
G G
Gn ⋅ r = (nb1 ) ⋅ ( xx ) = (2πn / a ) x
∴ ρ ( x) = ∑ ei (2π n / a ) x ρ n
n
important
Some useful formulas (for electrons in a lattice box)
1D
3D
L ⎛ k ⎞
δ (k ) =
δ⎜
⎟
2π ⎝ 2π / L ⎠
L
→
δ k ,0
2π
∴ Lδ k ,0 ↔ 2πδ (k )
δ (k ) =
∑
e
ik ⋅na
k∈1st BZ
∑e
ik ⋅na
G
∴
∑
= N δ n ,0
G
k ∈1st BZ
V
3
e
∑e
G G
ik ⋅ R
G G
ik ⋅ R
G
R
∫ d re
= aδ G ,0
3
⎛
kx
⎝ 2π / Lx
δ⎜
⎞ ⎛ kx
⎟ δ ⎜⎜
⎠ ⎝ 2π / Ly
3
δ kG ,0
= N δ RG ,0
= N δ kG ,0
G G
iG ⋅r
Restrict k to the 1st BZ,
Cf. Marder, App A.4
= vδ GG ,0
cell
cell
note:
⎞ ⎛ kx ⎞
⎟⎟ δ ⎜
⎟
L
2
π
/
⎝
⎠
z
⎠
( 2π )
G
3
V δ kG ,0 ↔ ( 2π ) δ (k )
n
∫ dx e
( 2π )
→
= N δ k ,0
iGx
Lx Ly Lz
∫d re
3
all
GG
ik ⋅r
A+M, App. D
⎧⎪ V δ kG ,0
G
=⎨
3
⎪⎩( 2π ) δ k
( )
for discrete k
for continuous k
important
How do we determine uk(x) from the potential U(x)?
Schrodinger equation
⎡ pG + =kG
⎢
⎢ 2m
⎢⎣
(
)
2
⎤
G ⎥ G G
G
+ U (r ) uk (r ) = ε kG ukG (r )
⎥
⎥⎦
Keypoint: go to k-space to simplify the calculation
Fourier transform
1. the lattice potential
G G
G
iG ⋅r
U (r ) = ∑
U GG e
G
G
2. the wave function
Schrod. eq. in k-space
aka. the central eq.
G − iGG ⋅rG
G
G
u (r ) = ∑
Ck (G )e
G
G
k
(ε
0
G G
k −G
G=2πn/a
k=2πn/L
G
2 2
G
G
=
k
0
G G C G (G ') = 0, ε ≡
− ε k CkG (G ) + ∑
U
k
G ' −G k
G
2m
G'
)
Matrix form of the central eq. (in 1D)
%
⎛ ε k0+ 2 g − ε k
⎜
⎜ Ug
⎜ U2g
⎜
⎜ U3g
⎜
⎝ U4g
G=ng (g=2π/a)
#
⎞ ⎛ Ck (−2 g ) ⎞
⎟⎜
⎟
C
g
−
(
)
k
⎟⎜
⎟
0
Ug
εk − εk
U−g
U −2 g ⎟ ⎜ Ck (0) ⎟ = 0 for a
⎟⎜
⎟
particular k
0
U2g
Ug
ε k−g − ε k
U − g ⎟ ⎜ Ck ( g ) ⎟
U3g
U2g
Ug
ε k0− 2 g − ε k ⎟⎠ ⎜⎝ Ck (2 g ) ⎟⎠
%
#
• For a given k, there are many eigen-energies εnk, with eigen-vectors Cnk.
U−g
ε k0+ g − ε k
U −2 g
U−g
U −3 g
U −2 g
ε
U(x)=0
C(-g)=1
k
-2g
-g
U −4 g
U −3 g
0
ε3k
ε2k
ε1k
C(0)=1
g
C(g)=1
2g
k
1st BZ
• when U(x) ≠0, for a particular k, unk is a linear combination of
plane waves, with coefficients Cnk :
unk ( x) = ∑ Cnk (G )e − iGx
G
Also valid in
higher dim
• From the central eq., one can see
Ck +G ' (G + G ') = Ck (G )
unk ( x) = ∑ Cnk (G )e − iGx
G
⇒ un ,k +G ( x) = e − iGx unk ( x)
ψ n ,k +G ( x) = ψ nk ( x)
• Bloch energy εn,k+G = εnk
Example.
U(x) = 2U cos2πx/a
= U exp(2πix/a)+U exp(-2πix/a) (Ug=U-g=U)
Matrix form of the central eq.
%
⎛ ε k0+ 2 g − ε k
⎜
⎜ U
⎜
0
⎜
0
⎜
⎜
0
⎝
#
U
ε k0+ g − ε k
U
0
0
0
U
ε k0 − ε k
U
0
0
0
U
ε k0− g − ε k
U
0
0
0
U
ε k0− 2 g
⎞ ⎛ C k ( −2 g ) ⎞
⎟⎜
⎟
−
C
g
(
)
⎟⎜ k
⎟
⎟ ⎜ Ck (0) ⎟ = 0
⎟⎜
⎟
⎟ ⎜ Ck ( g ) ⎟
− ε k ⎟⎠ ⎜⎝ Ck (2 g ) ⎟⎠
%
#
A solvable model in 1-dim: The Kronig-Penny model (1930)
(not a bad model for superlattice)
a
U ( x ) = A a ∑ δ ( x − sa ) = ∑ U n eiGn x =U 0 + 2 ∑ U n cosGn x
s
a /2
1
U n = ∫ dx U ( x ) e − iGn x = A
a − a /2
(ε
0
k − ng
− ε k ) C n + ∑ U m − n Cm = 0
m
⇒ Cn =
A
ε k − ε k0− ng
⇒1= ∑
n
n >0
n
∑C
A
ε k − ε k0− ng
m
m
=2
=∑
2mA n
=
1
2K
=
a
4K
1
2π n ⎞
⎛
K −⎜k −
⎟
a ⎠
⎝
2
2
, K2 ≡
2mε k
=2
⎛
⎞
⎜
⎟
1
1
+
∑n ⎜
2π n
2π n ⎟
⎜K +k−
⎟
K −k +
a
a ⎠
⎝
a
a⎤
⎡
cot
K
k
cot
K
k
+
+
−
(
)
(
)
⎢⎣
2
2 ⎥⎦
1
∑n x + nπ = cot x
...
=2
a
sin Ka
⇒
=
2mA 2 K cos ka − cos Ka
P
⇒
sin Ka + cos Ka = cos ka
Ka
K has a real solution when
P
sin Ka + cos Ka
Ka
⎛
mAa 2 ⎞
⎜P =
⎟
=2 ⎠
⎝
ε k = ε n ,k '+G (k ' ∈1st BZ)
=ε n ,k '
Sometimes it is convenient to
extend the domain of k with the
following requirement
Fig from Dr. Suzukis’ note (SUNY@Albany)
1st
Brillouin
zone
First Brillouin zone for 2D reciprocal lattice
ε n ,kG +GG = ε n ,kG
G
G
ψ n ,kG +GG (r ) = ψ n ,kG (r )
Wigner-Seitz cell of
the reciprocal lattice
Ex: Hexagonal lattice
direct lattice
reciprocal lattice
BZ
The first BZ of FCC lattice (its reciprocal lattice is BCC lattice)
4π/a
The first BZ of BCC lattice (its reciprocal lattice is FCC lattice)
z
4π/a
y
x
Momentum and velocity of an electron in a Bloch state
• momentum
Not a momentum eigen-state
=
∇ψ nk
i
GG
G
ik ⋅r =
= =kψ nk + e
∇unk
i
Crystal (quasi)-momentum
• velocity
=
1
G G
vn (k ) = ψ nk ∇ ψ nk
m
i
G
1
=
=
unk ∇ + k unk
m
i
G
∂H (k )
G unk
= unk
=∂k
G
∂ε n (k )
G
=
=∂k
~ group velocity
ψ nk ( x) = ∑ Cn ,k −G ei ( k −G ) x
G
∴ for a Bloch state with crystal
momentum ħk, the actual momentum
can have, with various probabilities,
an infinity of values.
GG
GG
G
− ik ⋅r
ik ⋅r
H (k ) = e He
2
=2 ⎛ ∇ G ⎞
G
=
⎜ + k ⎟ + U (r )
2m ⎝ i
⎠
G
G
H (k ) unk = ε n (k ) unk
• In a perfect lattice, the
motion is unhindered
• velocity is zero at
Brillouin zone boundary
εn(k)
k
vn(k)
k
A. Wilson (1932)
… Bloch, in showing that tightly bound electrons could in fact move
through the lattice, had "proved too much"—that all solids should be
metals. Were insulators simply very poor conductors? However,
implicit in Peierls's papers on the Hall effect lay the clue, not carried
further by Peierls, that a filled band would carry no current.
A filled band does not carry current (Peierls, 1929)
z Electric current density
G 1
j=
V
G
∂
ε
(
k
d k 1 n )
G
G
(
−
ev
)
=
−
e
∑G
3
∫
π
=
(2
)
∂
k
filled k
3
z Inversion symmetry,εn(k)= εn(-k)
Æ electrons with momenta ħk and -ħk have opposite velocities
Æ no net current in equilibrium
Æ a filled band carries no current even in an external field
Hoddeson L – Out of the Crystal Maze, p.120
Difference between conductor and insulator (Wilson, 1931)
• There are N k-points in an energy band, each k-point can be
occupied by two electrons (spin up and down).
conductor
∴ each energy band has 2N “seats” for electrons.
E
• If a solid has odd number of valence electron per primitive
cell, then the energy band is half-filled (conductor).
For example, all alkali metals are conductors
k
• If a
even
solidnumber
has even
of electrons
number ofper
valence
primitive
electron
cell, then
per primitive
there are 2
possibilities:
cell,
then the(a)
energy
no energy
band is
overlap
filled (insulator).
or (b) energy overlap.
Wilson
recalls
Bloch's
replycan
after
his or
arguments:
E.g., alkali
earth
elements
behearing
conductor
insulator.
"No, it's quite wrong, quite wrong, quite wrong, not possible at all."
insulator
conductor
E
E
k
k
A nearly-filled band
• “charge” of a hole:
G
e
j =−
V
=−
G
v
∑G
filled k
e⎛
G
G⎞
v
v
−
∑G ⎟
⎜ ∑
V ⎝ kG∈1st BZ
unfilled k ⎠
e
=+
V
∑
G
v
G
unfilled k
∴ unoccupied states
behave as +e charge
carriers
• “momentum” of a hole:
If an electron of wavevector k0 is missing, then the sum over filled k, Σk= -k0.
Alternatively speaking, a hole with wavevector kh (= -k0). is produced.
important
DOS for a Bloch band εnk
Surface ε+dε =const
dSε
dk⊥
Surface ε=const
⎛ L ⎞
D (ε )d ε = ⎜
⎟
⎝ 2π ⎠
3
∫
Shell
d 3k
d 3 k = dSε dk ⊥
d ε = ∇ kG ε dk ⊥
∴ d 3 k = dSε
dε
∇ kG ε
⎛ L ⎞
⇒ D (ε ) = ⎜
⎟
⎝ 2π ⎠
3
∫
dSε
∇ kG ε
If vk = ∇ kG ε = 0, then there is
"van Hove singularity" (1953)
For example, DOS of 1D energy bands
Van Hove
singularity