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1
Ch. 8: Electron Levels In Periodic Potential
What are the consequences of a periodic potential on the form of the wave functions, irrespective of the exact form of the potential and irrespective of whether the material is a conductor or an insulator? U is the effective one-
)()( rURrU rrr=+
electron potential.
ψεψψ =⎟⎟⎠
⎞⎜⎜⎝
⎛+∇−= )(
22
2
rUm
H rh
Bloch’s Theorem: The eigenstates of the one-electron Hamiltonian, with a periodic potential in a Bravais lattice, can be chosen to have the form of a plane wave times a function with the periodicity of the Bravais lattice.
Or, equivalently
)()( ruer knrki
kn
rrr
rr
r⋅=ψ )()( Rruru knkn
rrrrr +=
)()( reRr knRki
kn
rrrr
rr
r ψψ ⋅=+
1st Proof Of Bloch’s Theorem
Define translation operator TR:
)()( RrfrfTRrrr
r +=
Hamiltonian is periodic,
)()()()( RrRrHrrHTRrrrrrr
r ++= ψψ
RR HTHT rr =
RRRRRR TTTTT rrrrrr +′′′ ==But, also
Since H and all TR’s commute with one another, eigenstate of H can be chosen to be eigenstate of all TR’s.
ψεψ =H )()()( rRcrTRrrr
r ψψ =
a number
2
1st Proof Of Bloch’s Theorem
ψεψ =H )()()( rRcrTRrrr
r ψψ =
)()()( RcRcRRc ′′+rrrr
)()()( RcRcRRc =+
Let iixi eacπ2)( =r
332211 anananRrrrr
++=)()( reRr Rki r
rr rr ψψ ⋅+
primitive vectora (complex) number
332211 222)( nixnixnix eeeRc πππ=r
RkieRcrrr⋅=)(
332211 bxbxbxkrrrr
++=
)()( reRr knkn rr ψψ =+
To see the constraints on k, we need to introduce boundary conditions.
Born – von Karman Boundary Condition
3,2,1)()( ==+ iraNr iirrr ψψ PBC
i t
3,2,1)()( ==+ ⋅ ireaNr knakiN
iiknii
rrrr
rr
r ψψ
12 =ii xNie π332211 bxbxbxkrrrr
++=
Nmx / 33
22
11 b
mbmbmk
rrrr++=mi integer
volume per allowed state
iii Nmx /= 332
21
1
bN
bN
bN
k ++
( )3213
3
2
2
1
1 1 bbbNN
bNb
Nb
krrr
rrrr
×⋅=⎟⎟⎠
⎞⎜⎜⎝
⎛×⋅=Δ
cellVNk
3)2( π=Δr
same as free electron gas!
3
2nd Proof of Bloch’s Theorem (Plane Wave Expansion)
Working within the Born-von Karman periodic boundary condition, we expand both the one-electron wave function and the effective potential in plane waves
rKi
KK eUrU
rr
rr
r ⋅∑=)( k’s are all the wave vectors allowed by the PBC, but K’s are only reciprocal lattice vectors because the potential is periodic.)(1 rUerd
VU rKi
cellcell
K
rr rrr ⋅−∫=
Plugging into Schroedinger’s eq. to look for eigen states
rki
kk ecr
rr
rr
r ⋅∑=)(ψ
ψεψψ =⎟⎟⎠
⎞⎜⎜⎝
⎛+∇−= )(
22
2
rUm
H rh
∑∑ ∑ ⋅⋅+⋅ =⎬⎫
⎨⎧
+ rkirKkirki ceecUeckrrrrrrrh ε)(
22
*KK UU rr −=
∑∑ ∑ ⋅−⋅ =⎭⎬⎫
⎩⎨⎧
+k
krki
k KKkKk
rki cecUcmke
rr
rr
r rrrrr
rr h ε2
22
Since k’s form a complete orthogonal set, coefficient for a specific plane wave on LHS must equal coefficient for the same k on the RHS.
∑∑ ∑ =⎭⎬
⎩⎨ +
kk
k KkKk ceecUecm r
rr r
rrr ε)(2
k k’-K
Plane Wave Expansion (cont.)
02
22
=+⎟⎟⎠
⎞⎜⎜⎝
⎛− ∑ −
KKkKk cUcm
kr
rrrrh ε
The C coefficient for plane wave with a specific k is coupled to the coefficients for all wave vectors k’ related to k by k’=k+K, where K is a
i i i⎠⎝ K reciprocal lattice vector and UK is non-zero.
Note that U can always be chosen (shifted) so that UK=0=0.
Plane waves not differing by a reciprocal lattice vector are unrelated. Or, if a plane wave is to couple, it can only couple to other plane waves with wave vectors differing by K from its own. Grab all plane waves with wave vectors related to k and form a subset (Hilbert space) that is independent of other subsets. Now a plane wave can only couple to waves within its own subset. To distinguish subsets we can identify them by
For a specific ε, this equation has non-trivial solutions only for a few k’s (or for none).
What does it mean physically?
(1) If UK vanishes for large K’s, say, K>G ?
(2) If UK vanishes for a series of K’s (e.g. nG, where n is an odd integer)?
coup e o w ves w s ow subse . o d s gu s subse s we c de y e bythe one wave vector, q, in each subset that belongs to the first Brillouin zone.
4
Plane Wave Expansion (cont.)
02
22
=+⎟⎟⎠
⎞⎜⎜⎝
⎛− ∑ −
KKkKk cUcm
kr
rrrrh ε
qnqnqnH rrr ψεψ =rKqi
Kqnqn ecrrrr
rrrr ⋅+
+∑= )(,)(ψ
There is an infinite number of plane waves in each subset (because the number of reciprocal lattice points is infinite). However, the number of subsets is finite ( number of allowed q = number of unit cells in the Born von Karman box). Within each subset, the above matrix can be diagonalized to yield an infinite number of discreteeigenvalues εn(q), n=0, 1, 2,.... Each eigenstate is composed of a (sub)set of plane waves from one subset, with their proper C coefficients, which satisfies the Schroedinger equation
qnqnqn qKr ,
What can we say about the C’s, if the periodic potential is weak?
Eigenstates with same q but different n can be made to be orthogonal to each other. Perturbation theory tells us that for an incremental change in q, (i.e. neighboring subsets) there must be allowed eigenvalues that are incremental changed from each other. These states can be viewed as forming bands and n is then the band index.
Plane Wave Expansion (cont.)
∑∑ ∑ ⋅−⋅ =⎭⎬⎫
⎩⎨⎧
+k
krki
k KKkKk
rki cecUcmke
rr
rr
r rrrrr
rr h ε2
22rki
kk ecr
rr
rr
r ⋅∑=)(ψ
Looking at the Schroedinger equation and requesting that Ψ be an eigenstate with eigenvalue ε, we see that only a very small number of subsets could have ε as one of their eigenvalues. A specific Ψ can be expanded in plane waves from these few independent subsets with a specific, allowed, eigenvalue.
a’s are independent and arbitrary
)()(),( ,,,
, qnqnqn
qn rar rrr
rrr εεδψεψ −=∑
few terms
rKqiKqn
Kqn ecr
rrrrr
rrr ⋅+
+∑= )(,)(ψ )(),( ,,,
rar qnqnqn
rrrr
r
ψεψεε
∑=
=
We have as many choices as there are terms in the sum to define eigenstates. Certainly, any eigenfunction can be chosen to have only contribution(s) from one q.
5
2nd Proof of Bloch’s Theorem
rKqiKqn
Kqn ecr
rrrrr
rrr ⋅+
+∑= )(,)(ψ any eigen function of a periodic lattice
all terms periodic with lattice
K
rKiKqn
K
rqiqn ecer
rr
rrr
rrrr ⋅
+⋅ ∑= ,)(ψ
)()( Rruru knknrrr
rr +=)()( ruer knrki
kn
rrr
rr
r⋅=ψ
Velocity of Bloch Electrons
22 rrh ⎟
⎞⎜⎛
)()()(2
22
rrrUm knknkn
rrrrhrrr ψεψ =⎟⎟
⎠
⎞⎜⎜⎝
⎛+∇
−
222
2)( q
mkiq
mHH kqk
hrrrhrrr ++∇−⋅+=
+
...)()()(2
* ++∇−⋅+=+ ∫ nknknn ukiqmurdkqkrrrhrrrr εε
)()()()(2
)( 2 rururUkim
ruH kkkkkrrrrrhr
rrrrr ε=⎟⎟⎠
⎞⎜⎜⎝
⎛++∇−=
)()( ruer knrki
kn
rrr
rr
r⋅=ψ
1)()(2
+∂
+∇++ ∑ nkk εεεε rrrr
But, we also know that
)()( Rruru kkrrr
rr +=
...2
)()(,
+∂∂
+∇⋅+=+ ∑ji
jiji
nnknn qqkk
qkqk εεε r
nknknkn vri
mrrd
krrr
hrr
rh
=∇−=∂∂
∫ )()(1)(* ψψε
know that
6
About Bloch’s Theorem
1. Allowed wave vector k follow same PBC as free l t b t k i l t /h belectrons, but k is no longer momentum/h bar.
2. k can always be confined to the first Brilloun zone.
3. band index n
4 repeated zone representation
)()( Rruru kkrrr
rr +=
)()()()(2
)( 22
rururUkim
ruH kkkkkrrrrrhr
rrrrr ε=⎟⎟⎠
⎞⎜⎜⎝
⎛++∇−=
Kkk rr rrrrr
+= )()( ψψ4. repeated zone representation
5. velocity of electron
6. Filled bands, empty bands, partially filled bands and Fermi surface.
Kknkn
Kknkn
rrr+
+
= ,,,, )()(
εεψψ
Density of Levels
)(8
2)(2 3 kQkdVkQQ nn
rrr∫∑∑ ==
One is often interested in summing quantities from occupied individual electronic levels. In the limit of large crystals, the summation can be
l d b i t l
over Brillouin zone
8 3,n
nknnr ∫∑∑ πreplaced by an integral.
If the interested quantity only depends on k through energy, then the integral can be carried out
)()( εεε QgdVQq ∫== ))((4)()( 3 k
kdgg nn
nn
rr
εεδπ
εε −== ∫∑∑Consider equi-energy surfaces in reciprocal space. For small enough energy difference, we get
)()( kkdSdgr
δεε ∫= )(4)( )( 3 kkdg nSn δπεε ε∫=
)(|)(| kkkd nrr
δεε ∇=
|)(|1
4)(
)( 3 kdSg
nSn n
rεπ
εε ∇
= ∫
7
van Hove Singularities
van Hove singularities in 3D: gn is finite. However, slope divergesdiverges.
Ch. 9: Electrons In Weak Periodic Potential
Assuming weak periodic potential in some metals, a reasonable starting point for eigenstates of conduction electrons is plane waves.
Why do we even expect the periodic potential experienced by some conduction (outer-shell) electrons to be weak in some metals?
(1) Outer shell electrons are forbidden from coming close to the ion cores.
(2) Outside the cores, potential is effectively screened by mobile electrons.
Bloch’s theorem rKki
KKkk ecr
rrr
rrrr
r ⋅−−∑= )()(ψ
2h ⎞⎛
02
)( 22=+⎟⎟
⎠
⎞⎜⎜⎝
⎛−
− ∑′
′−−′−K
KkKKKk cUcmKk
rrrrrrr
rrh ε
Weak Potential: UK’s ~ U
8
Extreme Case: Free Electrons
In the case of ultimately weak potential (free electron gas),
02
22
=⎟⎟⎠
⎞⎜⎜⎝
⎛− qcm
qr
h ε
rqiq e
rrr
⋅∝ψ222
qmqh
r =ε
22
0
2q
mqh
r =ε
00 ....1 mKqKq
rrrr++ == εε
For free electrons or nearly free electrons, energy is related to the plane wave vector in an obvious way. Inconvenient to use band index. Convenient to use extend zone scheme, i.e. full wave vector.
degenerate case
rKim
jj
rqiq
jeaerrrr
r⋅
=
⋅ ∑=1
ψ
a’s are independent and arbitrary
Nearly Free Electrons: Non-degenerate
22
0
2q
mqh
r =ε∑∑ ∑ ⋅−⋅ =⎭⎬⎫
⎩⎨⎧
+k
krki
k KKkKk
ok
rki cecUcer
rrr
r rrrrrr
rr
εε
First, non-degenerate cases, by which we mean
( ) 0111
0 =+− ∑ −−−−K
KkKKKkKk cUc r rrrrrrrr εε
KKallforUKkKkrr
rrrr ≠′>>− ′−− ||00 εε
Concentrate on eigenstate “derived” from k-K1 plane wave
Expect energies and wave functions to be only slightly modified from FEG case.
factor of U/εsmaller than ck-K1
summing over all K (U0 = 0)
∑∑ ∑ −⋅−′
′−−′−−⋅− =
⎭⎬⎫
⎩⎨⎧
+k
KkrKki
k KKKkKKk
oKk
rKki cecUcer
rrrrr
r rrrrrrrrr
rrr )()( εεTo find
we rewrite the topmost eq.
eigenvalue “derived” from k-K1, not k-K
Kkc rr−
∑≠′
′−−′−−−−+=−
111
)(KK
KkKKKkKKKko
Kk cUcUc rr rrrrrrrrrrrrεε
9
Non-Degenerate NFE
∑′
′−−′−−
− −+
−= 11 00
KK
KkKKKkKKKk
cUcUc
rr r
rrrr
r
rrrr
rr
εεεε
To order of U2,
≠′ −− 1KK KkKkrrrr εεεε
)( 2UO
( ) 0111
0 =+− ∑ −−−−K
KkKKKkKk cUc r rrrrrrrr εε
...1 000 111
1+
−+= ∑ −−−−
K KkKk
KkKKKK
KkKk
cUU
c r rrrr
rrrrrr
rr
rr
εεεε
copied from last page
small
)(||
300
20
1
1
1UO
U
K KkKk
KKKk +−+= ∑
−−
−
− r rrrr
rr
rr
εεεε
Shift in energy is to order of U2. Repelled by other energy levels.
11 −−− K KkKkKk
Perturbation Theory
Equivalently, we could have used perturbation theory and would have gotten the answer quicker
22
0 2∇−=
mH h )(1 rUH
r=
21
20 )(
2Kk
mrrh
−=ε
rKkierrr⋅−∝ )(0 1ψ
...||||||2
111
0 +−
>−−<+>−−−−
+>−=>−−−
−∑ KkU
KkKkKkKk
KK
K
newrrrrrr
rrrr
rr
r εε
1/ KkKk cc rrrr −−=
1KKU rr−=
10
Nearly Free Electrons: m-Fold Near-Degeneracy
Near degeneracy means:
mjiU 1|| 00 =≤εε
mKkKk KKKmiUirrr
rrrr ,...,...,,1,|| 100 ≠=>>−−−
εε
mjiUij KkKk
...,,1,,|| =≤−−−rrrr εε
∑ ′−−′−− += 11 00 KkKKKkKK
Kk
cUcUc
rrrrrrrr
rrIn non-degenerate case the c’s for the plane wave component to the ∑
≠′ −−− −−
1
00KK KkKk
Kk rr rrrr εεεεthe plane wave component to the state derived from k-K1 is
In case of near-degeneracy, c’s are not necessarily small. Eigen states are not (necessarily) dominated by one plane wave component, even for very weak potential.
Degenerate NFE
( ) 0111
0 =+− ∑ −−−−K
KkKKKkKk cUc r rrrrrrrr εεfrom plane wave expansion
di ti i h t
( ) micUcUcm
ijijiiKKK
KkKK
m
jKkKKKkKk ,...,1,
...1
0
1
=+=− ∑∑≠
−−=
−−−− rrrrrrrrrrrrrrrεε
mKKK
KkKK
m
jKkKK
KkKk KKKcUcUc
mjj
rrr
rrrrrrrrrrr
rr
rr ,...,,1 1...1
01
≠⎟⎟⎠
⎞⎜⎜⎝
⎛+
−= ∑∑
≠′′−−′
=−−
−− εε
distinguish two groups
m
m
jKkKK
KkKk KKKUOcUc jj
rrrrrrr
rr
rr ,...,,)(1 12
10 ≠+−
= ∑=
−−−
− εε
11
Degenerate NFE
( ) miUOUUccUc m KKKKKkm
KkKKKkKkji
jjijii,...,1,)( 30
0 =+⎟⎟⎠
⎞⎜⎜⎝
⎛+=− ∑ ∑∑ −−−−−−−
rrrr
rrrrrrrrrr
εεεε
( ) miUOcUc mj
KkKKKkKk jijii,...,1,)( 2
1
0 =+=− ∑=
−−−−rrrrrrrrεε
( )j KKK Kk
Kkj
KkKkKkm
jjijii1 ...1 1
⎟⎠
⎜⎝ −
∑ ∑∑= ≠ −=
rrr rrεε
To first order in U only need to consider the nearly degenerate statesTo first order in U, only need to consider the nearly degenerate states.
Eigenvalue evolves with wave vector. When degeneracy occurs for certain k-K1, it occurs for a small volume around that point in the k-space.
Splitting Near Single Bragg Plane
Only two vectors K1 and K2 leads to near-degeneracy (within U) of the k-K level near a single Bragg plane. Namely, not 3-or more-fold degenerate.
Choose U such that Uq=0=0.
0,|| 00rrr
rrr KKforUKqq ≠′>>− ′−εεUqKq
12
Single Bragg Plane
For non-trivial solutions to be found, 00*
0
=−−−−
KqK
Kq
UU
rrr
rr
εεεε
−KqK
( ) 2200
00
221
KKqq
Kqq U rrrr
rrr +⎟⎟⎠
⎞⎜⎜⎝
⎛ −±+= −−
εεεεε
( ) 0|| 200002 =−++− −− KqKqqKq U rrrrrrr εεεεεε
⎠⎝
Kq U rr ±=0εεFor q lying in the Bragg plane
Additional Comments: Single Bragg Plane
( ) ( ) 22000041
21
KKqqKqq U rrrrrrr +−±+= −− εεεεε 42
⎟⎠⎞
⎜⎝⎛ −=
∂∂ Kq
mqrrh
r 212ε Gradient lies in the Bragg plane.
Constant energy surface perp. to plane.
stationary state (standing wave)
( ) KqKqq cUc rrrrr −=− 0εε
KqKq cUc rrrr −±= )sgn( Assume UK to be real (potential has inversion symmetry)
( ) KqKqqhigher band
lower band
13
Nature of States
rrrrrrrrr
KqKq cUc rrrr −±= )sgn(
0>KU r ⎪⎩
⎪⎨⎧
−=⋅∝
+=⋅∝
Kq
Kq
UrKrUrKr
rr
rr
rrr
rrr
02212
02212
)(sin)()(cos)(
εεψ
εεψIon cores at a/2.
p-like
2/)cos(~ 21)2/()()( rKeee rKqirKqirqi r
rrrrrrr⋅=+ ⋅−⋅−⋅+ψ
2/)sin(~ 21)2/()()( rKieee rKqirKqirqi r
rrrrrrrrr⋅=− ⋅−⋅−⋅−ψ
Direction of propagation?
0
14
3D Examples
FEG energy bands. NoteFEG energy bands. Note degeneracy.
Energy gaps at Bragg planes.
There are as many k-states in a Brillouin zone as there are cells in the Born von Karman box. Two electrons (spin up and down) to each k-state. At 1 electron per unit cell, the lowest band should be exactly half-filled or, in some extreme cases, about half-filled.
Brillouin Zones And Fermi Surfaces
||22
2||
22
Kinner UkK
mE rh −
⎥⎥⎦
⎤
⎢⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛=
||22
2||
22
Kouter UkKE rh +
⎥⎥⎤
⎢⎢⎡
+⎟⎠⎞
⎜⎝⎛=
(1) Draw the free electron Fermi surface
(2) Deform it slightly near Bragg plane
(3) Translate the portion within n-th Brillouin zone through all reciprocal lattice vectors. This constructs a repeated zone Fermi surface.
||22 || Kouter m ⎥
⎥⎦⎢
⎢⎣
⎟⎠
⎜⎝
15
Shape of N-th Brillouin Zones
Free Electron Fermi Surface
valence 4
16
FCC Valence 2 and 3
Lattices with Basis (Identical Atoms)
)()( jdRrrUrrrr
−−= ∑∑ φ
∑∫ −−= ⋅−jR
jrKi
cellKdRrerd
vU
,
)(1r
rr
r
rrrr φ
Geometrical
Structure Factor
)()( jjR
r ∑∑ φ
*)(1 KK SKvU rr
rφ=
r r
∫
∑∫ −= ⋅−j
jrKi
spacedrerd
v)(1
rrr rr φ
∑ ⋅−= dKiK jeSrr
r
Structure Factor)()( rerdK rKispace
rrr r φφ ⋅−∫= j
No splitting on Bragg planes with vanishing structure factors, e.g. basal planes of hcp metal. Convenient to use “Jones zones”: zero-gap planes ignored.
17
Other Contributions To Degeneracy Removal
22
∇H h 20 2∇−=
mH
)(1 rUHr
=
SLHrr⋅=ξ2
Spin-orbit coupling
Spin-orbit coupling is important for heavy elements and for degeneracy not removed by weak periodic potential.
Homework Chapter 8 - 10
Chapter 8: Problems 1
Chapter 9: Problems 1, 3
Chapter 10: Problem 1
Due 3/16/2012