5.73 Lecture #37
37 - 1
Infinite 1-D Lattice
CTDL, pages 1156-1168
LAST TIME:
hole ( h + ) vs. e – configurations:
l
N
(
↔ l2 2
l
+ 1) − N
for N > 2l + 1
2
e rij unchanged
[ζ
ζ ( NLS ) → −ζ ( NLS )
nl
unchanged
]
N
Hund’s 3rd Rule (Lowest L - S term of l only)
E MIN for J = L − S
N < 2l + 1
( 2 l +1) +1
N = 2l + 1
S
J=
2 l +1
2
E MIN for J = L + S
N > 2l + 1
regular
S state: no fine structure
inverted
Zeeman Effect
Wigner-Eckart Theorem used to define gJ
EZeeman = −µ0 M J g J Bz
equal spacings
gJ as L,S,J diagnostic
gJ = 1 +
Confirm by H
J (J + 1) + S(S + 1) − L( L + 1)
Zeeman
2J (J + 1)
in Slater determinantal basis
TODAY:
1. H2+ as example of localization, delocalization, tunneling
2. ∞ secular equation for simplified 1- D lattice
3. eigenvectors by equal probability trick
4. restrict k to k < π
l
5. E ( k) = E0 − 2 A cos kl
next
lecture
6.
7.

8.
9.
: 1st Brillouin Zone
(all of the allowed states?)
Bloch functions ψ k ( x ) = e ikxuk ( x )
wavepackets,motion, group velocity
transitions – energy bands and intensity profiles
conductivity
updated September 19, 2003 @ 2:22
5.73 Lecture #37
37 - 2
Start with H2+ , a lattice with only 2 equivalent sites.
qualitative picture:
atomic energy levels
tunneling between identical localized states
slow behind big barrier (small splitting)
fast behind small barrier (large splitting)
levels → bands, of width related to tunneling rate
−R 2
H2+
+R 2
x
•
)
E(n0+1
2 atoms ∅ 2 states
E(n0 )
R >> a0
Elowest = −
ℜ
12
and doubly degenerate
for exact degeneracy, can choose any linear combination
Localized basis set
(0)
(0)
ψ localized = ψ left
or ψ right
Delocalized basis set
(0)
(0)
ψ delocalized = 2−1/2 ψ left
± ψ right
[
]
updated September 19, 2003 @ 2:22
5.73 Lecture #37
37 - 3
If initially in localized state, tunneling rate depends on
(
)
* height relative to E(0)
of barrier
n
* width of barrier
(0)
(0)
* size of overlap between exponential tails of ψ left
and ψ right
clear that tunneling rate (i.e. splitting) increases
* as n↑ at constant R (internuclear separation)
* as R↓ at constant n
E
0
E1
∆
double degeneracy
at R → ∞
R
∆ is tunneling splitting—gets larger as R↓
N ATOMS ALONG A STRAIGHT LINE
wide
narrow
♦
♦ N atoms, N states
each electronic state of isolated atom becomes band of states for ∞ lattice.
Energy width of each band increases as the principal q.n. increases because
atomic states require more room: ⟨r⟩n ∝ a0n2. Tunneling gets faster.
Greater sensitivity to world outside one atom.
updated September 19, 2003 @ 2:22
5.73 Lecture #37
37 - 4
Simplified Model for ∞ 1–Dimensional Lattice: basis for qualitative insights and
early time predictions.
1. Each ion, called q, has one bound state, νq⟩
at E0 = ⟨νqHνq⟩ [diagonal element of H]
(actually 2 spin-orbitals)
2. permit orbitals only on adjacent ions to interact [simplifying assumption] like
Hückel theory.
3. symmetry: all ions are equally spaced, xq+1 – xq = l, and all adjacent-orbital
interaction matrix elements are identical
[off-diagonal elements of H]
⟨νqHνq+1⟩ ≡ −Α
(ℑAℑ would increase as l → 0)
[reasons for – A sign choice later.]
 E0
 −A

so H = 




−A
O
O
0
O
E0
−A




−A

O O


O
0
tridiagonal infinite
matrix
since this is infinite, need a trick to diagonalize it.
general variational function
ϕ =
∞
∑
cq ν q
superposition of AO’s at
each site
q = −∞
get requirements on cq by plugging this into Schrödinger
equation
Hϕ = E ϕ
left multiply by ν q
updated September 19, 2003 @ 2:22
5.73 Lecture #37
37 - 5
q’th
position
LHS
(0…1…0)
picks out q-th row of H
ν q Hϕ = E ν q ϕ
(0…− AE 0 − A…0 )
RHS
[
E νq ϕ
∴
] = E [c ]
q
[
 c−∞ 
 M 
  = − Acq − 1 + E0cq − Acq + 1
 M 
 
 c+∞ 
]
0 = cq E0 − E − cq − 1 A − cq + 1 A
comes from the
assumed simple
form of model
TRICK: probability of finding e– on each lattice site should be the same
for all sites (complex amplitudes might differ but probabilities
will be constant)
let cq = e
ikql
2
cq = 1
for all q
This choice of cq is a good guess that is consistent with expectation of
equal probabilities on each lattice site.
is distance between adjacent atoms
q is integer
ql is the coordinate of the q-th site: looks like eikx plane wave
k is of dimension l–1
problem reduces to finding allowed values of k.
l
periodicity of lattice provides the important result that if k is replaced by k′ ,
where k′ = k +
2π
l
, the wavefunction does not change (translational symmetry)
updated September 19, 2003 @ 2:22
5.73 Lecture #37
37 - 6
cq′ = e ik ′q = e
l

2π 
ql 
 ikql + i


l
i 2 πq
= e ikq e{
= cq
l
=1
Since all distinguishable ϕ may be generated by choosing k in the interval
−
π
l
≤k<
π
l
, restrict k to this range: called “First Brillouin zone”.
Return to question about what happens when k is not in 1st Brillouin Zone next
time [get another part of the band structure using qualitative perturbation
theory rather than a matrix diagonalization calculation].
Plug
cq = e ikql into Schrödinger Equation
0 = cq ( E0 − E ) − A(cq +1 + cq −1 )
ik q +1 l
ik q −1 l
0 = e ikql ( E0 − E ) − A(e ( ) + e ( ) )
divide by e ikq and rearrange
l
− ik
]
E = E0 − A[1
e ik42
+ e4
3
l
2 cos kl
l
This is the condition on
E,k that must be satisfied
for all eigenfunctions of
the Schrödinger equation
E = E0 − 2 A cos kl
updated September 19, 2003 @ 2:22
5.73 Lecture #37
37 - 7
E varies continuously over finite interval E0 ± 2 A
E(k)
E0 + 2 A
E0 —
E0 − 2 A
−π
l
0
+π
l
k
The choice ν q H ν q +1 = − A leads to minimum E at k = 0.
Are these all of the allowed energy levels that arise from a single orbital at
each lattice site? Apparently not — see next time. Only half of the states.
[One orbital per atom → two spin-orbitals per atom. Antisymmetrization gives
another separate band.]
Could repeat calculation for a higher energy state at each site. Would get a
broader band centered at higher energy.
updated September 19, 2003 @ 2:22
5.73 Lecture #37
37 - 8
closer look at spatial form of ϕ k ( x ) ≡ x ϕ k
ϕ k ( x) = x ϕ k =
+∞
∑
e ikq x νq
123
l
q =−∞
νq ( x)
goal is to replace infinite sum by single term:
This is called
a Bloch function
6
474
8
ϕ k ( x ) ~ e ikxuk ( x )
show that:
periodicity
of lattice
plane wave
(Free particle)
begin by requiring that ϕ k ( x) =
∞
∑
q =−∞
e ikql νq ( x)
Translational symmetry imposes a relationship
between νq(x) and ν0
each νq(x) is localized at site q.
ν q ( x ) = ν 0 ( x − ql )
ϕ k( x ) =
∑ e ν0 ( x − ql)
∞
ikql
q = −∞
ϕ k( x + l ) =
x + l − ql )
∑ e ikq ν10 (4
4244
3
∞
shift x by
–ql to get
from site q
to site 0
l
q = −∞
= e ik
l
( (
))
=ν 0 x − q −1
l
ik( q − 1)
ν 0 ( x − (q − 1)l )
∑ e
l
updated September 19, 2003 @ 2:22
5.73 Lecture #37
37 - 9
re-index sum (replace q–1 by q)
ϕ k( x + l ) = e ϕ k( x )
ikl
translation
by l!
This form of φ k has all of the symmetry properties we will need. This form
is sufficient to satisfy the symmetry requirements (boundary conditions).
This means, instead of writing ϕ k ( x ) as sum over atom - localized
ν q ( x )’s, it is possible to write ϕ k ( x ) as product of 2 factors
ϕ k ( x ) = e ikxuk ( x )
1st factor conveys translational symmetry of a plane wave with wavevector k, 2nd
factor builds in translational symmetry of lattice with spacing l. This is a more
general expression that incorporates all of the properties of the original definition
of ϕk(x) as a sum over localized orbitals.
u k ( x + l ) = uk ( x )
note that
[
]
ϕ k ( x + l ) = e ikx e ik uk ( x + l ) = e ik e ikxuk ( x )
l
l
= e ik ϕ k ( x )
l
as required.
Note also that ℑϕk(x + nl)ℑ2 = ℑϕk(x)ℑ2 implies that, as required, e– has
equal probability of being found on each site.
updated September 19, 2003 @ 2:22