9 Handout Application of LCAO to Energy Bands in Solids and the Tight Binding Method
In this lecture you will learn:
An approach to energy bands in solids using LCAO and the tight binding method
Energy
Es
4Vss
a
a
k
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Example: A 1D Crystal with 1 Orbital per Primitive Cell
Consider a 1D lattice of atoms:
r a1
r r Rm = m a1
a
Each atom has the energy levels as shown The electrons in the lowest energy level(s) are well localized and do not take Energy levels part in bonding with neighboring atoms The electrons in the outermost s-orbital participate in bonding The crystal has the Hamiltonian: 0
r Va (r )
x
Es
s (r )
r
0
r
r V (r )
rr h2 2 H= + Va r Rm 2m m
(
)
Potential in a crystal
0
a
x
ECE 407 Spring 2009 Farhan Rana Cornell University
1
Tight Binding Approach for a 1D Crystal r
r a1 a
rr h2 2 H= + Va r Rm 2m m
r Rm = m a1
x
Periodic potential
(
)
We assume that the solution is of the LCAO form: (r ) = cm s r Rm
m
r
(r
r
)
If we have N atoms in the lattice, then our solution is made up of N different sorbitals that are sitting on the N atoms In principle one can take the assumed solution, as written above, plug it in the Schrodinger equation, get an NxN matrix and solve it (just as we did in the case of molecules). But one can do better .. We know from Blochs theorem that the solution must satisfy the following:
(r + R ) = (r ) 2
r r
2
r
(r + R ) = e i k . R (r )
r r r
ECE 407 Spring 2009 Farhan Rana Cornell University
r
r
Tight Binding Approach for a 1D Crystal r
r a1
r Rm = m a1
a
Consideration 1: For the solution: (r ) = cm s r Rm m to satisfy:
x
r
(r
r
)
(r + R ) = (r ) 2
r r
2
r
one must have the same value of same weight).
cm
2
for all m (i.e. all coefficients must have the
So we can write without loosing generality:
cm =
ei m N
r 2 3r (r ) d r = 1
Consideration 2: For the solution: to satisfy:
rr ei m s r Rm N m rr rr r i k .R r +R =e (r )
(r ) =
r
(
)
r r
(
)
one must have the phase value equal to: m = k . Rm
ECE 407 Spring 2009 Farhan Rana Cornell University
2
Tight Binding Approach for a 1D Crystal r
r a1
r Rm = m a1
a
Consideration 2 (contd): Proof:
x
(r ) =
r
rr ei m s r Rm N m
(
)
)
r r
For the Bloch condition we get:
r r rr rrv r e i k . Rm e i k . Rm r +R = s r + R Rm = s r Rm R N N m m
(
)
r
r
(
((
))
Let:
r rr Rm R = R p
i k . Rp i k . (R p + R ) rr rr rr rr e e r + R = s r R p = e i k . R s r R p N N p p rr r i k .R =e (r )
(
)
r
r
(
)
r
r
(
)
ECE 407 Spring 2009 Farhan Rana Cornell University
Tight Binding Approach for a 1D Crystal r
r a1 a
Finally we can write the solution as:
v k
r Rm = m a1 x
(r ) = e
r
m
rr i k . Rm
N
r r
s (r Rm )
r r
And we know that it is a Bloch function because:
r r k (r + R ) = e i k . R k (r )
r
r
r
All that remains to be found is the energy of this solution so we plug it into the Schrodinger equation:
m
e
rr i k . Rm
r r r rr H k (r ) = E k k (r ) rr H s r Rm
()
N
(
)
r rr e i k . Rm =E k s r Rm N m
()
r
r
(
)
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Tight Binding Approach for a 1D Crystal
r r Rm = m a1
a
m
e
rr i k . Rm
0 rr H s r Rm
N
(
)
r e =E k
m
()
rr i k . Rm
x
Multiply this equation with s (r ) and: keep the energy matrix elements for orbitals that are nearest neighbors and assume that the orbitals on different atoms are orthogonal
r
N
s (r Rm )
r r
r rr r r r rr e i k . R1 e i k . R 1 1 s (r ) H s r R1 + s (r ) H s (r ) + s (r ) H s r R 1 N N N r1 r r =E k s (r ) s (r ) N
r
r
(
)
r
r
(
)
()
r e i k . a1 1 e i k . a1 Vss + Vss = E k Es N N N
rr
rr
()
1 N
r rr E k = E s 2Vss cos k . a1
()
(
)
ECE 407 Spring 2009 Farhan Rana Cornell University
Tight Binding Approach for a 1D Crystal r
r a1 a r rr E k = E s 2Vss cos k . a1
r Rm = m a1 x
()
(
)
Energy levels in an isolated atom
Energy
0
r Va (r )
Es
Energy levels
s (r )
r
Es
4Vss
0
r
a
a
k
r V (r )
4Vss
Energy band
Energy levels in a crystal
0
ECE 407 Spring 2009 Farhan Rana Cornell Binding University
a
x
4
Tight Approach for a 1D Crystal
r a1 a r rr E k = E s 2Vss cos k . a1 x
Number of quantum states at the starting point = 2 x number of orbitals used in the LCAO solution = 2N Number of quantum states at the ending point = 2 x energy levels per band for an N atom crystal = 2N Initial number of quantum states = Final number of quantum states
()
(
)
Energy
Es
4Vss
a
a
k
N : Es
4Vss
A band of N energy levels 2N quantum states
ECE 407 Spring 2009 Farhan Rana Cornell University
Tight Binding vs NFEA for a 1D Crystal
r rr E k = E s 2Vss cos k . a1
Energy Would have also obtained the higher energy bands in LCAO if higher energy atomic orbitals were also included in the LCAO solution
()
LCAO Tight Binding
(
)
Nearly Free Electron Approach (NFEA) Energy
Es
4Vss
~
Vo
h2 2m a
2
a
a
k
a
a
h2 1 m a2
k
The energy matrix elements are of the order of: Vss ~
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Example: A 1D Crystal with 2 Orbitals per Primitive Cell r
s p
r r a1 Rm = m a1 a x
Each atoms now has a s-orbital and a p-orbital that contributes to energy band formation
s (r ) r p (r )
r r
r
Es Ep
We write the solution in the form:
v k
i k . Rm s r r rr rr (r ) = e cs k s r Rm + c p k p r Rm N m
[ () (
r
)
() (
)]
Verify that it satisfies:
r r k (r + R ) = e i k . R k (r )
r
r
r
r
And plug it into the Schrodinger equation:
r r r rr H k (r ) = E k k (r )
ECE 407 Spring 2009 Farhan Rana Cornell University
()
Tight Binding Approach for a 1D Crystal
x r r Rm = m a1
0 r r r rr H k (r ) = E k k (r )
()
Step 1: r Multiply the equation with s (r ) and: keep the energy matrix elements for orbitals that are nearest neighbors and assume that the orbitals on different atoms are orthogonal
rr rr r r r v [ Es 2Vss cos(k .a1) ] cs (k ) + 2i Vsp sin(k .a1) cp (k ) = E (k ) cs (k )
Step 2: r Multiply the equation with p (r ) and: keep the energy matrix elements for orbitals that are nearest neighbors and assume that the orbitals on different atoms are orthogonal
rr r rr r r r [ E p + 2Vpp cos(k .a1) ] cp (k ) 2i Vsp sin(k .a1) cs (k ) = E (k ) cp (k )
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Tight Binding Approach for a 1D Crystal
r r r R = m a1 a1 m a x
We can write the two equations in matrix form:
rr E s 2Vss cos k .a1 rr 2i Vsp sin k .a1
(
(
)
)
rr 2i Vsp sin k .a1 rr E p + 2Vpp cos k .a1
(
(
)
)
r r c c s k r=E k s c p k c p
() ()
()
(k ) r (k )
r
Energy
For each value of wavevector one obtains two eigenvalues corresponding to two energy bands For k = 0 we get:
r
r E k = 0 = E p + 2Vpp r c s k = 0 0 r = c p k = 0 1 r E k = 0 = E s 2Vss r cs k = 0 1 r = c p k = 0 0
(
( (
)
4Vpp
Bloch function is made of only p-orbitals
) )
(
( (
)
4Vss
) )
Bloch function is made of only s-orbitals a
0
a
k
ECE 407 Spring 2009 Farhan Rana Cornell University
Tight Binding Approach for a 1D Crystal
r r r Rm = m a1 a1
a
Energy
r For k = x we get: 2a r E k = x = ? 2a r c s k = 2a x ? = r ? c p k = x Bloch function is made 2a
x
4Vpp
of both s- and p-orbitals
r E k = =? 2a r c s k = 2a x ? = Bloch function is made r c p k = x ? of both s- and p-orbitals 2a
4Vss
a
0
a
k
ECE 407 Spring 2009 Farhan Rana Cornell University
7
Tight Binding Approach for a 1D Crystal
r r r R = m a1 a1 m
a
Energy
x
r For k = x we get: a
r E k = x = E p 2Vpp a r c s k = a x 0 = r Bloch function is made c p k = x 1 of only p-orbitals a r E k = x = E s + 2Vss a r cs k = a x 1 Bloch function is made of only s-orbitals r = 0 c p k = x a
4Vpp
4Vss
a
0
a
k
ECE 407 Spring 2009 Farhan Rana Cornell University
Tight Binding Bands For Germanium
Germanium: Atomic number: 32 Electron Configuration: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p2 Number of electrons in the outermost shell: 4 Tight Binding Bands for Ge
Energy (eV)
1 2 1 2 1 1 2 1 1
2
FBZ (for FCC lattice)
1
1
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