Towards Electrons in Solids
Review
2
2

 
i ( x, t ) 
( x, t )  V ( x, t )( x, t )
2
t
2m  x
Schroedinger wave equation
Arrived at the time independent version by solving for  ( x, t ) through
separation of variables and taking V = V(x) only
 ( x, t )   ( x)T (t )
T (t )  ei ( E / )t  ei t
2
E ( x) 
 ( x)  V ( x) ( x)
2
2m  x

2
Time independent wave equation
Electrons about an atomic nucleus: Coulombic interaction

Angular momentum
ψ(r , ,  )  R(r ) ( ) ( )


Exact analytical solution for one electron

 Ze 2
V
 V (r )  f ( ,  )
4o r
Energy
n
l
ml
Electrons in Atoms
n2
Single electron atom or ion (arbitrary Z) rn 
Z
orbitals known exactly
Hydrogen: Z = 1
 13.6Z 2
En 
n2
 1 0 0 (r, ,  )  Ae Zr / ao
1s, 2s, 2p, 3s, 3p…
n l ml
Multi-electron atoms: (Z > 1)
1s orbital
1

π ao3 / 2
• Pauli exclusion principle
– only one electron with given set of quantum numbers
– can’t all be in the lowest energy state; fill lowest first, then go up
• Electron-electron interaction
– removes degeneracy with respect to l
– Es < Ep
• Filling order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s
Electrons in Molecules
H2+ ionic molecule: 2 protons and 1 electron
Take new molecular*symmetry
orbital about
to bex =a0linear
atomic
orbitals
requirescombination
coefficients to be of
equal
in magnitude
Orbitals*
 1 0 0 (r, ,  )  Ae Zr / ao

~ S  ψ a  ψb
ψ
~ A  ψ a  ψb
ψ
a
b
- ½R
½R
a
b
position
Probability densities: * = ||2
||2
Bonding
Antibonding
electron between nuclei
electron on either side of nuclei
a
b
- ½R
½R
a
position
intuitively: this orbital has lower coulombic energy
b
“node” high energy
Types of MOs from LCAOs
s
s
+
pz
pz
+
s bonds
s
pz
+
px (py)
+
 bond
Types of MOs from LCAOs
s
s
+
Showed: symmetric = bonding
s bonds
Symmetric
pz

+
Anti-bonding
pz
+
+

Antisymmetric
+

+
+

+

Bonding

+
+

+
+
The H2 Ion
Vary inter-nucleus distance, R, between the two protons
R0
R
E
anti-bonding
electron sees only one
proton, no interaction
-13.6 eV
bonding
E
 0  R0
R
To find equilibrium bond distance, R0
R
The H2 Molecule
• Introduce 2nd electron
• Solution has perturbations due to electron-electron interactions
• Ignore these and place 2nd electron in ‘same’ bonding orbital but
with opposite spin
alternative
depictions
1s
R
E

H  H
2 anti-bonding states
-13.6 eV
2 protons

2 bonding states
• Result: H2 covalent bond
• Directional; typical of molecules
two 1s states each
 4 states total
1s