Wolfram Alpha:

```Band Theory of Solids
---------------------

Linear Combination of Atomic Orbitals
-------------------------------------

The linear combinations of atomic orbitals (LCAO)
approach can be used to estimate the molecular
orbitals that are formed upon bonding between a
molecule’s constituent atoms. Similar to an atomic
orbital, a Schrödinger equation can be constructed
for a molecular orbital as well.  Linear combinations
of atomic orbitals, or the sums and differences of
the atomic wavefunctions, provide approximate solutions
to these molecular Schrodinger equations.  These sums
and differences represent constuctive and destuctive
interference between the individual atomic wavefunctions.

For simple diatomic molecules, the obtained wavefunctions
are represented mathematically by the equations:

ΨConstructive = ψ1 + ψ2
ΨDestructive = ψ1 - ψ2

Where the atomic orbitals ψi are the combined spatial
and spin wavefunctions.

If we compute the probabilities for each situation
we see that |ψConstructive|2 ≠ 0 whereas |ψDestructive|2 = 0.
This indicates that there is high electron density
between the nuclei in the constructive case that
results in a net attractive force between the atoms
(see more on this below).

The net result is the splitting of orbital energy
levels into SINGLET states as shown below.

antibonding MO (destructive) σ+
- spend most of their time away
from regions between 2 nuclei.
|
E                     v
^                  --------
|                 /        \
|  AO atom 1 -----          ------- AO atom 2
\        /
--------
^
|
bonding MO (constructive) σ
- spend most of their time
between 2 nuclei.

The orbitals get filled starting at the lowest
energy.  The bonding orbitals get filled first
followed by the anti-bonding orbitals.  The molecular
stability is determined by the BOND ORDER which is
defined as:

Bond Order = (1/2){No. of electrons in the BO

- No. of electrons in the ABO}

The higher the BO, the greater the stability.
A BO of 0 indicates the molecule is completely
unstable.

Consider 2 H atoms.  The individual AOs combine
to form 2 MOs.  The Pauli Exclusion Principle
dictates that a state can only contains 2 electrons
with opposite spins.  Thus, the lowest energy
configuration for the system is when an H2 molecule
is formed.  The Bond Order is 1.

The physical interpretation of this is as follows:

When the hydrogen atoms are brought together,
two new forces of attraction appear because of
the attraction between the electron on one atom
and the proton on the other.  But two forces of
repulsion are also created because the two
negatively charged electrons repel each other,
as do the two positively charged protons.

The force of repulsion between the protons can
be minimized by placing the pair of electrons
between the two nuclei. The distance between
the electron on one atom and the nucleus of
the other is now smaller than the distance
between the two nuclei. As a result, the force
of attraction between each electron and the
nucleus of the other atom is larger than the
force of repulsion between the two nuclei, as
long as the nuclei are not brought too close
together.

If the nuclei are close enough together to share
the pair of electrons, but not so close that
repulsion between the nuclei becomes too large
then the net result of pairing the electrons
and placing them between the two nuclei is a
system that is more stable than a pair of isolated
atoms (). The hydrogen atoms in an H2 molecule
are therefore held together (or bonded) by the
sharing of a pair of electrons.

In the case of 3 H atoms, the atomic orbitals
combine to form 3 molecular orbitals.  Again,
each orbital can hold 2 electrons.  However,
there are only 3 electrons in the system, so the
lowest energy situation is for 2 electrons to be
in the lowest energy state, 1 electron in the next
highest state and 0 in the highest energy state.
However, the overall energy is still higher than
the H2 system.  That is why we don't get H3 molecules
in nature.  In this case the Bond Order is
(1/2){2 - 1} = 1/2 which indicates that H3 is not
that stable.

Likewise, we can extend this idea to a 2 He atom
system.  There are 2 resulting MOs each containing
2 electrons.  At a simplifed level, there is no
energy advantage of forming a molecule so the He
atoms don't combine.   The Bond Order is (1/2){2 - 2}
= 0 which confirms this.

In general, we can summarize as follows:

When N atoms are brought together, each individual
energy state splits into a band of N energy levels.
There will be bands corresponding to the 1s, 2s etc.
atomic orbitals.  The width of the band depends on the
strength of the interaction and the overlap between
neighboring atoms.  Two or more energy bands may
coincide in energy at special positions or there may
be gaps between bands. This leads to the distinction
between metals, insulators and semiconductors.

Degenerate Orbitals - Hund's Rule
---------------------------------

In the case of degenerate orbitals (i.e. orbitals
with the same energy) Hund's rule is applied.
Hund's rule basically says that the electrons
don't pair up until they have to.  They space
out with one in each orbital.  Only when they
run out of orbitals with equal energy do they
begin to pair up.  A good analogy is an empty
bus.  Assume that everyone boarding the bus is
travelling alone.  Because they would rather
sit by themselves than next to a stranger,
passengers will individually fill the rows even
if there is space next other passengers.  It is
not until all the empty rows are filled that
passengers will begin sitting next to strangers
and fill the row.

Quantitative Band Theory
---------------------------

Density of States Function:

It is impractical to solve the Schrodinger equation
for energy for every molecular orbital.  Fortunately,
there is an easier way which provides a good
approximation in most situations.  If we consider
the electrons being confined inside a potential well
where the dimensions of the well are equal to the
dimensions of the metal we can use the FREE ELECTRON
THEORY to derive a density of states function g(E).
The density of states equals the density per unit
volume and energy of the  number of solutions to
Schrodinger's equation.

Free Electron Theory:

In the FE theory a metal is described in terms
of a potential well.

------------           --------- V(x)
|         |
|         |
|         |
---------
0         L

Inside the well, V(x) = 0

The time independent Schrodinger equation is:

∂2ψ/∂x2 + 2mEψ/h2 = 0

or,

∂2ψ/∂x2 = -2mEψ/h2

= -k2ψ

Where k = √(2mE)/h

This has the solution:

ψ(x) = Asinkx + Bcoskx

≡ Cexp(ikx) + Dexp(-ikx)

The boundary conditions are:

ψ(0) = ψ(L) = 0

ψ(0) = Asink0 + Bcosk0 = 0 => B = 0

and,

ψ(L) = AsinkL = 0 => A = 0 or sinkL = 0

In the latter case, this implies that k = nπ/L

Now, E = p2/2m

= h2k2/2m

= n2π2h2/2mL2

Thus, only certain values of energy are allowed
which correspond to n = 0, 1, 2, 3 ...

This result can be generalized to 3D as follows:

E = π2h2|n|2/2mL2 where |n|2 = nx2 + ny2 + nz2

Density of States:

E = π2h2|n|2/2mL2

|n| = √(2mL2E/π2h2)

Number of states in n-space with energy less
than E is given by the volume of a sphere of

R = √(2L2mE/π2h2)

Therefore,

Total number of states = (4π/3)(2mL2E/π2h2)3/2

Real space occupies 1/8 of sphere.  Thus,

Number of real states = (4π/3)(2mL2E/π2h2)3/2
which must equal ∫g(E)dE.

E
∫g(E)dE = (1/6π2h3)(2m)3/2E3/2
0

Differentiate this to obtain:

g(E) = (1/4π2h3)(2m)3/2E1/2

Accounting for the Pauli Exclusion Principle
we get:

g(E) = (1/2π2h3)(2m)3/2E1/2

What happens if the PE inside the well is not
0 but is periodic as would be the case in a
normal metal?

Bloch Function
--------------

The Bloch function is a wavefunction for a particle
in a periodically-repeating environment, most
commonly an electron in a crystal. Bloch's theorem
states that the energy eigenfunction for such a
system may be written as the product of a plane
wave envelope function and a periodic function
(periodic Bloch function), uk(x), that has the
same periodicity as the potential (uk(x) = uk(x + a)).
The Bloch function is the solution to Schrodinger
Equation with a potential V(x).  Thus:

(p2/2m + V(x))ψ = Eψ

ψ = uk(x)e-ikx

Kronig-Penney Model
-------------------

Model crystal as a series of potential wells.

<---- a ----->
-----   -----------   -----------     ---------
|  |          |  |          |  |
V|  |          |  |          |  |
--            --            --
b

The solution to the time independent Schrodinger
Equation is:

cos(ka) = Psin(αa)/αa + cos(αa)

where P = mbaV/h2 and α = √(2mE/h2)

Note:
Plotting the RHS against αa reveals that the
electron in a periodic potential can only have
energies that lie in certain bands.  If P = 0,
α = k and we get the free electron situation:

E = p2/2m = h2k2/2m.

When P is finite there are discontinuities at
k = nπ/a.

E-k curve
---------

E
\             |               /
.           |             .
.         |           .
|             ^
|             | Forbidden energy
.        |         .   v <-- dE/dk = 0
.      |       .
\     |     /<-- point of inflection d2/dk2 =0
\    |    /
.  |  .
--------+--------. .---------+---------------- k
-π/a                    +π/a

<-- 1st Brillouin -->
zone

The resulting curve is no longer parabolic.  However,
if the forbidden energy gap is relatively large,
the parabolic free-electron (E = h2k2/2m + V)
approximation can be used at the extrema of the
bands (the .. regions).  The slope of the curve
at the discontinuity is 0.

However, if the gap is small this proves to be
ineffective because the interaction between bands
cannot be ignored.  In this case it is necessary
to use k.p perturbation theory to determine the
electron energy dispersion.

The periodic function uk satisfies the following
Schrodinger-type equation:

Hk(x) = Ekuk(x)

We can write Hk(x) as the sum of an unperturbed
Hamiltonian and a perturbation term:  Hk = H0 + Hk'
where H0 = (p2/2m + V) and Hk' = hk.p/m + h2k2/2m

This combines to:

Hk = p2/2m + V + hk.p/m + h2k2/2m

were k.p = kx(-ih∂/∂x) +  ky(-ih∂/∂y) +  kz(-ih∂/∂z)

Which can be solved to derive and expression
for the dispersion of E:

E = En0 + h2k2/2m + (h2/m2)Σm≠n|<ψn0|k.p|ψm0|2/(En0 - Em0)

Effective Mass
-------------

vg = ∂ω/∂k = (1/h)∂E/∂k since E = hω

The work done by an electric field, ε, is dE:

dE = eεvgdt

= eε((1/h)dE/dk)dt

∴ dE.dk/dt = eε(1/h)dE

∴ dk/dt = eε(1/h)

dvg/dt = (1/h)(d2E/dkdt)

= (1/h)(d2E/dk2)(dk/dt)

= (1/h)(d2E/dk2)eε(1/h)

= (1/h2)(d2E/dk2)eε

Now F = ma = mdvg/dt = eε.  Therefore,

dvg/dt = (1/h2)(d2E/dk2)mdvg/dt

1 = (1/h2)(d2E/dk2)m

Define:  m* = h2/{d2E/dk2}