• Contact Us
• Programming Blog
• Tutoring Services
• Bio

  Search by keyword:  

Last modified: November 18, 2021 ✓

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 
radius,

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}

For free electron E = h2k2/2m so d2E/dk2 = h2/m 
which yields m* = m as expected.

The electron velocity increases from 0 at k = 0 
(accelerating so d2E/dk2 is positive).  The maximum 
velocity is at the point of inflection at k = π/2a 
(d2E/dk2 = 0). The electron velocity then decreases 
to 0 again at k = π/a (decelerating so d2E/dk2 is 
negative).  Thus,  we have the situation where m* 
changes from positive to negative as k increases.  
The negative mass corresponds to a HOLE.

Newsletter Sign-up:

Copyright © 2022 Redshift Academy

website built by Redshift Academy using the Codeigniter MVC framework