 We have a band gap at the
Brillouin zones, of course. It's magnitude is directly given by the amplitude of the periodic
potential, i.e. we have 
 

  The last thing we need to
notice is that the available states will be occupied right up to the 1st Brillouin
zone, because we have one "free" electron per lattice constant. 

What we see now is that the total energy of the "free"
electrons is smaller if there is a band gap. But we don't get it for free  we have to pay
some elastic energy E_{mech} to get a reduction in the electronic energy.

  In order to find out out
if "pays" to invest some elastic energy E_{mech} with respect
top the return of electronic energy, we have to look eat the total energy E_{total}
(e) = E_{mech}(e) + E_{
elec}( e) and see if it has a minimum , and if yes, for
what kind of elastic deformation . 
  Doing
that is not exactly easy  even for our rather simple minded approximations used so far. 
 
But it is not so difficult either, so that we cannot understand
the gist of the argumentation here 
 So let's not
go onto details, but just discuss what needs to be done. 


First, we need some formula for the dispersion curve for the
case with a band gap. Whatever this formula will be, it depends on the magnitude of the band
gap, i.e. on the amount of strain e and on the constant
A , which describes how much an elastic deformation influences electrons. 
 
This formula, as always, also defines the available states for
the electrons in kspace. 
  The total electronic energy of the system then is simply the integral
over this curve up to the last occupied states, i.e. up to the Fermi wave vector or Fermi
energy. 
  To this integral
we add the elastic energy that we have to invest to produce
the periodic potential in the first place. This gives us the total energy 
 
E_{total}(e)
= E_{elec}(e) + E_{mech}(e) 
 
  Then
we find the minimum of that total energy by differentiating it with respect to the strain
e and setting the resulting differential quotient to zero,
i.e. we do 
  ¶ E_{total}(e ) ¶e  =  ¶ ¶e  æ ç è 
k_{F} ó õ 0 
E(k,e, A
) + Y · e^{2}  ö ÷ ø  = 0   

This will be our master equation. It looks pretty formidable even for the most simple model
and approximation we can chose. We can easily make it more complicated by looking at more
sophisticated models, but it always will have always one
basic property: It either has a solution or it doesn't. 


If it has a solution, it means that a Peierls instability does
occur because it is energetically favorable. An inherent symmetry will be broken by some elastic
deformation, and a band gap in electronic states will open up. This gap might be so small
that at room temperature it will not be noticed, but it will be there nonetheless. 
 
If it does not have a solution,
it means that there is no Peierls instability  bond lengths are the same, the band structure
does not have a gap, and the (model) material is a conductor. 
 As
it turns out, for out problem the master equation does have a solution, and that is also true
if you use more sophisticated models or math. 
  In
the most simple form, one obtains for the strain e that
minimizes the total energy something like 
 
e 
=  ^{2} · k_{F}^{2} m · A^{ } 
·  1^{ }_{ }
sinh – (^{2}
· k_{F} · p · Y / 4 ·
m · A^{2})   

It's not so obvious, what this means. Look up the hyperbolic
sinus here if you are unsure what
it looks like . 
  Playing
around with numbers a bit (which means making some educated guesses about the range of possible
values for A), one realizes that the argument of the hyperbolic sinus tends
to be >>1, which means we can approximate the equation from above by 
 
e 
»  2 ·
^{2} · k_{F}^{2} m · A^{ } 
· exp –  ^{2} · k_{F} · p ·
Y 4 · m · A^{2} 
 
  That is an interesting equation, because it comes up
in similar form for various problems, most noteworthy, perhaps, for superconductivity , which also owns its existence to some
kind of Peierls instability 

Looking back, we now can draw some more conclusions: 
  For conjugated
carbon chains, the Peierls instability has a large effect. For the most simple real conjugated
polymer which is poly ..... , it causes the transition from a onedimensional metal to a semiconductor
with a rather large band gap of » 1.6 eV. 
  But any periodic arrangement of atoms, ions or whatever, might undergo
some kind of Peierls instability. It might be so small, however, that it is not noticeable
at finite temperatures. 
  The effect is not limited to onedimensional
chains. As long as we can consider the x, y and zdimension
of the electronic energy separately, as we do in the free electron gas model, we will have
a Peierls instability in three dimensions, too. 
 However,
let's not get too general at this point. The Peierls instability results from the coupling
of phonons and electrons, and this is but a first step into a complicated world of collective
phenomae in solids. 
  It
may happen, and if energetically favorable, it will happen,
causing large effects on occasion, as for conjugated polymer chains. But other effects might
happen, too, and it would be too simple minded to invoke the Peierls instability for everything
out there not yet understood by us. 
  