
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. 


