*Please wait for the animation to completely load.*

In Section 11.4 we considered
what happens when we have two finite wells nearby each other. What happens
when there are even more finite wells side by side? Such a situation is
called a *finite lattice* of square wells. This finite lattice is
modeled by a set of *N* finite square wells (*V*_{0 }< 0) each of width *b* = 2*a* and a distance
*D*
apart from each other. In addition, for a finite lattice, the
boundary condition on the energy eigenfunction is such that it is zero at the edges of
the lattice, ψ_{edges }= 0.

In the first animation (** ħ = 2m = 1**), you can change the number of wells in the
lattice from 1 to 3 to 5, while maintaining the individual
well's width and depth. Notice what happens to the energy
level diagram. For these particular wells, there are just two bound
states possible. What happens when we increase the lattice to
include three finite wells? Five finite wells? What you should
notice is that the number of bound states increases as the number
of wells increases. There are still two groups of states, but now each group
has N individual states, where N is the number of finite wells. Therefore with three wells there are 6 bound
states (three and three) and for five wells there are 10 bound
states (five and five). As the number of wells increases, the
number of bound states, therefore, will also increase. As the
number of wells approaches the number in a metal, on the order of
10

In order to consider a more quantitative model, we consider the
Kronig-Penney model. In the Kronig-Penney
model, the finite nature of the lattice is removed by using periodic boundary
conditions: requiring the energy eigenfunction at the edges of the lattice match, ψ_{left edge }= ψ_{right
edge}. This is clearly different than the
condition we considered above. For such a periodic potential, one
which repeats every *D*, the periodicity of the potential can be expressed by V(*x*) = V(*x* + *D*). Bloch's theorem^{5} tells us that
the solution to the time-independent Schrödinger equation for
such a periodic potential energy function is an energy eigenfunction of the form:

ψ(*x* + *D*) = exp(*iKD*) ψ(*x*),
(11.8)

for a constant *K*. Since a solid does not go on forever, we apply
the periodic boundary condition such that the energy eigenfunction matches after
it has gone through all *N* wells:

ψ(*x *+* ND*) = ψ(*x*),
(11.9)

and since ψ(*x* + *D*) = exp(*iKD*) ψ(*x*), we have that

ψ(*x* + *ND*) = exp(*iKD*) ψ(*x*)
= ψ(*x*),
(11.10)

and therefore *NKD* = 2π*n*, with *n* = 0, ±1, ±2, .... In order to solve
this problem we must match wave functions and in doing so we get a transcendental equation for the bound states:

cos(*KD*) = cos((|*E*|α)^{1/2}) cosh(η)
− [(2*E* −* V*_{0})/(2(−*E*^{2 }+ |*E*||*V*_{0}|)^{1/2})] sin((|*E*|α)^{1/2}) sinh(η),
(11.11)

where η = [(|*V*_{0}| − |*E*|)β]^{1/2}. The left-hand side
(shown in teal in the animation) varies from 1 to −1 in tiny little steps
since *KD* = 2π*n*/*N*, where *N* is a large number (the number of finite wells in the
lattice). Only for certain values of the right-hand side (shown in red in
the animation), between 1 to −1, are there valid solutions in the form of bands
of allowable energies. This can be seen in the animation by varying the values of
*b*, *D*, and |*V*_{0}|.

___________________

^{5}For more
details see pages 289-306 of R. Liboff, *Introductory
Quantum Mechanics*, Addison Wesley (2003) and the original paper, F. Bloch,
*Z. Physik*, **52** (1928).