Section 11.5: Finite and Periodic Lattices
Choose the # of wells:
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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 (V0 < 0) each of width b = 2a 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, ψedges = 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 108, the individual states form an almost continuous band of states, while the energies between these bands are called gaps.
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, ψleft edge = ψright 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 theorem5 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 wave function 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 = 2πn, with n = 0, ±1, ±2, .... In order to solve this problem we must match energy eigenfunctions and in doing so we get a transcendental equation for the bound states:
cos(KD) = cos((|E|α)1/2) cosh(η) − [(2E − V0)/(2(−E2 + |E||V0|)1/2)] sin((|E|α)1/2) sinh(η), (11.14)
where η = [(|V0| − |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 = 2πn/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 |V0|.
5For 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).