Kronig-penny modal

 The Kronig–Penney model (named after Ralph Kronig and William Penney^[2]) is a simple, idealized quantum-mechanical system that consists of an infinite periodic array of rectangular potential barriers.

The potential function is approximated by a rectangular potential:

Using Bloch's theorem, we only need to find a solution for a single period, make sure it is continuous and smooth, and to make sure the function u(x) is also continuous and smooth.

Considering a single period of the potential:
We have two regions here. We will solve for each independently: Let E be an energy value above the well (E>0)

${\displaystyle \mathrm {For} \quad 0<x<(a-b)}$ :

${\displaystyle {-\hbar ^{2} \over 2m}\psi _{xx}=E\psi }$

${\displaystyle \Rightarrow \psi =Ae^{i\alpha x}+A'e^{-i\alpha x}\quad \left(\alpha ^{2}={2mE \over \hbar ^{2}}\right)}$

${\displaystyle \mathrm {For} \quad -b<x<0}$ :

${\displaystyle {-\hbar ^{2} \over 2m}\psi _{xx}=(E+V_{0})\psi }$

${\displaystyle \Rightarrow \psi =Be^{i\beta x}+B'e^{-i\beta x}\quad \left(\beta ^{2}={2m(E+V_{0}) \over \hbar ^{2}}\right).}$

To find u(x) in each region, we need to manipulate the electron's wavefunction:

${\displaystyle \psi (0<x<a-b)=Ae^{i\alpha x}+A'e^{-i\alpha x}=e^{ikx}\cdot \left(Ae^{i(\alpha -k)x}+A'e^{-i(\alpha +k)x}\right)\,\!}$

${\displaystyle \Rightarrow u(0<x<a-b)=Ae^{i(\alpha -k)x}+A'e^{-i(\alpha +k)x}.\,\!}$

And in the same manner:

${\displaystyle u(-b<x<0)=Be^{i(\beta -k)x}+B'e^{-i(\beta +k)x}.}$

To complete the solution we need to make sure the probability function is continuous and smooth, i.e.:

${\displaystyle \psi (0^{-})=\psi (0^{+})\qquad \psi '(0^{-})=\psi '(0^{+}).}$

And that u(x) and u′(x) are periodic:

${\displaystyle u(-b)=u(a-b)\qquad u'(-b)=u'(a-b).}$

These conditions yield the following matrix:

${\displaystyle {\begin{pmatrix}1&1&-1&-1\\\alpha &-\alpha &-\beta &\beta \\e^{i(\alpha -k)(a-b)}&e^{-i(\alpha +k)(a-b)}&-e^{-i(\beta -k)b}&-e^{i(\beta +k)b}\\(\alpha -k)e^{i(\alpha -k)(a-b)}&-(\alpha +k)e^{-i(\alpha +k)(a-b)}&-(\beta -k)e^{-i(\beta -k)b}&(\beta +k)e^{i(\beta +k)b}\end{pmatrix}}{\begin{pmatrix}A\\A'\\B\\B'\end{pmatrix}}={\begin{pmatrix}0\\0\\0\\0\end{pmatrix}}.}$

For us to have a non-trivial solution, the determinant of the matrix must be 0. This leads us to the following expression:

${\displaystyle \cos(ka)=\cos(\beta b)\cos[\alpha (a-b)]-{\alpha ^{2}+\beta ^{2} \over 2\alpha \beta }\sin(\beta b)\sin[\alpha (a-b)].}$

To further simplify the expression, we perform the following approximations:

${\displaystyle b\to 0;\quad V_{0}\to \infty ;\quad V_{0}b=\mathrm {constant} }$

${\displaystyle \Rightarrow \beta ^{2}b=\mathrm {constant} ;\quad \alpha ^{2}b\to 0}$

${\displaystyle \Rightarrow \beta b\to 0;\quad \sin(\beta b)\to \beta b;\quad \cos(\beta b)\to 1.}$

The expression will now be:

${\displaystyle \cos(ka)=\cos(\alpha a)+P{\frac {\sin(\alpha a)}{\alpha a}},\qquad P={\frac {mV_{0}ba}{\hbar ^{2}}}.}$

For energy values inside the well (E < 0), we get:

${\displaystyle \cos(ka)=\cos(\beta b)\cosh[\alpha (a-b)]-{\beta ^{2}-\alpha ^{2} \over 2\alpha \beta }\sin(\beta b)\sinh[\alpha (a-b)],}$

with ${\displaystyle \alpha ^{2}={2m|E| \over \hbar ^{2}}}$ and ${\displaystyle \beta ^{2}={2m(V_{0}-|E|) \over \hbar ^{2}}}$.

Following the same approximations as above (${\displaystyle b\to 0;\,V_{0}\to \infty ;\,V_{0}b=\mathrm {constant} }$), we arrive at

${\displaystyle \cos(ka)=\cosh(\alpha a)+P{\frac {\sinh(\alpha a)}{\alpha a}}}$

with the same formula for P as in the previous case ${\displaystyle \left(P={\frac {mV_{0}ba}{\hbar ^{2}}}\right)}$