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# Kronig-Penney model

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 Title: Kronig-Penney model Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:

### Kronig-Penney model

In quantum mechanics, the particle in a one-dimensional lattice is a problem that occurs in the model of a periodic crystal lattice. The potential is caused by ions in the periodic structure of the crystal creating an electromagnetic field so electrons are subject to a regular potential inside the lattice. This is an extension of the free electron model that assumes zero potential inside the lattice.

## Problem definition

When talking about solid materials, the discussion is mainly around crystals - periodic lattices. Here we will discuss a 1D lattice of positive ions. Assuming the spacing between two ions is a, the potential in the lattice will look something like this:

The mathematical representation of the potential is a periodic function with a period a. According to Bloch's theorem, the wavefunction solution of the Schrödinger equation when the potential is periodic, can be written as:

$\psi \left(x\right) = e^\left\{ikx\right\} u\left(x\right). \,\!$

Where u(x) is a periodic function which satisfies:

$u\left(x+a\right)=u\left(x\right) \,\!$

When nearing the edges of the lattice, there are problems with the boundary condition. Therefore, we can represent the ion lattice as a ring following the Born-von Karman boundary conditions. If L is the length of the lattice so that L >> a, then the number of ions in the lattice is so big, that when considering one ion, its surrounding is almost linear, and the wavefunction of the electron is unchanged. So now, instead of two boundary conditions we get one circular boundary condition:

$\psi \left(0\right)=\psi \left(L\right). \,\!$

If N is the number of Ions in the lattice, then we have the relation: aN = L. Replacing in the boundary condition and applying Bloch's theorem will result in a quantization for k:

$\psi \left(0\right) = e^\left\{ik \cdot 0\right\} u\left(0\right) = e^\left\{ikL\right\} u\left(L\right) = \psi \left(L\right) \,\!$
$u\left(0\right) = e^\left\{ikL\right\} u\left(L\right)=e^\left\{ikL\right\} u\left(N a\right) \rightarrow e^\left\{ikL\right\} = 1 \,\!$
$\Rightarrow kL = 2\pi n \rightarrow k = \left\{2\pi \over L\right\} n \qquad \left\left( n=0, \pm 1, \pm 2, ..., \pm \left\{N \over 2\right\} \right\right). \,\!$

## Kronig–Penney model

The Kronig–Penney model (named after Ralph Kronig and William Penney) 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:

$\mathrm\left\{For\right\} \quad 0 < x <\left(a-b\right)$ :
$\left\{-\hbar^2 \over 2m\right\} \psi_\left\{xx\right\} = E \psi \,\!$
$\Rightarrow \psi = A e^\left\{i \alpha x\right\} + A\text{'} e^\left\{-i \alpha x\right\} \quad \left\left( \alpha^2 = \left\{2mE \over \hbar^2\right\} \right\right) \,\!$

$\left\{-\hbar^2 \over 2m\right\} \psi_\left\{xx\right\} = \left(E+V_0\right)\psi \,\!$
$\Rightarrow \psi = B e^\left\{i \beta x\right\} + B\text{'} e^\left\{-i \beta x\right\} \quad \left\left( \beta^2 = \left\{2m\left(E+V_0\right) \over \hbar^2\right\} \right\right). \,\!$

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

And in the same manner:

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

$\psi\left(0^\left\{-\right\}\right)=\psi\left(0^\left\{+\right\}\right) \qquad \psi\text{'}\left(0^\left\{-\right\}\right)=\psi\text{'}\left(0^\left\{+\right\}\right). \,\!$

And that u(x) and u'(x) are periodic

$u\left(-b\right)=u\left(a-b\right) \qquad u\text{'}\left(-b\right)=u\text{'}\left(a-b\right). \,\!$

These conditions yield the following matrix:

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

For us not to have the trivial solution, the determinant of the matrix must be 0. This leads us to the following expression:

$\cos\left(k a\right) = \cos\left(\beta b\right) \cos\left[\alpha\left(a-b\right)\right]-\left\{\alpha^2+\beta^2 \over 2\alpha \beta\right\} \sin\left(\beta b\right) \sin\left[\alpha\left(a-b\right)\right]. \,\!$

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

The expression will now be:

$\cos\left(k a\right) = \cos\left(\alpha a\right)-P\left\{\sin\left(\alpha a\right) \over \alpha a\right\} \qquad \left\left( P=\left\{m V_0 ba\over \hbar^2\right\} \right\right). \,\!$

## Kronig–Penney model: Alternative Solution

An alternative treatment to a similar problem is given. Here we have a delta periodic potential: $V\left(x\right) = A\cdot\sum_\left\{n=-\infty\right\}^\left\{\infty\right\}\delta\left(x-n\cdot a\right)$. $A$ is some constant, and $a$ is the lattice constant (the spacing between each site). Since this potential is periodic, we could expand it as a Fourier series: $V\left(x\right) = \sum_K \tilde\left\{V\right\}\left(K\right)\cdot e^\left\{i\cdot K \cdot x\right\}$, where $\tilde\left\{V\right\}\left(K\right) = \frac\left\{1\right\}\left\{a\right\}\int_\left\{-a/2\right\}^\left\{a/2\right\}dx\,V\left(x\right)\,e^\left\{-i\cdot K\cdot x\right\} = \frac\left\{1\right\}\left\{a\right\}\int_\left\{-a/2\right\}^\left\{a/2\right\}dx\sum_\left\{n=-\infty\right\}^\left\{\infty\right\}A\cdot \delta\left(x-na\right)\,e^\left\{-i\,K\,x\right\} = \frac\left\{A\right\}\left\{a\right\}$.

The wave-function, using Bloch's theorem, is equal to $\psi_k\left(x\right) = e^\left\{i k x\right\} u_k\left(x\right)$ where $u_k\left(x\right)$ is a function that is periodic in the lattice, which means that we can expand it as a Fourier series as well: $u_k\left(x\right)=\sum_\left\{K\right\} \tilde\left\{u\right\}_k\left(K\right)e^\left\{i K x\right\}$. Thus the wave function is: $\psi_k\left(x\right)=\sum_\left\{K\right\}\tilde\left\{u\right\}_k\left(K\right)\,e^\left\{i\left(k+K\right)x\right\}$.

Putting this into the Schroedinger equation, we get:

$\left\left[\frac\left\{\hbar^2\left(k+K\right)^2\right\}\left\{2m\right\}-E_k\right\right]\cdot\tilde\left\{u\right\}_k\left(K\right)+\sum_\left\{K\text{'}\right\}\tilde\left\{V\right\}\left(K-K\text{'}\right)\,\tilde\left\{u\right\}_k\left(K\text{'}\right)=0$

or rather:

$\left\left[\frac\left\{\hbar^2\left(k+K\right)^2\right\}\left\{2m\right\}-E_k\right\right]\cdot\tilde\left\{u\right\}_k\left(K\right)+\frac\left\{A\right\}\left\{a\right\}\sum_\left\{K\text{'}\right\}\tilde\left\{u\right\}_k\left(K\text{'}\right)=0$

Now we define a new function:

$f\left(k\right):=\sum_\left\{K\text{'}\right\}\tilde\left\{u\right\}_k\left(K\text{'}\right)$

Plug this into the Schroedinger equation:

$\left\left[\frac\left\{\hbar^2\left(k+K\right)^2\right\}\left\{2m\right\}-E_k\right\right]\cdot\tilde\left\{u\right\}_k\left(K\right)+\frac\left\{A\right\}\left\{a\right\}f\left(k\right)=0$

Solving this for $\tilde\left\{u\right\}_k\left(K\right)$ we get:

$\tilde\left\{u\right\}_k\left(K\right)=\frac\left\{\frac\left\{2m\right\}\left\{\hbar^2\right\}\frac\left\{A\right\}\left\{a\right\}f\left(k\right)\right\}\left\{\frac\left\{2mE_k\right\}\left\{\hbar^2\right\}-\left(k+K\right)^2\right\}=\frac\left\{\frac\left\{2m\right\}\left\{\hbar^2\right\}\frac\left\{A\right\}\left\{a\right\}\right\}\left\{\frac\left\{2mE_k\right\}\left\{\hbar^2\right\}-\left(k+K\right)^2\right\}\,f\left(k\right)$

We sum this last equation over all values of $K$ to arrive at:

$\sum_\left\{K\right\}\tilde\left\{u\right\}_k\left(K\right)=\sum_\left\{K\right\}\frac\left\{\frac\left\{2m\right\}\left\{\hbar^2\right\}\frac\left\{A\right\}\left\{a\right\}\right\}\left\{\frac\left\{2mE_k\right\}\left\{\hbar^2\right\}-\left(k+K\right)^2\right\}\,f\left(k\right)$

Or:

$f\left(k\right)=\sum_\left\{K\right\}\frac\left\{\frac\left\{2m\right\}\left\{\hbar^2\right\}\frac\left\{A\right\}\left\{a\right\}\right\}\left\{\frac\left\{2mE_k\right\}\left\{\hbar^2\right\}-\left(k+K\right)^2\right\}\,f\left(k\right)$

Conveniently, $f\left(k\right)$ cancel outs and we get:

$1=\sum_\left\{K\right\}\frac\left\{\frac\left\{2m\right\}\left\{\hbar^2\right\}\frac\left\{A\right\}\left\{a\right\}\right\}\left\{\frac\left\{2mE_k\right\}\left\{\hbar^2\right\}-\left(k+K\right)^2\right\}$

Or:

$\frac\left\{\hbar^2\right\}\left\{2m\right\}\frac\left\{a\right\}\left\{A\right\}=\sum_\left\{K\right\}\frac\left\{1\right\}\left\{\frac\left\{2mE_k\right\}\left\{\hbar^2\right\}-\left(k+K\right)^2\right\}$

To save ourselves some unnecessary notational effort we define a new variable:

$\alpha^2:=\frac\left\{2mE_k\right\}\left\{\hbar^2\right\}$

and finally our expression is:

$\frac\left\{\hbar^2\right\}\left\{2m\right\}\frac\left\{a\right\}\left\{A\right\}=\sum_\left\{K\right\}\frac\left\{1\right\}\left\{\alpha^2-\left(k+K\right)^2\right\}$

Now, $K$ is a reciprocal lattice vector, which means that a sum over $K$ is actually a sum over integer multiples of $\frac\left\{2\pi\right\}\left\{a\right\}$:

$\frac\left\{\hbar^2\right\}\left\{2m\right\}\frac\left\{a\right\}\left\{A\right\}=\sum_\left\{n=-\infty\right\}^\left\{\infty\right\}\frac\left\{1\right\}\left\{\alpha^2-\left(k+\frac\left\{2\pi n\right\}\left\{a\right\}\right)^2\right\}$

We can juggle this expression a little bit to make it more suggestive:

$\frac\left\{\hbar^2\right\}\left\{2m\right\}\frac\left\{a\right\}\left\{A\right\}=\sum_\left\{n=-\infty\right\}^\left\{\infty\right\}\frac\left\{1\right\}\left\{\alpha^2-\left(k+\frac\left\{2\pi n\right\}\left\{a\right\}\right)^2\right\}=-\frac\left\{1\right\}\left\{2\alpha\right\}\sum_\left\{n=-\infty\right\}^\left\{\infty\right\}\left\left[\frac\left\{1\right\}\left\{\left(k+\frac\left\{2\pi n\right\}\left\{a\right\}\right)-\alpha\right\}-\frac\left\{1\right\}\left\{\left(k+\frac\left\{2\pi n\right\}\left\{a\right\}\right)+\alpha\right\}\right\right]=-\frac\left\{a\right\}\left\{4\alpha\right\}\sum_\left\{n=-\infty\right\}^\left\{\infty\right\}\left\left[\frac\left\{1\right\}\left\{\pi n + \frac\left\{k a\right\}\left\{2\right\}-\frac\left\{\alpha a\right\}\left\{2\right\}\right\}-\frac\left\{1\right\}\left\{\pi n +\frac\left\{k a\right\}\left\{2\right\}+\frac\left\{\alpha a\right\}\left\{2\right\}\right\}\right\right]=$
$=-\frac\left\{a\right\}\left\{4\alpha\right\}\left\left[\left\left(\sum_\left\{n=-\infty\right\}^\left\{\infty\right\}\frac\left\{1\right\}\left\{\pi n + \frac\left\{k a\right\}\left\{2\right\}-\frac\left\{\alpha a\right\}\left\{2\right\}\right\}\right\right)-\left\left(\sum_\left\{n=-\infty\right\}^\left\{\infty\right\}\frac\left\{1\right\}\left\{\pi n +\frac\left\{k a\right\}\left\{2\right\}+\frac\left\{\alpha a\right\}\left\{2\right\}\right\}\right\right)\right\right]$

If we use a nice identity of a sum of the cotangent function (Equation 18) which says:

$cot\left(x\right)=\sum_\left\{n=-\infty\right\}^\left\{\infty\right\}\frac\left\{1\right\}\left\{n\pi+x\right\}$

and plug it into our expression we get to:

$\frac\left\{\hbar^2\right\}\left\{2m\right\}\frac\left\{a\right\}\left\{A\right\}=-\frac\left\{a\right\}\left\{4\alpha\right\}\left\left[\cot\left\left(\frac\left\{k a\right\}\left\{2\right\}-\frac\left\{\alpha a\right\}\left\{2\right\}\right\right)-\cot\left\left(\frac\left\{k a\right\}\left\{2\right\}+\frac\left\{\alpha a\right\}\left\{2\right\}\right\right)\right\right]$

We use the sum of $\cot$ and then, the product of $\sin$ (which is part of the formula for the sum of $\cot$) to arrive at:

$\cos\left(k a\right)=\cos\left(\alpha a\right)+\frac\left\{m A\right\}\left\{\hbar^2 \alpha a\right\}\sin\left(\alpha a\right)$

This equation shows the relation between the energy (through $\alpha$) and the wave-vector, $k$, and as you can see, since the left hand side of the equation can only range from $-1$ to $1$ then there are some limits on the values that $\alpha$(and thus, the energy) can take, that is, at some ranges of values of the energy, there is no solution according to these equation, and thus, the system will not have those energies: energy gaps. These are the so called band-gaps, which can be shown to exist in any shape of periodic potential (not just delta or square barriers).