# Materials in Electronics/Confined Particles/1D Infinite Wells

## The 1D Infinite Well

Often, it is convenient to consider an electron "trapped" in a closed space, for example in a pice of semiconductor with a large bandgap between it and its surroundings. This kind of electron trap is called a 'potential well. The theoretically most striaghtforward case is the infinite potential well in one dimension, where an electron is confined completely and can never escape, as to do so would require an infinte amount of energy.

We will later expand this to more dimensions and the more realistic finite potential well, where the electron can escape.

For now, consider an electron trapped in an infintie potenital well, of width L (right). The potential of this systems is given by:

$V\left( x \right) = \begin{cases} \infty ,\quad x < 0 \\ 0, \quad \,\,\, 0 \le x \le L \\ \infty , \quad x > L \end{cases}$

The probability of finding an electron outside this region is zero. As the wavefunction of the electron in this well must be continuous according to the condtions on the wavefuction, we are seeking a solution to ψ(x) that is zero at x=0 and x=L. if this were not true, we would have a discontinuity here.

## Solving for the Wavefunction

We need to solve Schrödingers Equation to find ψ(x):

$- {\hbar^2 \over {2m}}{{d^2 \psi \left( x \right)} \over {dx^2 }} + V\left( x \right)\psi \left( x \right) = E\psi \left( x \right), \quad \quad 0 \le x \le L$.

Since V(x) is zero on this inteval, we can write:

$- {\hbar^2 \over {2m}}{{d^2 \psi \left( x \right)} \over {dx^2 }}= E\psi \left( x \right) \quad \quad 0 \le x \le L$.

Rearranging,

${{d^2 \psi \left( x \right)} \over {dx^2 }} = - {{2mE} \over \hbar^2 }\psi \left( x \right)$

Recall that the energy is given by

 $E\,$ $= T\left( x \right) + V\left( x \right)\,$ $= {{\hbar ^2 k^2 } \over {2m}}$

Substituting for E in our differential equation gives:

${{d^2 \psi \left( x \right)} \over {dx^2 }} = - k^2 \psi \left( x \right)$

This is now a second-order differential equation with a standard form. The general solution is given by:

[General Solution]

$\psi \left( x \right) = A \sin{kx} + B \cos{kx}$

## Applying Boundary Conditions

Applying the first boundary condition $\left( \psi \left( 0 \right) =0 \right)$:

$\psi \left( 0 \right) = A\sin \left( 0 \right) + B\cos \left( 0 \right) = 0$

Therefore, B=0.

Applying the second boundary condition $\left( \psi \left( L \right) =0 \right)$:

$\psi \left( L \right) = A\sin \left( k L \right) = 0$

Therefore, kL must be a multiple of π:

$kL=n \pi \quad n=\{ 0,1,2,\ldots \}$
$k=\frac{n \pi}{L} \quad n=\{ 0,1,2,\ldots \}$

We call A the normalisation coefficient, ψ0. This exists to ensure the probability of finding an electron in all of space is 1. Also, since when n=0 the whole equation goes to zero, the proability of an electron existing is zero, and thus, the solutions to the Schrödinger Equation for a particle in a 1D box start at n=1:

$\psi _{n_x } \left( x \right) = \psi _0 \sin \left( {{{n_x \pi } \over L} }x \right), \quad n_x = \left\{ {1,2,3, \ldots } \right\}$

Becasue we a working in the x-direction, we have called our indexing number nx. This is called the wavenumber. The energies associated with the above solutions are given by:

 $E_{n_x }$ $= {{\hbar ^2 k^2 } \over {2m}}$ $= {{\hbar ^2 n_x^2 \pi ^2 } \over {2mL^2 }}, \quad n_x = \left\{ {1,2,3, \ldots } \right\}$

There are two thing to note here.

• A wavefunction solving the Schrödinger Equation exists only for integer wavenumbers, n.
• Each wavefunction (eigenfunction) has a specific energy (eigenvalue) associated with it. This is proportional to the square of the wavenumber and the energy of the wavefunction with n=1, E1.

The diagram below shows the wavefunctions with wavenumbers 1, 2 and 3, plotted against the associated energy relative to E1:

The probability densities can be found easily once the wavefunctions have been found - simply take the modulus (in the case of complex wavefunctions) and square. This results in proability densities as shown below:

## Normalisation Coefficient

We can find the normalisation coefficient, ψ0 using the knowlegde that the electron must be somewhere in the quantum well, and therefore the area under the proabability density function is one:

$\int\limits_{ - \infty }^\infty {\left| {\psi \left( x \right)} \right|^2 dx} = \int\limits_0^L {\psi \left( x \right)^*\psi \left( x \right)} dx = 1$

So, for our wavefunction, $\psi _{n_x } \left( x \right) = \psi _0 \sin \left( {{{n_x \pi } \over L} }x \right)$,

$\int\limits_0^L {\psi_0^2 \sin^2 \left(\frac{n_x \pi}{L}x \right) dx} = 1$
$\psi_0^2 \int\limits_0^L {\frac{1}{2} \left(1-\cos \left(\frac{2 n_x \pi}{L} \right)x \right) dx} = 1$
$\psi_0^2 \left[ \int\limits_0^L {\frac{1}{2} dx} - \frac{1}{2}\int\limits_0^L {\cos \left(\frac{ 2 n_x \pi}{L} x \right) dx} \right] = 1$
$\psi_0^2 \left[\frac{L}{2} - \frac{L}{4 n_x \pi}\left[\sin \left(\frac{ 2 n_x \pi}{L} x\right) \right]_0^L \right]= 1$

Because the area under the sine curve over a whole number of periods in zero, we have

$\psi_0^2 \frac{L}{2} = 1$
$\psi_0 = \sqrt{\frac{2}{L}}$

It turns out that this is independent of the wavenumber, so our normalised wavefunction for an electron in a 1-D infinite potential well is zero outside the well and given by the equation:

[Wavefunction of Particle 1D Infinite Potential Well]

 $\psi _{n_x } \left( x \right) = \sqrt{\frac{2}{L}} \sin \left( {{{n_x \pi } \over L}} x \right), \quad n_x = \left\{ {1,2,3, \ldots } \right\}$

Inside it, the associated energies are given by:

[Associated Energies of Wavefunctions]

 $E_{n_x} = {{\hbar ^2 \pi ^2 } \over {2mL^2 }}n_x^2, \quad n_x = \left\{ {1,2,3, \ldots } \right\}$