📂Quantum Mechanics

Finding the Wave Function Eigenfunctions and Energy Eigenvalues in an Infinite Potential Well

Summary

When the potential has the form of an infinite rectangular box, the energy (eigenvalue) of the wavefunction is

$$E_{n}=\frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}} \quad (n = 0, 1, 2, \dots)$$

and the wavefunction (eigenstate) corresponding to each energy is as follows.

$$\psi_{n}{(x)} = \sqrt{\frac{2}{a}}\sin \frac{n\pi}{a}x$$

Explanation

Such a form of potential is called an infinite potential well. It is also referred to as the particle in a box model. It is a model that describes a situation where particles cannot escape a specific interval. Though it is a very simplistic model, it shows notably different results from classical mechanics. In classical mechanics, the position of the particle found within the interval is the same, but in quantum mechanics, the probability of finding the particle varies based on its position.

Proof

Let us assume the potential $U$ is given as follows.

$$-\frac{\hbar ^{2}}{2m}\frac{d^{2} \psi{(x)}}{dx^{2}}+U\psi{(x)}=E\psi{(x)},\ \ U=\begin{cases} \infty, & -\infty < x <0 \\ 0, & 0<x<a \\ \infty, & a<x<\infty \end{cases}$$

$E \lt 0$

Since the potential is always greater than or equal to $0$, the wavefunction does not exist when the energy is negative.

$E \gt 0$

$0 \lt x \lt a$

In the interval $[0, a]$, the potential is $U = 0$, and the Schrödinger equation is as follows.

$$\frac{d^{2} \psi}{dx^{2}} + \frac{2m}{\hbar^{2}}E\psi=0$$

Since $E$ is positive, $\dfrac{2m}{\hbar^{2}}E$ is also positive, therefore let us denote it as $k^{2}$. Then the Schrödinger equation is as follows.

$$\frac{d^{2} \psi}{dx^{2}} + k^{2}\psi = 0$$

The solution of this differential equation is as follows.

$$\psi = A\sin kx + B\cos kx$$

By applying the boundary condition $( \mathrm{boundary \ condition})$, $\psi{(0)}=\psi{(a)}=0$ and $\sin$ functions satisfy the boundary condition but $\cos$ function does not, thus let’s find $k$ using the boundary condition $\psi{(x)}=A \sin kx$. Since the sine function becomes $0$ at the integer multiples of $\pi$, $\displaystyle ka=n\pi \implies k=\frac{n\pi}{a}$ and $\psi{(x)}=A \sin \frac{n\pi}{a}x$, finally let’s find $A$ through normalization. Since the probability of the particle must be $1$ throughout the entire interval, \displaystyle \begin{align*} 1 &= \int_{0}^a |A|^{2} \sin^{2} \frac{n\pi}{a}x dx \\ &= |A|^{2} \int_{0}^a \frac{1}{2}(1-\cos \frac{2n\pi}{a}x)dx \\ &= |A|^{2}\frac{1}{2} \left[x-\frac{a}{2n\pi}\sin \frac{2n\pi}{a}x\right]_{0}^a \\ &= |A|^{2} \frac{a}{2} \end{align*} and $\implies A=\sqrt{\frac{2}{a}}$. Therefore, the wavefunction (eigenfunction) in the infinite potential well is $\displaystyle \psi{(x)}=\sqrt{\frac{2}{a}}\sin \frac{n\pi}{a}x$. The lowest energy level among the wavefunctions is called the ground state $\mathrm{Ground\ state}$. The states other than the ground state are called excited states $\mathrm{Excited\ state}$. That is, the ground state in the infinite potential well is $\psi_{1}(x)=\sqrt{\dfrac{2}{a}}\sin\left( \dfrac{\pi}{a}x \right)$. The first excited state is $\psi_2(x)=\sqrt{\dfrac{2}{a}}\sin\left( \dfrac{2\pi}{a}x \right)$, and the second excited state is $\psi_{3}(x)=\sqrt{\dfrac{2}{a}}\sin\left( \dfrac{3\pi}{a}x \right)$. Now, let’s calculate the eigenvalues (energies). Since $\displaystyle \frac{2m}{\hbar^{2}}E=k^{2},\ k=\frac{n\pi}{a}$ is given, it is $\displaystyle E=\frac{\hbar^{2}k^{2}}{2m}=\frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}}$. Energy $E$ is quantized based on integer $n$. This means that not all energies are allowed, but only those defined by $n$, and that it is proportional to $n^{2}$. Using the subscript $n$, it is denoted as follows: $\displaystyle E_{n}=\frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}}$.