Finding the Wave Function Eigenfunctions and Energy Eigenvalues in an Infinite Potential Well
Proposition
When the potential takes the form of an infinite well over the interval $[0, a]$, the energy (eigenvalue) $E_{n}$ and the wave function (eigenstate) $\psi_{n}$ of the wave function are as follows.
$$ \begin{align*} E_{n} &=\frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}} \\[1em] \psi_{n}{(x)} &= \textstyle \sqrt{\frac{2}{a}}\sin \left( \frac{n\pi}{a}x \right) \end{align*} \qquad\qquad n = 0, 1, 2, \dots \tag{0} $$
Explanation
$$ V(x) = \begin{cases} \infty, & -\infty \lt x \lt 0 \\ 0, & 0 \lt x \lt a \\ \infty, & a \lt x \lt \infty \end{cases} $$
A potential of the form shown above $U$ is called an infinite potential well. This system models a situation where a particle cannot leave a certain interval, also known as the particle in a box model. It is a very simple model but serves as an important example where it demonstrates results significantly different from those in classical mechanics. In classical mechanics, the position where a particle may be found is uniform within the interval, whereas in quantum mechanics, the probability of finding the particle varies with its position.
The state with the lowest energy level ($=n$) among the wave functions is known as the ground state. Any state that is not the ground state is referred to as an excited state.
$$ \begin{align*} \text{ground state: } & \textstyle \psi_{1}(x)=\sqrt{\frac{2}{a}}\sin\left( \frac{\pi}{a}x \right) \\[1em] \text{first excited state: } & \textstyle \psi_2(x)=\sqrt{\frac{2}{a}}\sin\left( \frac{2\pi}{a}x \right) \\[1em] \text{second excited state: } & \textstyle \psi_{3}(x)=\sqrt{\frac{2}{a}}\sin\left( \frac{3\pi}{a}x \right) \\ \vdots \end{align*} $$
The wave function cannot have arbitrary energy, but only energies of the form depicted in equation $(0)$. Energy is proportional to $n^{2}$, and it appears in discrete rather than continuous values. This is called quantization.
Derivation
The time-independent 1-dimensional Schrödinger equation is given as follows.
$$ \left(-\frac{\hbar^{2}}{2m}\frac{ d ^{2} }{ d x^{2} }+V(x)\right)\psi(x) = E\psi(x) $$
$$ V(x) = \begin{cases} \infty, & -\infty \lt x \lt 0 \\ 0, & 0 \lt x \lt a \\ \infty, & a \lt x \lt \infty \end{cases} $$
$E \lt 0$
Since the potential is always greater than or equal to $0$, a wave function does not exist when the energy is negative.
$E \gt 0$
[1] For $|x| \gt a$
In this interval, since $0 \lt E \lt V = \infty$, it follows that $\psi(x) = 0$.
[2] For $0 \lt x \lt a$
In the interval $[0, a]$, the potential is $V = 0$, and the Schrödinger equation is given by:
$$ \frac{d^{2} \psi}{dx^{2}} + \frac{2m}{\hbar^{2}}E\psi=0 $$
Since $E$ is positive, $\dfrac{2m}{\hbar^{2}}E$ is also positive. Let us define this as $k^{2}$. Then the Schrödinger equation becomes:
$$ \frac{d^{2} \psi}{dx^{2}} + k^{2}\psi = 0 $$
The solution to this differential equation is as follows.
$$ \psi(x) = A\sin kx + B\cos kx $$
[3] Boundary Conditions
Given that the wave function $\psi$ is continuous, the values of the functions obtained in [1] and [2] must match at the boundary. Thus, when $x = 0$, the following must hold:
$$ 0 = \psi (0) = A\sin 0 + B\cos 0 = B \implies B=0 $$
The boundary condition must also hold when $x = a$, therefore:
$$ 0 = \psi (a) = A\sin ka \qquad (\because B = 0) \tag{1} $$
The sine function is zero for integer $n$, meaning that solutions exist for all integer $n$, leading to:
$$ A \sin ka = 0 \implies ka = n\pi \implies k = \frac{n\pi}{a} \tag{2} $$
$$ \implies \psi_{n}(x) = \textstyle A\sin \left( \frac{n\pi}{a}x \right) $$
The wave function must be square integrable such that $1$.
$$ \begin{align*} \int\limits_{-\infty}^{\infty} \psi_{n}^{\ast} \psi_{n} dx = 1 &= \int_{0}^{a}\textstyle |A|^{2} \sin^{2} \frac{n\pi}{a}x dx \\ &= |A|^{2} \int_{0}^{a}\textstyle \frac{1}{2}(1-\cos \frac{2n\pi}{a}x)dx \\ &= |A|^{2}\textstyle \frac{1}{2} \left[x-\frac{a}{2n\pi}\sin \frac{2n\pi}{a}x\right]_{0}^a \\ &= |A|^{2}\textstyle \frac{a}{2} \end{align*} $$
$$ \implies |A|^{2} = \frac{2}{a} \implies A = \sqrt{\frac{2}{a}} \tag{3} $$
In the third equality, the half-angle formula was used. Therefore, the wave function is as follows.
$$ \psi_{n}(x)= \textstyle \sqrt{\frac{2}{a}}\sin \left( \frac{n\pi}{a}x \right) $$
Also, given that $\dfrac{2m}{\hbar^{2}}E=k^{2}$ was defined earlier, the energy corresponding to the wave function $\psi_{n}$ is given by:
$$ E_{n} = \frac{\hbar^{2}k^{2}}{2m} = \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}} $$
■