Question:
There is an infinite potential well of width \(a\) for a quantum system. Find the ground state energy.
Answer:
The potential energies for each domain are
\[ V(x) = \infty \quad (x \leq 0, \ x \geq a) \]
\[ V(x) = 0 \quad (0 < x < a) \]
For the region of \(V(x) = \infty\), there is no particle since it gives zero for the wave function. For inside the well, the hamiltonian is
\[ H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} \]
Therefore the Schroedinger equation becomes
\[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \]
\[ \frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi = 0 \]
\[ \frac{d^2\psi}{dx^2} + k^2\psi = 0 \]
where \(k^2=\frac{2mE}{\hbar^2}\).
This differential equation has the general solution as follows:
\[ \psi(x) = A\sin kx + B\cos kx \]
The wave function is a continuous function; and at both edges it must be zero due to the infinite potential.
\[ \psi(0) = \psi(a) = 0 \]
This indicates that the wave function must be odd because \(0<x<a\). Thus, from \(\psi(x) = A\sin kx + B\cos kx\), only the sine function survives, so the wave function should be \(\psi(x)=A\sin kx\). From the above boundary condition \( \psi(0) = \psi(a) = 0 \),
\[ A\sin ka = 0 \]
This gives
\[ k_n a = n\pi, \quad n=0,1,2,… \]
Since \(k=\frac{n\pi}{a}\), the wave function becomes
\[ \psi(x) = A\sin \left(\frac{n\pi x}{a}\right) \]
Let us now normalize this function.
\[ A^2\int^a_0 \sin^2\left(\frac{n\pi x}{a}\right) dx = 1 \]
Put \(\theta=\frac{n\pi x}{a}\), so \(dx=\frac{a}{n\pi}d\theta\) and \(0<\theta<n\pi\).
\[ \frac{A^2a}{n\pi}\int^{n\pi}_0 \sin^2\theta d\theta = 1 \]
\[ \frac{A^2a}{n\pi}\left[\frac{\theta}{2}-\frac{\sin 2\theta}{4}\right]^{n\pi}_0 = 1 \]
\[ \frac{A^2a}{n\pi}\frac{n\pi}{2} = 1 \]
\[ A = \sqrt{\frac{2}{a}} \]
Hence, the wave function is completed as
\[ \psi(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right) \]
Now, we obtain the energy eigen values. From equations \(k^2=\frac{2mE}{\hbar^2}\) and \(k=\frac{n\pi}{a}\), we have
\[ E_n = \frac{\hbar^2 k^2_n}{2m} = \frac{\hbar^2 n^2 \pi^2}{2ma^2} \]
\(n=0\) corresponds to \(E=0\), so the ground state energy is when \(n=1\).