Analytical Solution of the 1D Finite Square Potential Well

UPSC CSE / IFoS Paper II — Quantum Mechanics

1. Problem Definition

Consider a particle of mass m trapped in a symmetric finite square well of width 2a and depth V_0:

V(x) = \begin{cases} -V_0 & \text{for } |x| \le a \\ 0 & \text{for } |x| > a \end{cases}

We seek bound state solutions where energy E < 0. Let us define: \alpha = \frac{\sqrt{2m(E + V_0)}}{\hbar}, \quad \beta = \frac{\sqrt{-2mE}}{\hbar}

2. Parity and Eigenfunctions

Since the Hamiltonian possesses reflection symmetry (V(-x) = V(x)), the eigenfunctions must exhibit definite parity.

Symmetric (Even Parity) States

\psi(x) = \begin{cases} C e^{\beta x} & x < -a \\ A \cos(\alpha x) & -a \le x \le a \\ C e^{-\beta x} & x > a \end{cases}

Matching boundary conditions of continuity for \psi(x) and \psi'(x) at x = a: \alpha \tan(\alpha a) = \beta

Introducing dimensionless parameters \xi = \alpha a and \eta = \beta a: \eta = \xi \tan\xi, \quad \text{with } \xi^2 + \eta^2 = \frac{2m V_0 a^2}{\hbar^2} \equiv R^2

The bound state energies are the intersections of the curve \eta = \xi \tan\xi with the circle of radius R in the first quadrant.

3. UPSC Key Takeaway

1. Always At Least One Bound State: No matter how shallow (V_0 \to 0) or narrow (a \to 0) a one-dimensional symmetrical potential well is, there always exists at least one even bound state.
2. For odd states (\eta = -\xi \cot\xi), a critical minimum potential depth V_0 \ge \frac{\pi^2 \hbar^2}{8 m a^2} is required.