The Finite Square Well and Evanescent Tails
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You should be comfortable solving the time-independent Schrödinger equation in piecewise-constant potentials and applying continuity conditions for the wavefunction and its first derivative at boundaries.
Why this matters
The finite square well is the simplest model that captures the essential physics of bound states in real systems.
Unlike the infinite well, where the particle is strictly confined, the finite well allows the wavefunction to penetrate the classically forbidden region. This "evanescent tail" is not a mathematical artefact but a physical reality responsible for phenomena such as nuclear alpha decay and quantum tunnelling. Understanding how the number of bound states depends on the well's depth and width provides a direct link between microscopic parameters and observable spectral lines.
The Physical Picture
Imagine a particle trapped in a box with soft, permeable walls. Inside the box, where the potential energy is lower, the particle behaves like a standing wave, oscillating back and forth. Outside the box, where the potential energy is higher than the particle's total energy, the wave does not oscillate; instead, it decays exponentially into the barrier. The particle is not "inside" or "outside" in a classical sense; its probability density is spread out, with the bulk of the probability inside the well and a small but non-zero probability finding the particle in the barrier. The condition for a stable bound state is that the wave inside must fit perfectly with the decaying wave outside, creating a self-consistent standing wave pattern that does not radiate energy away.
Parity and Symmetry
Because the potential \( V(x) \) is symmetric about \( x = 0 \), the Hamiltonian commutes with the parity operator. This allows us to classify solutions into even and odd parity states. For an even parity state, the wavefunction is symmetric, \( \psi(-x) = \psi(x) \), and the derivative is zero at the origin. For an odd parity state, the wavefunction is antisymmetric, \( \psi(-x) = -\psi(x) \), and the wavefunction itself is zero at the origin. This symmetry simplifies the boundary conditions significantly, as we only need to solve the equation for \( x > 0 \) and then extend the solution to \( x < 0 \) using the appropriate symmetry property.
The Transcendental Conditions
Applying the matching conditions at the boundary \( x = a \) leads to the quantisation conditions derived in D-144. For even parity states, the condition is \( u \tan u = v \). For odd parity states, the condition is \( -u \cot u = v \). Here, \( u = ka \) and \( v = \kappa a \) are dimensionless parameters. The relationship \( u^2 + v^2 = R^2 \) defines a circle in the \( u \)-\( v \) plane, where \( R^2 = \frac{2mV_0 a^2}{\hbar^2} \) is a measure of the "strength" of the well. Bound states exist only where the curves \( u \tan u = v \) and \( -u \cot u = v \) intersect this circle. Since the tangent and cotangent functions have asymptotes, the number of intersections is finite and depends on the radius \( R \).
Counting Bound States
A key result from the graphical analysis is that a one-dimensional finite square well always has at least one bound state, regardless of how shallow or narrow the well is. This is because the first even-parity solution always intersects the circle near the origin. As the well strength \( R \) increases, more bound states appear. Even states need \( \tan u \ge 0 \), so one appears in each interval \( [0, \pi/2), [\pi, 3\pi/2), \dots \) that the circle reaches; odd states need \( \cot u \le 0 \), one in each of \( [\pi/2, \pi), [3\pi/2, 2\pi), \dots \). A new state therefore appears each time \( R \) passes a multiple of \( \pi/2 \), alternating even and odd, and the number of bound states is \( N = \lceil 2R/\pi \rceil \) (the integer part of \( 2R/\pi \), plus one). This discrete spectrum is a hallmark of quantum mechanics, contrasting sharply with the continuous range of energies a classical particle in the well can have.
The derivation
Finite Square Well: Transcendental Quantisation
For a one-dimensional finite square well of depth V0 and width 2a, a bound state (energy 0 < E < V0) exists only when the interior wavenumber k = √(2mE)/ℏ and the exterior decay constant κ = √(2m(V0−E))/ℏ satisfy the transcendental matching conditions k tan(ka) = κ (even parity) or k cot(ka) = −κ (odd parity). Equivalently, with dimensionless u = ka, v = κa, the states are the intersections of u2 + v2 = R2 (with R2 = 2mV0a2/ℏ2) with the branch curves v = u tan u and v = −u cot u, giving a finite number of levels with exponentially decaying (evanescent) tails outside the well.
Two opposite mistakes are common. One is to ignore the tails altogether: for a shallow or narrow well (small \( R \)) the ground state can have most of its probability outside the well. The other is to think that penetration means the particle "has negative kinetic energy" there: a position measurement that finds the particle in the forbidden region disturbs it enough (\( \Delta p \gtrsim \hbar\kappa \)) that energy conservation is never seen to fail. For deep wells, \( \kappa \) is large, the tails are short, and the levels approach those of the infinite well. C
Worked example
Problem. An electron (\( m = 9.11 \times 10^{-31}\ \mathrm{kg} \)) is in a finite square well of width \( 2a = 1.0 \times 10^{-9}\ \mathrm{m} \) and depth \( V_0 = 1.0\ \mathrm{eV} \). Calculate the strength parameter \( R \) and the number of bound states, with their parities. Use \( \hbar = 1.055 \times 10^{-34}\ \mathrm{J\,s} \) and \( 1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J} \).
- Depth in joules: \( V_0 = 1.602 \times 10^{-19}\ \mathrm{J} \); half-width \( a = 5.0 \times 10^{-10}\ \mathrm{m} \).
- \( R^2 = \frac{2mV_0 a^2}{\hbar^2} = \frac{2 (9.11 \times 10^{-31})(1.602 \times 10^{-19})(5.0 \times 10^{-10})^2}{(1.055 \times 10^{-34})^2} = \frac{7.30 \times 10^{-68}}{1.113 \times 10^{-68}} \approx 6.56 \), so \( R \approx 2.56 \).
- Count: \( 2R/\pi = 5.12/3.1416 \approx 1.63 \), so \( N = 2 \). The circle passes \( u = 0 \) (first even state) and \( u = \pi/2 \approx 1.57 \) (first odd state) but not \( u = \pi \approx 3.14 \).
- For comparison, a well ten times deeper (\( V_0 = 10\ \mathrm{eV} \)) has \( R \approx 8.1 \) and \( 2R/\pi \approx 5.2 \): six bound states, three of each parity.
Units check. \( mV_0a^2/\hbar^2 \) is \( \mathrm{kg \cdot J \cdot m^2/(J^2\,s^2)} = \mathrm{kg\,m^2/(J\,s^2)} \), and since \( \mathrm{J} = \mathrm{kg\,m^2/s^2} \) this is dimensionless, as \( R \) must be.
Three quick tests
- Does a finite square well always have at least one bound state?
Answer
Yes, due to the symmetry and the behaviour of the transcendental equations near the origin, the first even state always exists for any \( V_0 > 0 \). - How does the number of bound states change if the width \( 2a \) is doubled?
Answer
Since \( R \propto a \), doubling the width doubles \( R \). This generally increases the number of bound states, as the circle in the \( u \)-\( v \) plane grows larger, intersecting more branches of the tangent/cotangent curves. - What is the physical significance of the evanescent tail?
Answer
It represents the non-zero probability of finding the particle in the classically forbidden region, a purely quantum mechanical effect that enables phenomena like tunnelling.
Problem set for PU-202 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.