The Finite Square Well

The infinite square well is the workhorse of introductory quantum mechanics: walls at $x = 0$ and $x = L$ force $\psi = 0$ there, the Schrödinger equation inside is free-particle, and the energy levels come out in a clean closed form. But infinite walls are an idealization. Real quantum systems — semiconductor quantum dots, electrons bound in an atomic-scale potential, neutrons in a nucleus — all have finite walls. Removing the infinity changes the wave function, the energy spectrum, and — crucially — introduces a new phenomenon: the wave function can penetrate into regions that are classically forbidden.

The two potentials, side by side

Infinite well (width $a$, centered at the origin):

The boundary condition $\psi(\pm a/2) = 0$ is exact and forces the wave function to vanish identically outside the well.

Finite well (same width $a$, depth $V_0 \gt 0$):

Here $V_0$ is finite and the walls are penetrable. We look for bound states with energy $-V_0 \lt E \lt 0$.

Solving the Schrödinger equation in each region

Inside the well ($|x| \leq a/2$) the kinetic energy $E - V = E + V_0$ is positive, so the solution oscillates. Define

which is real and positive. The time-independent Schrödinger equation inside the well is

with general solution $A\cos(kx) + B\sin(kx)$.

Outside the well ($|x| \gt a/2$) the energy $E \lt 0$ while $V = 0$, so the kinetic energy $E - V = E$ is negative. Define

also real and positive. The equation outside is

with general solution $Ce^{\kappa x} + De^{-\kappa x}$. Normalizability requires the wave function to decay as $|x| \to \infty$, so we keep only the decaying exponential:

Even and odd parity states

Because the potential is symmetric ($V(x) = V(-x)$), the Hamiltonian commutes with the parity operator. Stationary states can therefore always be chosen to be even (cosine inside) or odd (sine inside). This halves the algebra.

Even states inside: $\psi_\text{in}(x) = A\cos(kx)$.

Matching $\psi$ and $\psi'$ at $x = +a/2$:

Dividing the second equation by the first eliminates $A$ and $G$ and yields the even-parity transcendental equation:

Odd states inside: $\psi_\text{in}(x) = B\sin(kx)$.

The same matching procedure gives the odd-parity transcendental equation:

Neither equation can be solved in closed form for general $V_0$. The energy eigenvalues must be found graphically or numerically.

How the energy levels compare to the infinite well

For the infinite well of width $a$, the exact energies are

These grow without bound: every positive integer $n$ contributes a level, and there are infinitely many of them.

For the finite well the situation is fundamentally different in two ways.

1. Finite number of bound states. A bound state exists only if the decaying exponential outside is matched to an oscillating solution inside — and there is only enough "room" to fit a limited number of oscillations when $V_0$ is finite. A standard result is that the number of bound states $N$ is

where $\lceil\cdot\rceil$ denotes the ceiling function (the smallest integer not less than its argument). No matter how shallow the well, there is always at least one bound state (this follows because the even-parity equation always has a solution with $ka/2 \in (0, \pi/2)$, as can be seen from the graphical solution).

2. Lower energy eigenvalues. Every bound level of the finite well lies below the corresponding level of the infinite well of the same width:

Intuitively, the wave function in the finite well is not forced to zero at the walls; instead it has a nonzero tail that "extends" slightly beyond $x = \pm a/2$. A longer effective wavelength means a lower kinetic energy. Equivalently, the ground-state probability density is slightly broader in the finite well, consistent with the uncertainty principle: a less confined particle has a smaller momentum spread and hence lower kinetic energy.

The evanescent tail: penetration into classically forbidden regions

The exponential decay $e^{-\kappa|x|}$ outside the well is the finite-well counterpart of the hard zero in the infinite well. Its characteristic penetration depth is

A particle in the ground state has the largest $|E|$ relative to the barrier height (its binding is deepest), so $\kappa$ is largest and the tail decays fastest. Higher excited states have energies closer to zero (less tightly bound) so $\kappa$ is smaller and the tail extends further — the least bound state has the most penetration.

This penetration is not a curiosity: it is the physical origin of quantum tunneling. If the well wall has finite thickness rather than extending to infinity, the evanescent wave can reach the far side before decaying completely. That scenario — the rectangular barrier — has been the subject of earlier lessons in this module. The finite square well is the bound-state limit of exactly the same mathematics.

Summary: key contrasts

| Property | Infinite well | Finite well | |---|---|---| | Wave function at wall | Exactly zero | Exponential decay | | Number of bound states | Infinitely many | Finite ($\geq 1$) | | Energy levels | $E_n \propto n^2$ (exact) | $E_n \lt E_n^\infty$ (transcendental) | | Penetration depth | Zero | $\hbar/\sqrt{2m|E|}$ |

The infinite well is a useful and tractable model; the finite well is the physically realistic one. As $V_0 \to \infty$ the two descriptions converge: $\kappa \to \infty$, $\delta \to 0$, and $E_n^\text{finite} \to E_n^\infty$. Every result of the infinite well can be understood as a limit of the finite well.

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