Counting the Bound States

The infinite well is solved by a rule: fit a half-integer number of wavelengths between the walls, and every state is bound. Give the walls a finite height $V_0$ and neither half survives. The wavefunction no longer has to vanish at the edge, so the allowed energies stop being a clean arithmetic sequence — and because a particle with enough energy simply escapes, only finitely many states remain bound at all.

There is no closed-form formula for those energies. What there is instead is a construction that turns the problem into a picture, and the picture answers "how many states?" at a glance.

Two families, not one

Take the well to be $V(x) = -V_0$ for $|x| < a$ and $0$ outside, and look for a bound state, $-V_0 < E < 0$. Inside, the particle has kinetic energy to spare and oscillates; outside, it is in the classically forbidden region and decays:

The well is symmetric about the origin, so $V(-x) = V(x)$ and the Hamiltonian commutes with the parity operator. The two can be diagonalized together, which means every bound state can be chosen either even or odd — there is no third possibility, and no state that mixes the two. Inside the well that fixes the form completely: $\cos kx$ for the even states, $\sin kx$ for the odd ones.

Requiring $\psi$ and $\psi'$ to join smoothly at $x = a$ — equivalently, requiring $\psi'/\psi$ to match, which drops the unknown amplitudes — gives one condition per family:

One circle to hold them all

Those are transcendental: no rearrangement solves them for $E$. But writing $X = ka$ and $Y = \beta a$ makes them plottable, and adds one more relation for free. The definitions of $k$ and $\beta$ above give $k^2 + \beta^2 = 2mV_0/\hbar^2$ regardless of $E$, so

Every bound state of the well lies on a circle of radius $R$ — and $R$ is the only thing about the well that matters. Depth and width enter solely through the combination $V_0a^2$. The two matching conditions become curves in the same plane:

A bound state is an intersection. Drag $R$ below and watch the circle grow through the branches.

Why the count is what it is

The branches are what make the picture answer the counting question. Branch $n$ leaves the $X$ axis at $X = n\pi/2$ and climbs to infinity before the next one starts, so the quarter-circle of radius $R$ crosses it exactly once — if it reaches that far at all. It does precisely when $R > n\pi/2$, giving

Two consequences are worth stating plainly.

A 1D well always binds at least one state. The even branch starts at the origin, so however small $R$ — however shallow or narrow the well — the circle crosses it. There is no minimum depth. (This is special to one dimension; in three dimensions a shallow enough well binds nothing.)

The first odd state needs $R > \pi/2$. Odd states are not born alongside even ones. An odd wavefunction must pass through zero at the centre, which costs curvature and therefore energy, and a weak well cannot pay. This is why parity strictly alternates as states appear: even, odd, even, odd.

Nudge $R$ to just above a threshold and the figure will sometimes report one state fewer than $N$. That is not a rounding slip. A state born at a threshold has $Y \to 0$, so its decay length $1/\beta = a/Y$ diverges: it is bound, but spread over hundreds of well widths. The numerical solver works in a finite box, and once the tail is longer than the box the state no longer fits inside it. The construction and the solver disagree exactly where "bound" stops being a practical description.

The states themselves

The construction gives the energies; the wavefunctions follow from them. Each state is drawn on its own energy level below — solid for even, dashed for odd.

Two features have no counterpart in the infinite well. The wavefunctions do not stop at the walls: each decays outside as $e^{-\beta|x|}$, over a length $1/\beta = a/Y$. And that leakage is not uniform across the ladder — the lower a state sits, the larger its $Y$ and the tighter it is held, while the topmost state, barely bound, can have a substantial fraction of its probability outside the well entirely. Push $R$ down toward a threshold and watch the highest state spread out and let go.

Recovering the infinite well

Let $R \to \infty$. The circle becomes enormous, and it meets each branch far up where the curve is nearly vertical — that is, close to the asymptote $X \to (n+1)\pi/2$. Then $ka \to (n+1)\pi/2$, and with $L = 2a$,

the infinite-well spectrum, measured from the bottom of the well. The finite well's states are the infinite well's, pulled down and pushed out by the finite wall — and $R$ measures how much of that idealization survives.