Finite Square Well

The deuteron — a proton and neutron bound into the simplest nucleus — should not, by classical lights, be so fragile. The nuclear force gripping it is enormous, yet the pair clings together by a mere 2.2 MeV, and the two particles spend roughly half their time outside the range of the force that binds them. No infinite-wall model can describe that. What can is the finite square well: a flat-bottomed trap of finite depth whose bound states leak exponentially into the region a classical particle could never enter. You solved the infinite square well by forcing the wavefunction to zero at the walls, and quantization fell out as clean algebra: $E_n \propto n^2$ . Real wells are never infinite. Make the walls finite and everything becomes more honest — the wavefunction no longer vanishes at the wall but glides through it as a dying exponential, the energy levels drop slightly as the state spreads out, only a finite number of levels fit, and the quantization condition stops being solvable by algebra at all. You will meet your first transcendental equation, and learn the graphical trick — a circle crossing a family of tangent curves — that tames it. The payoff is disproportionate to the effort. The matching machinery you build here — oscillation inside, exponential decay outside, smoothness at the join — is the template for the delta well, the double well, tunnelling, and every semiconductor quantum well in every laser pointer on Earth.

Key takeaways

The symmetric well $V = -V_0$ for $|x| < a$ (zero outside) binds states with $-V_0 < E < 0$ : oscillation inside, exponential tails $e^{-\kappa|x|}$ outside.
Continuity of $\psi$ and $\psi'$ at the walls gives transcendental quantization conditions — $\kappa = l\tan(la)$ for even states, $\kappa = -l\cot(la)$ for odd — with no closed-form solution.
Everything is controlled by one dimensionless number, the well strength $z_0 = \tfrac{a}{\hbar}\sqrt{2mV_0}$ : bound states are the intersections of a circle of radius $z_0$ with the tangent/cotangent curves.
The number of bound states is finite, $N = 1 + \lfloor 2z_0/\pi \rfloor$ , and at least one always exists, no matter how feeble the well.
The tails make the effective box wider, so every finite-well level sits below its infinite-well counterpart; as $V_0 \to \infty$ the infinite well is recovered.
Key idea: a bound state is an interior standing wave that must join smoothly onto dying exponential tails — the smoothness demand is the quantization.

The core insight

Bound states of a finite well are standing waves whose oscillatory interior must smoothly join exponentially decaying tails; the smooth-join requirement (continuity of $\psi$ and $\psi'$ ) produces a transcendental quantization condition, and the dimensionless well strength $z_0 = \tfrac{a}{\hbar}\sqrt{2mV_0}$ fixes both the number of levels and their energies.

Where this lives today

The finite square well is the most manufactured quantum system in existence. Every semiconductor quantum-well laser — in DVD players, fiber-optic transceivers, laser pointers — is a thin sheet of low-bandgap material (say GaAs) between higher-gap barriers (AlGaAs), and the emitted photon energy is set by exactly the level positions you computed in the worked examples: engineers dial $V_0$ with alloy composition and $2a$ with atomic-layer deposition, then read the laser's color off the transcendental equation. Quantum cascade lasers chain dozens of coupled finite wells so that a single electron emits a photon at every step down the staircase — infrared spectrometers built from solved matching conditions. The barely-bound limit runs modern nuclear physics too: halo nuclei like $^{11}$ Li, whose outermost neutrons orbit far beyond the core in exactly the long-tailed, weakly-bound states of a near-threshold well, are mapped at rare-isotope facilities such as FRIB. And in every scanning tunnelling microscope, the $e^{-\kappa d}$ sensitivity that Gamow first exploited resolves individual atoms: a 0.1 nm change in tip height changes the current by an order of magnitude.

The physical picture

Think of the well as a resonant cavity with leaky walls. Inside, where $E + V_0 > 0$ , the particle has positive kinetic energy: the wavefunction curves toward the axis and oscillates, exactly like a free wave. Outside, the energy sits below the potential — kinetic energy would be negative — and the equation flips sign: $\psi$ now curves away from the axis, and the only behavior that doesn't explode is exponential decay. A bound state exists when the interior oscillation arrives at the wall with exactly the right value-to-slope ratio to hand off, seamlessly, to a falling exponential. Too little phase accumulated inside and the slope cannot turn over; the matching fails, and that energy is simply not allowed.

Figure

Leaky confinement — three bound states of a well with z₀ = 4

Read the figure from the bottom up. The deep state barely notices the walls are finite: its tail dies within a fraction of the well width, and it looks almost like an infinite-well state. The shallow state is another creature entirely — bound by a sliver of energy, it spills far into the forbidden region, living a substantial fraction of its life where a classical particle could not exist. That spilling is not a curiosity. It is the seed of quantum tunnelling, the mechanism of alpha decay, and the working principle of the scanning tunnelling microscope. One more habit to build: symmetry first. The well is symmetric under $x \to -x$ , so its bound states come in definite parities — even (cosine-like), odd (sine-like) — alternating up the ladder. Parity halves the algebra before you write a single matching equation.

The math: match, then intersect

Set $V(x) = -V_0$ for $|x| < a$ , zero outside, and look for bound states, $-V_0 < E < 0$ . Inside the well the Schrödinger equation sees a positive kinetic energy $E + V_0$ and delivers oscillation; outside it sees the deficit $-E$ and delivers exponential decay. Each region hands the problem its own characteristic inverse length, and everything that follows is a negotiation between the two:

l = \frac{\sqrt{2m(E + V_0)}}{\hbar}, \qquad \kappa = \frac{\sqrt{-2mE}}{\hbar} \qquad (1)

where: $l$ — oscillation wavenumber inside the well; $\kappa$ — decay rate of the tails outside; $1/\kappa$ — the penetration depth into the forbidden region. Notice the trade-off built into the definitions: a deep-lying state (very negative $E$ ) has small $l$ and large $\kappa$ — it barely wiggles and its tail dies almost immediately — while a state near the rim ( $E \to 0^-$ ) has $\kappa \to 0$ and a tail stretching far beyond the well. Those are the short-tailed deep state and long-tailed shallow state of the figure above, now in symbols. The solution is built piecewise — and symmetry does the first round of work. The Hamiltonian commutes with parity and one-dimensional bound states are non-degenerate, so every bound state is either even or odd, and one interior amplitude vanishes before any matching begins:

\psi(x) = \begin{cases} F\,e^{\kappa x}, & x < -a \\ A\cos lx \;\text{ or }\; B\sin lx, & |x| < a \\ \pm F\,e^{-\kappa x}, & x > a \end{cases}

where: the growing exponentials are discarded because the state must be normalizable, and parity ties the two tails together ( $+$ for even states, $-$ for odd). Only the ratio of interior to tail amplitude remains free — and the matching conditions are about to fix it.

Derivation AThe matching conditions

The target is the finite well's replacement for the infinite well's clean fitting rule. The plan: write the even solution, demand that value and slope both agree at the wall where oscillation hands off to decay, then divide the two conditions so the arbitrary amplitudes cancel. That division is the heart of the method — quantization must come from the shape of the state, never from its normalization. Watch parity do half the work, and watch the answer refuse to be algebraic.

Step 1. Impose continuity of $\psi$ at $x = a$ for the even state: $A\cos(la) = Ge^{-\kappa a}$ . Why: a discontinuous wavefunction would have infinite kinetic energy; the value must agree at the join.

Step 2. Impose continuity of $\psi'$ at $x = a$ : $-Al\sin(la) = -\kappa Ge^{-\kappa a}$ . Why: the potential jump at the wall is finite, so the slope is continuous too — only an infinite spike (a delta function) can kink $\psi'$ .

Step 3. Divide the two equations. The amplitudes $A$ and $G$ cancel entirely:

\kappa = l\tan(la) \qquad \text{(even states)} \qquad (2)

Why: quantization cannot depend on normalization — dividing strips away everything but the shape constraint.

Step 4. Repeat with $B\sin lx$ inside. The same division gives

\kappa = -\,l\cot(la) \qquad \text{(odd states)} \qquad (3)

Why: since $\kappa > 0$ , even states need $\tan(la) > 0$ and odd need $\cot(la) < 0$ — each parity is confined to its own windows of $la$ , which is why even and odd levels alternate. *Verified: equations (2)–(3) reproduce Griffiths §2.6; re-derive by dividing the value and slope matching equations.*

Equations (2) and (3) contain the energy on both sides, buried inside $l$ and $\kappa$ — trigonometric on one side, square root on the other. No algebra will unbury it. This is a transcendental equation: its solutions exist, are discrete, and must be found graphically or numerically.

Go deeper · Derivation B · 1 minThe circle construction

Derivation BThe circle construction

Equations (2) and (3) hide the energy inside two different functions — trigonometric on one side, square root on the other — and no algebra will dig it out. The rescue is geometric: find a combination of $l$ and $\kappa$ from which the energy drops out entirely, so the matching conditions become fixed curves in a plane and bound states become intersections. We want three payoffs — a picture that solves the equation, a count of how many states a given well holds, and a guarantee that at least one always survives. Watch a single dimensionless radius do all the budgeting.

Step 1. Eliminate $E$ . From the definitions (1), $l^2 + \kappa^2 = 2mV_0/\hbar^2$ — the energy cancels. Why: $E$ enters $l^2$ and $\kappa^2$ with opposite signs, so their sum depends only on the well itself.

Step 2. Scale by $a$ . With $z = la$ and the well-strength parameter $z_0 = \tfrac{a}{\hbar}\sqrt{2mV_0}$ ,

z^2 + (\kappa a)^2 = z_0^2 \qquad (4)

Why: in the $(z, \kappa a)$ plane every allowed state must sit on a circle whose radius is fixed by depth and width alone.

Step 3. Intersect. The even condition (2) reads $\kappa a = z\tan z$ and the odd condition (3) reads $\kappa a = -z\cot z$ : fixed families of curves. Bound states are exactly the intersections of the circle with those curves — every crossing is one level. Why: a state must satisfy the energy bookkeeping (the circle) and the smooth-matching demand (the curves) simultaneously.

Step 4. Count. The even branch through the origin always cuts the circle once — so every well, however shallow, binds at least one state. Each time $z_0$ grows past a multiple of $\pi/2$ , a new branch (alternating parity) enters the circle:

N = 1 + \left\lfloor \frac{2z_0}{\pi} \right\rfloor \qquad (5)

Why: the branches start at $z = 0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}, \dots$ ; the circle reaches a new one exactly when its radius crosses the next threshold. *Verified: thresholds at $z_0 = n\pi/2$ match the graphical analysis in Cohen-Tannoudji Complement H_I.*

Figure

The whole spectrum in one picture — circle meets tangent curves

Once a root $z_n$ is in hand, the energy follows from $E_n = \hbar^2 z_n^2 / (2ma^2) - V_0$ : the phase that fits inside the well is the energy, measured from the well bottom.

Go deeper · Derivation C · 1 minThe shallow well and the delta-well limit

Derivation CThe shallow well and the delta-well limit

Two claims to pin down for the barely-binding well: the lone bound state's energy scales as the square of the depth, and the coefficient must match the delta-function well exactly. The strategy is a small- $z_0$ expansion of the even matching condition, and it deserves care — a sloppy expansion produces a wrong factor of 4, and that wrong factor sat in this article's own source material. The safety rail is the circle: $z$ stays pinned near $z_0$ , and the tiny gap between them is where all the binding lives.

Step 1. Expand the even condition. For $z_0 \ll 1$ the circle cuts only the first even branch, so $z \le z_0 \ll 1$ and $\tan z \approx z$ : the even condition $\kappa a = z\tan z$ collapses to $\kappa a \approx z^2$ . Why: the branch through the origin is the only curve inside so small a circle — this is the always-bound even state, alone.

Step 2. Feed it to the circle. In $z^2 + (\kappa a)^2 = z_0^2$ the term $(\kappa a)^2 \approx z^4$ is fourth order, so to leading order $z^2 \approx z_0^2$ , and therefore $\kappa a \approx z^2 \approx z_0^2$ . Why: almost all the well strength is spent on interior phase; the second-order leftover is the binding — why weak binding is so very weak.

Step 3. Restore units. With $\kappa \approx z_0^2/a = 2maV_0/\hbar^2$ , the binding energy is

E_0 = -\frac{\hbar^2\kappa^2}{2m} \approx -\frac{2ma^2V_0^2}{\hbar^2}

Why: quadratic in the depth — halve $V_0$ and the binding drops fourfold. Beware the tempting mis-expansion $\kappa a \approx z_0^2/2$ , which lands on $-ma^2V_0^2/2\hbar^2$ , off by that factor of 4.

Step 4. Check against the delta well. Collapse the well keeping its area $\alpha = 2aV_0$ fixed: $E_0 \approx -m\alpha^2/2\hbar^2$ — precisely the delta well's lone bound state. The shallow finite well is a delta well that hasn't finished collapsing. Why: a weak short-range 1D attraction binds via its integrated strength alone; shape has stopped mattering. *Verified: the numerical root at z₀ = 0.05 gives E₀ over (−2ma²V₀²/ħ²) equal to 0.997, tending to 1 as z₀ shrinks; the mis-expanded value is off by 4.*

Limiting cases

Case	Result	Meaning
$V_0 \to \infty$ ( $z_0 \to \infty$ )	$E_n + V_0 \to \dfrac{n^2\pi^2\hbar^2}{2m(2a)^2}$	tails vanish, $\psi \to 0$ at the walls — the infinite square well returns
shallow well, $z_0 \ll 1$	one bound state, $E_0 \approx -\dfrac{2ma^2V_0^2}{\hbar^2}$	binding quadratic in $V_0$ — a barely-bound state, mostly living outside the well
$a \to 0$ , $V_0 \to \infty$ , $2aV_0 = \alpha$ fixed	$E_0 \to -\dfrac{m\alpha^2}{2\hbar^2}$	the delta-function well: one bound state survives the collapse
any $z_0 > 0$	$N \geq 1$	a 1D well is never empty — contrast 3D, where too-shallow wells bind nothing
high levels, $z_n \to z_0$	$\kappa \to 0$ , tail length $1/\kappa \to \infty$	states near the rim barely hold on, spreading far beyond the well

Common trap“The wavefunction must vanish at the walls”

It must not — that habit belongs to the infinite well and dies here. At $x = \pm a$ the potential jump is finite, so both $\psi$ and $\psi'$ are continuous and $\psi(\pm a)$ is generally nonzero: the state leaks into the forbidden region with penetration depth $1/\kappa$ . Demanding $\psi(\pm a) = 0$ would give you back the infinite-well energies, which are measurably wrong — the true levels sit lower, because the leaking tails widen the effective box. And a second habit to unlearn: do not hunt for closed-form energies. The conditions $z\tan z = \sqrt{z_0^2 - z^2}$ are transcendental; the circle diagram or a numerical solver is not a fallback, it is the method.

Worked examples

1 · How many states fit? (confidence builder)

An electron sits in a quantum well of depth $V_0 = 1.0\ \mathrm{eV}$ and width $2a = 2.0\ \mathrm{nm}$ . How many bound states does the well hold, and with what parities?

Step 1. Compute the well strength: $z_0 = \tfrac{a}{\hbar}\sqrt{2mV_0}$ . With $a = 1.0\times10^{-9}\ \mathrm{m}$ and $2mV_0 = 2(9.11\times10^{-31})(1.602\times10^{-19}) = 2.92\times10^{-49}$ : $\sqrt{2mV_0} = 5.40\times10^{-25}\ \mathrm{kg\,m/s}$ , so $z_0 = (10^{-9})(5.40\times10^{-25})/(1.055\times10^{-34}) = 5.12$ .

Step 2. Count with (5): $2z_0/\pi = 3.26$ , so $N = 1 + \lfloor 3.26 \rfloor = \boxed{4}$ bound states.

Step 3. Parities alternate from the even ground state: even, odd, even, odd. (Check against the thresholds: $z_0 = 5.12$ exceeds $\pi/2, \pi, 3\pi/2$ but not $2\pi \approx 6.28$ — three thresholds crossed, plus the guaranteed ground state.) Squeeze the well to $2a = 0.4\ \mathrm{nm}$ and $z_0$ drops to $1.02 < \pi/2$ : only the lone even state survives. Width and depth budget the spectrum through $z_0$ alone.

2 · Pinning down the ground state (exam level)

For the same well ( $V_0 = 1.0\ \mathrm{eV}$ , $a = 1.0\ \mathrm{nm}$ , $z_0 = 5.12$ ), find the ground-state energy and the penetration depth of its tails, and compare with the infinite-well prediction.

Step 1. Solve the even condition $z\tan z = \sqrt{z_0^2 - z^2}$ in the first window $(0, \pi/2)$ . Bisect: at $z = 1.25$ the left side is $3.76$ , the right $4.97$ — left too small; at $z = 1.35$ , left $= 6.02$ , right $= 4.94$ — too big; converging: $z_1 = 1.312$ .

Step 2. Energy above the well bottom: $E + V_0 = \dfrac{\hbar^2 z_1^2}{2ma^2} = \dfrac{(1.055\times10^{-34})^2(1.312)^2}{2(9.11\times10^{-31})(10^{-9})^2} = 1.05\times10^{-20}\ \mathrm{J} = 0.066\ \mathrm{eV}$ , so $E_1 = \boxed{-0.934\ \mathrm{eV}}$ .

Step 3. Tail decay: $\kappa a = \sqrt{z_0^2 - z_1^2} = \sqrt{26.2 - 1.72} = 4.95$ , so the penetration depth is $1/\kappa = a/4.95 = \boxed{0.20\ \mathrm{nm}}$ — the electron's presence extends a fifth of a nanometer into the barrier.

Step 4. Compare with the infinite well of the same width: $E_1^{\infty} = \dfrac{\pi^2\hbar^2}{2m(2a)^2} = 0.094\ \mathrm{eV}$ above the bottom. The finite well's $0.066\ \mathrm{eV}$ is 30% lower. The leaking tails widen the effective box, soften the wavefunction's curvature, and cheapen the kinetic energy — hard walls always overestimate confinement energies.

Origins · 1 minGeorge Gamow and how this was discovered

George Gamow

1904–1968 · Göttingen, then Copenhagen

In 1928, weeks after arriving in Göttingen, the 24-year-old Gamow read Rutherford's data on alpha decay and saw what everyone had missed: the alpha particle does not climb out of the nuclear well — it leaks through the wall, riding exactly the kind of exponential tail you just computed. His tunnelling formula explained the staggering range of alpha-decay lifetimes (nanoseconds to billions of years) as the exquisite sensitivity of $e^{-\kappa d}$ to barrier width, and it was the first time the wavefunction's forbidden-region tail did real physical work. Gurney and Condon reached the same idea independently that year. Gamow went on to co-found Big Bang nucleosynthesis — and to write the Mr Tompkins books that taught generations to imagine quantum wells.

History

1926

— Schrödinger's wave equation makes piecewise potentials solvable: continuity of

\psi

and

\psi'

becomes the master tool.

1928

— Gamow, and independently Gurney & Condon, explain alpha decay as leakage through a finite barrier — forbidden-region tails become physics, not formalism.

1935

— Bethe and Peierls model the deuteron as a shallow finite well: one barely-bound state, its wavefunction mostly outside the force's range.

1949

— Mayer and Jensen's nuclear shell model builds on finite-well-like potentials, ordering the magic numbers of nuclear stability.

1970

— Esaki and Tsu propose semiconductor superlattices: finite square wells become a manufacturable technology.

Going further

Texts. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §2.6 · Shankar, Principles of Quantum Mechanics, Ch. 5 · Cohen-Tannoudji, Diu & Laloë, Vol. 1, Complement H_I · Townsend, A Modern Approach to Quantum Mechanics.

Provenance. Equations and limits cross-checked against Griffiths 3rd ed. §2.6, Cohen-Tannoudji Vol.1 Complement HI, and Shankar Ch.5. A verifier should (1) re-derive the even/odd conditions by dividing the value and derivative matching equations; (2) confirm $z^2+(\kappa a)^2=z_0^2$ from the definitions of $l,\kappa$ ; (3) check the count formula $N=1+\lfloor 2z_0/\pi\rfloor$ against the graphical threshold pattern at $z_0=n\pi/2$ ; (4) verify the shallow-well result $E_0\approx-2ma^2V_0^2/\hbar^2$ and its consistency with the delta-well limit $E_0=-m\alpha^2/2\hbar^2$ with $\alpha=2aV_0$ ; (5) confirm the deep-well limit reproduces the infinite square well of width $2a$ ; plus the worked numbers ( $z_0 = 5.12$ , $N = 4$ , $z_1 = 1.312$ , $E_1 = -0.934\ \mathrm{eV}$ , $1/\kappa = 0.20\ \mathrm{nm}$ , $E_1^{\infty} = 0.094\ \mathrm{eV}$ ). Merged with the v3 pilot article (v3-finite-square-well) at Step-2; AllQFT depth grafted onto the v3 narrative spine. Correction at merge: the AllQFT base and the v3 pilot both quoted the shallow-well energy as $-ma^2V_0^2/2\hbar^2$ , inconsistent by a factor of 4 with their own delta-well limit; corrected to $E_0 \approx -2ma^2V_0^2/\hbar^2$ and verified numerically.

Prerequisites. Time-independent Schrödinger equation in 1D · bound vs. scattering states · the infinite square well · free-particle plane waves. Leads to. The delta-function well (the collapse limit) · quantum tunnelling & barrier transmission · double wells and level splitting · the quantum harmonic oscillator (a well with soft walls).

How can the particle exist where E is less than V — negative kinetic energy?

No measurement ever catches it with negative kinetic energy. To confirm the particle is in the forbidden region you must localize it within a tail length $1/\kappa$ , and that measurement injects momentum uncertainty of order $\hbar\kappa$ — kinetic energy of exactly the size needed to cover the deficit. The honest statement: the amplitude out there is nonzero, probability genuinely leaks, tunnelling and level shifts see the tails — and every completed energy audit stays consistent. Negative kinetic energy is a property of no measurement outcome.

Why does every 1D well bind a state, but a shallow 3D well can bind nothing?

Write the 3D s-wave problem for $u(r) = r\psi(r)$ and it becomes exactly a 1D well problem with one extra demand: $u(0) = 0$ . That boundary condition admits only the odd-type solutions — and odd states switch on only once $z_0$ exceeds $\pi/2$ . The 1D privilege lives entirely in the even branch through the origin, which the 3D constraint kills. This is why the deuteron needed a deep enough well to exist at all, and why "at least one bound state" must never be quoted in three dimensions.

Is there a quantitative version of the wider effective box idea?

Yes, and it is a serviceable estimator. The tails add roughly one penetration depth on each side, so compare with an infinite well of width $2a + 2/\kappa$ : $E + V_0 \approx \pi^2\hbar^2/[2m(2a + 2/\kappa)^2]$ . For the worked well ( $1/\kappa = 0.20$ nm on a 2.0 nm box) this predicts about $0.065$ eV against the exact $0.066$ eV. It is only an estimate — $\kappa$ depends on the energy you are computing — but it turns "levels sit lower" into a number and shows why deep levels barely shift while rim states shift hard.

What exactly happens at a threshold like z₀ equal to π/2?

A state is born at zero binding. At threshold the circle just touches the next branch where $\kappa = 0$ : that marginal state has $E = 0^-$ , an infinitely long tail, and sits at the edge of normalizability. Just above it is genuinely bound, with tiny binding and enormous extent; just below, it dissolves into the continuum — often surviving as a low-energy scattering resonance. The count $N = 1 + \lfloor 2z_0/\pi \rfloor$ is for generic $z_0$ ; exactly at a threshold, do not count the marginal state.

My textbook's finite-well formulas look different — width L instead of 2a?

Almost certainly a convention mismatch — the most common factor-of-2 factory in this topic. Griffiths and this article put the well between $-a$ and $+a$ (width $2a$ ) and define $z_0 = \tfrac{a}{\hbar}\sqrt{2mV_0}$ with the half-width. Books using the full width $L = 2a$ carry strength parameters twice as large and thresholds that look shifted. Translate both sources to one convention before comparing, and in an exam state yours before computing.

If the deuteron spends half its time outside the force's range, what holds it together out there?

Nothing does — and nothing needs to. Outside the well the wavefunction is a pure decaying exponential: no force acts there, the amplitude simply persists, financed by the binding accumulated inside. With binding energy 2.2 MeV the tail length $1/\kappa \approx 4.3$ fm exceeds the roughly 2 fm range of the nuclear force, so most of the probability sits outside. That is this article's barely-bound regime at its physical extreme — and why the deuteron is fragile and halo nuclei are its exaggerated cousins.