Delta-Function Potential Well and Bound State

Take the finite square well you just solved and abuse it: squeeze the width toward zero while cranking the depth toward infinity, keeping the product — depth times width — fixed. The excited states are shed one by one as the well narrows, but the ground state rides the collapse all the way down. What remains is the strangest trap in one-dimensional quantum mechanics: a potential that exists at a single point, $V(x) = -\alpha\,\delta(x)$ , and yet holds a particle. The particle cannot live inside a well of zero width — there is no inside — so it lives in the free space on either side, its wavefunction a pair of exponential tails meeting in a sharp kink at the origin. That kink is the whole story. Everywhere else the particle is free; the entire potential acts through one boundary condition, a prescribed jump in the slope of $\psi$ , obtained by integrating the Schrödinger equation across the singularity. Matching two dying exponentials through that jump has exactly one solution — not one family, one state — with energy $E = -m\alpha^2/2\hbar^2$ fixed entirely by the strength $\alpha$ . Where the infinite well gave you infinitely many levels and the finite well a tunable few, the delta well gives you precisely one, and a two-line dimensional argument will tell you why it never could have been otherwise. The payoff is leverage. This is the fastest nontrivial bound state in physics — a favorite exam problem — and the working model for any attraction much narrower than the wavelengths it traps: impurities in crystals, contact interactions between ultracold atoms, the deuteron in its long-tailed, barely-bound reality.

Key takeaways

An attractive delta potential $V(x) = -\alpha\,\delta(x)$ ( $\alpha > 0$ ) supports exactly one bound state — no more, no less, for any strength.
The bound state is $\psi(x) = \sqrt{\kappa}\,e^{-\kappa|x|}$ with $\kappa = m\alpha/\hbar^2$ : two exponential tails joined by a cusp at the origin.
Its energy is $E = -m\alpha^2/2\hbar^2$ — the only energy you can build from $m$ , $\hbar$ , and $\alpha$ .
The singular potential trades the usual smoothness rule for a jump condition: $\psi$ stays continuous, but $\psi'$ jumps by $-(2m\alpha/\hbar^2)\,\psi(0)$ .
The same well scatters positive-energy waves with transmission $T = [1 + (m\alpha/\hbar^2k)^2]^{-1}$ — bound state and scattering controlled by one parameter.
Key idea: integrate the Schrödinger equation across the singularity and the potential becomes a boundary condition — the cusp that exactly one state can satisfy.

The core insight

Integrating the Schrödinger equation across the delta singularity converts the singular potential into a boundary condition — a fixed jump in $\psi'$ ; matching the two decaying exponentials through this jump yields exactly one bound state, with $\kappa = m\alpha/\hbar^2$ and $E = -\hbar^2\kappa^2/2m$ .

Where this lives today

The delta potential's modern career is as the universal stand-in for any interaction much shorter-ranged than every other length in the problem. Ultracold-atom physics runs on it: at nanokelvin temperatures, atomic de Broglie wavelengths dwarf the actual interatomic potential, so theorists replace it wholesale with a contact (delta) interaction whose one strength parameter — the scattering length — is tunable in the lab via Feshbach resonances. Entire many-body models built from delta interactions, like the Lieb–Liniger gas of bosons on a line, are exactly solvable and are now engineered atom-by-atom in one-dimensional optical traps, letting experiments check exact solutions of the very idealization you just solved. In the solid state, the shallow donor level of a doped semiconductor — the electron loosely bound to a phosphorus atom in silicon — is a delta-well problem dressed in an effective mass, and the Kronig–Penney chain of deltas remains the first band-structure calculation every physicist performs. The weak-binding limit ( $1/\kappa$ far beyond the well's range) is the effective-range picture of the deuteron and of halo nuclei — the same long exponential tail, drawn at femtometre scale.

The physical picture

Every smooth attractive potential bends the wavefunction gently toward the axis, trading potential depth for reduced curvature energy over some region of space. The delta well has no region of space. All its attraction is concentrated at one point, so it does the only thing a point can do: it kinks the wavefunction. On either side of the origin the particle is free with negative energy, and the only free solutions that die at infinity are exponentials $e^{-\kappa|x|}$ . Two such tails, falling away symmetrically, meet at the origin with slopes $-\kappa$ and $+\kappa$ — a built-in kink of just the kind the delta demands. Binding happens when the kink the tails have equals the kink the well supplies, and that single matching selects a single $\kappa$ .

Figure

The delta well's lone bound state — two free tails and one cusp

Now run the dimensional argument, because it explains the "exactly one" better than any equation-grinding. The strength $\alpha$ carries units of energy × length (it multiplies a delta function, whose integral is 1 over a length). From $m$ , $\hbar$ , and $\alpha$ you can build exactly one inverse length, $\kappa = m\alpha/\hbar^2$ , and exactly one energy, $m\alpha^2/\hbar^2$ (up to a number). A second bound state would need a second scale to live at — and there is none. Contrast the finite well, whose width and depth are independent knobs: their ratio furnishes the dimensionless parameter $z_0$ that counts its levels. Collapse the well and that knob is welded shut. Here is that collapse drawn as three snapshots. Narrow a finite well while holding its area $2aV_0 = \alpha$ fixed: the excited levels are squeezed out of existence one by one, and the ground state alone rides through to the singular limit.

Figure

Where the delta well comes from — the collapse limit of a finite well

One more habit from this problem travels far: in one dimension, any attraction binds. As $\alpha \to 0$ the state survives — it just becomes absurdly shallow and absurdly wide, $E \to 0^-$ while the tails stretch as $1/\kappa \to \infty$ . A barely-bound state lives almost entirely outside its well, which is precisely the situation of the real deuteron. The point's influence does not end at negative energy. Send in a wave with $E > 0$ and the same singularity partially reflects it — and the fraction that gets through is governed by the very scale the bound state fixed, the binding energy $|E_b| = m\alpha^2/2\hbar^2$ :

Figure

Transmission through the delta well — the bound state's fingerprint

Slow waves mostly bounce off this attractive well — pure quantum reflection — while fast ones barely notice it, and the fifty-fifty point sits exactly at $E = |E_b|$ . One parameter controls both faces of the problem; the scattering section picks the curve back up.

The math: one jump condition does everything

Now make the collapse limit official. All of the squeezed well's attraction survives as a single number $\alpha$ — the area (depth × width) held fixed during the squeeze — multiplying a delta function that plants it at the origin. The eigenvalue problem is the usual one, with the singular potential written in:

V(x) = -\alpha\,\delta(x), \qquad -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} - \alpha\,\delta(x)\,\psi = E\,\psi \qquad (1)

where: $\alpha > 0$ — well strength, units of J·m; $\delta(x)$ — Dirac delta, zero everywhere except $x = 0$ , with $\int\delta(x)\,dx = 1$ ; $E < 0$ for the bound state we seek. Read the structure before solving: away from $x = 0$ this is the free-particle equation, so the delta cannot shape the wavefunction anywhere — it can only impose a condition at the one point it inhabits. The derivations below extract that condition, then cash it in.

Derivation AThe jump condition at the singularity

The delta potential is infinite at one point, so the familiar rule that $\psi'$ is continuous cannot be trusted there — we need to find what replaces it. The strategy is the standard cure for singular terms: integrate the full Schrödinger equation (1) over a shrinking interval $[-\epsilon, +\epsilon]$ straddling the singularity and watch which terms survive as $\epsilon \to 0$ . Whatever survives is the boundary condition the delta imposes. Watch the asymmetry: the kinetic and delta terms stay finite while the energy term dies with the interval — that imbalance is the entire result.

Step 1. Integrate equation (1) across the singularity, from $-\epsilon$ to $+\epsilon$ : $-\dfrac{\hbar^2}{2m}\big[\psi'(\epsilon) - \psi'(-\epsilon)\big] - \alpha\,\psi(0) = E\displaystyle\int_{-\epsilon}^{\epsilon}\psi\,dx$ . Why: the delta is infinitely tall but integrably so — integration is the tool that tames it, and $\int\delta(x)\psi(x)\,dx = \psi(0)$ .

Step 2. Shrink the interval. As $\epsilon \to 0$ the right side vanishes ( $\psi$ is bounded, the interval isn't there), leaving

\Delta\psi' \equiv \left.\frac{d\psi}{dx}\right|_{0^+} - \left.\frac{d\psi}{dx}\right|_{0^-} = -\frac{2m\alpha}{\hbar^2}\,\psi(0) \qquad (2)

Why: a finite quantity survived the shrinking interval only because the potential was infinite there. For any finite potential the same integral vanishes and $\psi'$ is continuous — this jump is the delta's signature.

Step 3. Note what does not jump. $\psi$ itself stays continuous: $\psi'$ has a step, so its integral $\psi$ has only a kink, not a break. Why: each integration smooths by one degree — $\psi''$ contains a delta, $\psi'$ a step, $\psi$ a cusp. *Verified: equation (2) matches Griffiths §2.5.2 and Cohen-Tannoudji Complement K_I.*

Go deeper · Derivation B · 1 minThe unique bound state

Derivation BThe unique bound state

With the delta reduced to a boundary condition, the eigenvalue problem can be solved completely, and we do it here: energy, wavefunction, normalization. The route is to solve the free equation on each side of the origin keeping only the normalizable pieces, glue the branches with continuity of $\psi$ , then let the jump condition from the previous card select the one decay constant the well can support. Watch what does the selecting — the amplitude cancels out, so quantization comes entirely from matching the kink the tails have to the kink the delta demands, and one equation fixes one state.

Step 1. Solve where the potential vanishes. For $E < 0$ define $\kappa = \sqrt{-2mE}/\hbar > 0$ ; away from the origin $\psi'' = \kappa^2\psi$ , so keeping only normalizable pieces: $\psi = Ae^{\kappa x}$ for $x < 0$ and $\psi = Be^{-\kappa x}$ for $x > 0$ . Why: the growing exponentials would blow up at infinity; a bound particle must be findable somewhere finite.

Step 2. Impose continuity at the origin. $A = B$ — the state is symmetric, as it must be: the potential is even, and a lone bound state has no partner to share parities with. Why: the ground state of any symmetric 1D potential is even and nodeless.

Step 3. Feed the tails to the jump condition. The slopes are $\psi'(0^+) = -\kappa B$ and $\psi'(0^-) = +\kappa B$ , so (2) reads $-2\kappa B = -\dfrac{2m\alpha}{\hbar^2}B$ , giving

\kappa = \frac{m\alpha}{\hbar^2} \qquad (3)

Why: the amplitude $B$ cancels — quantization never depends on normalization — and one equation fixes one $\kappa$ . This is the entire spectrum.

Step 4. Read off the energy.

E = -\frac{\hbar^2\kappa^2}{2m} = -\frac{m\alpha^2}{2\hbar^2} \qquad (4)

Why: substitute (3) into the definition of $\kappa$ . One state, one energy, no free integers.

Step 5. Normalize. $\int|\psi|^2dx = 2B^2\!\int_0^\infty e^{-2\kappa x}dx = B^2/\kappa = 1$ gives $B = \sqrt{\kappa}$ :

\psi(x) = \sqrt{\kappa}\;e^{-\kappa|x|} \qquad (5)

Why: the tighter the binding, the taller and narrower the state — $\sqrt{\kappa}$ keeps the total probability at 1. *Verified against Griffiths §2.5.2 (Eqs. 2.129–2.132) and Shankar Ch. 5.*

Go deeper · Derivation C · 1 minThe energy audit

Derivation CThe energy audit

A bound state's energy should decompose sensibly into kinetic and potential parts, and for the delta well the audit is short enough to do exactly — so we do it as an independent check on the eigenvalue. The plan: compute $\langle V \rangle$ by letting the delta sample the probability density at the origin, get $\langle T \rangle$ from the slope integral after one integration by parts, and confirm the two add up to $E$ from the previous card. The split is worth remembering in its own right: it exposes the delta well's private virial relation, with $\langle V \rangle$ exactly twice the total energy.

Step 1. Potential energy. $\langle V \rangle = -\alpha\int\delta(x)|\psi|^2dx = -\alpha|\psi(0)|^2 = -\alpha\kappa = -\dfrac{m\alpha^2}{\hbar^2}$ . Why: the delta samples the probability density exactly where the well sits.

Step 2. Kinetic energy. Integrating by parts, $\langle T \rangle = \dfrac{\hbar^2}{2m}\int|\psi'|^2dx = \dfrac{\hbar^2\kappa^2}{2m} = \dfrac{m\alpha^2}{2\hbar^2}$ . Why: $|\psi'|^2 = \kappa^2|\psi|^2$ almost everywhere, and the norm is 1 — the cusp itself contributes nothing extra because $|\psi'|^2$ stays integrable.

Step 3. Add. $\langle T \rangle + \langle V \rangle = \dfrac{m\alpha^2}{2\hbar^2} - \dfrac{m\alpha^2}{\hbar^2} = -\dfrac{m\alpha^2}{2\hbar^2} = E$ ✓. Why: the checksum confirms (4) by an independent route. Note the pattern $\langle V \rangle = 2E$ , $\langle T \rangle = -E$ : the delta well's own virial relation. *Verified: cross-checks the eigenvalue via $\langle H \rangle$ ; matches Cohen-Tannoudji Complement K_I.*

The collapse limit pictured back in the physical-picture section is a computation worth doing once (Griffiths leaves it as a problem): the Finite Square Well's even-state condition $\kappa = l\tan(la)$ becomes, for $a \to 0$ with $2aV_0 = \alpha$ fixed, $\kappa \approx l^2a = 2mV_0a/\hbar^2 = m\alpha/\hbar^2$ — exactly (3). Every result of this article is the deep-narrow limit of the finite well with one bound state retained.

Scattering: the same point, seen by a traveler

For $E > 0$ the delta cannot bind, but it can deflect. A plane wave $e^{ikx}$ with $k = \sqrt{2mE}/\hbar$ arriving from the left partially reflects and partially transmits; the same jump condition (2), applied to oscillating instead of decaying exponentials, gives

T(E) = \frac{1}{1 + \dfrac{m\alpha^2}{2\hbar^2E}} = \frac{1}{1 + |E_b|/E}, \qquad R = 1 - T \qquad (6)

where: $|E_b| = m\alpha^2/2\hbar^2$ — the binding energy of the bound state, resurfacing uninvited in a pure scattering formula. The wave is mostly reflected when $E \ll |E_b|$ and passes almost freely when $E \gg |E_b|$ ; at $E = |E_b|$ the coin is fair, $T = \tfrac12$ . The bound state at negative energy leaves its fingerprint on physics at all positive energies — a first glimpse of a deep scattering-theory fact (bound states appear as poles of the transmission amplitude continued to $E < 0$ ). Equation (6) is exactly the transmission curve previewed in the physical-picture section: monotonic, resonance-free, with its fifty-fifty point pinned at $E = |E_b|$ . The honest bookkeeping behind such formulas — reflection and transmission defined as ratios of probability current, and why they must sum to one — is owned by Transmission and Reflection Coefficients; the repulsive twin of this calculation, same $|t|^2$ but nothing to bind, belongs to Delta-Function Barrier and Scattering.

Limiting cases

Case	Result	Meaning
$\alpha \to 0^+$	$E \to 0^-$ , $1/\kappa \to \infty$	the state survives arbitrarily weak attraction — in 1D any well binds — but grows huge and shallow
$\alpha \to \infty$	$E \to -\infty$ , $1/\kappa \to 0$	a spike pinned to the origin; binding grows quadratically in strength
$V_0 \to \infty$ , $a \to 0$ , $2aV_0 = \alpha$	$E \to -m\alpha^2/2\hbar^2$	the finite square well collapses onto the delta well, keeping one state
$E > 0$	$T = [1 + \|E_b\|/E]^{-1}$	scattering states form the continuum; no second bound level hides anywhere
$\alpha < 0$ (repulsive)	no bound state	a repulsive kink pushes the slopes the wrong way; only scattering, with the same $\|T\|$

Common trap“The slope of ψ must be continuous — it always was before”

At every finite potential you have met — steps, finite wells — both $\psi$ and $\psi'$ are continuous, and it is tempting to promote that to a law of nature. It is not a law; it is a theorem about finite potentials, and the delta breaks its hypothesis. Here $\psi'$ jumps by $-(2m\alpha/\hbar^2)\psi(0)$ , and if you smooth the kink away you lose the bound state entirely — the cusp is not a blemish, it is where all the binding happens. Keep the hierarchy straight: $\psi$ continuous always (probability current must make sense); $\psi'$ continuous unless the potential is infinitely singular; and at the infinite square well's hard walls, $\psi'$ likewise jumps, for the same reason. One more trap: do not go hunting for the delta well's excited states. There is no $n$ anywhere in equations (3)–(5) — the spectrum is one state deep, and the dimensional argument says it could never be otherwise.

Worked examples

1 · Sizing up a point trap (confidence builder)

An electron is bound by a delta well of strength $\alpha = 2.0\ \mathrm{eV\cdot\mathring{A}} = 3.20\times10^{-29}\ \mathrm{J\cdot m}$ . Find the binding energy, the size of the state, and the probability of finding the electron beyond one decay length.

Step 1. Decay constant from (3): $\kappa = \dfrac{m\alpha}{\hbar^2} = \dfrac{(9.11\times10^{-31})(3.20\times10^{-29})}{(1.055\times10^{-34})^2} = 2.62\times10^{9}\ \mathrm{m^{-1}}$ , so the state's size is $1/\kappa = \boxed{0.38\ \mathrm{nm}}$ — a few atomic diameters from a zero-width well.

Step 2. Energy from (4): $E = -\dfrac{\hbar^2\kappa^2}{2m} = -\dfrac{(1.055\times10^{-34})^2(2.62\times10^9)^2}{2(9.11\times10^{-31})} = -4.2\times10^{-20}\ \mathrm{J} = \boxed{-0.26\ \mathrm{eV}}$ — a typical shallow-impurity scale in a semiconductor.

Step 3. Tail probability: $P(|x| > 1/\kappa) = 2\displaystyle\int_{1/\kappa}^{\infty}\kappa\,e^{-2\kappa x}dx = e^{-2} = \boxed{0.135}$ .

Step 4. Pause on Step 3: for a bound state at $E < 0$ , every point with $V = 0$ is classically forbidden — a classical particle with negative total energy and zero potential energy cannot exist anywhere. The quantum particle spends its entire life in the forbidden region, anchored to a single allowed point of measure zero. No other textbook problem states the case so baldly.

2 · The bound state meets a beam (exam level)

A beam of electrons is fired at the same delta well ( $|E_b| = 0.26\ \mathrm{eV}$ ). Find the transmission at beam energies $E = 0.065$ , $0.26$ , and $2.6\ \mathrm{eV}$ , and the energy at which 90% of the beam passes.

Step 1. The single control ratio is $|E_b|/E$ . At $E = 0.065\ \mathrm{eV}$ : $|E_b|/E = 4$ , so $T = \dfrac{1}{1+4} = \boxed{0.20}$ — a slow beam mostly bounces off a well (attractive!), pure quantum reflection.

Step 2. At $E = |E_b| = 0.26\ \mathrm{eV}$ : $T = \dfrac{1}{1+1} = \boxed{0.50}$ . The binding energy is exactly the fifty-fifty point of the scattering problem.

Step 3. At $E = 2.6\ \mathrm{eV}$ : $|E_b|/E = 0.1$ , so $T = \dfrac{1}{1.1} = \boxed{0.91}$ ; fast beams barely notice the point. Check the trend against (6): $T$ rises monotonically — a single delta has no resonances to make it wiggle.

Step 4. For $T = 0.90$ : solve $0.90 = [1 + |E_b|/E]^{-1}$ , giving $|E_b|/E = \tfrac{1}{9}$ and $E = 9|E_b| = \boxed{2.3\ \mathrm{eV}}$ .

Step 5. Conservation check: at each energy $R = 1 - T$ (0.80, 0.50, 0.09) — reflection plus transmission exhausts the beam, as the probability current demands.

Origins · 1 minPaul Dirac and how this was discovered

Paul Dirac

1902–1984 · Cambridge

The delta "function" that names this well entered physics through Dirac, who deployed it unapologetically in his 1927 papers and enshrined it in The Principles of Quantum Mechanics (1930) — a function that is zero everywhere, infinite at a point, and integrates to one. Mathematicians winced; Dirac, trained as an electrical engineer, knew a working idealization when he saw one, having met its ancestor in Heaviside's impulse calculus. He used it to write the orthonormality of continuum states, $\langle x|x'\rangle = \delta(x - x')$ , making position eigenstates usable at all. Twenty years later Laurent Schwartz's theory of distributions made the delta rigorous — and earned Schwartz the Fields Medal, a mathematician's prize for legalizing a physicist's shortcut. Dirac himself, characteristically, had never doubted it.

History

1926

— Schrödinger's wave equation makes potentials into boundary-value problems; singular limits become thinkable.

1930

— Dirac's Principles of Quantum Mechanics systematizes the delta function as the working tool of the continuum formalism.

1931

— Kronig and Penney chain deltas into a periodic potential and produce the first exactly solvable model of energy bands in crystals.

1936

— Fermi introduces the delta pseudopotential for slow-neutron scattering — real short-range forces replaced by a contact interaction, the delta well's scattering face.

1950

— Schwartz's distribution theory makes the delta rigorous, earning the Fields Medal and retroactive mathematical honesty for two decades of physics.

Going further

Texts. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §2.5 · Shankar, Principles of Quantum Mechanics, Ch. 5 · Cohen-Tannoudji, Diu & Laloë, Vol. 1, Complement K_I · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed.

Provenance. A verifier should recompute: (1) the jump condition by integrating $-\tfrac{\hbar^2}{2m}\psi'' - \alpha\delta\psi = E\psi$ over $[-\epsilon, \epsilon]$ ; (2) $\kappa = m\alpha/\hbar^2$ and $E = -m\alpha^2/2\hbar^2$ from continuity plus the jump; (3) the normalization $A = \sqrt{\kappa}$ ; (4) $\langle T \rangle = m\alpha^2/2\hbar^2$ and $\langle V \rangle = -m\alpha^2/\hbar^2$ ; (5) the transmission $T = [1 + (m\alpha/\hbar^2k)^2]^{-1}$ ; (6) the finite-well limit with $2aV_0 = \alpha$ ; and the worked numbers ( $\kappa = 2.62\times10^9\ \mathrm{m^{-1}}$ , $1/\kappa = 0.38$ nm, $E = -0.26$ eV, $P = e^{-2} = 0.135$ ; $T = 0.20, 0.50, 0.91$ and $E_{90\%} = 2.3$ eV). All match Griffiths & Schroeter §2.5.2, Shankar Ch. 5, and Cohen-Tannoudji Complement K_I.

Merge note. Merged with the v3 pilot article (v3-delta-function-well) at Step-2; AllQFT depth grafted onto the v3 narrative spine.

Prerequisites. Time-independent Schrödinger equation in 1D · bound vs. scattering states · free-particle plane waves · the finite square well (whose collapse limit this is). Leads to. Quantum tunnelling and barrier transmission · the Kronig–Penney model and energy bands · wave packets scattering off localized potentials · contact interactions in cold-atom physics.

Is a delta function even a legitimate potential? V is infinite at a point — why is the energy still finite?

The delta is an idealization with a precise meaning: it is the limit of a narrow, deep well whose area (depth × width) is held at $\alpha$ while the width shrinks below every other length in the problem. Every physical quantity — $\kappa$ , the energy, the transmission — depends only on that area and converges smoothly in the limit, which is what makes the idealization safe. The infinity never appears in an observable; it is bookkeeping for "much narrower than anything else around."

Why doesn't the uncertainty principle forbid binding by a zero-width well?

It would, if the particle actually had to live inside the well. It doesn't. The bound state spreads over $1/\kappa = \hbar^2/m\alpha$ — nanometres in the worked example — so its momentum spread is a modest $\sim\hbar\kappa$ , costing kinetic energy $\hbar^2\kappa^2/2m$ , which is exactly what the audit found. The particle borrows localization from the region around the well, not from the point itself; squeezing $\psi$ down to the point would indeed cost infinite kinetic energy, and the compromise between attraction and confinement cost is precisely what fixes $\kappa$ .

If every point except x = 0 has V = 0 and E is negative, isn't the particle always in a classically forbidden region?

Yes — and that is the cleanest statement of how non-classical a bound state can be. A classical particle with $E < 0$ and $V = 0$ is a contradiction; the quantum particle spends all its time there, anchored to one allowed point of measure zero. The exponential tails are the same evanescent behavior you meet under a potential step, just glued back-to-back. Probability still balances: $e^{-2}$ of it lies beyond one decay length, and nothing is flowing anywhere in a stationary state.

How do I remember why there's exactly one bound state, without re-deriving anything?

Count scales. Binding energies must be built from $m$ , $\hbar$ , and $\alpha$ , and those three make exactly one energy, $m\alpha^2/\hbar^2$ , up to a pure number. A spectrum of levels needs a dimensionless knob to index it — the infinite well has $n$ via its width, the finite well has $z_0$ via depth-times-width — and the delta well's knob was welded shut when depth and width fused into one constant. One scale, one state.

What changes if the delta is repulsive?

Flip $\alpha \to -|\alpha|$ and the jump condition pushes the slopes apart instead of pulling them together, so no pair of decaying tails can match: no bound state, for any strength. Scattering survives — the transmission magnitude is the same function of $|\alpha|$ , so a measurement of $T$ alone cannot even tell you the sign of the potential. The bound state is what distinguishes attraction, which is exactly how it shows up as a pole of the scattering amplitude at negative energy.

Is the kink in ψ′ physically real, or an artifact of the idealization?

Both, in a precise sense. In any real narrow well, $\psi'$ changes continuously but very fast — across the well's actual width. The delta limit compresses that rapid turn into a point, exactly as it compresses the well itself. So the cusp is as real as the statement "the well is much narrower than $1/\kappa$ ": zoom out and every prediction agrees with the smooth problem to corrections of order (width × $\kappa$ ). It is the same license you use when you call a real string's bend at a knot a "corner."

Delta-Function Potential Well and Bound State

Where this lives today

The physical picture

The math: one jump condition does everything

Scattering: the same point, seen by a traveler

Limiting cases

Worked examples

1 · Sizing up a point trap (confidence builder)

2 · The bound state meets a beam (exam level)

Going further

Before this

Where this leads