Admissibility Conditions on Wavefunctions

Solve the Schrödinger equation as a pure exercise in differential equations and it will hand you far more functions than nature uses. For every energy you try — every energy, not just the famous quantized ones — the mathematics produces a two-parameter family of solutions. Most of them are frauds: they blow up exponentially, or refuse to fade at infinity, or carry a kink that would cost infinite energy to manufacture. Quantum mechanics runs a door policy. Before a function $\psi$ may call itself a state, it must show three credentials: it must be single-valued (one amplitude per place), continuous (no teleporting amplitude), and square-integrable (a finite total $\int|\psi|^2$ , so that Born's rule can normalize it to a probability). These are not extra postulates bolted on for tidiness — each one is the Born interpretation or the Schrödinger equation itself, enforcing its own consistency. This article is the door policy written out: what each condition says, why physics demands it, and what it costs to violate one. The centerpiece arguments are quantitative. A wavefunction that jumps pays an infinite kinetic-energy bill — we will derive the price and watch it diverge. A wavefunction with a lazy power-law tail fails the membership test at a sharp exponent — we will find the exact boundary. And the club these conditions guard has a name: the Hilbert space $L^2$ , whose formal architecture is built in Hilbert Space and State Vectors. The admissibility rules are also the engine of quantization — rejecting the growing exponential at infinity is precisely what makes bound-state energies discrete — and with them in hand, the foundations chapter is complete: you now know what a quantum state is.

Key takeaways

A solution of the Schrödinger equation becomes a physical state only if it is single-valued, continuous, and square-integrable — the admissibility conditions.
Square-integrability, $\int|\psi|^2\,dx < \infty$ , is the membership card of the Hilbert space $L^2$ : without it Born's rule has nothing to normalize. A power-law tail $\psi \sim 1/|x|^\alpha$ passes only for $\alpha > 1/2$ .
Continuity is an energy statement: a step discontinuity in $\psi$ makes $\int|\psi'|^2$ — and hence the mean kinetic energy — diverge. Smoothness is what finite energy looks like.
The derivative $\psi'$ is continuous wherever the potential is finite; it jumps only at genuine singularities — by $-(2m\alpha/\hbar^2)\psi(0)$ at a delta well, and freely at an infinite wall. The owners of those rules are the delta-well and matching-conditions articles.
Admissibility quantizes: demanding decay at $\pm\infty$ kills the growing exponential branch, and only discrete energies can satisfy both ends at once.
Plane waves and delta spikes fail the test yet remain indispensable — they are idealized basis elements, not states, handled rigorously by delta normalization.
Key idea: admissibility = single-valued + continuous + square-integrable. The first two make $|\psi|^2$ unambiguous and the energy finite; the third makes total probability finite and selects the physical spectrum.

The core insight

Admissibility = single-valued + continuous + square-integrable. The first two make the probability density unambiguous and the Schrödinger equation well-posed; the third — membership in $L^2$ — makes total probability finite and selects the physical spectrum.

Where this lives today

Every numerical quantum calculation is an admissibility engine. Shooting-method solvers for one-dimensional problems integrate the Schrödinger equation outward from both ends and hunt for the energies where the growing exponential cancels — the computer literally performs the kill-the-growing-branch argument, and the eigenvalues appear where the two admissible tails can be stitched smoothly in the middle. Quantum-chemistry codes bake the conditions into their basis sets: molecular orbitals are expanded in Gaussians precisely because Gaussians are smooth, single-valued, and die faster than any power law, so every trial state is admissible by construction and every matrix element is a convergent integral. The conditions also draw a live frontier. Delta-normalized continuum states — the plane waves this article turned away at the door — do real work in scattering theory, and making them rigorous citizens requires enlarging the architecture: the rigged Hilbert space, a nested triple in which genuine $L^2$ states sit in the middle and the idealized basis elements live in an outer shell of distributions. That construction, and the continuous-spectrum bookkeeping it licenses, arrive with the formalism chapter's continuous-spectra machinery. Meanwhile the single-valuedness clause has become laboratory hardware: the requirement that the quantum phase return to itself around a closed loop is why magnetic flux through a superconducting ring is quantized in exact units of $h/2e$ — a voltmeter-visible consequence of Condition 3.

The physical picture

Think of $|\psi|^2$ as probability paint. Admissibility says three things about the paint job: the painter used a finite bucket (square-integrability — you can rescale the total to exactly one), the coat is unbroken (continuity — the amplitude cannot jump across a point without an infinite energy bill), and each point got one coat (single-valuedness — two different amplitudes at the same place would make $|\psi|^2$ meaningless).

Figure

Four candidates at the door of Hilbert space — only one gets in unconditionally

The gallery is the subject at a glance. The growing exponential is the classic villain: it solves the equation perfectly and describes nothing, because no rescaling can make an infinite integral equal one. The plane wave is subtler — it is bounded, innocent-looking, and still inadmissible, because a constant $|\psi|^2$ over an infinite line is also infinite paint; we keep it anyway as an idealized basis element, the way a surveyor keeps an infinitely long ruler. The kinked function is legal (finite norm, finite energy) but sits one step from crime: sharpen the kink into a genuine break and the kinetic energy diverges. The smooth decayer is what actual states look like. Why should physics care about the tail of a function nobody measures? Because Born's rule is a zero-sum budget. The statement "the particle is somewhere with probability 1" — the bookkeeping built in Probability Density and Normalization and kept honest for all time in Conservation of Normalization — needs a finite total to divide by. The tail is where that total is won or lost.

Figure

How fast must the tail die — the α = 1/2 borderline

The charter, condition by condition

Condition 1 — square-integrability. The non-negotiable core, and the definition of the club:

\int_{-\infty}^{\infty} |\psi(x,t)|^2\, dx < \infty \qquad (1)

where: $\psi(x,t)$ — the candidate wavefunction; the integral — the total probability before normalization. Finiteness is exactly what lets you divide by the square root of this number and quote probabilities that sum to one. The set of all such functions is the Hilbert space $L^2(\mathbb{R})$ ; its axioms, its completeness, and the geometry that makes it the arena of quantum mechanics are the business of Hilbert Space and State Vectors. Note what the condition does not say: it does not literally demand $\psi \to 0$ at infinity — a function can be square-integrable while carrying ever-narrower spikes forever outward. Physically realizable states do decay, and we assume the decay whenever we discard boundary terms, but the honest statement is "finite total," not "vanishing tail." Once a state passes, it passes forever: the norm chosen at $t = 0$ is preserved by the dynamics. That theorem — the global conservation of the total probability, with its boundary-term honesty — was proved in Conservation of Normalization by integrating the local bookkeeping of Probability Current and the Continuity Equation, and we simply cash it here: admissibility is a one-time credential check, not a recurring fee.

Condition 2 — continuity. The amplitude cannot jump. The deep reason is energetic, and it is this article's first derivation: a discontinuous $\psi$ has infinite mean kinetic energy, because a jump is the limit of an ever-steeper ramp and the energy bill grows as the ramp sharpens. The companion rules for the derivative are bookkeeping at potential singularities, and they have owners already: where $V$ is finite, $\psi'$ is continuous too — proved by integrating the Schrödinger equation across the point in Matching Conditions and Wavefunction Continuity — while a delta well $V = -\alpha\,\delta(x)$ buys a prescribed kink, $\psi'(0^+) - \psi'(0^-) = -(2m\alpha/\hbar^2)\,\psi(0)$ , derived in Delta-Function Potential Well and Bound State, and an infinite wall forces $\psi = 0$ with the slope free to break, as at the walls of the Infinite Square Well. The rules assemble into a small table:

where you are	rule for $\psi$	rule for $\psi'$	owner of the proof
$V$ finite (even if it steps)	continuous	continuous	Matching Conditions and Wavefunction Continuity
delta potential $-\alpha\,\delta(x)$	continuous	jumps by $-(2m\alpha/\hbar^2)\psi(0)$	Delta-Function Potential Well and Bound State
infinite wall	$\psi = 0$ at the wall	may jump	Infinite Square Well
tail at $\pm\infty$ (bound state)	decays	decays	this article (membership test)

Condition 3 — single-valuedness. One configuration, one amplitude. On a line this is automatic; it grows teeth the moment the configuration space closes on itself. For a particle on a ring, coming home after $2\pi$ must reproduce the amplitude, $\psi(\phi + 2\pi) = \psi(\phi)$ , and imposing that on $e^{im\phi}$ forces $e^{2\pi i m} = 1$ — an integer $m$ , quantization purchased by pure topology before any equation is solved. That one-line miracle is the seed of the magnetic quantum number, and its full development belongs to Eigenfunctions of L_z and the Azimuthal Quantization in the angular-momentum chapter. And the conditions are not passive filters — they are the mechanism of quantization. At large $|x|$ in a classically forbidden region, solutions come in a dying branch and a growing branch (the pair met in Bound States versus Scattering States). Admissibility executes the growing branch at $+\infty$ and again at $-\infty$ ; generic energies cannot obey both death warrants at once, and the rare survivors are the discrete spectrum. Every transcendental quantization condition you will meet in the one-dimensional chapter is this charter, applied.

Derivation AA jump in the wavefunction costs infinite kinetic energy

We prove that a step discontinuity in $\psi$ is physically forbidden, not merely unaesthetic, by computing what it does to the mean kinetic energy. The strategy has two moves: first recast $\langle T\rangle$ as an integral of $|\psi'|^2$ — manifestly non-negative, and sensitive to steepness — and then price a jump by smoothing it into a linear ramp of width $\varepsilon$ and letting $\varepsilon \to 0$ . Watch the boundary term in the first move: discarding it is legitimate precisely for the decaying states this chapter admits, which is a pleasant consistency — the identity that polices admissibility is itself licensed by admissibility.

Step 1. Recast the kinetic-energy average. Starting from the operator form $\langle T\rangle = \int \psi^*\left(-\tfrac{\hbar^2}{2m}\psi''\right) dx$ and integrating by parts once:

\int_{-\infty}^{\infty} \psi^*(-\psi'')\, dx = \Big[-\psi^*\psi'\Big]_{-\infty}^{\infty} + \int_{-\infty}^{\infty} |\psi'|^2\, dx

Why the boundary term dies: for a normalizable state that decays at infinity, $\psi^*\psi' \to 0$ at both ends. What survives is beautiful:

\langle T\rangle = \frac{\hbar^2}{2m}\int_{-\infty}^{\infty} |\psi'|^2\, dx

— kinetic energy is literally the price of slope. Steep wavefunctions are expensive; a wavefunction is a budget allocation between localization and energy.

Step 2. Price a jump. Let $\psi$ climb by $\Delta$ over a ramp of width $\varepsilon$ . On the ramp $\psi' = \Delta/\varepsilon$ , so the ramp's contribution is

\int_{\text{ramp}} |\psi'|^2\, dx = \left(\frac{\Delta}{\varepsilon}\right)^2 \varepsilon = \frac{\Delta^2}{\varepsilon}

exactly — the bill scales as $1/\varepsilon$ .

Step 3. Take the limit. A true discontinuity is the ramp at $\varepsilon \to 0$ , and $\Delta^2/\varepsilon \to \infty$ : the mean kinetic energy diverges. No state of finite energy can carry a broken amplitude, which is Condition 2. The numerical ladder in the verification script watches the divergence happen: shrinking $\varepsilon$ tenfold raises the integral tenfold, every decade. Why a corner is still legal: a kink leaves $\psi'$ finite (merely discontinuous), so $\int|\psi'|^2$ stays finite. Kinks are exactly what delta potentials manufacture — expensive but affordable. Only the full break is priced out. *Verified: the integration-by-parts identity is checked symbolically on two independent states, the ramp integral is exact, and the $1/\varepsilon$ ladder is confirmed numerically over three decades in the verification script.*

Go deeper · Derivation B · 1 minHow fast must the tail die — the power-law membership test

Derivation BHow fast must the tail die — the power-law membership test

We find the exact decay rate that separates admissible tails from inadmissible ones. The strategy is to test the marginal family $\psi \sim C/|x|^{\alpha}$ for large $|x|$ and ask when the tail's contribution to $\int|\psi|^2$ converges. This one-parameter family is the right probe because any smooth tail is bracketed by power laws, so the family's threshold is the threshold. The result — $\alpha > 1/2$ — is startlingly permissive, and its permissiveness is the door through which the subtle counterexamples of this subject walk.

Step 1. Write the tail's bill. Beyond some $x = 1$ (units are irrelevant to convergence),

\int_{1}^{\infty} \left|\frac{C}{x^{\alpha}}\right|^2 dx = |C|^2 \int_{1}^{\infty} x^{-2\alpha}\, dx

Step 2. Apply the power-law convergence rule. The integral of $x^{-s}$ over $[1,\infty)$ converges if and only if $s > 1$ ; here $s = 2\alpha$ , so the tail is square-integrable iff $\alpha > 1/2$ . The borderline $\alpha = 1/2$ itself fails, logarithmically. Why the threshold is so low: probability weights the amplitude squared, so a tail may die far more slowly than you would guess — $1/x^{0.6}$ already passes, though such a state has bizarre moments. The Lorentzian-type state in Worked example 1, with $\alpha = 2$ , passes with room to spare.

Step 3. Register the two honesty clauses that the threshold exposes. First: square-integrable does not imply vanishing — one can build an $L^2$ function carrying unit-height spikes at every integer, thinned fast enough that the total area converges; it is admissible by the letter of Condition 1 and never goes to zero. Physical states do decay, and every boundary-term argument in this track quietly assumes it. Second: bounded does not imply admissible — the plane wave has $|\psi| = 1$ everywhere and fails — while unbounded does not imply inadmissible: $\psi \sim x^{-1/4}$ near the origin has $|\psi|^2 \sim x^{-1/2}$ , integrable, so a state may even spike to infinity at a point and still hold a membership card. Verified: convergence at $\alpha = 0.6$ , divergence at $\alpha = 0.4$ and at the $\alpha = 1/2$ borderline, and the integrable $x^{-1/4}$ spike are all checked symbolically in the verification script.

Limiting cases

case	result	meaning
$V$ finite everywhere, $E < V(\pm\infty)$	growing exponential branch forbidden	admissibility kills one solution per side; discrete energies are the survivors
power-law tail $\psi \sim x^{-\alpha}$	admissible iff $\alpha > 1/2$	the membership threshold; borderline $\alpha = 1/2$ fails logarithmically
step discontinuity in $\psi$ smoothed over $\varepsilon$	$\langle T\rangle \propto 1/\varepsilon \to \infty$	breaks in the amplitude are priced out by kinetic energy, not by decree
kink in $\psi$ ( $\psi'$ jumps, $\psi$ continuous)	$\langle T\rangle$ finite	legal; manufactured to order by delta potentials
plane wave $e^{ikx}$ , delta spike $\delta(x - x_0)$	not in $L^2$	idealized basis elements, delta-normalized; every physical state is a square-integrable superposition of them

Worked examples

Example 1 — a full admissibility audit (confidence builder). Audit the candidate state $\psi(x) = N/(x^2 + a^2)$ on the whole line, with $a = 0.1$ nm for an electron: check membership, normalize it, and price its energy.

Step 1. Tail check: for large $|x|$ , $\psi \sim N/x^2$ , so $\alpha = 2 > 1/2$ — admissible with room to spare. Continuity and single-valuedness are manifest.

Step 2. Normalize: $\int dx/(x^2 + a^2)^2 = \pi/2a^3$ , so $N^2 = 2a^3/\pi$ and $\boldsymbol{N = \sqrt{2a^3/\pi}}$ .

Step 3. Spread: $\langle x^2\rangle = N^2 \int x^2\,dx/(x^2+a^2)^2 = (2a^3/\pi)(\pi/2a) = \boldsymbol{a^2}$ — the particle lives within about $a$ of the origin, though this tail is fat enough that the fourth moment already diverges: admissible states can still have pathological statistics.

Step 4. Energy price via the slope identity: $\int |\psi'|^2 dx = 1/2a^2$ , so $\langle T\rangle = \frac{\hbar^2}{2m}\cdot\frac{1}{2a^2} = \frac{\hbar^2}{4ma^2}$ . For $a = 0.1$ nm and the electron mass, $\langle T\rangle = \boldsymbol{1.90\ \text{eV}}$ — a perfectly ordinary atomic-scale energy. Verdict: admitted, normalized, affordable.

Example 2 — the flat state's energy catastrophe (exam level). In an infinite well of width $L = 1$ nm, consider the deceptively innocent candidate $\psi(x) = 1/\sqrt{L}$ inside the well. It is bounded, positive, and exactly normalized: $\int_0^L |\psi|^2 dx = 1$ . Is it a good state?

Step 1. Check it against the wall rule. The well's admissible states must vanish at $x = 0$ and $x = L$ ; the flat state arrives at both walls at full height $1/\sqrt{L}$ — a step discontinuity against the vanishing exterior. By Derivation A, that should cost infinite kinetic energy. Let us watch it happen in the energy accounting.

Step 2. Expand in the well's own states $\psi_n = \sqrt{2/L}\sin(n\pi x/L)$ . The overlap integrals give $c_n = 2\sqrt{2}/n\pi$ for odd $n$ and $c_n = 0$ for even $n$ (the flat state is symmetric about the centre; the even modes are antisymmetric).

Step 3. Check the probability budget: $\sum |c_n|^2 = \frac{8}{\pi^2}\left(1 + \frac{1}{9} + \frac{1}{25} + \cdots\right) = \frac{8}{\pi^2}\cdot\frac{\pi^2}{8} = \boldsymbol{1}$ , exactly. The probability bookkeeping is flawless — normalization sees nothing wrong. (The completeness machinery licensing this expansion is the subject of Complete Orthonormal Bases.)

Step 4. Now check the energy budget: each odd mode contributes $|c_n|^2 E_n = \frac{8}{\pi^2 n^2}\cdot n^2 E_1 = \frac{8E_1}{\pi^2}$ — the same amount per mode, forever. With $E_1 = 0.376$ eV for $L = 1$ nm, that is $\boldsymbol{0.305\ \text{eV per odd mode}}$ , and the sum over infinitely many modes diverges: $\langle H\rangle = \infty$ . The flat state is normalizable but energetically fraudulent — its wall discontinuities hide an infinite kinetic bill, exactly as the ramp argument prices them. A state can pass Condition 1 and still be unpreparable; smoothness is a second, independent budget.

Origins · 1 minHenri Lebesgue and how this was discovered

Henri Lebesgue

1875–1941 · Nancy, Rennes, and the Collège de France

The membership test at the heart of this article — "is $\int|\psi|^2$ finite?" — is only a well-posed question because of Lebesgue. His 1902 doctoral thesis, Intégrale, longueur, aire, rebuilt integration from the ground up: where Riemann's integral chops the $x$ -axis and fails on wildly oscillating or spiky functions, Lebesgue's chops the $y$ -axis, measuring the set of points where the function is large. The payoff is exactly what quantum mechanics needs: the space $L^2$ of square-integrable functions, defined with his integral, is complete — limits of convergent sequences of members are still members, with no holes at the edges. (That completeness theorem, proved by Riesz and Fischer in 1907, is the load-bearing wall of Hilbert Space and State Vectors; the integral it stands on is Lebesgue's.) He was a schoolteacher's son who did much of the work while teaching at a lycée in Nancy, and his measure theory was initially attacked as pathology-hunting — functions "with no derivatives, monsters." The monsters turned out to be the arena of atomic physics: every wavefunction in this track is a Lebesgue-integrable object, and the door policy this article enforces is checked with his ruler.

History

1854

— Riemann formalizes the integral that bears his name; spiky and heavily oscillating functions expose its limits

1902

— Lebesgue's thesis rebuilds integration on measure theory, defining the integral that can handle limits of wild functions

1907

— Riesz and Fischer prove L² is complete under the Lebesgue integral — the future state space of quantum mechanics has no holes

1926

— Born's statistical interpretation makes square-integrability a physical requirement: the total probability must be finite

1927

— Kennard and others make the smoothness–energy tradeoff quantitative through the uncertainty relation

1932

— von Neumann's Mathematische Grundlagen enshrines L² as the official arena of quantum mechanics

Going further

References. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §1.4, §1.6, §2.5 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 1, Ch. 5 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. 1, Ch. I, Complement K-I · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., Ch. 2 · Townsend, A Modern Approach to Quantum Mechanics, 2nd ed., Ch. 6

Provenance. Upgraded in place from the AllQFT base article at Step-2. The base's delta-well bound-state derivation and its continuity-equation/norm-conservation derivation are not reproduced here — those proofs are owned by Delta-Function Potential Well and Bound State, Probability Current and the Continuity Equation, and Conservation of Normalization per chapter ownership, and are recalled by title. One base defect corrected: the base stated the delta-well jump condition as $\psi'(0^+)-\psi'(0^-) = +(2m\alpha/\hbar^2)\psi(0)$ for the attractive well $V = -\alpha\,\delta(x)$ ; the correct sign is $-(2m\alpha/\hbar^2)\psi(0)$ (integrating the TISE across the singularity gives $\Delta\psi' = (2mg/\hbar^2)\psi(0)$ for $V = g\,\delta(x)$ , and $g = -\alpha$ here), consistent with the owning articles; the base's final energy $E = -m\alpha^2/2\hbar^2$ was nonetheless correct. A verifier should: (1) reproduce the identity $\langle T\rangle = (\hbar^2/2m)\int|\psi'|^2 dx$ by parts; (2) confirm the ramp bill $\Delta^2/\varepsilon$ and its $1/\varepsilon$ divergence; (3) confirm the tail threshold $\alpha > 1/2$ ; (4) recompute the Lorentzian-type audit ( $N^2 = 2a^3/\pi$ , $\langle x^2\rangle = a^2$ , $\langle T\rangle = \hbar^2/4ma^2 = 1.90$ eV at $a = 0.1$ nm); (5) recompute the flat-state expansion ( $c_n = 2\sqrt{2}/n\pi$ odd, $\sum|c_n|^2 = 1$ exact, $8E_1/\pi^2 = 0.305$ eV per odd mode at $L = 1$ nm). All are asserted in this article's verification script.

Builds on: The Wavefunction · Born's Statistical Interpretation · Probability Density and Normalization · The Time-Dependent Schrödinger Equation

Leads to: Time-Independent Schrödinger Equation · Stationary States and Separation of Variables · de Broglie Matter Waves

Does a wavefunction have to go to zero at infinity?

Strictly, no — square-integrability is the actual requirement, and mathematicians can build L² functions carrying ever-narrower spikes out to infinity that never settle to zero. But such functions are unpreparable curiosities: physically realizable states decay, and the standard boundary-term arguments (norm conservation, the kinetic-energy identity) assume it. The honest hierarchy: the postulate demands finite ∫|ψ|², physics supplies the decay.

Why square-integrable rather than just integrable — why L² and not L¹?

Because Born's rule squares the amplitude: the probability density is |ψ|², so it is |ψ|² whose integral must be finite. The two conditions are genuinely different — 1/(1+|x|) is square-integrable but not integrable, and an integrable function with tall thin spikes can fail square-integrability. Quantum mechanics also needs the inner product ∫φ*ψ between any two states to converge, and L² is exactly the space where Cauchy–Schwarz guarantees that pairing always makes sense.

Is the plane wave a wavefunction or not? Textbooks seem to use it constantly.

It is a basis element, not a state. As a state it fails Condition 1 — infinite norm — and assigning it to a single particle produces contradictions (a uniform probability over an infinite line). But as an idealized building block it is indispensable: admissible states are superpositions ∫φ(k)e^{ikx}dk, and the plane waves are delta-normalized rather than unit-normalized. Free Particle and Plane Waves owns that bookkeeping; the formalism chapter's continuous-spectrum machinery makes it rigorous.

Does ψ have to be infinitely smooth, or just continuous?

Just continuous, with ψ′ continuous wherever the potential is finite. Beyond that, the wavefunction is exactly as smooth as the potential lets it be: the TISE says ψ″ = (2m/ħ²)(V−E)ψ, so ψ″ inherits any step in V, a delta in V kinks ψ′, and an infinite wall breaks ψ′ outright. Reading the smoothness of ψ is reading the potential — the second derivative is a map of where the forces live.

Can a wavefunction be infinite somewhere and still be a state?

Surprisingly, yes — the criterion is area, not height. A spike like ψ ~ x^(−1/4) near a point has |ψ|² ~ x^(−1/2), which integrates fine, so it passes Condition 1. Such states are exotic (their kinetic energy needs separate scrutiny) but they illustrate the discipline of this subject: check the integral, not the picture. The genuinely forbidden objects are those with infinite area — growing exponentials, non-decaying oscillations — however tame they look locally.

If admissibility kills the growing exponential, who enforces it during a real calculation?

The boundary conditions you impose. Solving the finite well, you keep only e^{−κ|x|} outside; solving hydrogen you will discard the divergent branch of the radial solution; a numerical shooting method tunes E until the growing branch's coefficient crosses zero. In every case "solve the Schrödinger equation" secretly means "solve it within the admissible class" — and the discreteness of bound spectra is the direct fingerprint of that restriction.

The flat state is normalized yet has infinite energy. Could I still use it as an initial condition?

As a mathematical initial condition, yes — it lives in L², so its expansion coefficients exist and its future evolution is perfectly well-defined. As physics, no laboratory can prepare it: infinite mean energy means every real preparation apparatus produces some smoothed version, with the corners rounded on some length scale ε and a large-but-finite ⟨T⟩ ~ 1/ε. It is the ideal gas of states — a useful fiction whose sharp corners you must remember are fictional.

Why does single-valuedness quantize anything? It sounds like a triviality.

On a line it is trivial — no path returns to its starting point. On a circle it is a genuine constraint: the amplitude at angle φ and at φ + 2π describe the same physical configuration, so they must be equal, forcing e^{2πim} = 1 and integer m. Quantization here comes from topology, not dynamics — no potential, no boundary, just the shape of the configuration space. The full story, including why this fixes the magnetic quantum number, belongs to the angular-momentum chapter.