From Single-Particle to Quantum Fields

For a decade after 1926, the best minds in physics tried to make a relativistic Schrödinger equation and kept breaking it. The Klein-Gordon equation, the honest quantization of $E^2=p^2c^2+m^2c^4$ , produced a conserved density that could go negative — useless as a probability. The Dirac equation fixed the density but paid for it with an infinite tower of negative-energy states reaching down to $-\infty$ , threatening every atom with radiative collapse into the abyss. Dirac's own patch — a filled sea of negative-energy electrons whose holes are positrons — is ingenious, but it works only for fermions, treats the vacuum as an infinite charged reservoir, and simply does not exist for a spin-0 particle. Something more fundamental was missing. The resolution is not a cleverer wave equation; it is a change in what the equation describes. We stop reading $\phi(x)$ and $\psi(x)$ as wavefunctions — amplitudes for one particle — and promote them to quantum operators acting on a space of many particles. The same equation, unchanged, now governs a field whose quanta are the particles. The negative-frequency solutions that terrified everyone become, in the operator theory, the creation operators for antiparticles: they add positive energy rather than subtracting it. There is no descent to $-\infty$ , because there is a lowest state — the vacuum — and every excitation costs positive energy. This article does not build the field-operator machinery; that is the work of Field Quantization of the Klein-Gordon and Dirac Fields. Here we make the argument for why that step is forced. We pin down three concrete, checkable failures of the single-particle picture — the indefinite density, the bottomless spectrum, and the impossibility of localizing a particle below its Compton wavelength — and show, at the level of signs and numbers, how promoting the wavefunction to a field cures all three at once. The price, and it is the correct price, is that particle number ceases to be conserved.

Key takeaways

Reading the Klein-Gordon or Dirac wavefunction as a single-particle probability amplitude fails: the density is not positive-definite and the energy spectrum is unbounded below.
The fix is a reinterpretation, not a new equation: promote $\phi(x)\to\hat\phi(x)$ , an operator on a many-particle Fock space, so its quanta are the particles.
A negative-frequency mode $e^{+iEt}$ is not a negative-energy state; its operator coefficient is a creation operator that adds a positive-energy antiparticle. The spectrum is bounded below.
The Klein-Gordon $\rho$ that goes negative was never a probability — it is a charge density, legitimately of either sign; $\rho\propto 2E$ flips sign between positive- and negative-frequency modes.
Relativity forbids localizing a particle below the Compton wavelength $\lambda_C=\hbar/mc$ (386 fm for the electron): the injected energy $\gtrsim mc^2$ makes pairs. So particle number cannot be fixed.
The pair-creation threshold is $2m_ec^2=1.022\ \mathrm{MeV}$ — a real, measured number that no single-particle theory can accommodate.
Key idea: do not quantize a single particle relativistically — quantize a field; the once-feared negative-frequency solutions become antiparticle creation operators, making energy positive-definite and replacing "probability of one particle" with a positive number operator on Fock space.

The core insight

Do not quantize a single particle relativistically; quantize a field. Promoting $\phi(x)\to\hat\phi(x)$ turns the negative-frequency solutions into antiparticle creation operators, so energy becomes positive-definite, the indefinite Klein-Gordon density becomes an honest charge density, and "probability of one particle" is replaced by a positive number operator on a many-particle Fock space.

Where this lives today

Every particle a modern experiment counts is a field quantum, and the collapse of the single-particle picture is not a historical curiosity but the daily working assumption of high-energy physics. At the LHC, collisions routinely turn kinetic energy into showers of particles that did not exist beforehand — the direct manifestation of a non-conserved particle number written into the field operator. The positron that Dirac's negative-frequency modes predicted now does useful work: positron-emission tomography images metabolism in the human brain by detecting the $511\ \mathrm{keV}$ annihilation photons, each carrying exactly $m_ec^2$ . And the vacuum energy that normal ordering discards is not always ignorable: when boundary conditions vary in space it produces the measurable Casimir force between conducting plates, a laboratory demonstration that the quantized field, not a single particle, is the real object. The equation that seemed broken in 1926 turned out to be describing something richer than anyone had asked it to.

The physical picture

The deepest reason single-particle relativistic quantum mechanics cannot exist is a collision between two facts you already trust: $E=mc^2$ and the uncertainty principle. To pin a particle to a region of size $\Delta x$ costs momentum $\Delta p\sim\hbar/\Delta x$ and therefore energy $\Delta E\sim\hbar c/\Delta x$ . Push $\Delta x$ down to the Compton wavelength $\lambda_C=\hbar/mc$ and that energy reaches $mc^2$ — enough to conjure a particle-antiparticle pair out of the vacuum. The moment you try hard enough to say "the particle is here," you have created more particles. A theory of exactly one particle is not approximately wrong at short distances; it is conceptually impossible there.

Figure

Localizing below the Compton wavelength creates pairs

A field handles this effortlessly. A classical field is an infinite collection of oscillators, one per mode; quantizing the field just means quantizing each oscillator, and the quanta of those oscillators are the particles. Add energy and you add quanta — the theory describes any number of particles automatically, from zero (the vacuum) upward. The old single-particle "wavefunction" does not disappear; it reappears as the one-particle sector of the larger space, the amplitude $\langle 0|\hat\phi(x)|\mathbf p\rangle$ to find a single quantum. Everything good about the plane wave survives; the diseases do not. The second picture is what happens to the negative-frequency solutions. Hermiticity forces the field to contain both $e^{-iEt}$ and $e^{+iEt}$ ; there is no throwing one away. In the single-particle reading the second is a state of energy $-E$ , and the spectrum plunges to $-\infty$ . In the operator reading the coefficient of $e^{+iEt}$ is a creation operator for an antiparticle, so that mode adds $+E$ . The bottomless spectrum of the wave equation becomes a spectrum bounded below by a unique vacuum.

Figure

The no-bottom problem, fixed

Finally, the negative density. The Klein-Gordon $\rho$ was never a probability. It is the time component of a conserved Noether current — a charge density. A complex field carries charge $+1$ for particles and $-1$ for antiparticles, so a negative density means nothing worse than a region net-populated by antiparticles. What looked like a fatal inconsistency was a category error: the object was mislabeled, not broken.

The core equations

The two relativistic wave equations are where the trouble starts, so we pin them and their conventions before quantizing anything. We work in natural units $\hbar=c=1$ with the mostly-minus metric $(+,-,-,-)$ , restoring $\hbar$ and $c$ only in the numeric examples.

\left(\Box+m^2\right)\phi=0,\qquad \Box=\partial_\mu\partial^\mu=\partial_t^2-\nabla^2 \qquad (1)

where: $\phi$ — the scalar (spin-0) field, about to be reinterpreted as an operator $\hat\phi(x)$ ; $\Box$ — the d'Alembertian (relativistic wave operator); $m$ — the mass of the quantum. Reading it physically: equation (1) is nothing but $E^2=\mathbf p^2+m^2$ with $E\to i\partial_t$ , $\mathbf p\to-i\nabla$ . It is manifestly Lorentz-covariant — which is exactly why it is second order in time, and that single fact is the root of the density problem. The Klein-Gordon Equation establishes this covariant structure; here we take it as given and follow its consequences. Because the equation is second order in time, its conserved current depends on the time derivative $\dot\phi$ , which is independent of $\phi$ . That is what lets the density change sign.

j^\mu=i\left(\phi^*\partial^\mu\phi-\phi\,\partial^\mu\phi^*\right),\qquad \rho\equiv j^0=i\left(\phi^*\dot\phi-\phi\,\dot\phi^*\right) \qquad (2)

where: $j^\mu$ — the conserved current, $\partial_\mu j^\mu=0$ ; $\rho=j^0$ — the density, which we will see is not sign-definite. Reading it physically: in the nonrelativistic Schrödinger theory the density is $|\psi|^2\ge0$ because the equation is first order in time and $\rho$ depends only on $\psi$ . Here $\rho$ depends on $\dot\phi$ and inherits the sign of the frequency — the fingerprint that Klein-Gordon Probability Current and Negative Densities identifies as tracking the energy, hence properly a charge. The cure is the reinterpretation. We do not solve (1) as a wavefunction; we read it as a field equation whose general solution superposes positive- and negative-frequency plane waves, and we promote the expansion coefficients to operators.

\hat\phi(x)=\int\!\frac{\mathrm d^3p}{(2\pi)^3}\,\frac{1}{\sqrt{2E_{\mathbf p}}}\left(a_{\mathbf p}\,e^{-ip\cdot x}+b_{\mathbf p}^\dagger\,e^{+ip\cdot x}\right),\qquad E_{\mathbf p}=+\sqrt{\mathbf p^2+m^2} \qquad (3)

where: $a_{\mathbf p}$ — annihilation operator for a particle of momentum $\mathbf p$ ; $b_{\mathbf p}^\dagger$ — creation operator for an antiparticle, sitting on the once-feared negative-frequency mode; $E_{\mathbf p}$ — always the positive root. Reading it physically: this is the entire idea in one line. The negative-frequency coefficient is not a state; it is $b^\dagger$ , an operator that manufactures a positive-energy antiparticle. We display (3) as the destination, not as a construction — building it, inverting it to get the mode commutators $[a_{\mathbf p},a_{\mathbf q}^\dagger]=(2\pi)^3\delta^3(\mathbf p-\mathbf q)$ , and computing the Hamiltonian is exactly the job of Field Quantization of the Klein-Gordon and Dirac Fields.

Derivation AThe Klein-Gordon density is not positive-definite

Our goal is to show, from equation (2) alone, that the Klein-Gordon density changes sign between positive- and negative-frequency plane waves — the concrete failure that dooms the single-particle probability interpretation and forces the charge reading. The strategy is direct substitution: evaluate $\rho$ on the two elementary solutions $e^{-iEt}$ and $e^{+iEt}$ and read off the coefficient. The one thing to watch is that the sign comes entirely from the time derivative, so the second-order-in-time character of the equation is doing all the work.

Step 1. Take a positive-frequency plane wave $\phi=N\,e^{-iEt}$ (suppressing the spatial part, which is common to both terms of $\rho$ ) and compute $\dot\phi=-iE\,N e^{-iEt}$ . Why: the positive-frequency solution is the "particle-like" branch; its density had better be positive if the naive interpretation is to survive.

Step 2. Substitute into (2): $\rho=i(\phi^*\dot\phi-\phi\dot\phi^*)=i\big(N^*e^{+iEt}(-iE)Ne^{-iEt}-\text{c.c.}\big)=+2E|N|^2$ . Why: the two terms of $\rho$ add rather than cancel because complex conjugation flips the sign of the exponent and of the explicit $i$ ; the result is manifestly positive for $E>0$ .

Step 3. Now the negative-frequency solution $\phi=N\,e^{+iEt}$ , with $\dot\phi=+iE\,N e^{+iEt}$ . The same substitution gives $\rho=-2E|N|^2$ . Why: flipping the sign of the frequency flips the sign of $\dot\phi$ , and since $\rho\propto\dot\phi$ , the density itself flips. There is no way to interpret $-2E|N|^2$ as a probability.

Step 4. Conclude: $\rho\propto 2E$ with the sign set by the frequency, so $\rho$ is not sign-definite. The consistent reading is that $\rho$ is a charge density, positive for particles and negative for antiparticles, and $\int\rho\,\mathrm d^3x$ is the conserved charge $Q$ , not a total probability. Why: charge is exactly the kind of quantity that comes in both signs; the single-particle theory tried to force a charge into the role of a probability, and the mismatch showed up as a negative density. Verified: `verify_aq-qm-single-particle-to-quantum-fields.py` computes $\rho$ symbolically on both plane waves and confirms $\rho(e^{-iEt})=+2E N^2$ , $\rho(e^{+iEt})=-2E N^2$ , differing in sign.

Go deeper · Derivation B · 2 minWhy promoting the coefficient to a creation operator bounds the energy below

Derivation BWhy promoting the coefficient to a creation operator bounds the energy below

Our goal is to show that the negative-frequency mode, which sinks the single-particle spectrum to $-\infty$ , instead raises the energy once its coefficient is an operator — so the quantized theory has a vacuum floor. The strategy is to track the sign of the energy carried by the negative-frequency mode through the promotion $\beta_{\mathbf p}\to b_{\mathbf p}^\dagger$ , using only the harmonic-oscillator commutator $[b,b^\dagger]=$ const. We stay at the level of a single mode and the sign of its energy; the full continuum Hamiltonian is assembled in the field-quantization article.

Step 1. In the single-particle reading, the negative-frequency solution $e^{+iEt}$ is an eigenstate of $\hat H=i\partial_t$ with eigenvalue $-E$ : the spectrum contains $-E_{\mathbf p}$ for every momentum, unbounded below. This is the crisis Negative-Energy Solutions and the Interpretive Crisis diagnoses. Why: $i\partial_t\,e^{+iEt}=-E\,e^{+iEt}$ ; the negative energy is a literal eigenvalue, not an artifact, so it cannot be dodged by a coordinate choice.

Step 2. Quantize instead. The negative-frequency coefficient becomes a creation operator, $\beta_{\mathbf p}\to b_{\mathbf p}^\dagger$ , and the mode contributes $E_{\mathbf p}\,b\,b^\dagger$ to the Hamiltonian rather than an eigenvalue $-E_{\mathbf p}$ . Why: this is the decisive reinterpretation of equation (3) — the amplitude of a classical mode has become an operator that acts on states, so "the energy of the mode" now means the energy it costs to excite it.

Step 3. Use the bosonic commutator $b\,b^\dagger=b^\dagger b+c$ , with $c=(2\pi)^3\delta^3(0)$ a (divergent) constant. Then $E_{\mathbf p}\,b\,b^\dagger=E_{\mathbf p}\,b^\dagger b+E_{\mathbf p}c$ , and normal ordering discards the c-number. Why: the constant $c$ is the zero-point energy; it is an additive shift that drops out of every energy difference, so removing it changes no physics and leaves $E_{\mathbf p}\,b^\dagger b\ge0$ .

Step 4. Conclude: the antiparticle mode enters the Hamiltonian with $+E_{\mathbf p}$ , so creating an antiparticle costs positive energy. The spectrum starts at the unique vacuum $a_{\mathbf p}|0\rangle=b_{\mathbf p}|0\rangle=0$ and rises. There is no abyss — for bosons and, with anticommutators, for fermions alike. Why: the same sign flip that saves the boson would fail for the Dirac field if one used commutators; there anticommutators are forced to keep $H\ge0$ , which is the field-theoretic origin of the spin-statistics connection taken up in Spin-Statistics and the CPT Theorem. Verified: the verifier confirms the single-particle spectrum contains $-E_p$ (unbounded below) while the field mode energy $\omega(n+\tfrac12)$ is minimized at $+\omega/2>0$ , and that the antiparticle mode enters with $+E$ after normal ordering.

Limiting cases

case	result	meaning
nonrelativistic, $E\ll mc^2$	number operator $N$ effectively conserved	pair creation suppressed; ordinary single-particle Schrödinger QM re-emerges
free field	$H=\sum_{\mathbf p}E_{\mathbf p}(a^\dagger a+b^\dagger b)\ge0$	unique vacuum, no negative-energy sea, no descent to $-\infty$
one-particle sector	$\langle 0\|\hat\phi(x)\|\mathbf p\rangle=e^{-ip\cdot x}$	the old relativistic plane wave lives here as a matrix element, not a fundamental object
neutral (Hermitian) field	$a_{\mathbf p}=b_{\mathbf p}$ , so $Q\equiv0$	the particle is its own antiparticle (e.g. the $\pi^0$ ); the conserved charge vanishes identically
$\Delta x\lesssim\lambda_C$	pair creation, $N$ ill-defined	the single-particle picture is not merely approximate but impossible at short distances

Worked examples

Example 1 — the Compton wavelength and the pair threshold: where single-particle QM must fail. Compute the electron's reduced Compton wavelength $\lambda_C=\hbar/m_ec$ , the energy needed to localize to that scale, and the pair-production threshold — the three numbers that pin the failure quantitatively.

Step 1. Reduced Compton wavelength. With $\hbar=1.0546\times10^{-34}\ \mathrm{J\,s}$ , $m_e=9.1094\times10^{-31}\ \mathrm{kg}$ , $c=2.9979\times10^8\ \mathrm{m/s}$ :

\lambda_C=\frac{\hbar}{m_ec}=\frac{1.0546\times10^{-34}}{(9.1094\times10^{-31})(2.9979\times10^8)}\ \mathrm m=\mathbf{3.862\times10^{-13}\ m=386\ fm}.

Step 2. Localization energy at that scale. Squeezing to $\Delta x=\lambda_C$ injects $\Delta E\sim\hbar c/\Delta x=\hbar c/(\hbar/m_ec)=m_ec^2$ . Numerically $m_ec^2=(9.1094\times10^{-31})(2.9979\times10^8)^2/(1.602\times10^{-19})\ \mathrm{eV}=\mathbf{0.511\ MeV}$ — the localization energy equals the rest energy, by construction.

Step 3. Pair threshold. Making a real particle-antiparticle pair costs $2m_ec^2=\mathbf{1.022\ MeV}$ . Below this energy pairs cannot be produced on-shell and the particle number is effectively fixed; above it, the single-particle description is dead. These are the same 386 fm and 1.022 MeV measured every day in electron-positron physics — a single-particle theory has no room for either.

Example 2 — the Klein-Gordon density on a negative-energy plane wave. Show explicitly that $\rho$ goes negative, and reinterpret the number.

Step 1. Take $\phi=N\,e^{+iEt}$ with $E=+\sqrt{\mathbf p^2+m^2}>0$ (a negative-frequency solution). Compute $\dot\phi=+iEN\,e^{+iEt}$ and $\dot\phi^*=-iEN^*e^{-iEt}$ .

Step 2. Insert into $\rho=i(\phi^*\dot\phi-\phi\dot\phi^*)$ :

\rho=i\big[N^*e^{-iEt}(iEN\,e^{+iEt})-N e^{+iEt}(-iEN^*e^{-iEt})\big]=i\big[iE|N|^2+iE|N|^2\big]=\mathbf{-2E|N|^2}.

Step 3. Reinterpret. As a probability, $-2E|N|^2<0$ is meaningless. As a charge density it is fine: with the convention that particles carry $+1$ , this mode is populated by antiparticles of charge $-1$ , so the net charge density is negative. The pathology was a label error — $\rho$ is a charge, and charges come in both signs, exactly as the Klein-Gordon current article insists.

Origins · 1 minquantum field and how this was discovered

quantum field

coined 1927 · Dirac quantizes the electromagnetic field — the first "second quantization" — turning wave amplitudes into operators and photons into quanta of the field

The idea that the fundamental object is a quantized field, not a wavefunction, was born in 1927 when Dirac quantized the electromagnetic field in "The Quantum Theory of the Emission and Absorption of Radiation," treating the mode amplitudes of the radiation field as creation and annihilation operators and deriving Einstein's $A$ and $B$ coefficients. This is the original "second quantization." The extension to matter fields — promoting the Klein-Gordon and Dirac wavefunctions themselves to operators — was carried out by Jordan, Klein, Wigner, and Pauli over 1927–1929, dissolving the negative-energy crisis of Dirac's own equation. The lesson took a decade to fully absorb: the electron is a quantum of the electron field, on exactly the same footing as the photon is a quantum of the electromagnetic field.

History

1927

— Dirac quantizes the electromagnetic field, the first second quantization

1928

— Dirac's relativistic wave equation for the electron; Jordan and Wigner introduce field anticommutators

1930

— Dirac proposes the filled negative-energy sea with holes as antiparticles (later superseded by field theory)

1932

— Anderson discovers the positron in cloud-chamber tracks, confirming antimatter

1934

— Pauli and Weisskopf quantize the charged scalar field, showing antiparticles arise for bosons too, with no sea

Going further

This article makes the case that single-particle relativistic quantum mechanics cannot exist and must be replaced by a field theory, using three concrete failures diagnosed elsewhere: the indefinite density of Klein-Gordon Probability Current and Negative Densities, the bottomless spectrum of Negative-Energy Solutions and the Interpretive Crisis, and the pair-creation reading of the transmission anomaly in The Klein Paradox. It supersedes the hole-theory picture of The Dirac Sea and Antiparticles with an interpretation that works for bosons too. The mechanics it deliberately leaves undone — the mode expansion, the field commutators, and the explicit Hamiltonian — is the subject of the next article, Field Quantization of the Klein-Gordon and Dirac Fields, and the statistics forced on those commutators is the theorem of Spin-Statistics and the CPT Theorem.

Provenance. Base article QUANT.RQM.SPQF.1104 (AllQFT "quantum-mechanics" track, Sakurai & Napolitano Ch. 8 / Peskin & Schroeder Ch. 2). Upgraded to v3 canon. The base built the full mode expansion and Hamiltonian; that material has been scoped out and forward-referenced to the field-quantization article, keeping this article to the conceptual argument. Every number — $\lambda_C=386\ \mathrm{fm}$ , $m_ec^2=0.511\ \mathrm{MeV}$ , $2m_ec^2=1.022\ \mathrm{MeV}$ — and every sign — $\rho\propto\pm2E$ , the energy bounded below — is recomputed in `verify_aq-qm-single-particle-to-quantum-fields.py`. One base overstatement was corrected: the Compton-wavelength value $3.8616\times10^{-13}\ \mathrm m$ is the reduced $\hbar/mc$ , not the full $h/mc$ (which is $2\pi$ larger, $2.426\times10^{-12}\ \mathrm m$ ); the article states "reduced" explicitly.

Prerequisites. The Klein-Gordon Equation · Klein-Gordon Probability Current and Negative Densities · Negative-Energy Solutions and the Interpretive Crisis · The Dirac Equation · The Dirac Sea and Antiparticles. Leads to. Field Quantization of the Klein-Gordon and Dirac Fields · Charge Conjugation and the Positron · Spin-Statistics and the CPT Theorem.

If the equation is unchanged, what exactly is different about "second quantization"?

Nothing in the equation; everything in the interpretation. In "first" quantization you take a classical particle, promote its position and momentum to operators, and the wavefunction $\phi(x)$ is a c-number amplitude. In second quantization you take that field $\phi(x)$ — already a solution of the wave equation — and promote it to an operator, so its Fourier coefficients become creation and annihilation operators on a many-particle space. The name is historical and slightly misleading: you are not quantizing twice, you are quantizing a field instead of a particle. The wave equation is the classical field equation; the quanta of the field are the particles.

Why does the second-order-in-time structure of Klein-Gordon cause the density problem?

A first-order equation like Schrödinger's determines $\psi$ at all times from $\psi$ at one time, and its conserved density $|\psi|^2$ depends only on $\psi$ — manifestly non-negative. The Klein-Gordon equation is second order, so specifying the state requires both $\phi$ and $\dot\phi$ independently, and the conserved density $\rho=i(\phi^*\dot\phi-\phi\dot\phi^*)$ necessarily involves $\dot\phi$ . Since $\dot\phi$ can point either way relative to $\phi$ , $\rho$ can be either sign. The Dirac equation is first order and does have a positive-definite density $\psi^\dagger\psi$ — but it pays with negative-energy states instead, so no single-particle relativistic equation escapes both problems at once.

Does the field theory really predict antimatter, or is it put in by hand?

It is forced, not inserted. Hermiticity (for a real field) or the requirement of a well-defined charge (for a complex field) obliges the field to contain both positive- and negative-frequency modes. Once you quantize, the negative-frequency coefficient must be a creation operator to keep the energy bounded below — and the object it creates has the same mass as the particle but opposite charge. That is an antiparticle. So antimatter is a theorem of relativistic quantum field theory, first seen in Dirac's electron theory and confirmed by Anderson's 1932 positron, and it holds equally for bosons (the $\pi^+/\pi^-$ ), where the Dirac-sea picture is unavailable.

What does it even mean to call an electron "a quantum of a field"?

It means the electron is an excitation of a field that pervades all of space, in exactly the way a photon is an excitation of the electromagnetic field. There is one electron field; a state with "two electrons here and one there" is a particular excited configuration of that single field, and the electrons are indistinguishable because they are excitations of the same underlying object — which is why identical-particle statistics is automatic rather than imposed. This is a genuine ontological shift, not just a computational convenience: the fundamental furniture of the world is a set of fields, and particles are the countable quanta those fields support. (A philosopher of physics would add that "particle" is a derived, and in curved spacetime observer-dependent, notion — but the field is primary.)

If number isn't conserved, how do we still talk about "the electron" in an atom?

Because in the nonrelativistic regime pair creation is energetically forbidden. Binding energies in atoms are electron-volts, while making a pair costs $1.022\ \mathrm{MeV}$ — six orders of magnitude more. The amplitude to create a virtual pair is suppressed by that huge energy gap, so to spectacular accuracy the electron number is fixed and ordinary single-particle quantum mechanics applies. Field theory does not forbid you from using the Schrödinger equation for hydrogen; it explains why you are allowed to. The tiny residual effects of the field — vacuum polarization, the Lamb shift, the anomalous magnetic moment — are exactly the corrections that single-particle theory cannot produce and that field theory predicts to ten significant figures.

Why can't we just fill the Dirac sea and be done, as Dirac originally proposed?

The sea works, but only barely and only for fermions. Filling every negative-energy electron state and invoking Pauli exclusion to block the cascade does stabilize the vacuum, and a hole in the sea correctly behaves as a positron — this is the picture of The Dirac Sea and Antiparticles. But it treats the vacuum as an infinite reservoir of charge and energy, it is not manifestly Lorentz-invariant, and it collapses completely for spin-0 particles, which cannot fill a sea because bosons have no exclusion principle. Yet the charged pion has an antiparticle just like the electron does. Field theory delivers antiparticles for bosons and fermions alike, with a finite, symmetric vacuum — which is why the sea is now taught as history rather than mechanism.

Is the "Compton wavelength" argument a rigorous bound or just a heuristic?

It is a heuristic scale, but a reliable one, and it is confirmed by the full theory. The estimate $\Delta E\sim\hbar c/\Delta x\to mc^2$ at $\Delta x=\lambda_C$ is order-of-magnitude, not a sharp inequality. The rigorous statement is that a relativistic particle cannot be localized to a sharp position at all: the position operator that would do so (the Newton-Wigner operator) is not Lorentz-covariant, and any attempt to build a localized single-particle state develops negative-frequency contamination on the Compton scale. So the heuristic correctly identifies $\lambda_C$ as the length below which "one particle here" stops making sense, and the field theory replaces the ill-defined position eigenstate with a genuine many-particle description.