Probability density & current
The Klein–Gordon equation, the first attempt at a relativistic wave equation, carried a fatal defect: the density it conserves, ρ = (i/2m)(ψ∂ₜψ − ψ∂ₜψ), is not sign-definite, and a probability that can turn negative is no probability at all. Paul Dirac traced the trouble to a single structural feature — the equation is second order in time, so ψ and ∂ₜψ are independent initial data, and any density built from both can be dialled negative. His response in 1928 was to demand an equation that is first order in time while still honouring the relativistic energy–momentum relation E² = p²c² + m²c⁴. Forcing linearity in ∂ₜ forces the wavefunction to become a four-component spinor and the coefficients to become the 4×4 matrices α and β. The pay-off is the subject of this article. Run the same conservation-law construction that Schrödinger and Klein–Gordon used — form the equation against its conjugate and collapse the result into total derivatives — and the conserved four-current comes out as ρ = ψ†ψ and j = c ψ†αψ. Because ψ is now a spinor, complex conjugation is promoted to Hermitian conjugation, and the density is ρ = ψ†ψ = |ψ₁|² + |ψ₂|² + |ψ₃|² + |ψ₄|²: a sum of four squared moduli, hence positive for every solution, at every point, regardless of the sign of the energy. That is exactly the object Born's statistical interpretation requires. This article builds the current from the Dirac equation, casts it in manifestly covariant form jᵘ = c ψ̄γᵘψ, and confronts the one subtlety that survives — that field theory ultimately reads this current as charge, not probability.
- The Dirac equation conserves a four-current with density ρ = ψ†ψ and spatial current j = c ψ†αψ, tied by the continuity equation ∂ₜρ + ∇·j = 0.
- The density ρ = ψ†ψ = |ψ₁|² + |ψ₂|² + |ψ₃|² + |ψ₄|² is a sum of squares — manifestly ≥ 0 — the very property Born's interpretation demands and the Klein–Gordon density lacks.
- The construction uses the Hermitian conjugate equation: because ψ is a four-spinor, ψ → ψ† = (ψ)ᵀ, and the Hermiticity of α, β makes the mass terms cancel.
- In covariant form jᵘ = c ψ̄γᵘψ = (cρ, j) with ψ̄ = ψ†γ⁰, and conservation is the single Lorentz-invariant statement ∂ᵤjᵘ = 0.
- For a plane wave the current is a flux j = ρv: the density streams at the particle velocity v = pc²/E, a genuinely relativistic result.
- Key idea: because the Dirac equation is first order in time, its conserved density ρ = ψ†ψ comes out as a sum of squares — automatically positive — so it can legitimately be read as a probability density.
Because the Dirac equation is first order in time, the conserved density ρ = ψ†ψ comes out as a sum of squares — automatically positive — so it can legitimately be read as a probability density, the feature the second-order Klein–Gordon equation could not supply.
Where this lives today
The Dirac four-current is not a museum piece — it is the electromagnetic vertex of the Standard Model. Coupling the current to the photon by the minimal replacement gives the interaction term e ψ̄γᵘψ Aᵤ, and jᵘ = e ψ̄γᵘψ is literally the electric charge–current that sources Maxwell's equations in quantum electrodynamics; every Feynman diagram with an electron emitting or absorbing a photon is one insertion of this bilinear. Split it by the Gordon decomposition and the current separates into a convection piece (the ψ̄∂ᵘψ motion of charge) plus a spin-magnetisation piece ∂ᵥ(ψ̄σᵘᵛψ), and that second piece is exactly the electron's magnetic moment — the g = 2 that Dirac got for free and that QED then corrects to g = 2.00231930436, agreement with experiment to twelve significant figures, the most precisely verified prediction in physics. The vector current ψ̄γᵘψ and its axial partner ψ̄γᵘγ⁵ψ together build the V−A structure of the weak interaction, so beta decay and neutrino scattering are written in the same language. And the conserved charge ∫j⁰d³x = N₊ − N₋, once you second-quantise, is what guarantees electric charge conservation exactly — the modern descendant of Dirac's insistence that the density behave.
The physical picture
A conserved probability is a bookkeeping promise: whatever probability drains out of a region has to reappear as flux through the region's walls, so that the total, integrated over all space, never changes. The pair (ρ, j) is exactly that ledger — ρ counts the density held in a volume, j counts the rate it streams across a surface, and the continuity equation ∂ₜρ + ∇·j = 0 is the statement that the books balance locally, point by point. For the Dirac field the entry ρ = ψ†ψ is a sum of four squared moduli, and this is what makes the ledger physically legitimate: no matter what spinor solution you write, the density it records is positive. The first figure sets Dirac's positive-definite density against Klein–Gordon's, which tracks the sign of the energy and goes negative for the negative-energy waves that any relativistic equation must contain.
Once the density is trustworthy, the current tells you where the probability is going. For a free particle the answer is as physical as it could be: the current is the density carried along at the particle's own velocity, j = ρv, with v = pc²/E the relativistic group velocity. The spinor expectation ⟨α⟩ that defines the current is precisely v/c, so a fast electron's probability streams forward at nearly the speed of light while a slow one's barely drifts. The second figure makes the continuity ledger concrete and attaches a real number to the flux: a γ = 2 electron of total energy 1.022 MeV moves at 0.866 c, and its probability current is that fraction of ρc.
The core equations
The starting point is the Dirac equation in its Hamiltonian (first-order-in-time) form. It is deliberately linear in ∂ₜ — that is the whole design goal — and linearity forces the coefficients α = (α₁, α₂, α₃) and β to be 4×4 Hermitian matrices obeying αᵢαⱼ + αⱼαᵢ = 2δᵢⱼ, αᵢβ + βαᵢ = 0, and αᵢ² = β² = 1, which is why ψ carries four components.
where: ψ — four-component spinor field ψ(x, t); α, β — Hermitian 4×4 matrices with α† = α, β† = β; m — particle rest mass; c — speed of light. Reading it: the Hamiltonian c α·p + β mc² is first order in the spatial derivatives too, which is the extra demand that made the matrices necessary. Running the conservation construction on equation (1) delivers the conserved four-current. The density and the spatial current are simple spinor bilinears, and the promotion from ψ* to the Hermitian conjugate ψ† is the one change from the Schrödinger case.
where: ρ — probability density (units of inverse volume once ψ is normalised); j — probability current density; ψ† = (ψ*)ᵀ — Hermitian conjugate (transpose-and-conjugate) of the spinor. Reading it: j is c times the expectation of the matrix α, and α plays the role of the velocity operator in the Dirac theory. These two are bound together by the continuity equation, the local statement that probability is neither created nor destroyed. Integrated over all space with fields that vanish at infinity, it guarantees d/dt ∫ρ d³x = 0, so a state normalised once stays normalised.
where: the divergence ∇·j is the net outflow per unit volume, exactly cancelling any local change ∂ₜρ. Reading it: this is the same continuity equation as in non-relativistic quantum mechanics and electromagnetism — only the entries (ρ, j) are new. The physically decisive line is the explicit component form of the density. Because ψ has four components and ψ†ψ sums each component against its own complex conjugate, ρ is a sum of squared moduli.
where: ψ₁…ψ₄ — the four complex spinor components. Reading it: every term is non-negative, so ρ is non-negative at every point and for every solution — the property that lets ρ serve as a probability density and the one the Klein–Gordon equation cannot match. Finally, the whole structure snaps into a single Lorentz-covariant object. Defining the gamma matrices γ⁰ = β and γⁱ = βαⁱ (so that {γᵘ, γᵛ} = 2gᵘᵛ) and the Dirac adjoint ψ̄ = ψ†γ⁰, the four-current is a bilinear that transforms as a genuine four-vector, and conservation becomes one covariant equation.
where: ψ̄ = ψ†γ⁰ — Dirac adjoint; γᵘ = (γ⁰, γⁱ) — gamma matrices; gᵘᵛ = diag(+, −, −, −) — Minkowski metric. Reading it: because jᵘ is a four-vector and ∂ᵤjᵘ is a Lorentz scalar, "probability is conserved" holds in every inertial frame at once. For contrast, the Klein–Gordon equation conserves its own four-current, but its density is built from a time derivative and is not sign-definite — the defect that motivated the entire Dirac programme.
where: ψ — a single complex scalar field (not a spinor). Reading it: ρ_KG depends on both ψ and ∂ₜψ, which are independent initial data for a second-order equation, so its sign is not fixed — for a plane wave it equals (E/m)|ψ|² and flips with the sign of E.
The goal is to construct the conserved four-current of the Dirac equation directly, by the same algebra Schrödinger used, and to see the positivity of the density emerge on the way. We take the Dirac equation in Hamiltonian form, build its Hermitian conjugate (the spinor analogue of complex-conjugating the Schrödinger equation), multiply the two so that the products become total derivatives, and subtract. The mass terms will cancel because β is Hermitian, leaving a clean continuity equation from which ρ = ψ†ψ and j = c ψ†αψ can simply be read off. Throughout we work in natural units ℏ = c = 1 to keep the algebra visible; the factors of ℏ and c reinstated in equations (1)–(2) do not affect the structure.
Step 1. Write the Dirac equation as i∂ₜψ = (−i α·∇ + βm)ψ. Every quantity here is a matrix acting on the four-spinor ψ, with α and β Hermitian by construction. Why: this Hamiltonian form isolates the single time derivative ∂ₜψ, which is what makes the density depend on ψ alone rather than on ψ and its time derivative separately.
Step 2. Take the Hermitian conjugate of the whole equation. Using (∂ₜψ)† = ∂ₜψ†, α† = α, β† = β, and remembering that Hermitian conjugation reverses the order of matrix–spinor products, this gives −i∂ₜψ† = i(∇ψ†)·α + mψ†β. Why: because ψ is a spinor, the correct conjugate is ψ† = (ψ)ᵀ, not ψ; the transpose is what keeps the matrix indices contracted properly in the next step.
Step 3. Left-multiply the Step 1 equation by ψ† and right-multiply the Step 2 equation by ψ. The first gives i ψ†∂ₜψ = −i ψ†α·∇ψ + m ψ†βψ; the second gives −i(∂ₜψ†)ψ = i(∇ψ†)·αψ + m ψ†βψ. Why: multiplying on the correct side keeps every term a scalar (a 1×1 spinor sandwich) and lines the two equations up so their mass terms are identical.
Step 4. Subtract the second from the first. The mass terms m ψ†βψ are identical and cancel exactly, leaving i(ψ†∂ₜψ + (∂ₜψ†)ψ) = −i(ψ†α·∇ψ + (∇ψ†)·αψ). Both sides are now total derivatives: the left is i ∂ₜ(ψ†ψ), the right is −i ∇·(ψ†αψ). Why: the cancellation of the mass term is the pay-off of β being Hermitian — it is precisely what leaves a pure divergence structure with no leftover source term.
Step 5. Divide through by i to obtain ∂ₜ(ψ†ψ) + ∇·(ψ†αψ) = 0. Comparing with the continuity equation ∂ₜρ + ∇·j = 0 identifies ρ = ψ†ψ and j = ψ†αψ (restore the c to get j = c ψ†αψ). Why: matching the standard continuity form is what licenses calling these expressions a density and a current rather than merely a conserved combination.
Step 6. Read the density in components: ρ = ψ†ψ = |ψ₁|² + |ψ₂|² + |ψ₃|² + |ψ₄|² ≥ 0. It is a sum of squared moduli, positive at every point for every solution. Why: this is the entire reason the construction was worth doing — the density is now something Born's rule can interpret, which the Klein–Gordon density was not.
Go deeper · Derivation B · 2 minThe covariant four-current and its manifest conservation
The aim here is to recast the density and current as a single four-vector jᵘ = ψ̄γᵘψ and to prove ∂ᵤjᵘ = 0 in one covariant line, which shows that "probability is conserved" is frame-independent. We introduce the gamma matrices, rewrite the Dirac equation and its adjoint in covariant form, and combine them. The key algebraic fact is that the Dirac-adjoint equation follows from the Hermitian conjugate through the relation γᵘ† = γ⁰γᵘγ⁰, which is exactly what makes ψ̄ the natural partner of ψ. Natural units ℏ = c = 1 are used.
Step 1. Define γ⁰ = β and γⁱ = βαⁱ. Multiplying the Dirac equation i∂ₜψ = (−iα·∇ + βm)ψ through by β = γ⁰ and using βαⁱ = γⁱ turns it into (iγᵘ∂ᵤ − m)ψ = 0. The gammas satisfy the Clifford algebra {γᵘ, γᵛ} = 2gᵘᵛ, with γ⁰ Hermitian and γⁱ anti-Hermitian. Why: the covariant form treats time and space derivatives on the same footing, which is what makes the resulting current a four-vector rather than a (density, current) pair glued together by hand.
Step 2. Hermitian-conjugate the covariant equation: −i∂ᵤψ†γᵘ† − mψ† = 0. Insert γᵘ† = γ⁰γᵘγ⁰ and right-multiply by γ⁰, using (γ⁰)² = 1. Defining the Dirac adjoint ψ̄ = ψ†γ⁰, this collapses to ψ̄(iγᵘ⃖∂ᵤ + m) = 0, where the derivative acts to the left. Why: the identity γᵘ† = γ⁰γᵘγ⁰ is precisely what makes ψ̄ = ψ†γ⁰ transform correctly and turns a messy conjugate into a clean adjoint equation.
Step 3. Multiply the Step 1 equation on the left by ψ̄ and the Step 2 equation on the right by ψ, then add. The mass terms (−mψ̄ψ and +mψ̄ψ) cancel, leaving ψ̄γᵘ(∂ᵤψ) + (∂ᵤψ̄)γᵘψ = 0. Why: adding rather than subtracting is what is needed here because the relative sign of the mass terms is opposite to the Hamiltonian-form derivation — the cancellation still happens, guaranteeing a pure divergence.
Step 4. Recognise the sum as a single divergence: ∂ᵤ(ψ̄γᵘψ) = 0. Hence jᵘ = ψ̄γᵘψ is conserved. Why: because γᵘ carries the only free Lorentz index, ψ̄γᵘψ is a bilinear that transforms as a four-vector, so ∂ᵤjᵘ is a scalar — conservation now holds identically in every frame.
Step 5. Check the components. The time component is j⁰ = ψ̄γ⁰ψ = ψ†γ⁰γ⁰ψ = ψ†ψ = ρ, using (γ⁰)² = 1. The space components are jⁱ = ψ̄γⁱψ = ψ†γ⁰γⁱψ = ψ†β(βαⁱ)ψ = ψ†αⁱψ, using β² = 1. So jᵘ = (ρ, ψ†αψ), reproducing equation (2). Why: recovering the same ρ and <strong>j</strong> confirms the covariant object is not a new current but the same one, now displayed as the four-vector it always was.
Go deeper · Derivation C · 1 minWhy the Klein-Gordon density changes sign
This derivation isolates the defect Dirac was solving. The goal is to construct the conserved current of the Klein–Gordon equation the same way, show that its density is forced to involve a time derivative, and see explicitly that this is what allows the density to be negative. We build the current, evaluate it on a plane wave, and contrast the result with the Dirac case, so that the advantage of first-order-in-time structure is quantitative rather than merely asserted. Natural units ℏ = c = 1.
Step 1. Start from the Klein–Gordon equation (∂ᵤ∂ᵘ + m²)ψ = 0 for a complex scalar ψ. Multiply it by ψ and subtract its complex conjugate multiplied by ψ, so the mass terms cancel: ψ∂ᵤ∂ᵘψ − ψ∂ᵤ∂ᵘψ = 0. Why: the same conjugate-and-subtract move as before, but now the equation is second order, so two derivatives survive.*
Step 2. Rewrite the left side as a total divergence, ∂ᵤ(ψ∂ᵘψ − ψ∂ᵘψ) = 0, which identifies the conserved four-current jᵘ ∝ i(ψ∂ᵘψ − ψ∂ᵘψ). Its time component is the density ρ_KG = (i/2m)(ψ∂ₜψ − ψ∂ₜψ), the normalisation chosen so ρ has the right units. Why: unlike the Dirac density, this ρ is built from ∂ₜψ, not from ψ alone — the direct fingerprint of the second-order equation.
Step 3. Evaluate on a plane wave ψ = N e^(−iEt + ip·x), for which ∂ₜψ = −iE ψ. Then ρ_KG = (i/2m)(|N|²(−iE) − |N|²(iE)) = (E/m)|N|². Why: because a relativistic equation admits both signs E = ±√(<strong>p</strong>² + m²), the density inherits that sign — positive for the E > 0 wave, negative for the E < 0 wave.
Step 4. Contrast with Dirac. There ρ = ψ†ψ is fixed by ψ alone and is a sum of squares, so no choice of initial data can make it negative; here ρ_KG scales with E and is negative for exactly half the free solutions. Why: this is the concrete statement of the article's thesis — first order in time ties the density to ψ alone and guarantees positivity; second order in time does not.
Limiting cases
| Case | Result | Meaning |
|---|---|---|
| Positive-energy plane wave | ρ = ψ†ψ > 0, j = ρv | normal probability flux; density streams at v = pc²/E |
| Negative-energy plane wave | ρ = ψ†ψ > 0 still | Dirac ρ stays positive where Klein–Gordon's goes negative — the core advantage |
| Particle at rest (p = 0) | j = 0, ρ = ψ†ψ constant | no current; density static and positive, as a bound state requires |
| Non-relativistic limit | upper spinor components dominate, ρ → ψ_upper†ψ_upper | reduces to the Schrödinger density ψ*ψ, recovering ordinary QM |
| Coupled to electromagnetism | eⱼᵘ = e ψ̄γᵘψ becomes the electric charge–current | the same conserved current sources Maxwell's equations in QED |
| Second-quantised (QFT) | jᵘ → :ψ̄γᵘψ: is charge, ∫j⁰d³x = N₊ − N₋ | conserved quantity is net charge, not particle number — probability picture retired |
Worked examples
1 · The Klein–Gordon density goes negative — with numbers (fully worked)
Take a free electron mass m = 0.511 MeV/c² and compare the Klein–Gordon and Dirac densities for a positive-energy plane wave and its negative-energy partner, both with total energy magnitude |E| = 1.022 MeV (that is, γ = 2) and unit amplitude |N|² = 1. Work in natural units ℏ = c = 1.
Step 1. Fix the kinematics. With γ = 2 the total energy is E = γm = 2 × 0.511 = 1.022 MeV, and the two relativistic branches are E = +1.022 MeV and E = −1.022 MeV, both satisfying E² = p² + m² with |p| = √(E² − m²) = √(1.044 − 0.261) = 0.885 MeV/c.
Step 2. Klein–Gordon density, positive branch. Using ρ_KG = (E/m)|N|² from Derivation C, ρ_KG = (+1.022)/(0.511) × 1 = +2.00. A sensible positive density.
Step 3. Klein–Gordon density, negative branch. Now E = −1.022 MeV, so ρ_KG = (−1.022)/(0.511) × 1 = −2.00. The density is negative — there is no way to read −2.00 as a probability per unit volume, which is exactly the defect Dirac set out to remove.
Step 4. Dirac density, both branches. Here ρ = ψ†ψ = |ψ₁|² + |ψ₂|² + |ψ₃|² + |ψ₄|². Whichever branch you pick, the spinor amplitudes enter as squared moduli. With the standard covariant normalisation u†u = v†v = 2E for the four-spinors, ρ = 2E = +2.044 on both branches (the antiparticle v-spinor carries its own positive norm). Dirac's density is positive for the negative-energy wave where Klein–Gordon's was negative — the entire point in a single number.
2 · The current is a flux: probability streams at the particle velocity (fully worked)
For the same γ = 2 electron plane wave, show that the spatial current is the density times the velocity, <strong>j</strong> = ρ<strong>v</strong>, and evaluate the velocity.
Step 1. Write the current. From equation (2), j = c ψ†αψ = ρ c ⟨α⟩, where ⟨α⟩ = (ψ†αψ)/(ψ†ψ) is the spinor expectation of the Dirac velocity matrix. So the whole problem is to evaluate ⟨α⟩.
Step 2. Use the free-spinor result. For a positive-energy plane wave u(p), the expectation of α is ⟨α⟩ = p/E (this is the momentum-space Ehrenfest statement that α is the velocity operator, and it holds because [α, H] generates the correct velocity). Numerically ⟨α⟩ = |p|/E = 0.885/1.022 = 0.866, dimensionless.
Step 3. Assemble the current and read the velocity. Then j = ρ c (0.866) = ρ (0.866 c), so the probability streams forward at v = pc²/E = 0.866 c — the same 0.866 c that γ = 2 gives from v = c√(1 − 1/γ²) = c√(0.75). The current is a flux, j = ρv, with v = 0.866 c, and it points along p as physical intuition demands.
3 · Positivity for an explicit spinor (quick check)
Confirm ρ ≥ 0 for the concrete normalised spinor ψ = (1/√2)(1, 0, i, 0)ᵀ (spatial factors suppressed).
Step 1. Apply equation (4): ρ = |ψ₁|² + |ψ₂|² + |ψ₃|² + |ψ₄|² = (1/2)(|1|² + |0|² + |i|² + |0|²) = (1/2)(1 + 0 + 1 + 0) = 1. The complex component ψ₃ = i/√2 contributes |i/√2|² = 1/2, not a negative amount — because it is squared in modulus, its phase cannot pull ρ down. ρ = 1 ≥ 0, as guaranteed for any spinor whatsoever.
Physics of the portrait
The probability current is Dirac's fingerprint on quantum mechanics: the demand for a positive density is what forced the electron to be a four-component spinor, and everything — spin, the magnetic moment, antimatter — followed from that one structural choice.
Origins · 1 minPaul Dirac and how this was discovered
Paul Adrien Maurice Dirac was, by temperament, an engineer turned mathematician who trusted the beauty of an equation over physical intuition, and the probability current is where that trust paid off first. Writing in 1928, he was troubled that the Klein–Gordon equation — the obvious relativistic upgrade of Schrödinger's — produced a density that could go negative and so could not carry Born's probability interpretation. He diagnosed the cause as its second-order time derivative and insisted on an equation linear in ∂ₜ. Demanding that this first-order equation still square to the relativistic energy relation E² = p²c² + m²c⁴ forced its coefficients to anticommute, which no ordinary numbers can do; Dirac was driven to 4×4 matrices and a four-component wavefunction. Out of that came the positive density ρ = ψ†ψ he had been chasing — and, unbidden, the electron's spin-½ and its magnetic moment g = 2, which he had not put in by hand. The same equation's negative-energy solutions, at first an embarrassment, led him in 1931 to predict the positron, discovered by Anderson the next year. Dirac shared the 1933 Nobel Prize with Schrödinger. He held Newton's Lucasian chair at Cambridge and was famously sparing with words; the four-line current derivation in this article is very much in his style — no motion wasted, the physics falling out of the algebra.
Going further
Texts. Thomson, Modern Particle Physics, §4.3 (probability density and current) · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.1–7.3 · Peskin & Schroeder, An Introduction to Quantum Field Theory, §3.2–3.4 (Dirac field, currents, Gordon identity) · Halzen & Martin, Quarks and Leptons, §5.3–5.4 · Schwartz, Quantum Field Theory and the Standard Model, Ch. 10 · Bjorken & Drell, Relativistic Quantum Mechanics, §3.
Provenance. The Hamiltonian-form Dirac equation, the matrices α, β, and their algebra are owned by aq-pp-dirac-equation and recalled here by name; the Klein–Gordon equation and its indefinite density are developed in aq-pp-klein-gordon-equation. The construction here reproduces Thomson's (4.14)–(4.17) and Griffiths's (7.43)–(7.68). A verifier should confirm: (1) the continuity equation ∂ₜρ + ∇·j = 0 with ρ = ψ†ψ, j = c ψ†αψ from the Hermitian-conjugate derivation; (2) the covariant form jᵘ = ψ̄γᵘψ with j⁰ = ρ and jⁱ = ψ†αⁱψ using γ⁰ = β, γⁱ = βαⁱ; (3) ρ_KG = (E/m)|N|² for a plane wave, negative for E < 0; (4) the γ = 2 numbers E = 1.022 MeV, |p| = 0.885 MeV/c, v = 0.866 c, and ⟨α⟩ = p/E = 0.866.
Prerequisites. The Dirac Equation · The Klein–Gordon Equation. Leads to. The Dirac Probability Current (covariant treatment) · Covariant Form of the Dirac Equation · Antiparticles and Negative-Energy Solutions.
Ask the tutor: Ask me to walk through the four-line continuity derivation on the board, contrast Dirac's ρ = ψ†ψ with the Klein–Gordon density on a negative-energy plane wave, or show how this same current becomes the charge current at the QED vertex.
Why did Dirac insist the equation be first order in time
Because Born's statistical interpretation needs a positive density, and the density's sign is decided by the equation's time-derivative structure. For a second-order equation like Klein–Gordon, the initial data are ψ and ∂ₜψ specified independently, so any conserved density built from them (and it must involve ∂ₜψ to be conserved) can be made either sign by choosing the data. A first-order equation fixes ∂ₜψ in terms of ψ itself, so the conserved density can be built from ψ alone — and ψ†ψ is then a sum of squares, automatically positive. First order in time is not a cosmetic preference; it is the structural requirement for a legitimate probability density.
Does positive ρ prove the Dirac equation is correct and Klein–Gordon wrong
No. It proves the single-particle Dirac theory is self-consistent as a probability theory, which the naive single-particle Klein–Gordon theory is not. But Pauli and Weisskopf showed in 1934 that Klein–Gordon's current is a perfectly good electric charge–current — allowed to be negative because antiparticles have opposite charge — and it correctly describes spin-0 particles like pions and the Higgs. In quantum field theory both equations are fine, and the Dirac density loses its probability meaning too: ∫ρ d³x becomes net charge, not particle number. So positive ρ was historically decisive but is not the final verdict on either equation.
What are the matrices α and β, and why must they be Hermitian
They are the 4×4 coefficients that make the Dirac equation linear in space and time derivatives while still reproducing E² = p²c² + m²c⁴. Squaring the Dirac Hamiltonian forces αᵢαⱼ + αⱼαᵢ = 2δᵢⱼ, αᵢβ + βαᵢ = 0, αᵢ² = β² = 1 — an algebra no ordinary numbers satisfy, which is why matrices (and a four-component ψ) are unavoidable. Their Hermiticity, α† = α and β† = β, is required for two reasons: it makes the Hamiltonian H = cα·p + βmc² Hermitian, so energies are real, and in the current derivation it is exactly what makes the mass terms cancel and the density come out as ψ†ψ.
Is the probability current literally the density times the velocity
For a free-particle plane wave, yes: j = ρv with v = pc²/E the relativistic group velocity, because the spinor expectation of the matrix α equals v/c. This is why α is called the Dirac velocity operator. There is a famous subtlety, though — the instantaneous eigenvalues of α are ±1 (that is, ±c), and a superposition of positive- and negative-energy states shows a trembling motion, Zitterbewegung, at frequency 2mc²/ℏ. For a pure positive-energy wave those oscillations average away and only the smooth drift v = pc²/E survives, which is what the current measures.
How does this probability current become the electric current in QED
By coupling it to the electromagnetic field. The minimal-substitution recipe pᵤ → pᵤ − eAᵤ introduces an interaction e ψ̄γᵘψ Aᵤ, in which jᵘ = e ψ̄γᵘψ — the same conserved bilinear, now multiplied by the charge e — appears as the source in Maxwell's equations. Every electron–photon vertex in a Feynman diagram is one factor of this current. Because ∂ᵤjᵘ = 0 was already guaranteed by the Dirac equation, electric charge conservation in QED is automatic rather than an extra assumption. The probability current and the charge current are the same object seen with and without the factor of e.
Why is ψ†ψ the time component of a four-vector rather than a scalar
Because of how the pieces transform. Under a Lorentz boost the spinor transforms as ψ → S(Λ)ψ, and S is not unitary (boosts are not), so ψ†ψ is not invariant — it transforms like the time component of the four-vector jᵘ = ψ̄γᵘψ, which is exactly right, since a density is the time component of a current (more density means more stuff per unit volume, and volumes Lorentz-contract). The genuine Lorentz scalar is ψ̄ψ = ψ†γ⁰ψ, a different bilinear, whose γ⁰ compensates the non-unitarity of S. Confusing ψ†ψ (density, a four-vector component) with ψ̄ψ (a scalar) is a common slip worth guarding against.
What happens to the probability interpretation in quantum field theory
It is retired and replaced by something sturdier. Once ψ is a quantum field, particle number is not conserved — pairs are created and annihilated — so ∫ρ d³x cannot be a fixed probability of finding one particle. Instead the conserved charge is Q = ∫:ψ̄γ⁰ψ: d³x = N₊ − N₋, the number of particles minus antiparticles, an integer that never changes. The current jᵘ = ψ̄γᵘψ survives intact and even gets more important — it is the charge current that couples to gauge fields — but its zeroth component now counts net charge, not probability. Single-particle probability was scaffolding; the conserved current is the permanent structure.