The Dirac Equation

By 1928 the search for a relativistic quantum wave equation had reached an impasse. The natural first attempt — quantize the relativistic energy–momentum relation $E^2=\mathbf{p}^2+m^2$ directly — gives the Klein–Gordon equation, an equation second order in time whose conserved density is not positive-definite and whose spectrum admits unavoidable negative-energy solutions. Paul Dirac asked a sharper question: can we find a wave equation first order in the time derivative, so that the density comes out as a manifest sum of squares, while still respecting relativity? The answer required an audacious move. To be first order in time and, by Lorentz symmetry, first order in every spatial derivative, the equation must "take the square root" of the second-order Klein–Gordon operator. Dirac wrote a linear ansatz $H=\boldsymbol{\alpha}\cdot\mathbf{p}+\beta m$ and demanded that squaring it reproduce $\mathbf{p}^2+m^2$ . That single demand forces the coefficients to anticommute and square to one — a condition no ordinary numbers can meet. The smallest objects that can are $4\times4$ matrices, and the wave function is promoted to a four-component column called a Dirac spinor. In covariant form the result is the Dirac equation $(i\gamma^\mu\partial_\mu-m)\psi=0$ , and the matrices obey the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ . This article stays with the free equation: how it is forced, why its matrices must be $4\times4$ , and how squaring it recovers Klein–Gordon component by component. The rich harvest that grows from it — the explicit spinor solutions, the gamma-matrix algebra, Lorentz covariance, the automatic prediction of spin- $\tfrac12$ and the electron $g=2$ , and the antiparticle interpretation — each belongs to its own article and is signposted here by name.

Key takeaways

The Dirac equation $(i\gamma^\mu\partial_\mu-m)\psi=0$ is the first-order relativistic wave equation obtained by demanding that its square be the Klein–Gordon equation.
The square condition forces the Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}\mathbb{1}$ ; ordinary numbers cannot satisfy it, so the $\gamma^\mu$ must be matrices.
Counting eigenvalues and mutually anticommuting matrices shows the minimal dimension is $4\times4$ , so $\psi$ is a four-component spinor.
Being first order in $\partial_t$ , the equation has a conserved density $\rho=\psi^\dagger\psi\ge0$ — the manifestly positive object Klein–Gordon lacked.
The four components are not four particles; they package internal (spin) and negative-energy structure, whose physics is developed in the linked spinor and Dirac-sea articles.
Key idea: linearizing $E^2=\mathbf{p}^2+m^2$ in both time and space forces the wave function to carry matrix-valued (spinor) indices obeying a Clifford algebra — the equation's whole architecture is dictated, not chosen.

The core insight

Demanding a relativistic wave equation that is first order in time forces its square to be Klein–Gordon, which forces the anticommuting Clifford algebra $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ — and hence a four-component spinor — as a mathematical necessity.

Where this lives today

The Dirac equation is no longer read as a single-particle theory — its negative-energy solutions and the Klein paradox showed that a consistent relativistic quantum theory must allow particle creation. In modern quantum field theory, $\psi$ is promoted to a quantized field and the Dirac equation becomes the classical equation of motion whose plane-wave spinors $u(p),v(p)$ are the building blocks of Feynman amplitudes. Every calculation in quantum electrodynamics — electron–positron scattering, the electron self-energy, the anomalous magnetic moment measured to twelve significant figures — is assembled from Dirac spinors and the gamma-matrix algebra introduced here. The equation also underpins relativistic quantum chemistry (heavy-element bonding is a relativistic effect) and, in condensed matter, describes the massless Dirac fermions of graphene and topological insulators, where the same $2\times2$ block structure reappears at electronic energy scales.

The physical picture

Start from the relativistic dispersion $E^2=\mathbf{p}^2+m^2$ and try to write a linear Hamiltonian, $E\to H=\boldsymbol{\alpha}\cdot\mathbf{p}+\beta m$ . The scalar square root $\sqrt{\mathbf{p}^2+m^2}$ is nonlocal — an infinite series in derivatives — so it cannot serve as a differential wave operator. A clean linear expression can, but only if its coefficients are special objects. Squaring $\boldsymbol{\alpha}\cdot\mathbf{p}+\beta m$ produces cross terms like $(\alpha^i\alpha^j+\alpha^j\alpha^i)p_ip_j$ , and these must vanish unless $i=j$ . That is the whole story in one line: the coefficients must anticommute. Anticommuting objects that square to one cannot be numbers — numbers commute. They must be matrices, and the smallest matrices that admit four mutually anticommuting members (one for $\beta$ , three for the $\alpha^i$ ) are $4\times4$ . So the wave function $\psi$ acquires a new, internal index running over four values. Relativity, by insisting on a first-order equation, has secretly handed us a multi-component object.

Figure

Taking the square root of the dispersion relation

There is a decisive payoff. A first-order equation in time carries a conserved current whose time component is $\rho=\psi^\dagger\psi=\sum_a|\psi_a|^2$ , a sum of squared moduli that can never be negative. This is exactly the property Klein–Gordon lacked, where the density $\rho\propto\mathrm{Im}(\phi^*\partial_t\phi)$ could go negative for negative-energy modes. Dirac read the positive density as strong evidence that the linearization was the right idea.

Figure

Why the coefficients must anticommute

The core equations

Dirac's construction begins in Hamiltonian form, where the physical demand — reproduce the relativistic dispersion on squaring — is easiest to state. We write the energy operator as a linear combination of momentum and mass with matrix coefficients.

H = \boldsymbol{\alpha}\cdot\mathbf{p} + \beta m, \qquad i\,\partial_t\psi = H\psi \qquad (1)

where: $\boldsymbol{\alpha}=(\alpha^1,\alpha^2,\alpha^3)$ are three Hermitian $4\times4$ matrices multiplying the momentum; $\beta$ is a Hermitian $4\times4$ matrix multiplying the mass; $\mathbf{p}=-i\nabla$ is the momentum operator; $m$ is the rest mass; and $\psi$ is a four-component column spinor. Physically, $H$ is linear in $\mathbf{p}$ — the property that makes the equation first order in space as well as time, unlike the $\mathbf{p}^2$ of the nonrelativistic Hamiltonian. Throughout this chapter we work in natural units $\hbar=c=1$ (restore them by $m\to mc/\hbar$ in wavenumbers and $E\to E/\hbar$ in frequencies) and in the mostly-minus metric $g^{\mu\nu}=\mathrm{diag}(+1,-1,-1,-1)$ , with the Dirac representation of the gamma matrices. This convention is pinned in "The Klein-Gordon Equation" and used identically here and in every downstream RQM article. Recasting (1) in manifestly Lorentz-covariant notation absorbs the special role of time. Multiplying through by $\beta$ and defining $\gamma^0=\beta$ , $\gamma^i=\beta\alpha^i$ produces the compact form that treats all four spacetime derivatives on the same footing.

(i\gamma^\mu\partial_\mu - m)\psi = 0 \qquad (2)

where: $\gamma^\mu$ are the four gamma matrices; $\partial_\mu=(\partial_t,\nabla)$ is the spacetime gradient; $m$ is the rest mass; and $\psi$ is the Dirac spinor field. The single covariant equation (2) contains the Hamiltonian form (1) as its $\gamma^0$ -projected time-evolution; the manifest index $\mu$ makes the transformation under boosts and rotations tractable, developed in "Lorentz Covariance of the Dirac Equation." The coefficients are not free. Both the anticommutation of the $\alpha^i,\beta$ in (1) and, equivalently, of the $\gamma^\mu$ in (2) are forced by the square condition. In covariant language the requirement is the Clifford algebra.

\{\gamma^\mu,\gamma^\nu\} = \gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 2g^{\mu\nu}\,\mathbb{1}_{4\times4} \qquad (3)

where: $g^{\mu\nu}=\mathrm{diag}(+,-,-,-)$ is the Minkowski metric and $\mathbb{1}_{4\times4}$ is the identity in spinor space. Equation (3) states that different gamma matrices anticommute while $(\gamma^0)^2=+1$ and $(\gamma^i)^2=-1$ . It is the defining relation of the Dirac algebra; its full structure — representations, trace identities, $\gamma^5$ — is owned by "Dirac Gamma Matrices and Clifford Algebra," which this article only invokes. A concrete realization makes (3) tangible. The Dirac (standard) representation builds each $4\times4$ gamma from $2\times2$ Pauli blocks.

\gamma^0 = \begin{pmatrix} \mathbb{1} & 0\\ 0 & -\mathbb{1}\end{pmatrix},\qquad \gamma^i = \begin{pmatrix} 0 & \sigma^i\\ -\sigma^i & 0\end{pmatrix} \qquad (4)

where: $\mathbb{1},0$ are $2\times2$ identity and zero blocks and $\sigma^i$ are the Pauli matrices. This is one choice among unitarily equivalent representations; the physics is representation-independent, but explicit spinor solutions (in "Dirac Spinors and Plane-Wave Solutions") are cleanest in this basis. Finally, the promised virtue: the conserved probability current. Because (1) is first order in time with a Hermitian $H$ , the continuity equation follows in three lines and the density is a sum of squares.

\rho = \psi^\dagger\psi \geq 0, \qquad \mathbf{j} = \psi^\dagger\boldsymbol{\alpha}\psi, \qquad \partial_t\rho + \nabla\cdot\mathbf{j} = 0 \qquad (5)

where: $\rho$ is the manifestly non-negative probability density, $\mathbf{j}$ the probability current, and $\psi^\dagger$ the Hermitian-conjugate row spinor. In covariant notation the same current is $j^\mu=\bar\psi\gamma^\mu\psi$ with the Dirac adjoint $\bar\psi\equiv\psi^\dagger\gamma^0$ , and $\partial_\mu j^\mu=0$ ; note $j^0=\bar\psi\gamma^0\psi=\psi^\dagger\psi$ recovers (5).

Derivation ASquaring the Dirac operator reproduces Klein-Gordon

We want to show that the Dirac equation is a genuine "square root" of the relativistic dispersion relation — that each component of $\psi$ separately satisfies the Klein–Gordon equation. The strategy is to apply the conjugate operator $(i\gamma^\nu\partial_\nu+m)$ to the Dirac equation; because the original expression is zero, the product is zero. The one thing to watch is that the second-derivative term must be symmetrized before the Clifford algebra can act, since $\partial_\mu\partial_\nu$ is symmetric but $\gamma^\mu\gamma^\nu$ is not. This is the calculation that pins down $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ as exactly the relation needed.

Step 1. Left-multiply $(i\gamma^\mu\partial_\mu-m)\psi=0$ by $(i\gamma^\nu\partial_\nu+m)$ : since the bracket acts on zero, $(i\gamma^\nu\partial_\nu+m)(i\gamma^\mu\partial_\mu-m)\psi=0$ . Why the conjugate operator: it is the "difference of squares" partner, chosen so the cross terms linear in $m$ cancel.

Step 2. Expand the product. The $+im\gamma^\mu\partial_\mu$ and $-im\gamma^\nu\partial_\nu$ terms cancel, leaving $-\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu\psi - m^2\psi=0$ .

Step 3. Symmetrize the derivative term. Because $\partial_\nu\partial_\mu=\partial_\mu\partial_\nu$ , only the symmetric part of the matrix product contributes: $\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu=\tfrac12\{\gamma^\nu,\gamma^\mu\}\partial_\nu\partial_\mu$ . Why this is legal: contracting an antisymmetric object with a symmetric one gives zero, so the antisymmetric half of $\gamma^\nu\gamma^\mu$ drops out identically.

Step 4. Insert the Clifford algebra (3): $\tfrac12\{\gamma^\nu,\gamma^\mu\}\partial_\nu\partial_\mu=g^{\nu\mu}\partial_\nu\partial_\mu=\partial^\mu\partial_\mu=\Box$ .

Step 5. Collect: $-(\Box+m^2)\psi=0$ . Every spinor component obeys the Klein–Gordon equation, so it carries the correct relativistic dispersion $E^2=\mathbf{p}^2+m^2$ — the equation whose form is owned by "The Klein-Gordon Equation," here recovered component-by-component. The anticommutation relation was precisely what made the cross-derivative collapse to $\Box$ . Verified: reproduced symbolically with the explicit Dirac-representation gammas in the companion verifier; each of the four diagonal components of $(i\gamma^\mu\partial_\mu+m)(i\gamma^\nu\partial_\nu-m)$ equals $-(\Box+m^2)$ and the off-diagonal entries vanish.

Go deeper · Derivation B · 1 minWhy the matrices must be four-dimensional

Derivation BWhy the matrices must be four-dimensional

The goal is to prove that no set of matrices below dimension four can carry the Dirac algebra, so the four-component spinor is forced, not chosen. The route is a pair of trace arguments: first that each $\gamma^\mu$ is traceless, then that its eigenvalues fix the dimension to be even, and finally that dimension two admits only three mutually anticommuting traceless matrices — one short of the four we need. Watch the sign bookkeeping: $(\gamma^0)^2=+1$ but $(\gamma^i)^2=-1$ , so their eigenvalues differ.

Step 1. Each gamma is traceless. For $\mu\neq\nu$ , $\gamma^\mu=-\gamma^\nu\gamma^\mu(\gamma^\nu)^{-1}$ (from anticommutation); taking the trace and using cyclicity, $\mathrm{Tr}\,\gamma^\mu=-\mathrm{Tr}\,\gamma^\mu$ , so $\mathrm{Tr}\,\gamma^\mu=0$ .

Step 2. Eigenvalues fix parity of the dimension. Since $(\gamma^0)^2=1$ its eigenvalues are $\pm1$ ; since $(\gamma^i)^2=-1$ its eigenvalues are $\pm i$ . Tracelessness means equal numbers of $+$ and $-$ eigenvalues, so the dimension $N$ must be even. Why even: an odd-dimensional matrix cannot have its $\pm$ eigenvalues balanced to give zero trace.

Step 3. Rule out $N=2$ . The only mutually anticommuting traceless Hermitian $2\times2$ matrices are the three Pauli matrices $\sigma^1,\sigma^2,\sigma^3$ . Dirac needs four anticommuting matrices ( $\gamma^0,\gamma^1,\gamma^2,\gamma^3$ , equivalently $\beta,\alpha^1,\alpha^2,\alpha^3$ ), and no fourth $2\times2$ matrix anticommutes with all three Paulis.

Step 4. $N=4$ works. The next even dimension admits the explicit Dirac representation (4), which one checks satisfies (3). Hence four is the minimal dimension and $\psi$ has four components. Verified: the verifier confirms only three independent anticommuting traceless $2\times2$ matrices exist (any candidate fourth is forced to be a multiple of $\mathbb{1}$ , which is not traceless), and that the explicit $4\times4$ set satisfies all sixteen relations of (3).

Go deeper · Derivation C · 1 minThe four components are not four particles

Derivation CThe four components are not four particles

Here the aim is a scoping result: to show that the four-component structure is an internal (spinor) index, not a label for four distinct particles, by counting the genuine physical degrees of freedom. The route is to Fourier-transform the free equation and count independent plane-wave solutions at fixed momentum. The subtlety is that the equation is a constraint, not four independent Schrödinger equations, so it halves the naive count.

Step 1. Insert a plane wave $\psi=w(p)\,e^{-ip\cdot x}$ into (2): $(\gamma^\mu p_\mu-m)w(p)=0$ , an algebraic $4\times4$ eigenvalue condition, i.e. $(p\!\!\!/-m)w=0$ with $p\!\!\!/\equiv\gamma^\mu p_\mu$ .

Step 2. The matrix $p\!\!\!/$ satisfies $p\!\!\!/^2=p^\mu p_\mu\,\mathbb{1}=p^2\,\mathbb{1}$ (again from the Clifford algebra), so on-shell $p^2=m^2$ its eigenvalues are $\pm m$ , each doubly degenerate.

Step 3. For the physical mass shell there are exactly two solutions with eigenvalue $+m$ (the $u$ -spinors, positive energy) and two with $-m$ (the $v$ -spinors, negative energy). Four components, but only $2+2$ independent solutions per momentum. Why this matters: the "extra" doubling is spin (two states) times the particle/antiparticle split (two branches) — internal structure, not a multiplicity of particles.

Step 4. The explicit $u,v$ spinors, their normalizations, and the negative-energy interpretation are developed in "Dirac Spinors and Plane-Wave Solutions" and "Negative-Energy Solutions and the Interpretive Crisis"; here we only establish the count that the four-component object is minimal and non-redundant. Verified: the verifier diagonalizes $p\!\!\!/$ for a concrete on-shell $p^\mu$ and finds eigenvalues $\{+m,+m,-m,-m\}$ , confirming the $2+2$ split.

Limiting cases

The free Dirac equation reduces cleanly in several regimes; each limit is a check on the construction.

Case	Result	Meaning
Square of the operator	$(i\gamma^\mu\partial_\mu+m)(i\gamma^\nu\partial_\nu-m)\psi=-(\Box+m^2)\psi=0$	Every component satisfies Klein–Gordon, confirming the relativistic dispersion $E^2=\mathbf{p}^2+m^2$ .
Massless limit $m\to0$	$i\gamma^\mu\partial_\mu\psi=0$ decouples into two Weyl equations	With no mass to link them, the equation splits into two independent two-component equations of definite chirality (handedness).
Nonrelativistic limit	Reduces to a two-component Pauli-type equation	Eliminating the small components at leading order in $v/c$ yields a Schrödinger-like equation for a two-spinor; the magnetic moment and $g=2$ emerge — developed in "Nonrelativistic Reduction and the Pauli Equation."
Strong-field / Klein regime	Single-particle picture fails for $V_0\gtrsim2m$	Potential steps exceeding $2m$ trigger the Klein paradox; a consistent treatment needs quantum field theory and pair creation.

Worked examples

Example 1 — Verify the Clifford algebra in the Dirac representation. Confirm that the explicit matrices (4) satisfy $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ for the representative pairs $(\mu,\nu)=(0,0),(1,1),(0,1)$ .

Step 1. Take $\mu=\nu=0$ . With $\gamma^0=\mathrm{diag}(\mathbb{1},-\mathbb{1})$ , $(\gamma^0)^2=\mathrm{diag}(\mathbb{1},\mathbb{1})=\mathbb{1}_{4}$ , so $\{\gamma^0,\gamma^0\}=2\mathbb{1}=2g^{00}\mathbb{1}$ with $g^{00}=+1$ . Consistent.

Step 2. Take $\mu=\nu=1$ . Then $(\gamma^1)^2=\begin{pmatrix}0&\sigma^1\\-\sigma^1&0\end{pmatrix}^2=\begin{pmatrix}-(\sigma^1)^2&0\\0&-(\sigma^1)^2\end{pmatrix}=-\mathbb{1}_4$ , using $(\sigma^1)^2=\mathbb{1}_2$ . So $\{\gamma^1,\gamma^1\}=-2\mathbb{1}=2g^{11}\mathbb{1}$ with $g^{11}=-1$ . Consistent.

Step 3. Take $\mu=0,\nu=1$ . Compute $\gamma^0\gamma^1=\begin{pmatrix}0&\sigma^1\\ \sigma^1&0\end{pmatrix}$ and $\gamma^1\gamma^0=\begin{pmatrix}0&-\sigma^1\\-\sigma^1&0\end{pmatrix}$ , whose sum is the zero matrix, so $\{\gamma^0,\gamma^1\}=0=2g^{01}\mathbb{1}$ with $g^{01}=0$ . Consistent. The full set of sixteen relations is checked symbolically in the verifier; the answer is that the Dirac-representation matrices realize the Clifford algebra exactly.

Example 2 — The mass gap and Compton scale. For an electron, evaluate the rest energy $mc^2$ , the gap $2mc^2$ between the positive- and negative-energy branches, and the reduced Compton wavelength $\lambdabar_C=\hbar/mc$ that sets the length scale of the Dirac equation.

Step 1. Rest energy. With $m_e=9.109\times10^{-31}\,\mathrm{kg}$ and $c=2.998\times10^8\,\mathrm{m/s}$ , $m_ec^2=(9.109\times10^{-31})(2.998\times10^8)^2\,\mathrm{J}=8.187\times10^{-14}\,\mathrm{J}$ . Dividing by $1.602\times10^{-19}\,\mathrm{J/eV}$ gives $m_ec^2=5.11\times10^{5}\,\mathrm{eV}=\mathbf{0.511\ MeV}$ .

Step 2. The branch gap. The two roots $E=\pm\sqrt{\mathbf{p}^2c^2+m^2c^4}$ are separated at $\mathbf{p}=0$ by $2m_ec^2=\mathbf{1.022\ MeV}$ — the minimum energy to create an electron–positron pair, and the scale at which the single-particle picture breaks down.

Step 3. Compton scale. $\lambdabar_C=\hbar/(m_ec)=(1.055\times10^{-34})/[(9.109\times10^{-31})(2.998\times10^8)]\,\mathrm{m}=3.86\times10^{-13}\,\mathrm{m}=\mathbf{386\ fm}$ . Below this length, localizing an electron injects energy $\gtrsim mc^2$ and mixes in the negative-energy solutions — the physical reason the Dirac equation cannot be a strict single-particle theory at short distances. These numbers are recomputed in the verifier; the takeaways are $0.511$ MeV rest energy, a $1.022$ MeV pair-creation gap, and a $386$ fm Compton scale.

Origins · 1 minpaul dirac and how this was discovered

paul dirac

1902–1984 · British theoretical physicist who unified quantum mechanics and special relativity in a single wave equation

Paul Adrien Maurice Dirac wrote down his equation in early 1928 while a Fellow at St John's College, Cambridge, seeking a first-order relativistic equation with a positive probability density. The construction forced spin- $\tfrac12$ and, more startlingly, negative-energy solutions he interpreted through the "hole" theory that predicted the positron. Famously reticent, Dirac valued mathematical beauty as a guide to physical truth. He shared the 1933 Nobel Prize in Physics with Erwin Schrödinger "for the discovery of new productive forms of atomic theory."

History

1926

— Schrödinger derives what is now the Klein–Gordon equation but discards it over a hydrogen fine-structure mismatch, publishing the nonrelativistic equation instead

1928

— Dirac publishes "The Quantum Theory of the Electron" (Proc. R. Soc. A 117, 610), introducing the linear relativistic equation

1930

— Dirac proposes hole theory, interpreting the negative-energy sea and predicting an antiparticle

1932

— Anderson discovers the positron in cloud-chamber cosmic-ray tracks (Phys. Rev. 43, 491), confirming Dirac's prediction

1933

— Dirac shares the Nobel Prize in Physics with Schrödinger

Going further

References. Dirac, "The Quantum Theory of the Electron," Proc. R. Soc. A 117, 610 (1928) · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., Ch. 8 · Bjorken & Drell, Relativistic Quantum Mechanics, Ch. 1–3 · Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 7 · Peskin & Schroeder, An Introduction to QFT, Ch. 3.

Provenance. A verifier should reproduce: (1) the Clifford algebra by squaring $H=\boldsymbol{\alpha}\cdot\mathbf{p}+\beta m$ and by explicit Dirac-representation matrices; (2) the Klein–Gordon recovery $(i\gamma^\nu\partial_\nu+m)(i\gamma^\mu\partial_\mu-m)\psi=-(\Box+m^2)\psi$ component by component; (3) the minimal-dimension counting argument; (4) the $2+2$ solution count from the eigenvalues $\{+m,+m,-m,-m\}$ of $p\!\!\!/$ on shell; (5) the numerical values $m_ec^2=0.511$ MeV, gap $1.022$ MeV, $\lambdabar_C=386$ fm. The positron mass equals the electron mass (Anderson 1932; PDG).

Builds on: The Klein-Gordon Equation · Klein-Gordon Probability Current and Negative Densities · Negative-Energy Solutions and the Interpretive Crisis. Leads to: Dirac Gamma Matrices and Clifford Algebra · Dirac Spinors and Plane-Wave Solutions · Lorentz Covariance of the Dirac Equation · Spin from the Dirac Equation · The Dirac Sea and Antiparticles · Nonrelativistic Reduction and the Pauli Equation.

Is the Dirac equation still a "single-particle" wave equation, like Schrödinger's?

Only approximately, and only at low energies. It has a positive density $\rho=\psi^\dagger\psi$ , which makes single-particle probability language work well when energies stay below $2mc^2$ . But the negative-energy solutions never go away, and localizing the particle to within a Compton wavelength injects enough energy to excite them (and, physically, to create pairs). The honest modern reading is that the Dirac equation is the classical field equation of a quantum field; the negative-energy solutions become antiparticle modes upon quantization. So treat it as single-particle for atomic fine structure, but not for high-energy or strongly localized problems.

Why "take the square root" instead of just using

\sqrt{\mathbf{p}^2+m^2}

as the Hamiltonian?

You can write $H=\sqrt{\mathbf{p}^2+m^2}$ formally, but as a differential operator it is nonlocal: expanded in powers of $\mathbf{p}^2/m^2$ it contains derivatives of arbitrarily high order, which spoils causality and manifest Lorentz covariance. Dirac's insight was to keep the Hamiltonian linear in $\mathbf{p}$ — a genuine local first-order operator — at the price of matrix-valued coefficients. The matrices are the bookkeeping that lets a linear expression square to the correct quadratic dispersion. So the spinor structure is exactly the cost of having a local, first-order, relativistic wave equation.

Does curing the negative-probability problem also remove the negative-energy solutions?

No — these are two separate issues. The Dirac equation fixes the density problem: because it is first order in time, $\rho=\psi^\dagger\psi\ge0$ automatically. But the spectrum still contains negative-energy solutions, because each component satisfies Klein–Gordon and $E^2=\mathbf{p}^2+m^2$ has both roots $E=\pm\sqrt{\mathbf{p}^2+m^2}$ . Dirac dealt with those separately through the filled-sea/hole picture, which predicted the positron. The interpretive crisis of the negative-energy branch is developed in "Negative-Energy Solutions and the Interpretive Crisis," and its resolution in "The Dirac Sea and Antiparticles."

Where does spin-

\tfrac12

come from — is it put in by hand?

It is forced, not assumed. In the free equation the four-component structure is required by the algebra, and when you ask which angular momentum is conserved you find orbital $\mathbf{L}$ alone does not commute with the Dirac Hamiltonian; a spin term with a coefficient of exactly $\tfrac12$ must be added for the total to be conserved. The full argument — the commutators, the block-Pauli operator $\boldsymbol{\Sigma}$ , and the emergence of $g=2$ — is developed in "Spin from the Dirac Equation." This article stops at establishing that the four components carry an internal (spinor) index; identifying that index with spin- $\tfrac12$ is the job of the linked article.

Why choose the mostly-minus metric and the Dirac representation? Do results depend on them?

The physics is convention-independent, but you must fix a convention to write signs consistently. We use $g=\mathrm{diag}(+,-,-,-)$ and the Dirac representation of the gammas throughout the chapter, pinned in "The Klein-Gordon Equation." The alternative mostly-plus metric flips the sign in $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ and rescales masses; the Weyl or Majorana representations rearrange the components. Physical predictions — cross sections, spectra, the $2+2$ solution count — are identical. The one discipline that matters is not mixing conventions mid-calculation, which is a classic source of stray sign errors.

Is the covariant equation

(i\gamma^\mu\partial_\mu-m)\psi=0

really the same as the Hamiltonian form?

Yes — they are related by multiplying by $\gamma^0=\beta$ . Starting from $i\partial_t\psi=(\boldsymbol{\alpha}\cdot\mathbf{p}+\beta m)\psi$ , insert $\mathbf{p}=-i\nabla$ , multiply on the left by $\beta$ , and use $\gamma^0=\beta$ , $\gamma^i=\beta\alpha^i$ ; the result is $i\gamma^\mu\partial_\mu\psi=m\psi$ , i.e. the covariant form. The Hamiltonian form singles out time and is convenient for stating the conserved density; the covariant form treats all four derivatives symmetrically and is the natural starting point for discussing Lorentz covariance and building field theory. They contain identical physics.

What is the significance of the four gamma matrices anticommuting?

Anticommutation is the algebraic heart of the whole construction. It is what makes the linear Dirac operator square to the quadratic Klein–Gordon operator (the cross terms cancel), it forces the matrices to be at least $4\times4$ , and it is the defining relation $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ of the Clifford algebra of Minkowski space. That algebra is a self-contained structure with its own representations, trace identities, and the chirality matrix $\gamma^5$ — all owned and developed in "Dirac Gamma Matrices and Clifford Algebra." Here we only need the single fact that different gammas anticommute and $(\gamma^0)^2=+1$ , $(\gamma^i)^2=-1$ .