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Quantum Mechanics

Covariant Dirac Equation

The free Dirac equation for a four-component spinor ψ(x)\psi(x) is

(iℏc γμ∂μ−mc2)ψ(x)=0.\boxed{ \left( i\hbar c\,\gamma^\mu\partial_\mu-mc^2 \right)\psi(x)=0 }.

It is first order in spacetime derivatives, yet its Clifford algebra ensures that every component satisfies the Klein–Gordon mass-shell equation. Lorentz covariance does not mean that ψ\psi is a four-vector: coordinates transform with Λ\Lambda, while spinor components transform with a matrix S(Λ)S(\Lambda) that intertwines Λ\Lambda with the gamma matrices. The resulting current has

j0=cψ†ψ≥0,j^0=c\psi^\dagger\psi\geq0,

which repairs the local-density problem of a one-particle Klein–Gordon amplitude. It does not remove the need for quantum fields when interactions can create particles and antiparticles.

Required background. Metric and Units, Four-Vectors, the Energy–Momentum Relation, Gamma Matrices, the Gamma-Matrix Conventions, and the Klein–Gordon Equation supply every convention and factorization used below.

Covariant operator and plane waves

Section titled “Covariant operator and plane waves”

Define the free Dirac operator

D=iℏc γμ∂μ−mc2.\mathcal D = i\hbar c\,\gamma^\mu\partial_\mu-mc^2.

For a constant spinor u(p)u(p) and the default plane wave,

ψ(x)=u(p)e−ip⋅x/ℏ,\psi(x)=u(p)e^{-ip\cdot x/\hbar},

the differential equation becomes the algebraic equation

(p ⁣ ⁣ ⁣/−mc)u(p)=0.(p\!\!\!/-mc)u(p)=0.

Multiplying from the left by p ⁣ ⁣ ⁣/+mcp\!\!\!/+mc gives

(p2−m2c2)u(p)=0.(p^2-m^2c^2)u(p)=0.

A nonzero plane-wave spinor is therefore on shell. The matrix equation also constrains its four components; they are not four independent scalar wave amplitudes.

The negative-frequency solutions are part of the free equation. In a fixed- particle treatment they signal the second energy sector. In the quantized Dirac field, the corresponding mode operators acquire the antiparticle interpretation without retaining an unbounded-below many-particle energy.

Squaring gives Klein–Gordon componentwise

Section titled “Squaring gives Klein–Gordon componentwise”

Multiply the equation by the conjugate algebraic factor:

(iℏc γν∂ν+mc2)(iℏc γμ∂μ−mc2)ψ=0.\left( i\hbar c\,\gamma^\nu\partial_\nu+mc^2 \right) \left( i\hbar c\,\gamma^\mu\partial_\mu-mc^2 \right)\psi=0.

The scalar mass commutes with the derivative operator, so the mixed terms cancel. Because partial derivatives commute,

γνγμ∂ν∂μ=12{γν,γμ}∂ν∂μ=□I4.\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu = \frac12\{\gamma^\nu,\gamma^\mu\} \partial_\nu\partial_\mu = \Box I_4.

The product is therefore

−ℏ2c2□−m2c4,-\hbar^2c^2\Box-m^2c^4,

and each spinor component obeys

(□+m2c2ℏ2)ψa=0.\left( \Box+\frac{m^2c^2}{\hbar^2} \right)\psi_a=0.

The converse is false. Four arbitrary Klein–Gordon solutions need not satisfy the first-order Dirac constraints. Squaring can introduce solutions of the second-order equation that the original factor does not annihilate.

Lorentz covariance as an intertwining relation

Section titled “Lorentz covariance as an intertwining relation”

Let a proper orthochronous Lorentz transformation act on coordinates as

x′μ=Λμνxν.x'^\mu=\Lambda^\mu{}_{\nu}x^\nu.

The spinor transforms as

ψ′(x′)=S(Λ)ψ(x),\psi'(x')=S(\Lambda)\psi(x),

where S(Λ)S(\Lambda) is chosen so that

S(Λ)−1γμS(Λ)=Λμνγν.S(\Lambda)^{-1}\gamma^\mu S(\Lambda) = \Lambda^\mu{}_{\nu}\gamma^\nu.

The derivative transforms with the inverse Lorentz matrix:

∂μ′=(Λ−1)νμ∂ν.\partial'_\mu = (\Lambda^{-1})^\nu{}_{\mu}\partial_\nu.

Since SS is constant for a global inertial-frame change,

γμ∂μ′ψ′(x′)=γμS(Λ−1)νμ∂νψ(x)=Sγν∂νψ(x).\begin{aligned} \gamma^\mu\partial'_\mu\psi'(x') &= \gamma^\mu S (\Lambda^{-1})^\nu{}_{\mu}\partial_\nu\psi(x) \\ &= S\gamma^\nu\partial_\nu\psi(x). \end{aligned}

Hence

D′ψ′(x′)=S(Λ)Dψ(x),\mathcal D'\psi'(x') = S(\Lambda)\mathcal D\psi(x),

so a solution in one inertial frame maps to a solution in every other.

Commuting diagram for Lorentz transformations of coordinates and spinors intertwined by the Dirac operator.

Lorentz covariance uses two linked representations. The vector matrix Λ\Lambda changes coordinates and derivatives; S(Λ)S(\Lambda) changes spinor components. The gamma-matrix intertwiner makes applying the Dirac operator before or after the transformation equivalent.

For transformations continuously connected to the identity, SS may be built from the generators σμν/2\sigma^{\mu\nu}/2. The exact exponential sign depends on the active/passive and parameter conventions, so the displayed intertwining identity is the authority used here.

The spinor representation is double valued over the Lorentz transformation: SS and −S-S induce the same Λ\Lambda. A 2π2\pi spatial rotation changes a spinor’s sign, while bilinears and physical rays return to themselves. This is the local algebraic trace of the double cover Spin+(1,3)≃SL(2,C)\mathrm{Spin}^+(1,3)\simeq SL(2,\mathbb C) of the proper orthochronous Lorentz group.

Adjoint equation and conserved current

Section titled “Adjoint equation and conserved current”

The Dirac adjoint is

ψˉ=ψ†γ0.\bar\psi=\psi^\dagger\gamma^0.

Take the Hermitian conjugate of the Dirac equation and use (γμ)†=γ0γμγ0(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0. The adjoint equation is

iℏc(∂μψˉ)γμ+mc2ψˉ=0.i\hbar c (\partial_\mu\bar\psi)\gamma^\mu +mc^2\bar\psi=0.

Multiplying the original equation on the left by ψˉ\bar\psi gives

iℏc ψˉγμ∂μψ−mc2ψˉψ=0.i\hbar c\,\bar\psi\gamma^\mu\partial_\mu\psi -mc^2\bar\psi\psi=0.

Multiplying the adjoint equation on the right by ψ\psi and adding cancels the mass terms:

∂μ(ψˉγμψ)=0.\partial_\mu(\bar\psi\gamma^\mu\psi)=0.

Define

jμ=cψˉγμψ.j^\mu=c\bar\psi\gamma^\mu\psi.

Then

∂μjμ=0,j0=cψ†γ0γ0ψ=cψ†ψ≥0.\partial_\mu j^\mu=0, \qquad j^0 = c\psi^\dagger\gamma^0\gamma^0\psi = c\psi^\dagger\psi \geq0.

The density ρ=j0/c=ψ†ψ\rho=j^0/c=\psi^\dagger\psi is pointwise nonnegative and the spatial current is

j=cψ†αψ.\mathbf j=c\psi^\dagger\boldsymbol\alpha\psi.

For solutions that decay sufficiently fast, or obey boundary conditions with zero net flux, integrating the continuity equation gives conservation of ∫d3x ψ†ψ\int d^3x\,\psi^\dagger\psi.

The matrix eigenvalues of each αi\alpha^i are ±1\pm1, and more generally ∣ψ†αψ∣≤ψ†ψ|\psi^\dagger\boldsymbol\alpha\psi|\leq\psi^\dagger\psi. Thus the current is causal: its local flow speed does not exceed cc.

Hamiltonian form and initial data

Section titled “Hamiltonian form and initial data”

Separate the time derivative and multiply by γ0\gamma^0:

iℏ∂tψ=HDψ,i\hbar\partial_t\psi = H_D\psi,

with

HD=cα⋅p+βmc2,p=−iℏ∇.H_D = c\boldsymbol\alpha\cdot\mathbf p +\beta mc^2, \qquad \mathbf p=-i\hbar\nabla.

This is the canonical owner of the free Dirac Hamiltonian used by the compact Dirac Hamiltonian card. The formula alone is not a self-adjoint operator: a Hilbert space, spatial domain, and boundary conditions must also be specified. On L2(R3,C4)L^2(\mathbb R^3,\mathbb C^4), the standard free realization is self-adjoint on H1(R3,C4)H^1(\mathbb R^3,\mathbb C^4). The detailed domain, spectral projectors, unitary evolution, and boundary-form analysis are developed in The Dirac Hamiltonian as an Operator.

Because the equation is first order in time, initial data consist of

ψ(t0,x)\psi(t_0,\mathbf x)

alone. Its time derivative is fixed by HDψH_D\psi. Specifying an independent ∂tψ(t0,x)\partial_t\psi(t_0,\mathbf x) would generally overdetermine the problem.

In momentum space, HD2=(p2c2+m2c4)I4H_D^2=(\mathbf p^2c^2+m^2c^4)I_4, so its free spectrum contains the two branches E±(p)E_\pm(\mathbf p). The positive density does not mean the negative-energy branch can be discarded locally without consequence; projection onto one branch is nonlocal in position space.

Fixed-particle scope and the QFT boundary

Section titled “Fixed-particle scope and the QFT boundary”

The free Dirac equation and weak prescribed external fields support highly accurate first-quantized calculations when pair creation is negligible. They describe spin, relativistic kinematics, currents, atomic fine structure, and the controlled Pauli limit.

They are not a complete interacting relativistic quantum theory. When appreciable pair-production channels are open, fixed particle number fails. An energy scale comparable to the particle–antiparticle gap is a warning, not by itself a sufficient criterion: field geometry, invariants, duration, and transition rates also matter. Quantum electrodynamics promotes ψ\psi to an operator-valued field; both frequency sectors are then necessary for locality, causal propagation, and antiparticle excitations.

Independently, the finite-dimensional Lorentz boost matrix need not be unitary in the ordinary Euclidean component norm, even in free one-particle theory. The conserved Hilbert-space norm is an integral on a chosen spacelike slice, while covariant bilinears use the Dirac adjoint.

Continue with Free Dirac Spinors for normalized modes and completeness, The Dirac Current for the Gordon decomposition and spin current, and Dirac Equation as Bridge for the additional structure required by quantum field theory.

Common pitfalls

Section titled “Common pitfalls”

Transforming a spinor as a four-vector. Four components do not determine the representation. Use S(Λ)S(\Lambda) and the gamma-matrix intertwiner.

Assuming S†S=IS^\dagger S=I for boosts. Spatial rotations are unitary in the finite spinor space; boosts are not. Covariant bilinears are preserved through ψˉ\bar\psi, not a Euclidean component norm at one event.

Forgetting the transformed argument. Covariance requires ψ′(x′)=Sψ(x)\psi'(x')=S\psi(x), not merely multiplication of components at an unchanged coordinate label.

Inferring the Dirac equation from Klein–Gordon alone. Squaring loses the first-order constraint. Every Dirac solution is Klein–Gordon componentwise, but not every four-tuple of Klein–Gordon solutions is a Dirac solution.

Exercises

Section titled “Exercises”

1. Operator factorization

Section titled “1. Operator factorization”

Multiply the two Dirac factors and recover the Klein–Gordon equation with the shared metric convention.

Solution

The mass cross terms cancel, and commuting derivatives remove the gamma-commutator contribution:

(iℏcγν∂ν+mc2)(iℏcγμ∂μ−mc2)=−ℏ2c2□−m2c4.\begin{aligned} &(i\hbar c\gamma^\nu\partial_\nu+mc^2) (i\hbar c\gamma^\mu\partial_\mu-mc^2) \\ &\qquad =- \hbar^2c^2\Box-m^2c^4. \end{aligned}

Dividing the resulting equation by −ℏ2c2-\hbar^2c^2 gives

(□+m2c2ℏ2)ψ=0.\left(\Box+\frac{m^2c^2}{\hbar^2}\right)\psi=0.

2. Current conservation

Section titled “2. Current conservation”

Derive the adjoint equation and use it to prove ∂μjμ=0\partial_\mu j^\mu=0.

Solution

Hermitian conjugation followed by right multiplication with γ0\gamma^0 gives

iℏc(∂μψˉ)γμ+mc2ψˉ=0.i\hbar c(\partial_\mu\bar\psi)\gamma^\mu+mc^2\bar\psi=0.

Multiplying by ψ\psi and adding the original equation multiplied by ψˉ\bar\psi cancels the mass terms and produces

iℏc ∂μ(ψˉγμψ)=0.i\hbar c\,\partial_\mu (\bar\psi\gamma^\mu\psi)=0.

3. Covariance check

Section titled “3. Covariance check”

Starting from the derivative transformation and intertwining identity, prove D′ψ′(x′)=SDψ(x)\mathcal D'\psi'(x')=S\mathcal D\psi(x).

Solution

Use

γμS=SΛμργρ\gamma^\mu S=S\Lambda^\mu{}_{\rho}\gamma^\rho

and contract with (Λ−1)νμ(\Lambda^{-1})^\nu{}_{\mu}:

γμS(Λ−1)νμ=Sγν.\gamma^\mu S(\Lambda^{-1})^\nu{}_{\mu} = S\gamma^\nu.

The mass term commutes with SS, so the full operator transforms in the stated way.

4. Hamiltonian form

Section titled “4. Hamiltonian form”

Derive HD=cα⋅p+βmc2H_D=c\boldsymbol\alpha\cdot\mathbf p+\beta mc^2 from the covariant equation.

Solution

Using ∂0=c−1∂t\partial_0=c^{-1}\partial_t, write

iℏγ0∂tψ+iℏcγi∂iψ−mc2ψ=0.i\hbar\gamma^0\partial_t\psi +i\hbar c\gamma^i\partial_i\psi -mc^2\psi=0.

Multiply by γ0\gamma^0 and use αi=γ0γi\alpha^i=\gamma^0\gamma^i, β=γ0\beta=\gamma^0, and pi=−iℏ∂ip_i=-i\hbar\partial_i to obtain

iℏ∂tψ=(cαipi+βmc2)ψ.i\hbar\partial_t\psi = (c\alpha^ip_i+\beta mc^2)\psi.

References

Section titled “References”