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Relativistic Green Functions

A relativistic Green function is specified by an operator, a delta source, and a boundary condition. In an external background it normally depends on two spacetime points separately. This page develops that two-point calculus and shows how it connects a forced Dirac equation to Hamiltonian evolution, energy-domain resolvents, and a background expansion. The explicit free scalar poles belong to Klein–Gordon Propagators.

Required background. Klein–Gordon Propagators fixes source and boundary conventions; The Dirac Hamiltonian as an Operator supplies the evolution operator. Helpful background. Resolvent Operator explains spectral inverses, and Minimal Coupling defines the background differential operator.

Two-point inverses and their source equations

Section titled “Two-point inverses and their source equations”

Use ℏ=c=1\hbar=c=1 and metric (+,−,−,−)(+,-,-,-). For a linear differential operator Dx\mathcal D_x acting on an rr-component column, a unit-source kernel obeys

DxKX(x,y)=δ(4)(x−y)Ir.\mathcal D_x K_X(x,y)=\delta^{(4)}(x-y)I_r.

The subscript XX specifies a boundary prescription. For suitable sources,

ψ(x)=ψhom(x)+∫d4y KX(x,y)J(y)\psi(x)=\psi_{\rm hom}(x)+\int d^4y\,K_X(x,y)J(y)

solves Dψ=J\mathcal D\psi=J, where Dψhom=0\mathcal D\psi_{\rm hom}=0. Initial data, incoming waves, or other boundary information determine the homogeneous part. An inverse chosen for one boundary problem need not implement another one.

For the scalar operator L=□+m2L=\Box+m^2, take r=1r=1 and denote the kernel by GXG_X. For the free Dirac operator D=iγμ∂μ−mD=i\gamma^\mu\partial_\mu-m, take r=4r=4 and write KXK_X. These unit-source definitions differ from some conventional field correlators: the scalar time-ordered DFD_F satisfies LDF=−iδ4LD_F=-i\delta^4, whereas the spinor time-ordered SFS_F satisfies DSF=+iδ4I4DS_F=+i\delta^4I_4. Multiplying a kernel by a constant changes its source normalization.

Translation invariance permits KX(x,y)=KX(x−y)K_X(x,y)=K_X(x-y) only when the operator and boundary choice share that invariance. A localized potential, time-dependent pulse, material boundary, or noninvariant state can destroy it. Even in a static potential, time translation may survive while spatial translation does not.

If two kernels solve the same unit-source equation, their difference satisfies

Dx(KX−KY)(x,y)=0.\mathcal D_x(K_X-K_Y)(x,y)=0.

The difference may still be nonzero: it records different homogeneous boundary data. The distributional inverse equation is therefore not enough to select a physical response.

The retarded Dirac inverse from unitary evolution

Section titled “The retarded Dirac inverse from unitary evolution”

Let a real prescribed background define a self-adjoint H(t)H(t) and a well-posed propagator U(t,t′)U(t,t'). For example, smooth bounded electromagnetic potentials on finite time intervals provide a useful controlled setting. Singular backgrounds require their own domain and existence analysis. Write

i∂tU(t,t′)=H(t)U(t,t′),U(t′,t′)=I.i\partial_tU(t,t')=H(t)U(t,t'), \qquad U(t',t')=I.

For an evolution source FF, (i∂t−H)ψ=F(i\partial_t-H)\psi=F, the retarded inverse is

RR(t,t′)=−iθ(t−t′)U(t,t′).R_R(t,t')=-i\theta(t-t')U(t,t').

Differentiating the step function yields (i∂t−H)RR=δ(t−t′)I(i\partial_t-H)R_R=\delta(t-t')I; the remaining terms cancel by the evolution equation. The spatial matrix elements of RRR_R include the spatial delta in this identity.

The covariant Dirac operator obeys DA=β(i∂t−H)D_A=\beta(i\partial_t-H), with β=γ0\beta=\gamma^0. Thus DAψ=JD_A\psi=J is the evolution equation with F=βJF=\beta J. Its spacetime inverse is consequently

KR(x,y)=−iθ(t−t′) ⟨x∣U(t,t′)∣y⟩β.K_R(x,y)=-i\theta(t-t')\, \langle\mathbf x|U(t,t')|\mathbf y\rangle\beta.

The β\beta is on the right. It acts on the source before propagation; moving it to the left is invalid unless the relevant propagator commutes with it. Across equal times the kernel has jump −iδ3(x−y)β-i\delta^3(\mathbf x-\mathbf y)\beta. The leading derivative iβ∂ti\beta\partial_t then gives the required +δ4I4+\delta^4I_4.

For t≥t0t\geq t_0, specified Cauchy data give

ψ(t)=U(t,t0)ψ(t0)−i∫t0tdt′ U(t,t′)βJ(t′).\psi(t)=U(t,t_0)\psi(t_0) -i\int_{t_0}^{t}dt'\,U(t,t')\beta J(t').

This formula distinguishes three objects: UU propagates initial data, RRR_R inverts the first-order evolution equation, and KRK_R inverts the covariant source equation. Only the last two contain a temporal step and delta-source jump. The second-order scalar problem needs two Cauchy data, as worked out in Solving Scalar Initial-Value Problems.

Static backgrounds and the spectral resolvent

Section titled “Static backgrounds and the spectral resolvent”

For time-independent self-adjoint HH, U(t,t′)=e−iH(t−t′)U(t,t')=e^{-iH(t-t')}. With transform R~(ω)=∫dt eiωtR(t,0)\widetilde R(\omega)=\int dt\,e^{i\omega t}R(t,0), the regulated retarded transform is

R~R(ω)=lim⁡ϵ↓0(ω+iϵ−H)−1.\widetilde R_R(\omega) =\lim_{\epsilon\downarrow0}(\omega+i\epsilon-H)^{-1}.

Before the limit, this is an ordinary bounded resolvent at a nonreal spectral parameter. On continuous spectrum its limit is generally a distributional or suitably weighted-space boundary value, not a bounded inverse on the original Hilbert space. Such limits require hypotheses. The advanced choice has ω−i0\omega-i0 instead.

For a static Dirac background,

K~R(ω;x,y)=⟨x∣(ω−H+i0)−1∣y⟩β.\widetilde K_R(\omega;\mathbf x,\mathbf y) =\langle\mathbf x|(\omega-H+i0)^{-1}|\mathbf y\rangle\beta.

A discrete bound state gives a pole, and a continuum gives a spectral discontinuity. This is the link to the canonical Resolvent Operator construction. A Feynman inverse imposes an additional positive/negative-frequency boundary selection; replacing every spectral denominator by the same +i0+i0 produces the retarded object instead.

Let D=D0+W\mathcal D=\mathcal D_0+W, where WW is a matrix multiplication operator for this discussion. Choose compatible inverses KK and K0K_0 with the same boundary problem. Algebraically,

K−K0=−K0WK,K=K0−K0WK.K-K_0=-K_0WK, \qquad K=K_0-K_0WK.

One way to obtain the identity is to apply D0\mathcal D_0 to the difference: D0(K−K0)=−WK\mathcal D_0(K-K_0)=-WK, then solve with the prescribed boundary inverse. This step requires the homogeneous boundary ambiguity to have been fixed. For retarded kernels and compactly supported sources, uniqueness of the causal problem supplies that condition.

Every product here is an ordered spacetime convolution. In components,

(K0WK)ab(x,y)=∫d4z (K0)ac(x,z)Wcd(z)Kdb(z,y).(K_0WK)_{ab}(x,y)= \int d^4z\,(K_0)_{ac}(x,z)W_{cd}(z)K_{db}(z,y).

For DA=iγμ∂μ−qγμAμ−mD_A=i\gamma^\mu\partial_\mu-q\gamma^\mu A_\mu-m, W=−qγμAμW=-q\gamma^\mu A_\mu, so iteration gives

KA=K0+qK0(γ⋅A)K0+q2K0(γ⋅A)K0(γ⋅A)K0+⋯ .\begin{aligned} K_A={}&K_0+qK_0(\gamma\cdot A)K_0\\ &+q^2K_0(\gamma\cdot A)K_0(\gamma\cdot A)K_0+\cdots. \end{aligned}

Matrix order matters. For retarded factors, only time-ordered chains y0≤z20≤z10≤x0y^0\leq z_2^0\leq z_1^0\leq x^0 contribute to the second correction. This is a source/background expansion; it does not quantize AμA_\mu or prove convergence of a scattering Born series.

For a bounded Hamiltonian perturbation V(t)V(t), the interaction-picture Dyson terms on a finite time interval are bounded by Λn/n!\Lambda^n/n!, where Λ=∫dt ∥V(t)∥\Lambda=\int dt\,\|V(t)\|. Thus the operator series for evolution converges there, and small Λ\Lambda controls a short truncation. An infinite-time Coulomb problem or an unbounded field is outside this simple estimate. Small charge alone is not a universal error bound.

  1. In a static problem replace H0H_0 by H0+vIH_0+vI, with real constant vv. Compare the exact retarded kernel with the first background correction.
Solution

U(t,t′)=e−iv(t−t′)U0(t,t′)U(t,t')=e^{-iv(t-t')}U_0(t,t'), so RR=RR,0[1−iv(t−t′)+O(v2(t−t′)2)]R_R=R_{R,0}[1-iv(t-t')+O(v^2(t-t')^2)] for t>t′t>t'. The energy resolvent is (ω−v−H0+i0)−1(\omega-v-H_0+i0)^{-1}. The expansion parameter on a finite interval is ∣v(t−t′)∣|v(t-t')|, not ∣v∣|v| without a time scale.

  1. Suppose JJ is supported after a time t0t_0. Can an arbitrary homogeneous solution be added to KRJK_RJ while retaining the same zero-past response condition?
Solution

No. For a well-posed hyperbolic Cauchy problem, a homogeneous solution with vanishing past data is zero. Homogeneous additions are available only when changing the boundary or initial data. The inverse equation alone misses precisely that uniqueness condition.

  1. Why is (D0+W)−1=D0−1(I+WD0−1)−1(D_0+W)^{-1}=D_0^{-1}(I+WD_0^{-1})^{-1} not an unconditional prescription on the real spectrum?
Solution

The inverses need compatible domains and boundary conditions. A real spectral value can lie in the spectrum; distributional boundary values require a specified limit, and their products may need further control. The identity is valid where the indicated inverses and products exist, or in a well-posed Green-operator setting. Formal notation does not establish those hypotheses.

  • Thaller, Bernd. The Dirac Equation. Springer, 1992. doi:10.1007/978-3-662-02753-0. Dirac operators, domains, and external fields.
  • Tong, David. Lectures on Quantum Field Theory. University of Cambridge, 2006, sections 2.7 and 5.4–5.5. Scalar kernels and spinor kernels.
  • Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995, chapter 6. Time ordering and perturbative evolution.