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Green's Functions

Source lecture(s): PHY621 Ch. 7

Intuition

The master key of the course, and the idea is disarmingly simple: if you know the response to a single sharp kick, you know the response to anything. Any source is a superposition of kicks, so by linearity the response is the same superposition of kick-responses.

\[\mathcal{L}G(x,x') = \delta(x - x') \qquad\Longrightarrow\qquad u(x) = \int G(x,x')f(x')\,dx' \ \text{ solves }\ \mathcal{L}u = f\]

Verify it in one line: \(\mathcal{L}u = \int\mathcal{L}G\,f\,dx' = \int\delta(x-x')f(x')dx' = f(x)\).

\(G\) is the inverse of a differential operator — \(\mathcal{L}^{-1}\) written as an integral kernel. That is the entire content, and it is why linear problems are the ones physics can solve.

Building one: the driven oscillator

For \(\mathcal{L} = d^2/dt^2 + \omega_0^2\) with causal (retarded) boundary conditions:

\[G(t,t') = \frac{\sin\omega_0(t-t')}{\omega_0}\,\Theta(t-t')\]

Read off the construction rules:

  • \(\Theta(t-t')\) enforces causality — no response before the kick.
  • \(G\) is continuous at \(t=t'\) (the mass cannot teleport).
  • \(\partial_t G\) jumps by 1 across \(t = t'\) — the impulse delivers momentum. Integrating \(\mathcal{L}G = \delta\) across the singularity gives exactly this jump, and it is what fixes the amplitude.

The response to any forcing is then \(u(t) = \int_{-\infty}^{t}G(t,t')f(t')\,dt'\) — a convolution, which is no accident: for a time-translation-invariant operator \(G\) depends only on \(t-t'\), and convolution is what translation invariance looks like.

Two routes to \(G\)

Eigenfunction expansion. If \(\mathcal{L}\phi_n = \lambda_n\phi_n\) with \(\{\phi_n\}\) complete and orthonormal,

\[G(x,x') = \sum_n \frac{\phi_n(x)\phi_n^*(x')}{\lambda_n}\]

which is just \(\mathcal{L}^{-1} = \sum\lambda_n^{-1}|n\rangle\langle n|\) — the spectral decomposition of the inverse. Note the danger signal: a zero eigenvalue makes this diverge, the analytic face of the null-space obstruction to inverting an operator. Resonance is a zero eigenvalue.

Transform methods. Fourier or Laplace transform the defining equation; \(\mathcal{L}\) becomes multiplication, so \(\tilde G = 1/\tilde{\mathcal{L}}\) and \(G\) is the inverse transform. The poles of \(\tilde G\) are the natural frequencies, and the contour you take around them is the boundary condition — retarded, advanced or Feynman. This is where PHY622's contour integration becomes indispensable.

Where it reappears

  • Electrostatics. \(G = 1/4\pi|\mathbf{r}-\mathbf{r}'|\) for the Laplacian: the potential of a point charge. Boundaries are handled by images — an image charge is a Green's function satisfying a boundary condition.
  • Quantum mechanics. The propagator \(\langle x|e^{-iHt/\hbar}|x'\rangle\) is a Green's function; perturbation theory expands in them; Feynman diagrams are pictures of their products.
  • Signal processing. The impulse response of a linear filter; its transform is the transfer function.
  • Plasma response. The dielectric function is the transform-space Green's function, and Landau damping is a statement about where its poles sit.

Common mistakes

  • Forgetting the boundary conditions. \(\mathcal{L}G = \delta\) has infinitely many solutions; the boundary conditions pick one. Retarded and advanced Green's functions solve the same equation and describe opposite physics.
  • Missing the derivative jump. Matching only the value across \(x'\) gives the wrong normalisation.
  • Using the eigenfunction expansion with a zero mode. It diverges — and that divergence is physical: the inhomogeneous problem is solvable only if the source is orthogonal to the zero mode (the Fredholm alternative).
  • Assuming \(G\) depends only on \(x - x'\). True only for translation-invariant operators; false as soon as there are boundaries.

Knowledge graph position

Prerequisites: Dirac delta, ODEs, transforms. Leads to: boundary-value problems, propagators, perturbation theory, scattering.

Quiz

Q1 (conceptual). Why does the derivative of \(G\) jump by 1 at \(x = x'\)?

Answer

Integrate \(\mathcal{L}G = \delta\) across an infinitesimal interval around \(x'\). For a second-order operator with unit leading coefficient only the highest derivative survives, giving \([\partial_xG] = 1\). Physically the impulse imparts a unit of momentum, so velocity changes discontinuously while position does not.

Q2 (conceptual). What goes wrong with \(G = \sum_n\phi_n\phi_n^*/\lambda_n\) when \(\lambda_k = 0\)?

Answer

It diverges — the operator is not invertible because it has a null vector. The inhomogeneous problem is then solvable only if the source is orthogonal to that zero mode (the Fredholm alternative), and the solution is unique only up to adding multiples of it. Driving a system exactly at resonance is this situation.

Q3 (MCQ). The retarded and advanced Green's functions for the wave equation differ by:

  • (a) the differential equation they satisfy
  • (b) the boundary conditions — equivalently, the contour taken around the poles
  • (c) a normalisation constant
  • (d) the source function
Answer

(b). Both satisfy \(\mathcal{L}G = \delta\). Which solution you get is fixed by how the inversion contour passes the poles, and that choice encodes whether effects follow causes or precede them.