Green Functions

A Green function is an integral kernel that solves an inhomogeneous linear problem by inverting an operator with specified boundary conditions. For a linear differential operator $L$, the Green function $G(x,\xi)$ is defined schematically by

$$L_xG(x,\xi)=\delta(x-\xi),$$

together with the boundary conditions in the $x$ variable. If

$$Lu=f,$$

then the solution is often written as

$$u(x)=\int G(x,\xi)f(\xi)\,d\xi,$$

with the integration region and measure determined by the problem.

The phrase “the Green function” is incomplete until the operator, domain, boundary conditions, and convention are specified.

Matrix Analogy

In finite dimensions, solving

$$A\mathbf u=\mathbf f$$

with invertible $A$ gives

$$\mathbf u=A^{-1}\mathbf f.$$

The entries of $A^{-1}$ are the finite-dimensional analogue of a Green function:

$$u_i=\sum_j (A^{-1})_{ij}f_j.$$

The continuous formula

$$u(x)=\int G(x,\xi)f(\xi)\,d\xi$$

is the same idea with a sum replaced by an integral and the inverse matrix replaced by an integral kernel. When $L$ is not invertible on the chosen space, one must modify the problem: add boundary conditions, project out zero modes, impose a radiation prescription, or choose a generalized inverse.

Boundary Conditions Are Part of the Definition

For differential operators, the same local formula can have different Green functions. The operator

$$L=-\frac{d^2}{dx^2}$$

has different Green functions on the full line, on a finite interval with Dirichlet boundary conditions, on a finite interval with Neumann boundary conditions, and on a circle with periodic boundary conditions.

For a boundary-value problem on an interval, $G(x,\xi)$ must satisfy:

• the inhomogeneous equation $L_xG(x,\xi)=\delta(x-\xi)$;
• the same boundary conditions in $x$ that $u(x)$ must satisfy;
• continuity and derivative-jump conditions implied by the delta source.

This is why Green functions are closely tied to Boundary Conditions and Eigenvalue Problems.

Dirichlet Interval Example

Consider

$$-u''(x)=f(x), \qquad 0<x<L,$$

with

$$u(0)=u(L)=0.$$

The Green function satisfies

$$-\frac{d^2}{dx^2}G(x,\xi)=\delta(x-\xi), \qquad G(0,\xi)=G(L,\xi)=0.$$

Away from $x=\xi$, the second derivative is zero, so $G$ is linear on each side. The result is

$$G(x,\xi) = \begin{cases} \dfrac{x(L-\xi)}{L}, & 0\leq x\leq \xi,\\ \dfrac{\xi(L-x)}{L}, & \xi\le x\le L. \end{cases}$$

It is continuous at $x=\xi$ and has derivative jump

$$\frac{\partial G}{\partial x}(\xi^+,\xi) - \frac{\partial G}{\partial x}(\xi^-,\xi) =-1.$$

That jump is exactly what produces the delta distribution after applying $-d^2/dx^2$. The solution is

$$u(x)=\int_0^L G(x,\xi)f(\xi)\,d\xi.$$

This example also shows why the Green function is not just a formula for an inverse differential expression. It encodes the endpoints.

The distributional rule behind derivative jumps is developed in Distributional Derivatives.

Spectral Representation

When an operator has a complete orthonormal set of eigenfunctions,

$$L\phi_n=\lambda_n\phi_n,$$

with nonzero $\lambda_n$, a Green function can often be written as

$$G(x,\xi) = \sum_n \frac{\phi_n(x)\phi_n^*(\xi)}{\lambda_n}.$$

This is the infinite-dimensional analogue of diagonalizing a matrix before inverting it. If an eigenvalue is zero, the inverse is singular unless the source is orthogonal to the zero mode or the problem is modified.

For Hamiltonians, the resolvent is the spectral inverse

$$R(z)=(z-H)^{-1},$$

defined for complex $z$ outside the spectrum of $H$. Its coordinate-space kernel is

$$G(x,\xi;z) = \langle x\rvert (z-H)^{-1}\lvert \xi\rangle.$$

For a purely discrete spectrum,

$$G(x,\xi;z) = \sum_n \frac{\psi_n(x)\psi_n^*(\xi)}{z-E_n}.$$

The poles occur at eigenvalues. In continuous spectra, the same idea becomes a spectral integral with branch cuts and boundary-value prescriptions.

Resolvent Intuition

The resolvent asks how the system responds to a source at a chosen spectral parameter. Formally, if

$$(z-H)u=f,$$

then

$$u=(z-H)^{-1}f.$$

Near an eigenvalue, the response can become large because the denominator $z-E_n$ is small. This is the source of poles in bound-state Green functions and resonant behavior after analytic continuation.

For real energies in the continuous spectrum, one usually writes

$$G^{(\pm)}(E) = \lim_{\epsilon\to0^+} \frac{1}{E-H\pm i\epsilon}.$$

The signs are not decorative. In scattering, they select outgoing or incoming boundary conditions. The Lippmann–Schwinger Equation uses exactly this resolvent prescription.

At real spectral points, the same $i0$ notation also has a distributional meaning: it splits the boundary value into a principal-value part and an on-shell delta part. See Principal Value Distributions for the identity behind this shorthand.

Fourier-Space Construction

For translation-invariant problems, Fourier transformation often constructs the Green function. On the real line, let

$$L=-\frac{d^2}{dx^2}+\kappa^2, \qquad \kappa>0.$$

The Green function for decay at infinity satisfies

$$\left( -\frac{d^2}{dx^2}+\kappa^2 \right)G(x)=\delta(x).$$

Fourier transformation gives

$$G(x) = \int_{-\infty}^{\infty} \frac{dk}{2\pi} \frac{e^{ikx}}{k^2+\kappa^2} = \frac{e^{-\kappa\lvert x\rvert}}{2\kappa}.$$

The pole locations in the complex $k$ plane and the choice of contour encode the decay condition. More advanced Green-function calculations therefore use Fourier Transform, Contour Integration, and Complex Analysis Essentials.

For translation-invariant problems, the solution can also be read as a Convolution of the Green function with the source.

Relation to Propagators

The time-evolution kernel

$$K(x,t;\xi,0) = \langle x\rvert e^{-iHt/\hbar}\lvert \xi\rangle$$

evolves wavefunctions in time. A resolvent Green function is an energy-domain inverse. They are related by transforms. With a common convention,

$$(E-H+i0)^{-1} = -\frac{i}{\hbar} \int_0^\infty e^{i(E-H+i0)t/\hbar}\,dt.$$

Taking coordinate-space matrix elements relates $G^{(+)}(x,\xi;E)$ to a one-sided time transform of the propagator kernel. The convention and sign of $i0$ determine whether the Green function is retarded, advanced, outgoing, incoming, or Feynman-like in a field-theory context.

For the time-domain kernel itself, see Propagator Kernel. The important distinction is:

• a propagator kernel evolves initial data in time;
• a Green function solves an inhomogeneous equation or resolvent equation with a specified boundary prescription.

They are related, but they should not be treated as interchangeable words without stating the convention.

Common Mistakes

• Writing $G$ without saying which operator and boundary conditions define it.
• Confusing an integral kernel with the null-space meaning of the word “kernel”.
• Forgetting that Neumann and periodic problems may have zero modes, so the inverse may not exist on all sources.
• Treating the $i0$ prescription as harmless notation rather than a boundary condition.
• Mixing time-domain propagators with energy-domain resolvents.
• Ignoring the measure when writing Green functions in curvilinear coordinates.
• Assuming a spectral sum over normalizable eigenfunctions covers continuous spectra.

Cross-Links

• Partial Differential Equations
• Boundary Conditions
• Eigenvalue Problems
• Delta Function
• Fourier Transform
• Contour Integration
• Branch Cuts
• Convolution
• Distributional Derivatives
• Principal Value Distributions
• Complex Analysis Essentials
• Spectral Theorem, Practical Version
• Propagator Kernel
• Lippmann–Schwinger Equation

References

• G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2013.
• G. F. Roach, Green’s Functions, 2nd ed., Cambridge University Press, 1982.
• E. N. Economou, Green’s Functions in Quantum Physics, 3rd ed., Springer, 2006.
• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.

Exercises

1. Let

$$A= \begin{pmatrix} 2 & 0\\ 0 & 5 \end{pmatrix}.$$

What is the finite-dimensional Green-function analogue for solving $A\mathbf u=\mathbf f$?

Solution

The inverse is

$$A^{-1} = \begin{pmatrix} 1/2 & 0\\ 0 & 1/5 \end{pmatrix}.$$

Thus

$$\mathbf u=A^{-1}\mathbf f.$$

The entries of $A^{-1}$ play the role of the Green-function kernel.

2. For the Dirichlet interval Green function above, verify the derivative jump at $x=\xi$.

Solution

For $x\lt\xi$,

$$G(x,\xi)=\frac{x(L-\xi)}{L}, \qquad \frac{\partial G}{\partial x} = \frac{L-\xi}{L}.$$

For $x>\xi$,

$$G(x,\xi)=\frac{\xi(L-x)}{L}, \qquad \frac{\partial G}{\partial x} = -\frac{\xi}{L}.$$

Therefore

$$\frac{\partial G}{\partial x}(\xi^+,\xi) - \frac{\partial G}{\partial x}(\xi^-,\xi) = -\frac{\xi}{L} -\frac{L-\xi}{L} =-1.$$

3. If $H\psi_n=E_n\psi_n$ with a discrete orthonormal basis, why does the resolvent Green function have poles at $z=E_n$?

Solution

The spectral representation is

$$G(x,\xi;z) = \sum_n \frac{\psi_n(x)\psi_n^*(\xi)}{z-E_n}.$$

Each term has a denominator $z-E_n$, so the resolvent is singular at the eigenvalues.

4. Why can two Green functions for the same differential expression differ?

Solution

A Green function is tied to an operator, not just a local differential expression. Changing boundary conditions, domains, or radiation prescriptions changes the inverse problem. The same expression, such as $-d^2/dx^2$, can therefore produce different Green functions on different domains or with different endpoint conditions.