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.

Green Functions as Inverse Matrices

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

• the inhomogeneous equation $L_xG(x,\xi)=\delta(x-\xi)$;
• the corresponding homogeneous boundary conditions in $x$ (nonzero boundary values require an additional homogeneous-equation solution);
• 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.

Neumann Boundary Conditions and the Zero Mode

Now keep the same equation $-u''=f$ on $[0,L]$ but impose $u'(0)=u'(L)=0$. A constant is a zero mode, so this operator has no inverse on the full space. Integrating the equation gives the compatibility condition

$$\int_0^L f(x)\,dx=-u'(L)+u'(0)=0.$$

When this condition holds, adding a constant to any solution gives another solution. We fix that freedom by requiring $\int_0^L u(x)\,dx=0$.

The normalized zero mode is $\phi_0=1/\sqrt L$, whose projector has kernel $1/L$. The reduced Green function inverts the operator on the subspace orthogonal to this mode:

$$-\partial_x^2G_N(x,\xi)=\delta(x-\xi)-\frac1L, \qquad \partial_xG_N(0,\xi)=\partial_xG_N(L,\xi)=0,$$

with $\int_0^L G_N(x,\xi)\,dx=0$. The compensating constant makes the integrated right-hand side vanish, as the boundary conditions require. This zero-mode removal and mean-zero normalization are the standard Neumann construction (DiBenedetto, 2010, Chapter 3, §7).

For $0<\xi<L$, the explicit kernel is

$$G_N(x,\xi)=\frac{x^2+\xi^2}{2L}-\max(x,\xi)+\frac L3.$$

To check it, its $x$ derivative is $x/L$ for $x<\xi$ and $x/L-1$ for $x>\xi$. The endpoint derivatives vanish, the derivative jumps by $-1$, and the ordinary second derivative is $1/L$ away from the source. Integrating the kernel over $x$ gives zero. It is also symmetric under $x\leftrightarrow\xi$.

Consequently,

$$u(x)=\int_0^L G_N(x,\xi)f(\xi)\,d\xi \quad\Longrightarrow\quad -u''(x)=f(x)-\overline f, \qquad \overline f=\frac1L\int_0^L f(\xi)\,d\xi.$$

For a compatible source, this is the unique mean-zero solution. For an incompatible source, the same integral solves the equation with the constant component removed; it does not solve the original equation. In particular, $f=1$ has no solution with homogeneous Neumann data, even though the reduced-kernel integral is well defined and gives zero.

This distinction is the continuous version of applying a matrix pseudoinverse: it inverts nonzero modes and projects away the null space. The Neumann operator domain specifies the corresponding self-adjoint realization.

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, an inverse on the full space does not exist. A source orthogonal to every zero mode may still admit solutions, but adding a zero mode makes the solution nonunique. A reduced inverse becomes unique after restricting both source and solution to the orthogonal complement of the zero modes.

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,

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

Isolated eigenvalues give operator-valued poles whose residues are the corresponding spectral projectors; an individual matrix element may have zero residue. Continuous spectra instead require spectral integrals and boundary-value prescriptions. Familiar free and scattering Green functions often express this structure through branch cuts.

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. Here the positive $i0$ gives the retarded time response and, for the usual stationary Schrödinger scattering problem, outgoing waves. A Feynman prescription involves additional time-ordering structure and should not be inferred from an unspecified sign alone.

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.

Related Topics

• Partial Differential Equations
• Delta Function
• Spectral Theorem, Practical Version

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.

5. On $[0,L]$, solve $-u''=2x/L-1$ with $u'(0)=u'(L)=0$ and zero mean. Then explain what the reduced Green function returns if the source is changed to $2x/L$.

Solution

The first source has zero integral. Integrating twice and imposing the endpoint conditions and zero mean gives

$$u(x)=-\frac{x^3}{3L}+\frac{x^2}{2}-\frac{L^2}{12}, \qquad u'(x)=x-\frac{x^2}{L}.$$

The derivative vanishes at both endpoints, $-u''=2x/L-1$, and the integral of $u$ is zero. The reduced-kernel integral therefore gives this same unique solution.

The second source has mean one. Subtracting its mean leaves $2x/L-1$, so the reduced Green function returns the same $u$. It solves the projected equation; the original equation $-u''=2x/L$ has no solution with these Neumann conditions because its source has nonzero integral. Adding a constant to $u$ cannot repair that incompatibility.

References

• G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2013.
• E. DiBenedetto, Partial Differential Equations, 2nd ed., Birkhäuser, 2010, Chapter 3, §7.
• E. N. Economou, Green’s Functions in Quantum Physics, 3rd ed., Springer, 2006.
• G. F. Roach, Green’s Functions, 2nd ed., Cambridge University Press, 1982.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.
• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.