In a Green's function \(G(x,x')\), the primed variable \(x'\) marks the source point — the location of a unit impulse — while \(x\) is the field point where the response is measured. This can be seen as a matrix element \(\langle x | L^{-1} | x' \rangle\) of the inverse operator. For the 1D Poisson equation \(-\frac{d^2 u}{dx^2}=f\), we obtain \(G(x,x')=-\frac12|x-x'|\) using \(\frac{d^2}{dx^2}|x-x'|=2\delta(x-x')\). Additional examples illustrate the same concept.
Operator notation as matrix elements
Dirac notation often writes the Green’s function as \[ G(x,x') = \langle x | L^{-1} | x' \rangle. \] That’s exactly the structure of a matrix element. In a finite‑dimensional space with an orthonormal basis \(\{|i\rangle\}\), a matrix \(M\) has entries \[ M_{ij} = \langle i | M | j \rangle. \] Here \(\langle i|\) is a bra (row vector) and \(|j\rangle\) is a ket (column vector); \(i\) selects the row, \(j\) the column.
Now imagine a continuous basis \(\{|x\rangle\}\) labelled by real numbers. A linear differential operator \(L\) and its inverse (assuming it exists and is well‑behaved) can be expressed as an integral kernel \(G(x,x')\): \[ (L^{-1} f)(x) = \int G(x,x')\, f(x')\, dx', \] and we identify \[ G(x,x') = \langle x | L^{-1} | x' \rangle. \] In this reading the left argument \(x\) plays the role of the row index — it’s where we evaluate the output. The right argument \(x'\) plays the role of the column index — it’s where the input is applied. Physically, \(x\) is the field point and \(x'\) is the source point.
Physical meaning: \(x'\) as source point
Consider a linear differential equation written as \[ L u(x) = f(x). \] The Green’s function is defined to be the solution when the right‑hand side is a point impulse: \[ L G(x,x') = \delta(x - x'). \tag{1} \] So \(G(x,x')\) is literally the system’s response at \(x\) to a unit source concentrated at the single point \(x'\). Because the equation is linear, the solution for a general \(f\) is just the superposition (integral) of all these individual responses: \[ u(x) = \int G(x,x') \, f(x') \, dx'. \tag{2} \] In (2), \(x'\) is a dummy variable that sweeps over the whole domain; after integration only \(x\) remains — the observation point.
The same logic extends to time‑dependent problems. There the source is labelled by a space–time pair \((x',t')\) and the response by \((x,t)\). Causality forces us to choose the retarded Green’s function (which vanishes for \(t<t'\)), but the fundamental meaning of the primed coordinates as the source location stays unchanged.
Construction for \(L = -\frac{d^2}{dx^2}\) on the real line
Let’s make all of this concrete with the simplest possible operator: \(L = -\frac{d^2}{dx^2}\) on the whole real line, with the boundary condition that \(u\) vanishes at infinity. We need to find \(G(x,x')\) satisfying \[ -\frac{d^2}{dx^2} G(x,x') = \delta(x - x'). \tag{3} \]
A useful distributional identity.
The absolute value function has a well‑known second derivative in the
sense of distributions: \[
\frac{d}{dx}|x| = \operatorname{sgn}(x), \qquad
\frac{d^2}{dx^2}|x| = 2\delta(x).
\] Shift the argument by a fixed \(x'\): \[
\frac{d^2}{dx^2}|x - x'| = 2\delta(x - x'). \tag{4}
\]
Finding \(G\).
From (4) we can isolate the delta by dividing by 2: \[
\frac{d^2}{dx^2}\Bigl( \tfrac12 |x - x'| \Bigr) = \delta(x -
x').
\] Now multiply by \(-1\) to
match the sign in (3): \[
-\frac{d^2}{dx^2}\Bigl( -\tfrac12 |x - x'| \Bigr) = \delta(x -
x').
\] Therefore \[
G(x,x') = -\tfrac12\,|x - x'|. \tag{5}
\]
Quick check via the jump condition.
Integrate (3) over an infinitesimal interval around \(x'\): \[
-\bigl[ G_x(x'^+) - G_x(x'^-) \bigr] = 1,
\qquad\Longrightarrow\qquad
G_x(x'^+) - G_x(x'^-) = -1.
\] For \(G = -\frac12|x -
x'|\), the derivative is \[
G_x = -\frac12 \operatorname{sgn}(x - x') =
\begin{cases}
+\frac12, & x < x', \\[4pt]
-\frac12, & x > x'.
\end{cases}
\] Thus \(G_x(x'^+) =
-\frac12\), \(G_x(x'^-) =
+\frac12\), and the jump is indeed \(-1\). If the operator were \(+\frac{d^2}{dx^2}\), the Green’s function
would be \(+\frac12|x - x'|\); the
operator’s sign completely determines the sign of \(G\).
Elementary examples
One-dimensional electrostatics
The equation \(-\frac{d^2 u}{dx^2} = f\) with \(u(\pm\infty)=0\) is just the 1D Poisson equation. Its Green’s function is again \(G(x,x') = -\frac12 |x - x'|\). So if we put a unit point charge at \(x'=3\), the potential at any \(x\) is \(-\frac12|x-3|\). Moving the charge to \(x'=5\) gives \(-\frac12|x-5|\). The primed coordinate plainly marks the source.
Heat equation (retarded Green’s function)
For the heat equation on an infinite line, \[ \frac{\partial u}{\partial t} - \kappa \frac{\partial^2 u}{\partial x^2} = f(x,t), \] the retarded Green’s function (the one that respects causality) is \[ G_R(x,t; x',t') = \begin{cases} \displaystyle \frac{1}{\sqrt{4\pi\kappa (t - t')}} \, \exp\!\Bigl[ -\frac{(x - x')^2}{4\kappa (t - t')} \Bigr], & t > t', \\[8pt] 0, & t < t'. \end{cases} \] Here \((x',t')\) is the point in space and time where we inject a very short, localized pulse of heat. The unprimed \((x,t)\) is where we later measure the temperature.
Discrete analogue: a mass–spring chain
Even the discrete version follows the same pattern. Suppose we have three masses connected by springs; the equilibrium displacement satisfies \(K \mathbf{u} = \mathbf{f}\), where \(K\) is the stiffness matrix. The inverse \(G = K^{-1}\) is the discrete Green’s function. Its entry \(G_{ij}\) is the displacement of mass \(i\) caused by a unit force applied only to mass \(j\). So \(j\) (the column index) is the source, and \(i\) (the row index) is the observation point — exactly the same logic as in the continuous case.
Summary
The whole story is an infinite‑dimensional version of matrix elements: - \(\langle x | L^{-1} | x' \rangle\) is a matrix element of the inverse operator. - \(x\) is the observation point (row index); \(x'\) is the source point (column index). - The defining equation is \(L G(x,x') = \delta(x - x')\): \(G\) is the response to a unit point source at \(x'\). - For \(-\frac{d^2}{dx^2}\), the identity \(\frac{d^2}{dx^2}|x - x'| = 2\delta(x - x')\) yields \(G(x,x') = -\frac12|x - x'|\) immediately; the sign follows from the operator. - When solving \(L u = f\), \(x'\) becomes a dummy integration variable — it scans through all source positions to build up the solution by superposition.