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Green’s functions and the integral equation · Quantum Mechanics | Math Infinitum

Intuition

To find $f$ , turn the Schrödinger equation into an integral equation with the boundary condition built in. Write it as $(\nabla^{2}+k^{2})\psi=U\psi$ with $U=2mV/\hbar^{2}$ , and read the right-hand side as a source. A point source at the origin radiates the outgoing spherical wave $G(\mathbf{r})=-e^{ikr}/4\pi r$ , the Green’s function, and any source is a sum of point sources. So the solution is the incident wave plus the waves radiated by $U\psi$ from every point where the potential acts: the Lippmann–Schwinger equation. Far away, all those waves travel in almost the same direction, and their sum is an outgoing spherical wave whose strength is an integral of $U\psi$ with the phases $e^{-i\mathbf{k}'\cdot\mathbf{r}'}$ — the scattering amplitude, in a formula exact for any short-range potential.

The sound heard far from a choir is the sum of every singer’s voice, each arriving with its own delay. The Green’s function is one singer’s voice; the integral equation adds them all, with the potential deciding how loudly each point sings.

A potential confined to the dashed circle, a point $\mathbf{r}'$ inside it and a distant detector at $\mathbf{r}$ . Far away, $|\mathbf{r}-\mathbf{r}'|$ is less the projection of on the direction of , which turns each point’s wave into the same outgoing wave times the phase .

Green’s functions and the integral equation

With $k^{2}=2mE/\hbar^{2}$ and $U=2mV/\hbar^{2}$ , the scattering state solves

\begin{aligned}\psi(\mathbf{r})&=e^{i\mathbf{k}\cdot\mathbf{r}}+\int G(\mathbf{r}-\mathbf{r}')\,U\psi(\mathbf{r}')\,d^{3}r'\\ G(\mathbf{r})&=-\frac{e^{ikr}}{4\pi r}\end{aligned}

Properties

$G$ solves $(\nabla^{2}+k^{2})G=\delta^{3}(\mathbf{r})$ and is an outgoing wave, ; the choice would describe waves converging on the target.

The scattering amplitude from the integral equation

f(\theta,\phi)=-\frac{m}{2\pi\hbar^{2}}\int e^{-i\mathbf{k}'\cdot\mathbf{r}'}\,V(\mathbf{r}')\,\psi(\mathbf{r}')\,d^{3}r'

The Green’s function turns the Schrödinger equation with its source into an integral equation. Far from the potential, the distance from a source point to the detector is $r$ less the source point’s projection on the direction of the detector; put that into $G$ , and the integral becomes an outgoing spherical wave times the amplitude.

Proof steps

(\nabla^{2}+k^{2})\psi=U\psi,\qquad U=\frac{2mV}{\hbar^{2}}

The Schrödinger equation at the energy $E=\hbar^{2}k^{2}/2m$ .

(\nabla^{2}+k^{2})\Bigl(-\frac{e^{ikr}}{4\pi r}\Bigr)=\delta^{3}(\mathbf{r})

Away from the origin $e^{ikr}/r$ solves the free equation; near it $-1/4\pi r$ is the potential of a point charge, whose Laplacian is the delta function.

\psi(\mathbf{r})=e^{i\mathbf{k}\cdot\mathbf{r}}+\int G(\mathbf{r}-\mathbf{r}')\,U\psi(\mathbf{r}')\,d^{3}r'

The free wave plus the waves radiated by the source $U\psi$ at every point.

\begin{aligned}|\mathbf{r}-\mathbf{r}'|&\approx r-\hat{\mathbf{r}}\cdot\mathbf{r}'\\ G(\mathbf{r}-\mathbf{r}')&\approx-\frac{e^{ikr}}{4\pi r}\,e^{-i\mathbf{k}'\cdot\mathbf{r}'}\end{aligned}

For $r\gg r'$ , with $\mathbf{k}'=k\hat{\mathbf{r}}$ ; in the slowly varying , alone is enough.

\psi\approx e^{i\mathbf{k}\cdot\mathbf{r}}+\frac{e^{ikr}}{r}\Bigl(-\frac{1}{4\pi}\int e^{-i\mathbf{k}'\cdot\mathbf{r}'}\,U\psi\,d^{3}r'\Bigr)

Read off $f$ , and put back $U=2mV/\hbar^{2}$ .

Applications

Practice

The Green’s Function

A point source radiates an outgoing spherical wave. It solves the free equation with a delta-function source.

G(\mathbf{r})=-\frac{e^{ikr}}{4\pi r}

Try it

Why is the Green’s function taken with $e^{+ikr}$ rather than $e^{-ikr}$ ?