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This is a continuation of yesterday’s post on wave function Green’s functions. We derived the 1D Green’s function, now it’s time for the 3D.

Next up after this will be the 2D Green’s function, which looks harder to evaluate than the 1D or 3D versions.

3D Green’s function.

The 3D Green’s function that we wish to try to evaluate is
\begin{equation}\label{eqn:helmholtzGreens:540}
G(\Br) = -\inv{(2 \pi)^3} \int \frac{e^{j \Bp \cdot \Br}}{\Bp^2 – k^2} d^3 p.
\end{equation}
We will have to displace the pole again, but we will get to that in a bit. First let’s make a spherical change of variables to evaluate the integral, with
\begin{equation}\label{eqn:helmholtzGreens:560}
\begin{aligned}
\Bp &= p \lr{ \sin\alpha \cos\phi, \sin\alpha \sin\phi, \cos\alpha } \\
\Br &= \Abs{\Br} \Be_3.
\end{aligned}
\end{equation}
This gives
\begin{equation}\label{eqn:helmholtzGreens:580}
G(\Br)
= -\inv{(2 \pi)^3} \int_0^\infty p^2 dp \int_0^\pi \sin\alpha d\alpha \int_0^{2 \pi} d\phi \frac{e^{j p \Abs{\Br} \cos\alpha}}{p^2 – k^2}.
\end{equation}
Let \( t = \cos\alpha \), to find
\begin{equation}\label{eqn:helmholtzGreens:600}
\begin{aligned}
G(\Br)
&= -\inv{(2 \pi)^2} \int_0^\infty p^2 dp \int_1^{-1} (-dt) \frac{e^{j p \Abs{\Br} t}}{p^2 – k^2} \\
&= \inv{(2 \pi)^2} \int_0^\infty p^2 dp \evalrange{\frac{e^{j p \Abs{\Br} t}}{\lr{p^2 – k^2} j p \Abs{\Br}}}{1}{-1} \\
&= \inv{j (2 \pi)^2 \Abs{\Br}} \int_0^\infty p dp \frac{e^{-j p \Abs{\Br}} – e^{j p \Abs{\Br}} }{p^2 – k^2} \\
&= -\inv{j (2 \pi)^2 \Abs{\Br}} \int_{-\infty}^\infty p dp \frac{e^{j p \Abs{\Br}} }{p^2 – k^2} \\
&\sim -\inv{j (2 \pi)^2 \Abs{\Br}} \int_{-\infty}^\infty p dp \frac{e^{j p \Abs{\Br}} }{p^2 – \lr{k + j \epsilon}^2}.
\end{aligned}
\end{equation}
In the last step, we’ve displaced the pole so that we can evaluate it using an infinite upper half plane semicircular contour, as illustrated in fig 3.

fig 3. Contours for 3D Green’s function evaluation

Which pole we choose depends on the sign we pick for the “small” pole displacement \( \epsilon \). For the \( \epsilon > 0 \) case, we find
\begin{equation}\label{eqn:helmholtzGreens:620}
\begin{aligned}
G(\Br)
&= -\inv{j (2 \pi)^2 \Abs{\Br}} \int_{-\infty}^\infty p dp \frac{e^{j p \Abs{\Br}} }{\lr{p – (k + j \epsilon)}\lr{p – (-k – j \epsilon)}} \\
&= -\frac{2 \pi j}{j (2 \pi)^2 \Abs{\Br}} \evalbar{\frac{p e^{j p \Abs{\Br}} }{p + k + j \epsilon}}{p = k + j \epsilon} \\
&= -\frac{1}{2 \pi \Abs{\Br}} (k + j \epsilon) \frac{e^{j (k + j \epsilon) \Abs{\Br}} }{2 (k + j \epsilon)} \\
&= -\frac{1}{4 \pi \Abs{\Br}} e^{j k \Abs{\Br}} e^{-\epsilon \Abs{\Br}} \\
&\rightarrow -\frac{e^{j k \Abs{\Br}} }{4 \pi \Abs{\Br}}.
\end{aligned}
\end{equation}
whereas for \( \epsilon < 0 \), we have
\begin{equation}\label{eqn:helmholtzGreens:640}
\begin{aligned}
G(\Br)
&= -\inv{j (2 \pi)^2 \Abs{\Br}} \int_{-\infty}^\infty p dp \frac{e^{j p \Abs{\Br}} }{\lr{p – (k + j \epsilon)}\lr{p – (-k – j \epsilon)}} \\
&= -\frac{2 \pi j}{j (2 \pi)^2 \Abs{\Br}} \evalbar{\frac{p e^{j p \Abs{\Br}} }{p – k – j \epsilon}}{p = -k – j \epsilon} \\
&= -\frac{1}{2 \pi \Abs{\Br}} (-k – j \epsilon) \frac{e^{j (-k – j \epsilon) \Abs{\Br}} }{2 (-k – j \epsilon)} \\
&= -\frac{1}{4 \pi \Abs{\Br}} e^{-j k \Abs{\Br}} e^{\epsilon \Abs{\Br}} \\
&\rightarrow -\frac{e^{-j k \Abs{\Br}} }{4 \pi \Abs{\Br}}.
\end{aligned}
\end{equation}

The Green’s function has the structure of either an outgoing or incoming spherical wave, with inverse radial amplitude:
\begin{equation}\label{eqn:helmholtzGreens:660}
\boxed{
G(\Bx, \Bx’) = -\frac{e^{\pm j k \Abs{\Br}} }{4 \pi \Abs{\Br}}.
}
\end{equation}