# Green function of $\left(\nabla^2+k^2\right)\!\psi=-4\pi\,\delta\!\left(\vec{r}\right)$?

I was revisiting an old paper by Leslie L Foldy where he states that the solution to $$\left(\nabla^2+k^2\right)\!\psi\!\left(\vec{r}\right)=-4\pi\,\delta\!\left(\vec{r}\right)$$ is $$\psi\!\left(\vec{r}\right)=\frac{e^{ikr}}{r}.$$ Here, $$\delta\!\left(\vec{r}\right)$$ is the (3D) Dirac delta and $$r=|\vec{r}|$$.

I can't find this identity anywhere. I've looked at a bunch of books on Green's functions without luck. Mathematical methods of physics by Mathews and Walker, for example, solves the very similar (in electrodynamics) $$\left(\nabla^2-\frac{1}{c}\,\partial_{tt}\right)\!\psi\!\left(\vec{r},t\right)=-4\pi\delta\!\left(\vec{r}\right)\delta(t)$$ ($$\partial_{tt}$$ is the second partial derivative with respect to $$t$$) and obtains $$\psi(\vec{r},t)=c\,\delta(r-ct),$$ which looks nothing like Foldy's outgoing spherical wave (an imaginary exponential divided by $$r$$).

I tried doing it myself but got stuck. I tried two methods:

1. If $$\Psi\!\left(\vec{k}\right)$$ is the (3D) Fourier transform of $$\psi\!\left(\vec{r}\right)$$, then we have $$\Psi\!\left(\vec{k}\right)=\frac{1}{(2\pi)^{3/2}}\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\psi\!\left(\vec{r}\right)e^{-i\vec{k}\cdot\vec{r}}\,\text{d}\vec{r},\\ \psi\!\left(\vec{r}\right)=\frac{1}{(2\pi)^{3/2}}\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\Psi\!\left(\vec{k}\right)e^{i\vec{k}\cdot\vec{r}}\,\text{d}\vec{k}.$$ Taking the Fourier transform of the equation we wish to solve, $$\left(-k_x\,\!^2-k_y\,\!^2-k_z\,\!^2+k^2\right)\!\Psi\!\left(\vec{k}\right)=-\frac{4\pi}{(2\pi)^{3/2}},$$ whereby $$\Psi\!\left(\vec{k}\right)=-\frac{4\pi}{(2\pi)^{3/2}\left(-k_x\,\!^2-k_y\,\!^2-k_z\,\!^2+k^2\right)}$$ and $$\psi\!\left(\vec{r}\right)=-\frac{1}{2\pi^2}\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{e^{i\vec{k}\cdot\vec{r}}}{-k_x\,\!^2-k_y\,\!^2-k_z\,\!^2+k^2}\,\text{d}\vec{k}.$$ This is where I get stuck; I have no idea how to solve that.

2. The equation we wish to solve is an inhomogeneous differential equation. The most general homogeneous solution without poles is $$\psi^\text{h}\,\!_{\ell,m}\!\left((\vec{r}\right)=j_\ell(kr)Y_{\ell,m}(\theta,\phi),$$ where $$j_\ell$$ is the $$\ell$$th spherical Bessel function of the first kind and $$Y_{\ell,m}$$ is the spherical harmonic of order $$(\ell,m)$$; however, something closer to what we're looking for is $$\psi^\text{h}\!\left(\vec{r}\right)=\frac{e^{ikr}}{r},$$ which is the general solution when $$\psi^\text{h}\!\left(\vec{r}\right)=\psi^\text{h}(r)$$. But then I have no idea what sort of particular solution $$\psi^\text{p}$$ to propose in order to arrive at $$\psi\!\left(\vec{r}\right)=\psi^\text{h}\,\!\!\left(\vec{r}\right)+\psi^\text{p}\,\!\!\left(\vec{r}\right)=\frac{e^{ikr}}{r}.$$

There's an article$$^1$$ which follows the second procedure I tried but then uses generalised functions to find $$\psi^\text{p}$$, and I got completely lost trying to follow it (I'm not familiar with generalised functions; that article was the first time I ever heard of them), so that's no use to me. The authors do reach the desired $$\psi\!\left(\vec{r}\right)=\frac{e^{ikr}}{r},$$ though, which makes me think Foldy wasn't wrong, just terrible at being clear in his writing.

How can I obtain the desired result without using magic— er, I mean, generalised functions?

$$^1$$ Schmalz J E, Schmalz G, Gureyev T E & Pavlov K M (2010): On the derivation of the Green's function for the Helmholtz equation using generalized functions, American Journal of Physics 78, 181–186

• Method 1 is fine: the integral over $d^3 k$ may be done in spherical $k$-co-ordinates
– Sal
Commented Mar 24, 2022 at 15:50
• @Sal Thanks for your comment. I tried that; I end up with (something proportional to) the integral of k_rexp(ik_r*r)/(k^2-k_r^2) with respect to k_r from 0 to infinity, which I couldn't solve even with a change of variable to a=sqrt(k^2-k_r^2) and which Mathematica says isn't finite. I'll keep trying, though.
– Rain
Commented Mar 24, 2022 at 17:16
• In the exponent $\mathbf{k}\cdot \mathbf{r}=kr\cos(\theta)$. Don't forget the volume element $d^3k=k^2\sin(\theta)dk \ d\theta \ d\phi$. The angular $\phi$ integral simply yields $2\pi$, and the $\theta$ integral may be done directly by substitution $u=\cos(\theta)$. Finally, it seems that the remaining $k$ integral will be simplest using contour integration
– Sal
Commented Mar 24, 2022 at 17:21
• I'm an idiot. I used $k_r\sin(k_\theta)\,\text{d}k_r\,\text{d}k_\theta\,\text{d}k_\varphi$ as the volume element instead of $k_r\,\!^2\sin(k_\theta)\,\text{d}k_r\,\text{d}k_\theta\,\text{d}k_\varphi$. Thanks again!
– Rain
Commented Mar 24, 2022 at 17:53