Greens function has me blue ....

[ognik]

Not following this example (PDE for Greens function) in my book:

Book states: $ \left( {\nabla}^{2} +{k}^{2}\right)G(r, r_2)=-\delta(r-r_2) = -\int \frac{e^{ip.(r-r_2)}}{\left(2\pi\right)^3} {d}^{3}p$

I recognised this as the Hemlmholtz eqtn, but cannot find where the 3rd term comes from? It looks like it could be the 3D Fourier Transform representation of the dirac-delta function? (if so a link to, or a derivation would be nice)

Then they say they solve the PDE in terms of a Fourier Integral $ G(r, r_2)=\int \frac{1}{\left(2\pi\right)^3}g_0(p) e^{ip.(r-r_2)} d^3p $

I know the Fourier Integral in 3D is $F(k)=\frac{1}{\left(2\pi\right)^{\frac{3}{2}}} \int f(r)e^{ik.r} \,d^3r $, (sqrt in denominator?) so I'm not sure what they are doing here?

 

[Jhoneine]

The third term in your example is indeed the 3D Fourier transform of the Dirac delta function. The 3D Fourier transform of a function $f(r)$ is given by $$\mathcal{F}[f(r)] = \frac{1}{(2\pi)^{\frac{3}{2}}} \int f(r) e^{i k \cdot r} d^3 r$$where $k \cdot r = k_x x + k_y y + k_z z$. This expression can be thought of as a generalization of the 2D Fourier transform, which uses a square root in the denominator instead of a cube root.The solution to the PDE that they give is a general solution to the homogeneous equation $\left( {\nabla}^{2} +{k}^{2}\right)G(r, r_2)=0$, which can be written as$$G(r, r_2) = \int \frac{1}{(2\pi)^3} g_0(p) e^{i p \cdot (r - r_2)} d^3 p.$$This is just the 3D Fourier transform of a function $g_0(p)$, which is some arbitrary function of momentum $p$.