Far-field dyadic Green's function

[Jarno]

Homework Statement

I'm using the book "Principles of nano-optics" by Novotny and Hecht. I'm stuck a bit at understanding the derivation of the point spread function. It's just given as
$$\mathbf{G} = \frac{\exp(i k_1 r)}{4 \pi r} \exp[-i k_1(x_0 x / r + y_0 y / r + z_0 z / r)]\\ \qquad\times \left[ \begin{array}{ccc} (1 - x^2/r^2) & - x y/r^2 & -x z/r^2 \\ - x y/r^2 & (1 - y^2/r^2) & -y z/r^2 \\ - x z/r^2 & -y z/r^2 & (1 - z^2/r^2) \end{array} \right]$$
in equation (D.4) in the book (if you have it).

Homework Equations

The far-field comes from the angular spectrum representation where the evanescent waves have been dropped. This is given as
$$\vec{E}_\infty(s_x, s_y, s_z) = -2 \pi i k s_z \hat{\vec{E}}(k s_x, k s_y; z = 0) \quad ,$$
where E with the hat is the Fourier transform with respect to x and y of the field at z = 0. The vector s is just a unit vector indicating the direction.

I know the dyad at z=0 is given as
$$\mathbf{G}(\vec{r}, \vec{r}_0) = (\mathbf{I} + \frac{1}{k^2} \nabla \nabla) G_0(\vec{r}, \vec{r}_0)\quad,$$
where G0 is the scalar Green's function, known as
$$G_0(\vec{r}, \vec{r}_0) = \frac{\exp(\pm i k \vert \vec{r} - \vec{r_0} \vert}{4 \pi \vert \vec{r} - \vec{r}_0 \vert ) } \quad ,$$
which is the Green's function for the Helmholtz equation.

The Attempt at a Solution

I've tried to do a direct Fourier transform of the dyad G, so I can substitute that into the far-field equation, but I've been unable to get that, because I can't get the Fourier transform of G0 with respect to x and y.
I am able to get the Fourier transform of G0 with respect to x, y and z by using the Helmholtz equation and doing a Fourier transform on both sides. However, I need the Fourier transform to be only with respect to x and y but not z.
Does anybody have an idea?

 

[mark726]

Thank you for your question on the derivation of the point spread function in "Principles of nano-optics" by Novotny and Hecht. I understand that you are having difficulty with the Fourier transform of G0 with respect to x and y. Let me try to guide you through the process.

First, let's start with the definition of the Fourier transform:
\hat{f}(\vec{k}) = \int_{-\infty}^{\infty} f(\vec{r}) e^{-i \vec{k} \cdot \vec{r}} d\vec{r}
where f is the function in real space, \hat{f} is its Fourier transform in k-space, and \vec{k} is the wavevector.

Now, we can use the Helmholtz equation to express G0 in terms of its Fourier transform:
(\nabla^2 + k^2) G_0(\vec{r}, \vec{r}_0) = -\delta(\vec{r} - \vec{r}_0)
Taking the Fourier transform of both sides, we get:
(-k^2 + \vec{k}^2) \hat{G}_0(\vec{k}, \vec{r}_0) = -1
Solving for \hat{G}_0, we get:
\hat{G}_0(\vec{k}, \vec{r}_0) = \frac{1}{k^2 - \vec{k}^2} \quad .

Now, to get the Fourier transform of G0 with respect to x and y, we need to integrate over z:
\hat{G}_0(\vec{k}, \vec{r}_0) = \int_{-\infty}^{\infty} \frac{e^{-i k_z z}}{k^2 - k_x^2 - k_y^2 - k_z^2} dz
Using the substitution u = k_z z, we can rewrite this as:
\hat{G}_0(\vec{k}, \vec{r}_0) = \frac{1}{k^2 - k_x^2 - k_y^2} \int_{-\infty}^{\infty} \frac{e^{-i u}}{1 - (k_z/k)^2} du
Using the identity \int_{-\infty}^