Evanescent field, condition of $z$ component of $k$?

Question

I am studying evanescent field and diffraction limit and I have one question.

Given a field $ U(x,y,0)$ we can decompose into 2D plane waves.

$U(x,y,0)= \int \int dk_x dk_y \tilde{U}(k_x,k_y) e^{+i(k_x x + k_yy)}$

if we want to have $U(x,y,z_0)$ we can propagate each wave to the plane $z_0$ in this way.

$U(x,y,z_0)= \int \int dk_x dk_y \tilde{U}(k_x,k_y) e^{+i(k_x x + k_yy)} e^{i k_z z_0}$

This is clear to me.

It is not clear why $k_z = \sqrt{ (\frac{2\pi}{\lambda})^2 - k_x^2 -k_y^2 }$.

Why suddenly the plane waves should be monochromatic with wavelength $\lambda$ ?

Answer 1

The wave $u(x,y,z,t)$ satisfies the Helmholtz wave equation $$\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}-\frac{1}{c^2}\frac{\partial^2u}{\partial t^2}=0 \tag{1}\label {1}.$$ This equation is solved by plane waves of the form $$U(x,y,z,t) = A e^{\mathfrak j (\omega t - (k_x x+k_yy+k_zz))}\tag{2}\label {2}.$$ The dispresion relation between $\omega$ and $\mathbf k(k_x,k_y,k_z)$ can be had by substituting $\eqref{2}$ in to $\eqref{1}$ and noting that $\frac{d^2 e^{\mathfrak jas}}{ds^2} = -a^2e^{\mathfrak jas}$ : $$-\frac{\omega^2}{c^2}+ k_x^2+k_y^2+k_z^2=0$$ or

$$\left(\frac{2\pi}{\lambda}\right)^2 = k_x^2+k_y^2+k_z^2 \tag{3}\label {3}.$$ The values of $\omega$ or $\mathbf k$ are not restricted to be real numbers, they just have to satisfy the dipersion relation $\eqref{3}$.