# Angular Spectrum Representation

The angular spectrum representation of a field can be somewhat difficult to understand at first, so don’t worry if you have a hard time getting through these next few paragraphs! However, the angular spectrum representation is a powerful tool to describing how laser light propagates and is focused, so it is important to try to understand it here. It will also pop up again in the diffraction section in the paraxial wave limit as equivalent to the Fourier optics depiction.

To begin, we can start again with the mode expansion of an arbitrary field:

$$E(r) = \hat{\epsilon}E_0\frac{1}{L^3}\sum_{k_j} a_{k}e^{ik_jj}\label{mode_exp}$$

where j = x, y,z. Then, converting this to an integral, we have:

$$E(r) = \hat{\epsilon}E_0(\frac{L}{2\pi})^3\int d^3k \frac{1}{\sqrt{L^3}}a(k)e^{-ik\cdot{r}}\label{int_mode_exp}$$

Now, say we are given the electric field $E(r)$ at a certain point in space. We already know now that we can represent this electric field as a summation of plane waves; we proved that in the Helmholtz section and it is stated again in Equation \eqref{mode_exp}. If we wanted to find the amplitude $a(k)$ of any of the plane wave components of this field corresponding to a certain $k$, then we could do this by getting rid of the exponential on the right hand side of Equation \eqref{int_mode_exp}. This would be accomplished by multiplying both sides of Equation \eqref{int_mode_exp} by:

$$u^*_{k’}(r) = \frac{1}{\sqrt{L^3}}e^{ik’\cdot{r}}$$

and integrating to yield:

\begin{align} \int d^3r E(r)\frac{1}{\sqrt{L^3}}e^{ik’\cdot{r}} & = \hat{\epsilon}E_0(\frac{L}{2\pi})^3\int d^3r\int d^3k \frac{1}{L^3}a(k)e^{-i(k-k’)\cdot{r}} \label{inv_trans}\\ & = \hat{\epsilon}E_0(\frac{1}{2\pi})^3\int d^3k~a(k)\int d^3r e^{-i(k-k’)\cdot{r}} \\ & = \hat{\epsilon}E_0(\frac{L}{2\pi})^3a(k)\end{align}

where the last step has utilized the property that we already discussed of delta function integrals:

$$\int d^3r e^{-i(k-k’)\cdot{r}} = L^3\delta(k-k’)$$

Now then, we can complete the remaining integral over $k$ (noting that $k = k’$) and find the amplitude of the plane wave component as:

$$a(k’) = (\frac{2\pi}{L})^3\int d^3r\frac{\hat{\epsilon}\cdot E(r)}{E_0}e^{ik’\cdot{r}} \label{free_space_mode}$$

Now, let’s breathe and reflect on what we just found, because it’s pretty cool if I do say so myself. The equation for $a(k’)$ is in the form the summation of plane waves over all space; this means that the modes of free space are plane waves, since $a(k’)$ is the equation for the modes. This is huge! This means that we can construct an arbitrary spatial field distribution $E(r)$ by summing over the modes, which again are just plane waves. Note too that we obtained this result by effectively taking the inverse Fourier transform of the field distribution (Equation \eqref{inv_trans}). If we want to construct the field $E(r)$ from $a(k’)$, we can just perform the inverse Fourier transform of Equation \eqref{free_space_mode} and find the field $E(r)$. This means that the summation over modes is equivalent to an inverse Fourier transform.

These results also mean, as stated earlier, that given a field distribution $E(r)$, we can find the amplitude of any specific plane wave component that makes up the field using Equation \eqref{free_space_mode}. Notice, though, that in order to find $a(k’)$, we need to know $E(r)$ at all points in space. Typically in optics, we are really only concerned with problems where we may know the field distribution in one plane and would like to determine the field distribution in another plane. If we wanted to do this using Equation{free_space_mode}, then we would not be able to as we would need to know the field at all points in space. Thus, we need a new solution…

Enter: the Angular Spectrum Representation! To find the angular spectrum, we will motivate our equation by assuming that the field is primarily propagating along the z-direction, so the phase would only be along the z-direction and the amplitude (defined as being in the plane transverse to propagation) would be only in the x- and y-planes. The equation representing this is:

$$E(r) = \script{E}(x,y)e^{-i\beta{z}}$$

Now, analogously to the inverse Fourier transform, which we said would be used to find the field distribution, we can find the field distribution at another xy-plane A also along the z-direction by integrating over only the $k_x$ and $k_y$ dimensions:

$$E_A(r) = \iint^{\infty}_{-\infty}A(k_x,k_y)e^{-i(k_xx+k_yy+\beta{z})}dk_xdk_y$$

which is the angular spectrum representation! The angular spectrum is the superposition of real waves and evanescent waves. We can find this explicitly by noting that E(r) must satisfy the Helmholtz equation, so then we obtain the condition:

$$(\nabla^2+k^2)E(r) = 0 = -k_x^2-k_y^2-\beta^2+k^2$$

so that

$$k^2 = k_x^2 + k_y^2 + \beta^2$$

Further Resources: This presentation from ETH Zurich gives much more information on the angular spectrum representation, albeit with a high level of tone than what I have here. Some of the stuff covered will be covered on this website in proceeding sections, but this is a good sneak peak!