Understanding the Fourier transform of a signal already with amplitude and phase information

Question

I'm from an image processing background and am working my way into optics. I'm working on phase retrieval problems, trying to understand the Gerchberg-Saxton algorithm. I found this video which is really great, but which made me realize I probably had some misconceptions regarding some of the physics going on. I would like to clear these up before going further, so apologies for the probably very confused question and maybe approximate vocabulary.

The setup in the video is as follows : some field (with amplitude and phase) is imaged on an image plane, where we get the image of the intensity of the field, but not the phase, and we would like to know that phase. So the field is also projected "in the far field", to get the image in the "Fourier plane".

I get the gist of the rest of the algorithm pretty well, which involves comparing images in the image plane and the Fourier plane, enforcing constraints on the amplitude to iteratively correct the phase.

but what I don't get - and I think this is probably my lack of knowledge regarding Fourier optics speaking - is what truly is the "Fourier transform" discussed here. From my image processing background, the Fourier transform decomposes an image, i.e. a real signal, function of space, into an amplitude and phase image in frequency space. Something like :

So, I first thought that the Fourier transform discussed here was decomposition happening between the "image plane"'s intensity and the "Fourier plane"'s intensity + amplitude, ie the red arrows I've added to the figure.

However, that doesn't seem right, because the phase information seems to be taken into account during the Fourier transform. But then, what does the transform indicated by the following red arrow mean ? The field has both phase and amplitude information : is it in frequency space ? does it carry the same information as the phase and amplitude of the Fourier transform (ie the two right images on the bottom row) ? What does it mean to take the Fourier transform of a such a signal ?

Answer 1

are you familiar with Fraunhofer diffraction? This is what is physically happening in your first picture as the light passes through the lens. The idea is that it is the new light field that is rigorously the 2D Fourier transform of the original one. However, in the end, you only measure intensity, i.e. the (square) amplitude.

Therefore, the first measurement is the amplitude of the field, the second measurement is the amplitude of the Fourier transform of the field. Combining both you can extract information on the phase of the initial field. Just to make it clear, this does not correspond to the amplitude of the Fourier transform of the amplitude of the initial field as you seem to suggest in the second picture (it would not even give any relevant information as you noted yourself).

Hope this helps and please tell me if there are any mistakes.

(Edit following @Solitus' comment)

I'll write explicitly the math behind my explication. Physically, your light is the EM field, a vector field, but in your case, you approximate it by a scalar field. Since it is regulated by a real linear second order in time differential equation, just as for the harmonic oscillator, you condense the field and its derivative by a single complex field $U(x,y)$ (complex amplitude). The intensity is therefore (up to a multiplicative constant) $I(x,y) = |U(x,y)|^2$ and the phase $\phi(x,y) = \arg(U(x,y))$.

Physically, you only measure $I$ with a photodetector, but you want $U$, so $\phi$. The trick is to transform $U$ and measure the new intensity in order to tease out information of $\phi$. The Fraunhofer diffraction transforms $U$ to $\mathcal F U$, the Fourier transform on a complex function. The second measurement is therefore of $|\mathcal F U|^2$ which is indirectly related to $\phi$. Heuristically, $U$ has $2$ real degrees of freedoms (per point), and you have two observables (per points), so you "should" be able to solve the equations to retrieve $U$. The Gerchberg-Saxton algorithm does this solve in a self-consistent loop.

To summarise, unlike image processing where the signals are real, in optics you are interested in complex signals.

Answer 2

A complex field $F$ is a function $F: \mathbb{R}^N \rightarrow \mathbb{C}$. i.e. it is a complex number defined at each point in $N$-dimensional space.

The Fourier transform takes a complex field $F(\mathbf{x})$ and maps it to a new complex field $\mathcal{FT}[F](\boldsymbol{\nu}) = \tilde{F}(\boldsymbol{\nu})$. It (and it's inverse) may be defined (up to various convention choices) as

\begin{align} \mathcal{FT}[F](\boldsymbol{\nu}) = \tilde{F}(\boldsymbol{\nu}) = \int_{x_1=-\infty}^{+\infty}\ldots\int_{x_N=-\infty}^{+\infty} F(\mathbf{x}) e^{-i2\pi \boldsymbol{\nu}\cdot\textbf{x}} dx_N\ldots dx_1\\ \mathcal{FT}^{-1}[\mathcal{FT}[F]](\textbf{x}) = \breve{F}(\textbf{x}) = \int_{x_1=-\infty}^{+\infty}\ldots\int_{x_N=-\infty}^{+\infty} F(\mathbf{x}) e^{+i2\pi \boldsymbol{\nu}\cdot\textbf{x}} dx_N\ldots dx_1\\ \end{align}

The content of the Fourier inversion theorem is that, for suitable functions $F$ (which includes many functions encountered in physics and signal processing) that $\breve{F}(\textbf{x}) = F(\textbf{x})$, i.e. that the inverse Fourier transform is in fact the inverse of Fourier transform.

In 2D we have

$$ \tilde{F}(\boldsymbol{\nu}) = \int_{x_1=-\infty}^{+\infty} \int_{x_2=-\infty}^{+\infty} F(\textbf{x})e^{-2\pi \boldsymbol{\nu}\cdot\textbf{x}}dx_1dx_2 $$

Ok, this is all well and good and already answers much of your question. It seems you are familiar with the Fourier transform of REAL functions or REAL fields and that the Fourier transform of a real field is a complex field, however, here, I have plainly told you that it is actually possible (and typical) to take the Fourier transform of complex fields. If $F$ is a real function then $\tilde{F}$ satisfies that $\tilde{F}(-\boldsymbol{\nu}) = \tilde{F}^*(\boldsymbol{\nu})$, but this relation does not hold for non-real input fields $F$.

Note that in general to determine $F(\textbf{x})$ from $\tilde{F}(\boldsymbol{\nu})$ it is necessary to perform the inverse Fourier transform which requires information about the phase of $\tilde{F}(\boldsymbol{\nu})$.

The Gerchberg-Saxton algorithm answers the question: Given only amplitude information about $\tilde{F}(\boldsymbol{\nu})$, i.e. $|\tilde{F}(\boldsymbol{\nu})|$, is it possible to determine $F(\textbf{x})$. It turns out the answer to this question is false as there are multiple fields $F$ that give rise to the same Fourier transform amplitude. But, the Gerchberg-Saxton algorithm gives us the somewhat surprising result that if we have knowledge of both $|F(\textbf{x})|$ and $|\tilde{F}(\boldsymbol{\nu})|$ that it IS indeed possible, using an iterative algorithm, to recover $F(\textbf{x})$. This result is very useful precisely because of the example given in the OP. It is often the case in both optical and electrical signal processing (especially optical) that it is easy to measure the amplitude of both a signal and it's Fourier transform while it is possible, but much more difficult, to measure both the amplitude and phase of a signal or it's Fourier transform.

I think this answers your main question about how you can take the Fourier tansform of a complex field and how that information plays into the Gerchberg-Saxton algorithm. There is a remaining question which is: "Why does allowing an optical field to propagate into the far field implement the complex Fourier transform on the input field?" However, (a) you did not ask that question and (b) that question is addressed elsewhere on this site and in other references so I will not address it here.