# Diffractionless Tomography

Classical tomography is based on the assumption that the illuminating light travels in straight lines through the object. While this assumption is only valid for waves of very short wavelength such as, for example, x-rays, it has also been extensively used as an approximation for the light in the visible spectrum.

#### Radon Transform

The Radon transform describes the relationship between a $D$-dimensional function and its projection onto a $D-1$ dimensional hyperplane. In practice, we are often interested in the cases when $D = 2$ or $D = 3$, which correspond to the projection of a two-dimensional (2D) surface onto a line or of a three-dimensional (3D) volume onto a plane, respectively. Generally, it is sufficient to consider the problem in 2D, since, the 3D case can be solved by decomposing it into a succession of 2D slices.

Consider the tomographic problem illustrated in the following figure.

Let $\Omega \in \mathbb{R}^2$ be a bounded, 2D domain which contains an object characterized by a function $f(\boldsymbol{x})$ with $\boldsymbol{x} = (x, y)$. We assume that $f$ varies within $\Omega$, but is zero outside of it. We probe the object by illuminating it with parallel, indefinitely thin beams of light that travel in the direction making an angle $\theta \in [0, \pi]$ with respect to the $y$-axis. We collect measurements along the $t$-axis, orthogonal to the direction of light propagation. Our objective is to reconstruct the object $f$ given the measurements $p_\theta(t)$ for $\theta \in [0, \pi]$ and $t \in \mathbb{R}$. This measurement problem can be described mathematically as

where $\delta$ is the Dirac distribution, $\boldsymbol{\theta} = (\cos \theta, \sin \theta)$ is a unit vector.  We can rewrite the equation by using the following variable change

as

The operator $\mathcal{R}: L_2(\mathbb{R}^2) \rightarrow L_2(\mathbb{R} \times [0, \pi])$ is called the Radon transform after Johann Radon (1887-1956) who introduced it in 1917. The data $\{p_\theta(t)\}_{\theta \in [0, \pi], t \in \mathbb{R}}$ generated by the transform is often called the sinogram, since it maps a point into a sinusoid. To see this consider a point $\boldsymbol{x}_0 = (x_0, y_0)$, whose projection onto the $t$-axis will be $t_0(\theta) = x_0 \cos \theta + y_0 \sin \theta$, which can be represented in polar coordinates $(r, \phi)$, with $x_0 = r \cos \phi$ and $y_0 = r \sin \phi$, as $t_0(\theta) = r \cos (\theta-\phi)$. Hence, the trajectory of the point $\boldsymbol{x}_0$ is a cosine of radius $r$ and shift $\phi$.

#### Fourier Slice Theorem

Fourier slice theorem is one of the most important results from classical tomography, which establishes the connection between the data $p_\theta(t)$ and the function $f(\boldsymbol{x})$ in Fourier space.

Theorem: The 1D Fourier transform of the projected data $\hat{p}_\theta(\omega) = \mathcal{F}_{\text{1D}}\{p_\theta(t)\}$ corresponds to a slice of the 2D Fourier transform of the function $\hat{f}(\boldsymbol{\omega}) = \mathcal{F}_{\text{2D}}\{f(\boldsymbol{x})\}$, with $\boldsymbol{\omega} = (\omega_x, \omega_y)$, taken along the angle $\theta$ through the origin, or more concisely

To establish this result, we consider the Fourier transform

\begin{align*}
\hat{p}_\theta(\omega) &= \int_\mathbb{R} p_\theta(t) \mathrm{e}^{-\mathrm{j} \omega t} \mathrm{d} t \\
&= \int_\mathbb{R} \left [\int_{\mathbb{R}^2} f(\boldsymbol{x}) \delta(\langle \boldsymbol{x}, \boldsymbol{\theta} \rangle - t) \mathrm{d} \boldsymbol{x} \right] \mathrm{e}^{-\mathrm{j} \omega t} \mathrm{d} t \\
&= \int_{\mathbb{R}^2} f(\boldsymbol{x}) \mathrm{e}^{- \mathrm{j} \omega \langle \boldsymbol{x}, \boldsymbol{\theta} \rangle } \mathrm{d} \boldsymbol{x} \\
&= \hat{f}(\omega \cos \theta, \omega \sin \theta),
\end{align*}

where in the second equality use used the definition of the Radon transform.

The remarkable aspect of the Fourier Slice Theorem is that it allows one to invert the Radon transform by (a) collecting measurements for all $\theta \in [0, \pi]$, (b) evaluating the 1D Fourier transform of these measurements and using them to fill the 2D Fourier space, and (c) evaluating inverse 2D Fourier transform to form the image.

#### Backprojection

We now define the backprojection operator $\mathcal{R}^\ast: L_2(\mathbb{R} \times [0, \pi]) \rightarrow L_2(\mathbb{R}^2)$

which is in fact the adjoint of the Radon transform, i.e., for all $p \in L_2(\mathbb{R} \times [0, \pi])$ and for all $f \in L_2(\mathbb{R}^2)$, we have that

Intuitively, for each point $\boldsymbol{x}$, backprojection collects all the projections that passed through that point.

#### Filtered Backprojection

Consider the following inverse Fourier transform

\begin{align*}
f(\boldsymbol{x})
&= \frac{1}{(2\pi)^2} \int_{\mathbb{R}^2} \hat{f}(\boldsymbol{\omega}) \mathrm{e}^{\mathrm{j} \langle \boldsymbol{x}, \boldsymbol{\omega} \rangle} \mathrm{d} \boldsymbol{\omega} \\
&= \frac{1}{(2\pi)^2} \int_0^\pi \int_\mathbb{R} \hat{f}(\omega \cos \theta, \omega \sin \theta) \mathrm{e}^{\mathrm{j} \omega \langle \boldsymbol{x}, \boldsymbol{\theta} \rangle} |\omega| \mathrm{d} \omega \mathrm{d} \theta\\
&= \frac{1}{(2\pi)^2} \int_0^\pi \int_\mathbb{R} |\omega| \hat{p}_\theta(\omega) \mathrm{e}^{\mathrm{j} \omega \langle \boldsymbol{x}, \boldsymbol{\theta} \rangle} \mathrm{d} \omega \mathrm{d} \theta \\
&= \int_0^\pi \left[\frac{1}{2\pi}\int_\mathbb{R} \frac{|\omega|}{2\pi} \, \hat{p}_\theta(\omega) \mathrm{e}^{\mathrm{j} \omega \langle \boldsymbol{x}, \boldsymbol{\theta} \rangle} \mathrm{d} \omega\right] \mathrm{d} \theta \\
&= \int_0^\pi q_\theta(x \cos \theta + y \sin \theta) \mathrm{d} \theta
\end{align*}
where we performed the variable change $\omega_x = \omega \cos \theta$ and $\omega_y = \omega \sin \theta$, which gives the Jacobian $J(\omega, \theta) = \omega$. This implies that it is possible to reconstruct the function $f$ by 1D filtering of the Radon data with

and then applying the backprojection operator $\mathcal{R}^\ast$ to the result $q_\theta(t) = (h \ast p_\theta)(t)$. To read the PDF version of this post click here.

# From Maxwell to Helmholtz

An accurate physical model is one of the main contributing factors towards the quality of  images that can be formed with an optical tomographic microscope. Our model in [1] was based on the beam propagation method, which splits the sample into layers, and computationally propagates the light layer-by-layer from the source all the way to the detector. The starting point for our derivations was the scalar Helmholtz equation

where $\boldsymbol{x} = (x, y, z)$ denotes a spatial coordinate, $u$ is what we called the total light-field at $\boldsymbol{x}$, $\Delta \triangleq (\partial^2/\partial x^2 + \partial^2/\partial y^2 + \partial^2/\partial z^2)$ is the Laplacian, $\mathrm{I}$ is the identity operator, and $k(\boldsymbol{x})$ is the wavenumber of the field at $\boldsymbol{x}$. The spatial dependence of the wavenumber $k$ is due to variations of the speed of light $c$ induced by the inhomogeneous nature of the sample under consideration. Once we have the Helmholtz equation, we can repeat the process described in [1], and obtain an algorithm for computationally forming an image from the measured data.

However, how does the Helmholtz equation above relate to the actual Maxwell's equations?

Maxwell's equations provide the full description of the optical waves, and there are four equations to consider. The first equation is often called the Faraday's law of induction

where $\mathbf{E}$ and $\mathbf{B}$ are electric and magnetic fields, respectively, and $\nabla \times$ is the curl operator. The second equation is sometimes called the Ampère's circuital law

where $\mathbf{J}$ is the is the total current density, $\mathbf{H}$ is a magnetic field, and $\mathbf{D}$ is the electric displacement field. The total current density $\mathbf{J}$ is related to the electric field $\mathbf{E}$ as $\mathbf{J} = \sigma \mathbf{E}$, where $\sigma(\boldsymbol{x})$ is the distribution of condictivity in the sample. The magnetic field $\mathbf{H}$ is related to $\mathbf{B}$ as $\mathbf{B} = \mu \mathbf{H}$, where $\mu(\boldsymbol{x})$ is the distribution of permeability in space. Similarly, the electric displacement field $\mathbf{D}$ is related to $\mathbf{E}$ as $\mathbf{D} = \epsilon \mathbf{E}$, where $\epsilon(\boldsymbol{x})$ is the distribution of permittivity in space. The third equation is the Gauss's law

where $\nabla \cdot$ is divergence operator and $\rho$ is the distribution of charge in the sample. Finally, the fourth equation is the Gauss's law for magnetism

To obtain the scalar Helmhotz equation, we assume that the sample is charge free and non-magnetic, i.e., $\rho(\boldsymbol{x}) = 0$ and $\mu(\boldsymbol{x}) = \mu_0$. We also limit ourselves to the electric fields that have transverse magnetic (TM) polarizations, i.e., $\mathbf{E}(\boldsymbol{x},t) = (0, 0, E_z(x,y,t))$ and $\mathbf{H}(\boldsymbol{x},t) = (H_x(\boldsymbol{x},t), H_y(\boldsymbol{x},t), 0)$, and that are thus divergence free $\nabla \cdot \mathbf{E} = 0$. Then by applying the identity $\nabla \times (\nabla \times \mathbf{E}) = \nabla (\nabla \cdot \mathbf{E}) - \Delta \mathbf{E}$, and limiting ourselves to the scalar $E_z$, we obtain

The usual Helmholtz equation is then obtained by taking a temporal Fourier transform form both sides, ignoring $\omega$, and rearranging the terms

where $k^2 = \omega^2 \mu_0 \epsilon(1 - \mathrm{j}\frac{\sigma}{\omega \epsilon})$. Typically, it is more convenient to write $k^2 = k_0^2 \epsilon$, where $k_0^2 = \omega^2 \epsilon_0 \mu_0$ corresponds to the wave vector in the air, and $\epsilon$ has been modified to incorporate $\sigma$, and is thus complex.

References:

[1] U. S. Kamilov, I. N. Papadopoulos, M. H. Shoreh, A. Goy, C. Vonesch, M. Unser, and D. Psaltis, “Optical tomographic image reconstruction based on beam propagation and sparse regularization,” IEEE Trans. Comp. Imag., vol. 2, no. 1, 2016,