Methods of Optical Coherence Tomography

Fig. 1

Schematic diagram of the light traveling in an OCT system. The laser beam emitted by the source is divided at the beam splitter into two light beams; one is reflected at a mirror, the other one backscattered from the sample. The superposition of the two reflected beams is then measured at the detector

There exist different variants of the OCT regarding the way the measurements are done:

Time and frequency domain OCT:

In time domain OCT, the position of the mirror is varied and for each position one measurement is performed. On the other hand, in frequency domain OCT, the reference mirror is fixed and the detector is replaced by a spectrometer. Both methods provide equivalent measurements which are connected by a Fourier transform.

Standard and full field OCT:

In standard OCT, the incoming light is focused through objective lenses to one spot in a certain depth in the sample and the backscattered light is measured in a point detector. This means that to obtain information of the whole sample, a transversal-lateral scan has to be performed (by moving the light beam over the frontal surface of the sample). In full field OCT, the entire frontal surface of the sample is illuminated at once and the single point detector is replaced by a two-dimensional detector array, for instance by a charge-coupled device (CCD) camera.

Polarization-sensitive OCT:

In classical OCT setups, the electromagnetic wave is simply treated as a scalar quantity. In polarization-sensitive OCT, however, the illuminating light beams are polarized and the detectors measure the intensity of the two polarization components of the interfered light.

There are also further modifications such as Doppler OCT and quantum OCT, which are not addressed here. In this chapter, the focus is mainly on time domain full field OCT, but also the others are discussed.

3 The Direct Scattering Problem

To derive a mathematical model for an OCT system, one has to describe on one hand the propagation and the scattering of the laser beam in the presence of the sample and on the other hand the way how this scattered wave is measured at the detectors. For the first part, the interaction of the incoming light with the sample can be modeled with Maxwell’s macroscopic equations.

Maxwell’s Equations

Maxwell’s equations in matter consist of the partial differential equations

$\displaystyle\begin{array}{rcl} {\mathrm{div}}_{x}D(t,x) = 4\pi \rho (t,x),\qquad \qquad \qquad \qquad t \in \mathbb{R},\;x \in \mathbb{R}^{3},& &{}\end{array}$

(1a)

$\displaystyle\begin{array}{rcl} {\mathrm{div}}_{x}B(t,x) = 0,\quad \qquad \qquad \qquad \qquad t \in \mathbb{R},\;x \in \mathbb{R}^{3},& &{}\end{array}$

(1b)

$\displaystyle\begin{array}{rcl} {\mathrm{curl}}_{x}E(t,x) = -\frac{1} {c} \frac{\partial B} {\partial t} (t,x),\quad \qquad \qquad \qquad \qquad t \in \mathbb{R},\;x \in \mathbb{R}^{3},& &{}\end{array}$

(1c)

$\displaystyle\begin{array}{rcl} {\mathrm{curl}}_{x}H(t,x) = \frac{4\pi } {c}J(t,x) + \frac{1} {c} \frac{\partial D} {\partial t} (t,x),\qquad \qquad t \in \mathbb{R},\;x \in \mathbb{R}^{3},& &{}\end{array}$

(1d)

relating the following physical quantities (at some time $t \in \mathbb{R}$ and some location $x \in \mathbb{R}^{3}$ ):

Speed of light	c	$\mathbb{R}$
External charge density	ρ(t, x)	$\mathbb{R}$
External electric current density	J(t, x)	$\mathbb{R}^{3}$
Electric field	E(t, x)	$\mathbb{R}^{3}$
Electric displacement	D(t, x)	$\mathbb{R}^{3}$
Magnetic induction	B(t, x)	$\mathbb{R}^{3}$
Magnetic field	H(t, x)	$\mathbb{R}^{3}$

Maxwell’s equations do not yet completely describe the propagation of the light (even assuming that the charge density ρ and the current density J are known, there are only 8 equations for the 12 unknowns E, D, B, and H).

Additionally to Maxwell’s equations, it is therefore necessary to specify the relations between the fields D and E as well as between B and H.

Let $\Omega \subset \mathbb{R}^{3}$ denote the domain where the sample is located. It is considered as a nonmagnetic, dielectric medium without external charges or currents, this means that for all $t \in \mathbb{R}$ and all $x \in \Omega$ the electric and magnetic fields fulfil the relations

$\displaystyle\begin{array}{rcl} D(t,x)& =& E(t,x) +\int _{ 0}^{\infty }\chi (\tau,x)E(t-\tau,x){\mathrm{d}}\tau,{}\end{array}$

(2a)

$\displaystyle\begin{array}{rcl} B(t,x)& =& H(t,x),{}\end{array}$

(2b)

$\displaystyle\begin{array}{rcl} \rho (t,x)& =& 0,{}\end{array}$

(2c)

$\displaystyle\begin{array}{rcl} J(t,x)& =& 0,{}\end{array}$

(2d)

where the function $\chi: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R}^{3\times 3}$ (for convenience, χ is also defined for negative times by χ(t, x) = 0 for t < 0, $x \in \mathbb{R}^{3}$ ) is called the (electric) susceptibility and is the quantity to be recovered. The time dependence of χ hereby describes the fact that a change in the electric field E cannot immediately cause a change in the electric displacement D. Since this delay is quite small, it is sometimes ignored and χ(t, x) is then replaced by δ(t)χ(x). Moreover, the medium is often considered to be isotropic, which means that χ is a multiple of the identity matrix.

The sample is situated in vacuum and the assumptions (2) are modified by setting for all $t \in \mathbb{R}$ and all $x \in \mathbb{R}^{3}\setminus \Omega$

$\displaystyle\begin{array}{rcl} D(t,x) = E(t,x),& &{}\end{array}$

(3a)

$\displaystyle\begin{array}{rcl} B(t,x) = H(t,x),& &{}\end{array}$

(3b)

$\displaystyle\begin{array}{rcl} \rho (t,x) = 0,& &{}\end{array}$

(3c)

$\displaystyle\begin{array}{rcl} J(t,x) = 0.& &{}\end{array}$

(3d)

This simply corresponds to extend the Eq. (2) to $\mathbb{R} \times \mathbb{R}^{3}$ and to assume χ(t, x) = 0 for all $t \in \mathbb{R}$ , $x \in \mathbb{R}^{3}\setminus \Omega$ .

In this case of a nonmagnetic medium, Maxwell’s equations result into one equation for the electric field E. To get rid of the convolution in (2a), it is practical to consider the Fourier transform with respect to time. In the following, the convention

$\displaystyle{\hat{f}(\omega,x) =\int _{ -\infty }^{\infty }f(t,x){\mathrm{e}}^{{\mathrm{i}}\omega t}{\mathrm{d}}t,}$

for the Fourier transform of a function f with respect to t is used.

Proposition 1.

Let E, D, B, and H fulfil Maxwell’s equations (1) . Moreover, let assumptions (2) and (3) be satisfied. Then the Fourier transform $\hat{E}$ of E fulfils the vector Helmholtz equation

$\displaystyle{ {\mathrm{curl}}_{x}{\mathrm{curl}}_{x}\hat{E}(\omega,x) - \frac{\omega ^{2}} {c^{2}}(\mathbb{1} +\hat{\chi } (\omega,x))\hat{E}(\omega,x) = 0,\quad \omega \in \mathbb{R},\;x \in \mathbb{R}^{3}. }$

(4)

Proof.

Applying the curl to (1c) and using (1d) with the assumptions B = H and J = 0, yields

$\displaystyle{ {\mathrm{curl}}_{x}{\mathrm{curl}}_{x}E(t,x) = -\frac{1} {c} \frac{\partial {\mathrm{curl}}_{x}B} {\partial t} (t,x) = -\frac{1} {c^{2}} \frac{\partial ^{2}D} {\partial t^{2}} (t,x). }$

(5)

The Fourier transform of (2a) and (3a) and the Fourier convolution theorem (recall that χ is set to zero outside $\Omega$ ) imply that

$\displaystyle{\hat{D}(\omega,x) = (\mathbb{1} +\hat{\chi } (\omega,x))\hat{E}(\omega,x),\quad \text{for all}\quad \omega \in \mathbb{R},\;x \in \mathbb{R}^{3}.}$

Therefore, the Eq. (4) follows by taking the Fourier transform of (5). □

Initial Conditions

The sample is illuminated with a laser beam described initially (before it interacts with the sample) by the electric field $E^{(0)}: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ which is (together with some magnetic field) a solution of Maxwell’s equations (1) with the assumptions (3) for all $x \in \mathbb{R}^{3}$ . Then, it follows from the proof of the 1, for χ = 0, that

$\displaystyle{ {\mathrm{curl}}_{x}{\mathrm{curl}}_{x}\hat{E}^{(0)}(\omega,x) - \frac{\omega ^{2}} {c^{2}}\hat{E}^{(0)}(\omega,x) = 0,\quad \omega \in \mathbb{R},\;x \in \mathbb{R}^{3}. }$

(6)

Moreover, it is assumed that E ⁽⁰⁾ does not interact with the sample until the time t = 0, which means that supp $E^{(0)}(t,\cdot ) \cap \Omega =\emptyset$ for all t ≤ 0.

The electric field $E: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ generated by this incoming light beam in the presence of the sample is then a solution of Maxwell’s equations (1) with the assumptions (2) and the initial condition

$\displaystyle{ E(t,x) = E^{(0)}(t,x)\quad \text{for all}\quad t \leq 0,\;x \in \mathbb{R}^{3}. }$

(7)

Since Maxwell’s equations for E in 1 are reformulated as an equation for the Fourier transform $\hat{E}$ , it is helpful to rewrite the initial condition in terms of $\hat{E}$ .

Proposition 2.

Let E (together with some magnetic field H) fulfil Maxwell’s equations (1) with the assumptions (2) and (3) and with the initial condition (7) .

Then the Fourier transform of E − E ⁽⁰⁾ fulfils that the function $\omega \mapsto \hat{E}(\omega,x) -\hat{ E}^{(0)}(\omega,x)$ , defined on $\mathbb{R}$ , can be extended to a square integrable, holomorphic function on the upper half plane $$$\mathbb{H} =\{\omega \in \mathbb{C}\mid \mathfrak{I}{\mathrm{m}}(\omega ) > 0\}$$” src=”/wp-content/uploads/2016/04/A183156_2_En_44_Chapter_IEq30.gif”></SPAN> <SPAN class=EmphasisTypeItalic>for every</SPAN> <SPAN id=IEq31 class=InlineEquation><IMG alt=$ .

Proof.

From the initial condition (7) it follows that E(t, x) − E ⁽⁰⁾(t, x) = 0 for all t ≤ 0. Thus, the result is a direct consequence form the Paley–Wiener theorem, which is based on the fact that in this case

$\displaystyle{\hat{E}(\omega,x) -\hat{ E}^{(0)}(\omega,x) =\int _{ 0}^{\infty }(E - E^{(0)})(t,x){\mathrm{e}}^{{\mathrm{i}}\omega t}{\mathrm{d}}t}$

is well defined for all $\omega \in \mathbb{H}$ and complex differentiable with respect to $\omega \in \mathbb{H}$ . □

Remark that the electric field E is uniquely defined by (4) and 2.

The Measurements

The measurements are obtained by the combination of the backscattered field from the sample and the back-reflected field from the mirror. In practice, see Fig. 1, the sample and the mirror are in different positions. However, without loss of generality, a placement of them around the origin is assumed in the proposed formulation, in order to avoid rotating the coordinate system. To do so, the simultaneously illumination of the sample and the mirror is suppressed and two different illumination schemes are considered. The gain is to keep the same coordinate system but the reader should not be confused with illumination at different times.

Thus, the electric field E, which is obtained by illuminating the sample with the initial field E ⁽⁰⁾ (that is E solves (4) with the initial condition (7)), is combined with E _rwhich is the electric field obtained by replacing the sample by a mirror and illuminating with the same initial field E ⁽⁰⁾.

The mirror is placed orthogonal to the unit vector e ₃ = (0, 0, 1) through the point re ₃. As in (7), it is assumed that supp E ⁽⁰⁾(t, ⋅ ) does not interact with the mirror for t < 0, so that

$\displaystyle{ E_{r}(t,x) = E^{(0)}(t,x)\quad \text{for all}\quad t < 0,\;x \in \mathbb{R}^{3}. }$

(8)

Then the resulting electric field $E_{r}: \mathbb{R} \times \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ is given as the solution of the same equations as E (Maxwell’s equations (1) together with the assumptions (2) and initial condition (8)) with the susceptibility χ replaced by the susceptibility χ _rof the mirror at position r. One sort of (ideal) mirror can be described via the susceptibility χ _r(t, x) = 0 for x ₃ > r and $\chi _{r}(t,x) = C\delta (t)\mathbb{1}$ for x ₃ ≤ r with an (infinitely) large constant C > 0.

The intensity I _rof each component of the superposition of the electric fields E and E _raveraged over all time is measured at some detector points. The detectors are positioned at all points on the plane

$\displaystyle{\mathcal{D} =\{ x \in \mathbb{R}^{3}\mid x_{ 3} = d\}}$

parallel to the mirror at a distance d > 0 from the origin. The mirror and the sample are both located in the lower half plane of the detector surface with some minimal distance to $\mathcal{D}$ . Moreover, the highest possible position R ∈ (δ, d − 2δ) of the mirror shall be by some distance δ > 0 closer to the detector than the sample, this means (see Fig. 2)

$\displaystyle{ \sup _{x\in \Omega }x_{3} < R -\delta \quad \text{and}\quad r \in (-\infty,R). }$

(9)

Fig. 2

The two scattering problems involved in OCT: On the left hand side the scattering of the initial wave on the sample $\Omega$ ; on the right hand side the reference problem where the initial wave E ⁽⁰⁾ is reflected by a perfect mirror at a tunable position r ∈ (−∞, R). The two resulting electric fields, E and E _r, are then combined and this superposition E + E _ris measured at the detector surface $\mathcal{D}$

To simplify the argument, let us additionally assume that the incoming electric field E ⁽⁰⁾ does not influence the detector after the time t = 0, meaning that

$\displaystyle{ E^{(0)}(t,x) = 0\quad \text{for all}\quad t \geq 0,\;x \in \mathcal{D}. }$

(10)

At the detector array, the data are obtained by measuring

$\displaystyle{ I_{r,j}(x) =\int _{ 0}^{\infty }\vert E_{ j}(t,x) + E_{r,j}(t,x)\vert ^{2}{\mathrm{d}}t,\quad x \in \mathcal{D},\;j \in \{ 1,2,3\}. }$

(11)

In standard OCT, the polarization is usually ignored. In this case, only the total intensity $I_{r} =\sum _{ j=1}^{3}I_{r,j}$ needs to be measured, see Sect. 5 for the reconstruction formulas in the isotropic case.

In this measurement setup, it is easy to acquire besides the intensity I _ralso the intensity of the two waves E and E _rseparately by blocking one of the two waves E and E _rat a time. Practically, it is sometimes not even necessary to measure them since the intensity of the reflected laser beam E _rcan be explicitly calculated from the knowledge of the initial beam E ⁽⁰⁾, and the intensity of E is usually negligible compared with the intensity I _r(because of the assumption (10), the field E contains only backscattered light at the detector after the measurement starts). Therefore, one can consider instead of I _rthe function

$\displaystyle{ M_{r,j}(x) = \frac{1} {2}\left (I_{r,j} -\int _{0}^{\infty }\vert E_{ j}(t,x)\vert ^{2}{\mathrm{d}}t -\int _{ 0}^{\infty }\vert E_{ r,j}(t,x)\vert ^{2}{\mathrm{d}}t\right ) }$

(12)

for r ∈ (−∞, R), j ∈ { 1, 2, 3}, and $x \in \mathcal{D}$ as the measurement data.

Proposition 3.

Let the initial conditions (7) and (8) and the additional assumption (10) be satisfied. Then, for all $x \in \mathcal{D}$ , r ∈ (−∞,R), and j ∈{ 1,2,3} the measurements M _r, defined by (12), fulfil

$\displaystyle\begin{array}{rcl} M_{r,j}(x)& =& \int _{-\infty }^{\infty }(E_{ j} - E_{j}^{(0)})(t,x)(E_{ r,j} - E_{j}^{(0)})(t,x){\mathrm{d}}t{}\end{array}$

(13a)

$\displaystyle\begin{array}{rcl} =\int _{ -\infty }^{\infty }(\hat{E}_{ j} -\hat{ E}_{j}^{(0)})(\omega,x)\overline{(\hat{E}_{ r,j} -\hat{ E}_{j}^{(0)})(\omega,x)}{\mathrm{d}}\omega,& &{}\end{array}$

(13b)

Proof.

Expanding the function I _r, j, given by (11), gives

$\displaystyle{I_{r,j}(x) =\int _{ 0}^{\infty }\big(\vert E_{ j}(t,x)\vert ^{2} + \vert E_{ r,j}(t,x)\vert ^{2} + 2E_{ j}(t,x)E_{r,j}(t,x)\big){\mathrm{d}}t.}$

Thus, by the Definition (12) of M _r, it follows that

$\displaystyle{M_{r,j}(x) =\int _{ 0}^{\infty }E_{ j}(t,x)E_{r,j}(t,x){\mathrm{d}}t,}$

which, using the assumption (10), can be rewritten in the form

$\displaystyle{M_{r,j}(x) =\int _{ 0}^{\infty }(E_{ j} - E_{j}^{(0)})(t,x)(E_{ r,j} - E_{j}^{(0)})(t,x){\mathrm{d}}t.}$

Then, since E and E _rcoincide with E ⁽⁰⁾ for t < 0, see (7) and (8), the integration can be extended to all times. This proves the formula (13a) for M _r. The second formula follows from Plancherel’s theorem. □

4 Solution of the Direct Problem

In this section the solution of the direct problem, to determine the measurements M _r, defined by (13a), from the susceptibility χ, is derived using Born and far field approximation for the electric field.

Proposition 4.

Let E be a solution of the Eqs. (4) and (7) . Then, the Fourier transform $\hat{E}$ solves the Lippmann–Schwinger integral equation

$\displaystyle{ \hat{E}(\omega,x) =\hat{ E}^{(0)}(\omega,x) + \left ( \frac{\omega ^{2}} {c^{2}}\mathbb{1} + {\mathrm{grad}}_{x}{\mathrm{div}}_{x}\right )\int _{\mathbb{R}^{3}}G(\omega,x - y)\hat{\chi }(\omega,y)\hat{E}(\omega,y){\mathrm{d}}y, }$

(14)

where G is the fundamental solution of the Helmholtz equation given by

$\displaystyle{G(\omega,x) = \frac{{\mathrm{e}}^{{\mathrm{i}} \frac{\omega } { c}\vert x\vert }} {4\pi \vert x\vert },\quad x\neq 0,\,\omega \in \mathbb{R}.}$

Proof.

Equation (4) can be rewritten in the form

$\displaystyle{{\mathrm{curl}}_{x}{\mathrm{curl}}_{x}\hat{E}(\omega,x) - \frac{\omega ^{2}} {c^{2}}\hat{E}(\omega,x) =\phi (\omega,x)}$

with the inhomogeneity

$\displaystyle{ \phi (\omega,x) = \frac{\omega ^{2}} {c^{2}}\hat{\chi }(\omega,x)\hat{E}(\omega,x). }$

(15)

Using that $\hat{E}^{(0)}$ solves (6), the difference $\hat{E} -\hat{ E}^{(0)}$ satisfies the inhomogeneous vector Helmholtz equation

$\displaystyle{ {\mathrm{curl}}_{x}{\mathrm{curl}}_{x}(\hat{E} -\hat{ E}^{(0)})(\omega,x) - \frac{\omega ^{2}} {c^{2}}(\hat{E} -\hat{ E}^{(0)})(\omega,x) =\phi (\omega,x). }$

(16)

The divergence of this equation, using that ${\mathrm{div}}_{x}{\mathrm{curl}}_{x}(\hat{E} -\hat{ E}^{(0)}) = 0,$ implies

$\displaystyle{ {\mathrm{div}}_{x}(\hat{E} -\hat{ E}^{(0)})(\omega,x) = -\frac{c^{2}} {\omega ^{2}} {\mathrm{div}}_{x}\phi (\omega,x). }$

(17)

Applying the vector identity

$\displaystyle{{\mathrm{curl}}_{x}{\mathrm{curl}}_{x}(\hat{E} -\hat{ E}^{(0)}) = {\mathrm{grad}}_{ x}{\mathrm{div}}_{x}(\hat{E} -\hat{ E}^{(0)}) - \Delta _{ x}(\hat{E} -\hat{ E}^{(0)})}$

in (16) and using (17) yields

$\displaystyle{\Delta _{x}(\hat{E} -\hat{ E}^{(0)})(\omega,x) + \frac{\omega ^{2}} {c^{2}}(\hat{E} -\hat{ E}^{(0)})(\omega,x) = -\frac{c^{2}} {\omega ^{2}} {\mathrm{grad}}_{x}{\mathrm{div}}_{x}\phi (\omega,x) -\phi (\omega,x).}$

This is a Helmholtz equation for $\hat{E} -\hat{ E}^{(0)}$ and the general solution which is (with respect to ω) holomorphic in the upper half plane (equivalent to (7) by 2) is given by, see [8]

$\displaystyle\begin{array}{rcl} (\hat{E} -\hat{ E}^{(0)})(\omega,x)& =& -\frac{c^{2}} {\omega ^{2}} \int _{\mathbb{R}^{3}}G(\omega,x - y)\left ( \frac{\omega ^{2}} {c^{2}}\mathbb{1} + {\mathrm{grad}}_{y}{\mathrm{div}}_{y}\right )\phi (\omega,y){\mathrm{d}}y {}\\ & =& -\frac{c^{2}} {\omega ^{2}} \left ( \frac{\omega ^{2}} {c^{2}}\mathbb{1} + {\mathrm{grad}}_{x}{\mathrm{div}}_{x}\right )\int _{\mathbb{R}^{3}}G(\omega,x - y)\phi (\omega,y){\mathrm{d}}y. {}\\ \end{array}$

For the last equality, integration by parts and ${\mathrm{grad}}_{x}G(\omega,x - y) = -{\mathrm{grad}}_{y}G(\omega,x - y)$ were used. The Lippmann–Schwinger equation (14) follows from the last expression inserting the expression (15) for ϕ. □

This integral equation uniquely defines the electric field E. The reader is referred to [1, 8] for the isotropic case and to [33] for an anisotropic medium.

Born and Far Field Approximation

To solve the Lippmann–Schwinger equation (14), the medium is assumed to be weakly scattering, which means that $\hat{\chi }$ is sufficiently small (implying that the difference $E -\hat{ E}^{(0)}$ becomes small compared to E ⁽⁰⁾) so that the Born approximation E ⁽¹⁾, defined by

$\displaystyle\begin{array}{rcl} \hat{E}^{(1)}(\omega,x)& =& \hat{E}^{(0)}(\omega,x) + \left ( \frac{\omega ^{2}} {c^{2}}\mathbb{1} + {\mathrm{grad}}_{x}{\mathrm{div}}_{x}\right ) \\ & & \int _{\mathbb{R}^{3}}G(\omega,x - y)\hat{\chi }(\omega,y)\hat{E}^{(0)}(\omega,y){\mathrm{d}}y, {}\end{array}$

(18)

is considered a good approximation for the electric field E, see [3]. To describe multiple scattering events, one considers higher order Born approximations. For different linearization techniques, the reader is referred to [1, 23]. Moreover, since the detector in OCT is typically quite far away from the sample, one can simplify the expression (18) for the electric field at the detector array by replacing it with its asymptotic behavior for | x | → ∞, that is replace the formula for E ⁽¹⁾ by its far field approximation (the far field approximation could also be applied to the solution E of the Lippmann–Schwinger equation (14)).

Proposition 5.

Consider, for a given function $\phi: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ with compact support and some parameter $a \in \mathbb{R},$ the function

$\displaystyle{g: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3},\quad g(x) =\int _{ \mathbb{R}^{3}} \frac{{\mathrm{e}}^{{\mathrm{i}}a\vert x-y\vert }} {\vert x - y\vert }\phi (y){\mathrm{d}}y.}$

Then, it follows, asymptotically for ρ →∞ and uniformly in $\vartheta \in S^{2},$ that

$$$\displaystyle{ (a^{2} + {\mathrm{grad}}_{ x}{\mathrm{div}}_{x})g(\rho \vartheta ) \simeq -\frac{a^{2}{\mathrm{e}}^{{\mathrm{i}}a\rho }} {\rho } \int _{\mathbb{R}^{3}}\vartheta \times (\vartheta \times \phi (y)){\mathrm{e}}^{-{\mathrm{i}}a\left <\vartheta,y\right >}{\mathrm{d}}y }$$” src=”/wp-content/uploads/2016/04/A183156_2_En_44_Chapter_Equ30.gif”></DIV></DIV><br /> <DIV class=EquationNumber>(19)</DIV></DIV></DIV></DIV><br /> <DIV class=$

Proof.

Consider the function

$\displaystyle{\Gamma: \mathbb{R}^{3} \rightarrow \mathbb{C},\quad \Gamma (x) = \frac{{\mathrm{e}}^{{\mathrm{i}}a\vert x\vert }} {\vert x\vert }.}$

Then

$\displaystyle\begin{array}{rcl} \frac{\partial ^{2}\Gamma } {\partial x_{j}\partial x_{k}}(x)& =& \frac{\partial } {\partial x_{j}}\left [\left ( \frac{{\mathrm{i}}a} {\vert x\vert ^{2}} - \frac{1} {\vert x\vert ^{3}}\right )x_{k}{\mathrm{e}}^{{\mathrm{i}}a\vert x\vert }\right ] {}\\ & =& \left [\left ( \frac{{\mathrm{i}}a} {\vert x\vert ^{2}} - \frac{1} {\vert x\vert ^{3}}\right )\delta _{jk} + \left ( \frac{{\mathrm{i}}a} {\vert x\vert ^{2}} - \frac{1} {\vert x\vert ^{3}}\right )\frac{{\mathrm{i}}ax_{j}x_{k}} {\vert x\vert } \right. {}\\ & & +\left.\left (-2 \frac{{\mathrm{i}}a} {\vert x\vert ^{3}} + 3 \frac{1} {\vert x\vert ^{4}}\right )\frac{x_{j}x_{k}} {\vert x\vert } \right ]{\mathrm{e}}^{{\mathrm{i}}a\vert x\vert }. {}\\ \end{array}$

Therefore, writing x in spherical coordinates: $x =\rho \vartheta$ with ρ > 0, $\vartheta \in S^{2}$ , for ρ → ∞ uniformly in $\vartheta,$ it can be seen that

$\displaystyle{ \frac{\partial ^{2}\Gamma } {\partial x_{j}\partial x_{k}}(\rho \vartheta ) = -\frac{a^{2}{\mathrm{e}}^{{\mathrm{i}}a\rho }} {\rho } \vartheta _{j}\vartheta _{k} + \mathcal{O}\left (\frac{1} {\rho ^{2}} \right ),}$

The approximation (locally uniformly in $y \in \mathbb{R}^{3})$

$$$\displaystyle{\vert \rho \vartheta - y\vert =\rho \sqrt{\vert \vartheta \vert ^{2 } - \frac{2} {\rho } \left < \vartheta,y\right > + \frac{1} {\rho ^{2}} \vert y\vert ^{2}} =\rho -\left < \vartheta,y\right > + \mathcal{O}\left (\frac{1} {\rho } \right ),}$$” src=”/wp-content/uploads/2016/04/A183156_2_En_44_Chapter_Equo.gif”></DIV></DIV></DIV>implies that (again uniformly in <SPAN id=IEq57 class=InlineEquation><IMG alt=$ )

$$$\displaystyle{ \frac{\partial ^{2}\Gamma } {\partial x_{j}\partial x_{k}}(\rho \vartheta -y) = -\frac{a^{2}{\mathrm{e}}^{{\mathrm{i}}a(\rho -\left <\vartheta,y\right >)}} {\rho } \vartheta _{j}\vartheta _{k} + \mathcal{O}\left (\frac{1} {\rho } \right ).}$$” src=”/wp-content/uploads/2016/04/A183156_2_En_44_Chapter_Equp.gif”></DIV></DIV></DIV>Now, considering the compact support of <SPAN class=EmphasisTypeItalic>ϕ</SPAN> and using that <SPAN id=IEq58 class=InlineEquation><IMG alt=$

$\displaystyle\begin{array}{rcl} ({\mathrm{grad}}_{x}{\mathrm{div}}_{x}g)_{j}(x)& =& \sum _{k=1}^{3} \frac{\partial } {\partial x_{j}}\int _{\mathbb{R}^{3}} \frac{\partial \Gamma } {\partial x_{k}}(x - y)\phi _{k}(y){\mathrm{d}}y {}\\ & =& \int _{\mathbb{R}^{3}}\sum _{k=1}^{3} \frac{\partial ^{2}\Gamma } {\partial x_{j}\partial x_{k}}(x - y)\phi _{k}(y){\mathrm{d}}y. {}\\ \end{array}$

Asymptotically for | x | → ∞ (again using the compact support of ϕ) one obtains

$$$\displaystyle{a^{2}g_{ j}(\rho \vartheta ) + ({\mathrm{grad}}_{x}{\mathrm{div}}_{x}g)_{j}(\rho \vartheta ) \simeq a^{2}\int _{ \mathbb{R}^{3}}\sum _{k=1}^{3}\frac{{\mathrm{e}}^{{\mathrm{i}}a(\rho -\left <\vartheta,y\right >)}} {\rho } \left (\delta _{jk} -\vartheta _{j}\vartheta _{k}\right )\phi _{k}(y){\mathrm{d}}y.}$$” src=”/wp-content/uploads/2016/04/A183156_2_En_44_Chapter_Equq.gif”></DIV></DIV></DIV>The approximation (<SPAN class=InternalRef><A href=$ 19) follows from the vector identity $$$\vartheta \times (\vartheta \times \phi ) = \left < \vartheta,\phi \right >\vartheta -\vert \vartheta \vert ^{2}\phi$$” src=”/wp-content/uploads/2016/04/A183156_2_En_44_Chapter_IEq59.gif”></SPAN> and <SPAN id=IEq60 class=InlineEquation><IMG alt=$ □

The application of both the far field and the Born approximation, this means 5 for the expression (18) of E ⁽¹⁾, that is setting a = ω∕c and $\phi = \tfrac{1} {4\pi }\hat{\chi }\hat{E}^{(0)}$ in 5, imply the asymptotic behavior

$$$\displaystyle{ \hat{E}^{(1)}(\omega,\rho \vartheta ) \simeq \hat{ E}^{(0)}(\omega,\rho \vartheta ) -\frac{\omega ^{2}{\mathrm{e}}^{{\mathrm{i}} \frac{\omega } {c}\rho }} {4\pi \rho c^{2}} \int _{\mathbb{R}^{3}}\vartheta \times \big (\vartheta \times (\hat{\chi }(\omega,y)\hat{E}^{(0)}(\omega,y))\big){\mathrm{e}}^{-{\mathrm{i}} \frac{\omega } {c}\left <\vartheta,y\right >}{\mathrm{d}}y. }$$” src=”/wp-content/uploads/2016/04/A183156_2_En_44_Chapter_Equ33.gif”></DIV></DIV><br /> <DIV class=EquationNumber>(20)</DIV></DIV></DIV></DIV><br /> <DIV id=Sec9 class=$

The Forward Operator

To obtain a forward model for the measurements described in Sect. 3, the (approximative) formula (20) is considered as a model for the solution of the scattering problem. To make this formula concrete, one has to plug in a function E ⁽⁰⁾ describing the initial illumination (recall that E ⁽⁰⁾ has to solve (6)).

The specific illumination is a laser pulse propagating in the direction − e ₃, orthogonal to the detector surface $\mathcal{D} =\{ x \in \mathbb{R}^{3}\mid x_{3} = d\}$ , this means

$\displaystyle{ E^{(0)}(t,x) = f(t + \tfrac{x_{3}} {c} )p, }$

(21)

which solves Maxwell’s equations (1) with the assumptions (3) for some fixed vector $p \in \mathbb{R}^{3},$ with $$$p_{3} = \left < p,e_{3}\right > = 0,$$” src=”/wp-content/uploads/2016/04/A183156_2_En_44_Chapter_IEq64.gif”></SPAN> describing the polarization of the initial laser beam.</DIV><br /> <DIV id=FPar6 class=$

Proposition 6.

The function E ⁽⁰⁾, defined by (21) with $$$\left < p,e_{3}\right > = 0$$” src=”/wp-content/uploads/2016/04/A183156_2_En_44_Chapter_IEq65.gif”></SPAN>, <SPAN class=EmphasisTypeItalic>solves together with the magnetic field H</SPAN> <SUP>(0)</SUP>, <SPAN class=EmphasisTypeItalic>defined by</SPAN><br /> <DIV id=Equr class=Equation><br /> <DIV class=EquationContent><br /> <DIV class=MediaObject><IMG alt=$

Maxwell’s equations (1) in the vacuum, that is with the additional assumptions (3) .

Proof.

The four equations of (1) can be directly verified:

$$$\displaystyle\begin{array}{rcl} {\mathrm{div}}_{x}E^{(0)}(t,x)& =& \frac{1} {c}f'(t + \tfrac{x_{3}} {c} )\left < e_{3},p\right > = 0, {}\\ {\mathrm{div}}_{x}H^{(0)}(t,x)& =& \frac{1} {c}f'(t + \tfrac{x_{3}} {c} )\left < e_{3},p \times e_{3}\right > = 0, {}\\ {\mathrm{curl}}_{x}E^{(0)}(t,x)& =& \frac{1} {c}f'(t + \tfrac{x_{3}} {c} )e_{3} \times p = -\frac{1} {c} \frac{\partial H^{(0)}} {\partial t} (t,x), {}\\ {\mathrm{curl}}_{x}H^{(0)}(t,x)& =& \frac{1} {c}f'(t + \tfrac{x_{3}} {c} )e_{3} \times (p \times e_{3}) = \frac{1} {c}f'(t + \tfrac{x_{3}} {c} )p = \frac{1} {c} \frac{\partial E^{(0)}} {\partial t} (t,x). {}\\ \end{array}$$” src=”/wp-content/uploads/2016/04/A183156_2_En_44_Chapter_Equ35.gif”></DIV></DIV></DIV> □ </DIV></DIV><br /> <DIV class=Para>To guarantee that the initial field <SPAN class=EmphasisTypeItalic>E</SPAN> <SUP>(0)</SUP> (and also the magnetic field <SPAN class=EmphasisTypeItalic>H</SPAN> <SUP>(0)</SUP>) does not interact with the sample or the mirror for <SPAN class=EmphasisTypeItalic>t</SPAN> ≤ 0 and neither contributes to the measurement at the detectors for <SPAN class=EmphasisTypeItalic>t</SPAN> ≥ 0 as required by (<SPAN class=InternalRef><A href=$ 8) and (10) the vertical distribution $f: \mathbb{R} \rightarrow \mathbb{R}$ should satisfy (see Fig. 2)

$\displaystyle{ {\mathrm{supp}}f \subset (\tfrac{R} {c}, \tfrac{d} {c}). }$

(22)

In the case of an illumination E ⁽⁰⁾ of the form (21), the electric field E _rproduced by an ideal mirror at the position r is given by

$$$\displaystyle{ E_{r}(t,x) = \left \{\begin{array}{@{}l@{\quad }l@{}} \big(f(t + \frac{x_{3}} {c} ) - f(t + \frac{x_{3}} {c} + 2\,\frac{r-x_{3}} {c} )\big)p\quad &\text{if}\;x_{3} > r,\\ 0 \quad &\text{if} \;x_{ 3} \leq r. \end{array} \right. }$$” src=”/wp-content/uploads/2016/04/A183156_2_En_44_Chapter_Equ37.gif”></DIV></DIV><br /> <DIV class=EquationNumber>(23)</DIV></DIV>This just corresponds to the superposition of the initial wave with the (orthogonally) reflected wave, which travels additionally the distance <SPAN id=IEq67 class=InlineEquation><IMG alt=$ . The change in polarization of the reflected wave (from p to − p) comes from the fact that the tangential components of the electric field have to be continuous across the border of the mirror.

The following proposition gives the form of the measurements M _r, described in Sect. 3, on the detector surface $\mathcal{D}$ for the specific illumination (21).

Proposition 7.

Let E ⁽⁰⁾ be an initial illumination of the form (21) satisfying (22) . Then, the equations for the measurements M _r from 3 are given by

$\displaystyle\begin{array}{rcl} M_{r,j}(x)& =& -p_{j}\int _{-\infty }^{\infty }(E_{ j} - E_{j}^{(0)})(t,x)f(t + \tfrac{2r-x_{3}} {c} ){\mathrm{d}}t,{}\end{array}$

(24a)

$\displaystyle\begin{array}{rcl} & =& -\frac{p_{j}} {2\pi } \int _{-\infty }^{\infty }(\hat{E}_{ j} -\hat{ E}_{j}^{(0)})(\omega,x)\hat{f}(-\omega ){\mathrm{e}}^{{\mathrm{i}} \frac{\omega } {c} (2r-x_{3})}{\mathrm{d}}\omega {}\end{array}$

(24b)

for all j ∈{ 1,2,3}, r ∈ (−∞,R), and $x \in \mathcal{D}$ .

Proof.

Since the electric field E _rreflected on a mirror at vertical position r ∈ (−∞, R) is according to (23) given by

$\displaystyle{E_{r}(t,x) =\big (f(t + \tfrac{x_{3}} {c} ) - f(t + \tfrac{2r-x_{3}} {c} )\big)p\quad \text{for all}\quad t \in \mathbb{R},\;x \in \mathcal{D},}$

the measurement functions M _r(defined by (12) and computed with (13a)) are simplified, for the particular initial illumination E ⁽⁰⁾ of the form (21), to (24a) for $x \in \mathcal{D}.$

Since formula (24a) is just a convolution, the electric field E − E ⁽⁰⁾ can be rewritten, in terms of its Fourier transform, in the form

$\displaystyle{M_{r,j}(x) = -\frac{p_{j}} {2\pi } \int _{-\infty }^{\infty }\int _{ -\infty }^{\infty }(\hat{E}_{ j} -\hat{ E}_{j}^{(0)})(\omega,x){\mathrm{e}}^{-{\mathrm{i}}\omega t}f(t + \tfrac{2r-x_{3}} {c} ){\mathrm{d}}\omega {\mathrm{d}}t.}$

Interchanging the order of integration and applying the Fourier transform $\hat{f}$ of f, it follows Eq. (24b). □

In the limiting case of a delta impulse as initial wave, that is for f(ξ) = δ(ξ −ξ ₀) with some constant $\xi _{0} \in (\frac{R} {c}, \frac{d} {c})$ satisfying (22), the measurements provide directly the electric field. Indeed, it can be seen from (24a) that

$\displaystyle{M_{r,j}(x) = -p_{j}(E_{j} - E_{j}^{(0)})(\tfrac{x_{3}-2r} {c} +\xi _{0},x).}$

By varying r ∈ (−∞, R), the electric field E can be obtained (to be more precise, its component in direction of the initial polarization) as a function of time at every detector position.

The following assumptions are made:

Assumption 1.

The susceptibility χ is sufficiently small so that the Born approximation E ⁽¹⁾ for the solution E of the Lippmann–Schwinger equation ( 14 ) can be applied.

Assumption 2.

The detectors are sufficiently far away from the object so that one can use the far field asymptotics ( 20 ) for the measured field.

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