# Complex Refractive Index


Generally, optics based on the electromagnetics is described with the refractive index, while such index does not barely appear in X-ray diffraction (XRD) from the kinematical approach, presumably because of the different historical background; however, it is of great importance to see their relationship to understand the diffraction anomalous fine structure (DAFS) method. For example, the energy dependence of $$f\pprime$$ term is not completely equivalent to a linear absorption coefficient, $$\mu$$; $$f\pprime$$ should be divided by the photon energy in order to treat it as the absorption spectrum equivalent to $$\mu$$ in the theoretical framework and analysis of the XAFS field. Thus, the following section briefly describes the relationship between the complex atomic scattering factor derived before and the conventional complex refractive index.

The refractive index is defined as the ratio of the wave numbers in a material and a vacuum as follows:

\begin{align}
\tilde{n} \equiv k / K = 1 – \delta + i\beta = n + i\beta,
\label{Eq:refractive_index}
\tag{1}
\end{align}

where $$K$$ and $$k$$ are the wave numbers in the vacuum and the material, respectively, $$n$$ is the real part of the refractive index and $$\delta$$ corresponds to the its discrepancy from 1, and $$\beta$$ is the imaginary part of the refractive index. Note that the sign in the imaginary part depends on the definition of the wavefunction; the above description is based on the wavefunction of $$\exp \left\{ i \left( \vec{k}\cdot \vec{r} – \omega t \right) \right\}$$. Thus, the wave number in the materials is written as

\begin{align}
k = \tilde{n} K,
\label{Eq:wave_number_in_material}
\tag{2}
\end{align}

and consequently the electric field in the material is calculated as

\begin{align}
E &= E_{0} \exp \left\{ i (kx – \omega t)\right\} \notag \\
%&= E_{0} \exp \left\{ i(\tilde{n} K x – \omega t)\right\} \notag \\
%&= E_{0} \exp \left\{ i (nKx – \omega t) \right\} \exp (-\beta K x) \notag \\
&= E_{0} \exp \left\{ i(Kx-\omega t) \right\} \exp (-i \delta K x) \exp (- \beta K x) .
\label{Eq:wave_in_material}
\tag{3}
\end{align}

The second and the third terms in the last line of the above equation indicate the phase shift and the absorption, respectively. The absorption is further described with $$\beta$$ in the intensity (i.e., proportional to the square of the electric field ) as

\begin{align}
I(x) = I(0) \exp (- 2\beta K x).
\label{Eq:absorption_due_to_refractive_index}
\tag{4}
\end{align}

Since the absorption is also written as $$I(x) = I(0) \exp (- \mu x)$$ with the linear absorption coefficient, $$\mu$$, the relationship between $$\beta$$ and $$\mu$$ is obtained by comparing the exponential parts as

\begin{align}
\beta = \frac{\mu}{2K} = \frac{\lambda} {4\pi} \mu.
\label{Eq:absorption_coefficient}
\tag{5}
\end{align}

The complex refractive index is also written in the form of

\begin{align}
\tilde{n} = \sqrt{\frac{\epsilon \mu_{m}}{\epsilon_{0} \mu_{m0}}},
\label{Eq:full_refractive_index}
\tag{6}
\end{align}

where $$\epsilon$$ and $$\mu_{m}$$ are dielectric constants and magnetic permeability, respectively. A subscript 0 denotes the values in the vacuum. The material is magnetically equivalent to the vacuum. Thus, since $$\mu_{m} = \mu_{m0}$$, the above index can be reduced as

\begin{align}
\tilde{n} = \sqrt{\frac{\epsilon }{\epsilon_{0}}}.
\end{align}

The dielectric constant, $$\epsilon$$, is also related to an electric susceptibility, $$\chi$$, with an equation of

\begin{align}
\epsilon = \epsilon_{0} (1 + \chi_{e}).
\end{align}

### Connection between the refractive index and the scattering factors

The refractive index is related to the atomic scattering factor as follows. Electric dipole moment, $$P_{e}$$, is written as $$P_{e}= \epsilon_{0} \chi_{e} E$$. At the same time, $$P_{e}$$ is also described as $$-n_{s} ex$$, where $$n_{s}$$ is the volume density of the dipoles. Then,

\begin{align}
P_{e} = – n_{s} e x_{0} = \epsilon_{0} \chi_{e} E_{0}.
\end{align}

Therefore, $$\chi_{e}$$ is further calculated with the amplitude of the forced oscillator described in Resonant Scattering as follows:

\begin{align}
\chi_{e} &= \frac{-n_{s} e}{\epsilon_{0} E_{0}} x_{0} \notag \\
&= \frac{-n_{s} e}{\epsilon_{0} E_{0}} \left(- \frac{eE_{0}}{m} \frac{1}{(\omega_{s}^{2} – \omega^{2} -i\omega\Gamma)} \right) \notag \\
&=\frac{n_{s} e^{2} }{\epsilon_{0} m} \frac{1}{(\omega_{s}^{2} – \omega^{2} -i\omega\Gamma)} \notag \\
& = \frac{n_{s} e^{2} }{\epsilon_{0} m \omega^{2}} \frac{- \omega^{2}}{(\omega_{s}^{2} – \omega^{2} -i\omega\Gamma)}.
\end{align}

Furthermore, the last term can be replaced by the atomic scattering factor of the single oscillator. Then,

\begin{align}
\chi_{e} & = – \frac{n_{s} e^{2} \lambda^{2}}{\epsilon_{0} m (2\pi c)^{2}} f_{s} \notag \\
& = – \frac{r_{0} }{\pi} n_{s} \lambda^{2} f_{s}. \notag \\
\end{align}

Thus, we obtain the relationship between the electric susceptibility and the dispersion correction term of the atomic scattering factor.

On the other hand, the complex refractive index is written with the electric susceptibility, $$\chi_{e}$$, by assuming $$\chi_{e} \ll 1$$ and consequently with the dispersion correction term as follows:

\begin{align}
\tilde{n} &= \left( \frac{\epsilon}{\epsilon_{0}} \right)^{\frac{1}{2}} \notag \\
&= \left( 1 + \chi_{e} \right)^{\frac{1}{2}} \notag \\
&\sim 1 + \frac{1}{2} \chi_{e} \notag \\
&= 1- \frac{r_0}{2\pi} \lambda^{2} n_{s} f_{s}.
\label{Eq:refractive_index_with_fs}
\tag{7}
\end{align}

In the scope of the single forced oscillator mode, this refractive index is expressed in the form of

\begin{align}
\tilde{n} = 1- \frac{2\pi r_{0} n_{s} c^{2}}{\omega^{2} – \omega_{s}^{2} + i\Gamma \omega}.
\label{Eq:refractive_index_of_oscillator_model}
\tag{8}
\end{align}

Again, the refractive index is also affected by the bound of the electron to the nucleus as seen in the atomic scattering factor.

Practically, $$f_{s}$$ is replaced by the complex atomic scattering factor determined from experiments, i.e., $$f_j (\vec{Q}, E) = f^{0}_{j}(\vec{Q}) + f’_{j}(E) + i f\pprime_{j}(E)$$. When assuming the forward scattering, i.e., $$\vec{Q} = 0$$, $$f^{0}$$ value is identical to the atomic number, $$Z$$, and then

\begin{align}
\tilde{n}= 1- \frac{r_0}{2\pi} \lambda^{2} \sum_j n_{j} \left( Z_{j} + f’_{j} + if”_{j} \right),
\label{Eq:refractive_index_with_scattering_factor}
\tag{9}
\end{align}

where $$n_{j}$$ denotes the number of atoms of element $$j$$ in a unit volume. If the material is a crystal, Eq. \eqref{Eq:refractive_index_with_scattering_factor} is also rewritten with the unit cell volume, $$v_{c}$$ as

\begin{align}
\tilde{n}= 1- \frac{r_0}{2\pi v_{c}} \lambda^{2} F(Q=0, E) =1- \frac{r_0}{2\pi v_{c}} \lambda^{2} \sum_j \left( Z_{j} + f’_{j} + if\pprime_{j} \right).
\label{Eq:refractive_index_with_structure_factor}
\tag{10}
\end{align}

Thus, the real and imaginary pars of the complex refractive index are obtained by comparing each pat in the above equation and Eq. \eqref{Eq:refractive_index} as follows

\begin{align}
\delta = \frac{r_0}{2\pi v_{c}} \lambda^{2} \sum_j \left( Z_{j} + f’_{j}\right),\ \ \beta = -\frac{r_0}{2\pi v_{c}} \lambda^{2} \sum_j f\pprime_{j}.
\label{Eq:each_parts_of_refractive_index_with_structure_factor}
\tag{11}
\end{align}

Furthermore, $$\mu$$ is linked with $$f\pprime$$ by Eq. \eqref{Eq:absorption_coefficient} by

\begin{align}
\mu = \frac{4\pi}{\lambda} \beta = \frac{2\lambda r_{0}}{v_{c}} \sum_j \left( -f\pprime_{j} \right).
\label{Eq:mu_and_f_double_prime}
\tag{12}
\end{align}

Therefore, $$f\pprime$$ obtained from the DAFS method is completely equivalent to $$\mu$$ by multiplying wavelength, $$\lambda$$, to $$f\pprime$$ (or divided by the photon energy). Importantly, $$f\pprime$$ is a negative value because $$\mu$$ is positive from Eq. \eqref{Eq:mu_and_f_double_prime}. Conventionally, $$f\pprime$$ appears as a positive value in many textbooks and/or tables. It causes no problem as long as we discuss the diffraction intensity, where only the square of $$f\pprime$$ appears in the intensity; however, it is of importance for the DAFS method to distinguish the sign of $$f\pprime$$ values because we need to extract the site- and/or phase dependent $$f\pprime$$ value itself by directly solving the simultaneous equation of the weighted $$f\pprime$$ values.

###### For further study…(This article was written based on the following books)
• J. Als-Nielsen and D. McMorrow, Elements of Modern X-ray Physics, 2nd ed., John Wiley & Sons, 2011.
• 菊田惺志(S. Kikuta), X線散乱と放射光科学 基礎編 (X-ray scattering and synchrotron radiation science -basics- (English title was translated by T.K.)), 東京大学出版 (University of Tokyo Press), 2011.