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Part 1: The wave vector
Part 2: The interface
Part 3: The s-polarization
Part 4: The p-polarization
Part 5: Multilayers

Here we will start to do things bit differently. For the p-polarization, the electric field vector lies in the plane of incidence and the magnetic field has only y-component. Consider the following figure:

The usual textbook definition of p-polarized Fresnel’s coefficients is \(r_p = E_{Rp}/E_{Ip} \) and \(t_p = E_{Tp}/E_{Ip} \), where the amplitudes \(E_{jp}\) are measured in the right-handed \(\boldsymbol\kappa,\mathbf{P},\mathbf{S}\) basis attached to the beams, so the reflected p-polarized direction is opposite from we have here (see Fresnel vs Verdet blog).

We will define here the \(r_p = E_{Rx}/E_{Ix} \) in terms of x-components, which is, in Fresnel’s convention,

\(
\begin{align*}
r_p = \frac{E_{Rx}}{E_{Ix}} \equiv \frac{E_{Rp}}{E_{Ip}},
\end{align*}
\)

since the angle of reflection is equal to angle of incidence. We will, however, define the transmission coefficient as

\(
\begin{align}
t_{px} = \frac{E_{Tx}}{E_{Ix}},
\end{align}
\)

and keep the index \(px\) to highlight that this definition differs from the textbook \(t_p\). We should not have any trouble with this definition, since in typical ellipsometry experiment, we do not measure the transmitted beam, and if we do, we do it back again in the ambient medium, so the angle of refraction is the same as angle of incidence, assuming of course we have planar and parallel interfaces on our sample.

We have two conditions for matching the in-plane field components:

\(
\begin{align}
\begin{aligned}
E_{Ix}+E_{Rx} &= E_{Tx},
\\
H_{Iy}+H_{Ry} &= H_{Ty},
\end{aligned}\label{eqFieldMatchP}
\end{align}
\)

So, dividing first with \(E_{Ix}\) we obtain

\( \begin{align}\label{eqDefCoefsP} 1+r_{p} &= t_{px}. \end{align} \)

where, thanks to the \(t_{px}\) definition and Fresnel’s convention, we’ve got condition analogical to the Part 3 eq. (2) valid for the s-polarization. We will also proceed in the same way, that is we will find the \(r_p\) and calculate \(t_{px}\) using \eqref{eqDefCoefsP}.

Using second Maxwell’s equation in the form of eqs. (9) in Part 1:

\(
\begin{align}
\varepsilon_j\mathbf{E}_j = – \boldsymbol\kappa_j \times \mathbf{H}_j = (\kappa_j H_{jy}, 0 ,-\xi H_{jy}),
\end{align}
\)

where again \( j = I,R,T \) and \( \kappa_I \equiv q \), \( \kappa_R \equiv -q \), \( \kappa_T \equiv \kappa \). We get the components explicitly

\(
\begin{align}
\begin{aligned}
E_{Ix} &= q H_{Iy},
& E_{Iz} &= – \xi H_{Iy},
\\
E_{Rx} &= – q H_{Ry},
& E_{Rz} &= – \xi H_{Ry},
\\
E_{Tx} &= \frac{\kappa H_{Ty}}{\varepsilon},
& E_{Tz} &= – \frac{\xi H_{Ty}}{\varepsilon},
\end{aligned}\label{eqHEcomponents}
\end{align}
\)

from where we get that \(r_p = E_{Rx}/E_{Ix} = – H_{Ry}/H_{Iy} \), so we take the first equation of \eqref{eqFieldMatchP}, replace the \(E_{jx} \) and have

\(
\begin{align*}
\varepsilon q H_I – \varepsilon q H_R = \kappa H_T,
\end{align*}
\)

where we dropped the \(y\) index. Finally, we use the second equation of \eqref{eqFieldMatchP} to get rid of the \(H_{T}\):

\(
\begin{align*}
\begin{aligned}
\varepsilon q H_I – \varepsilon q H_R &= \kappa H_I + \kappa H_R,
\\
(\varepsilon q – \kappa ) H_I &= (\varepsilon q + \kappa ) H_R.
\end{aligned}
\end{align*}
\)

Merrily getting the \(r_p\):

\( \begin{align} r_p = \frac{E_{Rx}}{E_{Ix}} = – \frac{H_{R}}{H_{I}} = \frac{ \kappa – \varepsilon q }{ \kappa + \varepsilon q }, \end{align} \)

and the \( t_{px} \) is obtained from \eqref{eqDefCoefsP}:

\( \begin{align} t_{px} = 1 + r_p = 1 + \frac{ \kappa – \varepsilon q }{ \kappa + \varepsilon q } = \frac{ 2 \kappa}{ \kappa + \varepsilon q }, \end{align} \)

And that’s it. We have derived the Fresnel’s coefficients in terms of the in-plane (\(x,y\)) components of the electric field amplitudes. We will see later that Berreman’s theory of light propagating in layers of very general electromagnetic media uses the same approach, so we will be rewarded with consistent framework.

---

Let me add some slightly confusing remarks here. Question is: what would happen, if we try to define the coefficients for p-polarization in terms of the \(H_{jy}\) field amplitudes, i.e. abusing the fact that \(\bf{H_j}\) have only y-components. We have a look on the second equation of \eqref{eqFieldMatchP}. Sure we can divide it by \(H_{Iy}\), getting:

\(
\begin{align*}
1 + \frac{H_{Ry}}{H_{Iy}} &= \frac{H_{Ty}}{H_{Iy}},
\\
1 + r_H &= t_H,
\end{align*}
\)

where we defined new pair of coefficients \( r_H \), \( t_H \). Using our results above, we get

\(
\begin{align*}
r_H = \frac{ \varepsilon q – \kappa}{ \varepsilon q + \kappa },\quad
t_H = \frac{ 2 \varepsilon q}{ \varepsilon q + \kappa }.
\end{align*}
\)

In this form we have pair of coefficients, which is completely symmetric with the \(r_s\), \(t_s\) from Part 3, if we exchange \( \varepsilon \) and \( \mu \). Which is nice, but would turn too disruptive later, when we will write the general Jones matrix with cross-polarization elements. Note that such \(r_H\) is also equivalent to \(r_p\) written in the Verdet’s convention, as it usually appears in textbooks.

So what is the textbook \(t_{pB}\) (here with index \(B\) as book)? It should be defined in terms of p-amplitudes of electric fields as \(t_{pB} = E_{Tp}/E_{Ip} \), so it assumes that we can define the

\(
\begin{align*}
E_{Tp} \equiv \sqrt{{E_{Tx}}^2 + {E_{Tz}}^2} = \frac{H_{Ty}}{\varepsilon} \sqrt{{\kappa}^2 + {\xi}^2} = \frac{H_{Ty}}{\varepsilon} \sqrt{\varepsilon \mu},
\end{align*}
\)

where we used the components from \eqref{eqHEcomponents}, while \(E_{Ip} = H_{Iy} \) in vacuum. So

\(
\begin{align*}
t_{pB} = \frac{E_{Tp}}{E_{Ip}} = \frac{H_{Ty}}{H_{Iy}} \frac{\sqrt{\varepsilon \mu}}{\varepsilon} = t_H \frac{\sqrt{\varepsilon \mu}}{\varepsilon} = \frac{ 2 \sqrt{\varepsilon \mu} q}{ \varepsilon q + \kappa }.
\end{align*}
\)

Madness. Mostly, I don’t like that here we no longer have the simple coupling \( 1+r=t\).

$\setCounter{0}$
Part 1: The wave vector
Part 2: The interface
Part 3: The s-polarization
Part 4: The p-polarization
Part 5: Multilayers

In the previous posts we’ve shown that Maxwell’s equations support plane waves, only restrict the wave vector. Then we’ve seen that on planar interface between two media, the direction of the wave vector must change, due to continuity of the field components. We work with isotropic medium, so these findings are independent on polarization of the fields.

What we want now, is to find – for given amplitude of the incident wave – the amplitudes of the reflected and transmitted waves. The incoming wave can be in any polarization state, which we can write as linear combination of two orthogonal polarization states. Due to the symmetry of the problem, the obvious choice is two linear polarizations: one with electric field perpendicular to the plane of incidence, called s-polarization (s is from German senkrecht ), or TE (transverse electric); second with electric field parallel to the plane of incidence, p-polarization, or TM (transverse magnetic).

In this blog we will have a look on the s-polarization. Following figure illustrates the situation:

One has to keep in mind that this picture is just schematic and corresponds to the case of simple dielectric with \( \varepsilon > 1 \). Important is that the s-polarized \(\bf{E}\) field has only y-component, so we will measure the field amplitudes in the positive direction of \(y\). We have plotted the magnetic field oriented so the vectors \(\mathbf{E},\mathbf{H},\boldsymbol{k}\) form a right-hand system, as we’ve seen previously. What is not captured in this figure, is the actual relative magnitude of the amplitudes, particularly the fact that the reflected amplitude should be much smaller than the incident and will have opposite sign. This we will see at the end of this post.

For all three waves, the electric field vector will be \( \mathbf{E}_j = (0,E_{j},0) \), where we will drop the index \(y\), and the magnetic field lies in the plane of incidence \( \mathbf{H}_j = (H_{jx},0,H_{jz}) \). The index \( j = I,R,T \) stands for the incident, reflected and transmitted wave.

So for given amplitude \( E_I \) we have two unknown amplitudes \( E_R \) and \( E_T \). The remaining components of the \( \mathbf{H} \) field can be resolved afterwards with the help of equations we already know from the Part 1 and Part 2. What we need here are two equations for conservation of the in-plane components of the \( \mathbf{E} \) and \( \mathbf{H} \), that is

\(
\begin{align}
\begin{aligned}
E_I+E_R &= E_T,
\\
H_{Ix}+H_{Rx} &= H_{Tx},
\end{aligned}\label{eqFieldMatch}
\end{align}
\)

We can divide the first eqation by \( E_I \), getting

\(
\begin{align*}
1+\frac{E_R}{E_I} &= \frac{E_T}{E_I},
\end{align*}
\)

or,

\( \begin{align}\label{eqDefCoefs} 1+r_s &= t_s. \end{align} \)

Here we defined the Fresnel’s coefficients for reflection, \( r_s = E_{Ry}/E_{Iy} \), and transmission, \( t_s = E_{Ty}/E_{Iy} \), for the s-polarization. For clarity I’ve added the index \(y\) back again.

In previous Part 2, equation (3), we have written the wave vectors \( \boldsymbol\kappa_j \). We will insert them into the third Maxwell’s equation in the form of eqs. (9) in Part 1:

\(
\begin{align}
\mu_j\mathbf{H}_j = \boldsymbol\kappa_j \times \mathbf{E}_j = (-\kappa_j E_{jy}, 0 , \xi E_{jy}),
\end{align}
\)

where the index \(j\) signifies that the formula is valid for all three waves, using the corresponding \( \kappa_j \), i.e. \( \kappa_I \equiv q \), \( \kappa_R \equiv -q \) and \( \kappa_T \equiv \kappa \). The components written explicitly read:

\(
\begin{align}
\begin{aligned}
H_{Ix} &= -q E_{Iy},
& H_{Iz} &= \xi E_{Iy},
\\
H_{Rx} &= q E_{Ry},
& H_{Rz} &= \xi E_{Ry},
\\
H_{Tx} &= -\frac{\kappa E_{Ty}}{\mu},
& H_{Tz} &= \frac{\xi E_{Ty}}{\mu},
\end{aligned}\label{eqEHcomponents}
\end{align}
\)

where we remember that \( q = \cos\varphi\) and \( \xi = \sin\varphi\). Then we plug the \( H_{jx} \) components to the second equation of \eqref{eqFieldMatch}, multiply by \( -\mu \), and drop the \(y\) index again, getting:

\(
\begin{align*}
\mu q E_I – \mu q E_R = \kappa E_T.
\end{align*}
\)

Then we replace the \( E_T \) on the right-hand side with first equation of \eqref{eqFieldMatch},

\(
\begin{align*}
\begin{aligned}
\mu q E_I – \mu q E_R &= \kappa E_I + \kappa E_R,
\\
(\mu q – \kappa ) E_I &= (\mu q + \kappa ) E_R.
\end{aligned}
\end{align*}
\)

So the desired \( r_s \) coefficient is

\( \begin{align} r_s = \frac{E_R}{E_I} = \frac{ \mu q – \kappa }{ \mu q + \kappa }, \end{align} \)

and the \( t_s \) is obtained from \eqref{eqDefCoefs}:

\( \begin{align} t_s = 1 + r_s = 1 + \frac{ \mu q – \kappa }{\mu q + \kappa} = \frac{ 2 \mu q }{ \mu q + \kappa }, \end{align} \)

Now we have solved the s-polarization on the interface between vacuum and isotropic \(\varepsilon\mu\) dielectric. The components of the \(\bf{H}\) field are given in eq. \eqref{eqEHcomponents}.
We should note that for simple transparent dielectric with \(\varepsilon > 1 \), \(\mu = 1\), as illustrated on the figure, the \( \kappa > q > 0 \), so the \(r_s\) will be negative and \(|r_s| < 1 \), while the \(t_s\) remains positive.

$\setCounter{0}$
Part 1: The wave vector
Part 2: The interface
Part 3: The s-polarization
Part 4: The p-polarization
Part 5: Multilayers

In this second part we will have a look what happens to electromagnetic waves at a planar interface between vacuum and isotropic material. First, without going into the textbook-level details here, we just remind ourselves that the Maxwell’s equations demand certain continuity of the \(\bf{E}\), \(\bf{H}\), \(\bf{D}\), \(\bf{B}\) fields at the interface. Particularly, the two equations with divergences will fix that the normal components of the \(\bf{D}\) and \(\bf{B}\) fields will be conserved across the interface. The two equations with rotations will then demand continuity of the in-plane components of the \(\bf{E}\) and \(\bf{H}\) fields.

\(
\begin{align}
\begin{aligned}
\mathbf{D}_{\perp 1}&=\mathbf{D}_{\perp 2},
&\mathbf{B}_{\perp 1}&=\mathbf{B}_{\perp 2},
\\
\mathbf{E}_{\parallel 1}&=\mathbf{E}_{\parallel 2},
&\mathbf{H}_{\parallel 1}&=\mathbf{H}_{\parallel 2},
\end{aligned}
\end{align}
\)

where the index 1, 2 denotes the first and the second medium.

Let’s choose our coordinate system. The interface will be the \(x-y\) plane at \(z=0\). We can freely orient our system so the incoming wave propagates in the \(z-x\) plane. We know that the field above the interface (in vacuum in our case) is composed of the incoming and the reflected wave, and there will also be the transmitted (refracted) wave in the second medium, which is isotropic material with both \( \varepsilon \) and \( \mu \) responses, i.e. generally frequency \( \omega \) dependent, but we excluded the points where \( \varepsilon \) or \( \mu \) are equal zero.

One important consequence of the continuity of the fields at the interface is that the x-components of the wave vectors of our three waves must be equal. And we remember that the magnitude of the wave vector was fixed by the eq. (10) in Part 1. This will lead to the fact that the angle of reflection is equal to angle of incidence, \(\varphi\) , and for the refracted wave, the Snell’s law is valid.

Here we should note that the Snell’s law in the textbook form, \( n_{1}\sin\varphi = n_{2}\sin\varphi’ \), in only meaningful for transparent materials with real \( n_1 \) and \( n_2 \). Otherwise, when the second medium is absorbing, with \( N_2=n_2+ik_2 \), we end up with complex \(\sin\varphi’ \).

In our picture, working with the geometrical components of the wave vectors, we will denote the x-component \( \xi = \sin\varphi \) (will always be non-negative real), and write the Snell’s law in general form

\(
\begin{align}
\xi=\text{const.}
\end{align}
\)

This is also valid for any layer in multilayer stack, as long as the interfaces are parallel. Here, we will also denote the z-component of the incoming wave as \( q = \cos\varphi \) and the z-component of the transmitted wave as plain \( \kappa \).
\(
\begin{align}
\begin{aligned}
\boldsymbol\kappa_{I}&=(\xi,0,q),
\\
\boldsymbol\kappa_{R}&=(\xi,0,-q),
\\
\boldsymbol\kappa_{T}&=(\xi,0,\kappa).
\end{aligned}
\end{align}
\)

Now, without writing it down, we remind ourselves that all the fields come with the phase factor in the form \( e^{i k_0 \kappa z} \cdot e^{i k_0 \xi x} \cdot e^{-i \omega t} \), where \( k_0 \equiv \frac{\omega}{c} \) is the physical magnitude of the wave vector in vacuum (with inverse length unit). So the problem is stationary in time \( t \), as well as in \( x \), we can cancel the the two exponential factors, and are left only with the z-propagation using the relevant \( \kappa \). Also, wherever possible, we cancelled the \( \frac{\omega}{c} \).

The geometrical z-component of the wave vector, \( \kappa \), is calculated from eq. (10) in Part 1

\(
\begin{align*}
\varepsilon\mu=\xi^{2} + \kappa^{2},
\end{align*}
\)

i.e.

\( \begin{align} \kappa = \sqrt{\varepsilon\mu – \xi^{2} }. \end{align} \)

From this generally complex square-root, we will choose the root with positive imaginary part.

The figure below illustrates the discussed configuration for the case of material with negative index of refraction, here \( n < -1 \) and \( k \) reasonably small, keeping the medium transparent. The right choice of \( \kappa \) is therefore with negative real part, so the transmitted wave travels apparently towards the interface, while its amplitude is decreasing with increasing distance from the interface. The Poyinting vector \( \bf{S} \) shows that the energy flows into the medium.

$\setCounter{0}$
Part 1: The wave vector
Part 2: The interface
Part 3: The s-polarization
Part 4: The p-polarization
Part 5: Multilayers

In a five-piece series of posts I will define and derive the Fresnel’s coefficients for reflection and transmission of plane waves on the interface between vacuum and isotropic material. It is going to be somewhat standard textbook explanation, but with three twists:

• we will use the HL units as defined in the previous blog
• we will keep isotropic permeability \(\mu\) – in this way we sneak in the possibility of negative index of refraction
• we will introduce the geometrical components of the wave vectors

So, let’s start from the Maxwell’s equations for HLU fields in the absence of external charges or currents

\(\DeclareMathOperator{\Div}{div}
\DeclareMathOperator{\Rot}{rot}
\newcommand{\parder}[2]{\frac{\partial {#1}}{\partial {#2}}}
\begin{align}
\begin{aligned}
\Div\mathbf{D} &=0,
&\Rot\mathbf{H} &=\frac{1}{c}\parder{\mathbf{D}}{t},
\\
\Rot\mathbf{E} &=-\frac{1}{c}\parder{\mathbf{B}}{t},
&\Div\mathbf{B} &=0.
\end{aligned}
\end{align}\)

The induced polarization and currents are captured in the material relations, here we will keep the \(\varepsilon\) and \(\mu\) as scalars:

\(\begin{align}
\begin{aligned}
\mathbf{D} &=\varepsilon\mathbf{E},
\\
\mathbf{B} &=\mu\mathbf{H}.
\end{aligned}
\end{align}\)

We consider the \(\bf{E}\) and \(\bf{H}\) to be the base fields, the \(\bf{D}\) and \(\bf{B}\) as linear response of the material, and we look for some solution in the form of plane wave:

\(\begin{align}
\begin{aligned}
&\mathbf{E}(\mathbf{r},t) = \mathbf{E}_{\mathbf{k},\omega} e^{i(\mathbf{kr}-\omega t)},
\\
&\mathbf{H}(\mathbf{r},t) = \mathbf{H}_{\mathbf{k},\omega} e^{i(\mathbf{kr}-\omega t)},
\end{aligned}\label{eqAmplitudes}
\end{align}\)

where \(\mathbf{E}_{\mathbf{k},\omega}\) \(\mathbf{H}_{\mathbf{k},\omega}\) are amplitude vectors for one chosen linear polarization. We will deal with this one particular component and rely on the linearity of Maxwell’s equations that the whole field can be written as some integral over the \(\bf{k}\) and \(\omega\) components and the two polarizations. We will also assume that the \(\varepsilon\) or \(\mu\) may depend on frequency \(\omega\), but not on \(\bf{k}\).

So let’s work on the Maxwell’s equations one by one:

\(\Div\mathbf{D} = \varepsilon\Div\mathbf{E} = \varepsilon\nabla\cdot\mathbf{E} = i\varepsilon\mathbf{k}\cdot\mathbf{E} = 0,\)

so unless the \(\varepsilon\) is zero – case which we will disregard for the moment – we get

\(\begin{align}\label{eqkE}
\mathbf{k}\cdot\mathbf{E} = 0.
\end{align}\)

Meaning that the electric field vector \(\mathbf{E}\), as well as the amplitude \(\mathbf{E}_{\mathbf{k},\omega}\), is always perpendicular to the direction of propagation.

Next:

\(\begin{align*}
\Rot\mathbf{H} = \nabla\times\mathbf{H} = i\mathbf{k}\times\mathbf{H}\; =\; \frac{1}{c}\parder{\mathbf{D}}{t} = \frac{\varepsilon}{c}\parder{\mathbf{E}}{t} = -i\varepsilon\frac{\omega}{c}\mathbf{E},
\end{align*}\)

which we will write for the moment as:

\(\begin{align}
\frac{\omega}{c}\mathbf{E} = -\dfrac{\mathbf{k}\times\mathbf{H}}{\varepsilon},
\end{align}\)

meaning that the \(\bf{E}\) and \(\bf{H}\) fields are proportional and orthogonal.

Third one:

\(\begin{align*}
\Rot\mathbf{E} = \nabla\times\mathbf{E} = i\mathbf{k}\times\mathbf{E}\; =\; -\frac{1}{c}\parder{\mathbf{B}}{t} = -\frac{\mu}{c}\parder{\mathbf{H}}{t} = i\mu\frac{\omega}{c}\mathbf{H},
\end{align*}\)

which we will again write as:

\(\begin{align}
\frac{\omega}{c}\mathbf{H} = \dfrac{\mathbf{k}\times\mathbf{E}}{\mu}.
\end{align}\)

And for \(\mu\) not equal to zero, analogically to \eqref{eqkE}, the last one simply yields:

\(\begin{align}
\mathbf{k}\cdot\mathbf{H} = 0.
\end{align}\)

Further, we define \(\boldsymbol\kappa\) as the geometrical component of the wave vector

\(\begin{align}
\mathbf{k} = \frac{\omega}{c}\boldsymbol\kappa,
\end{align}\)

and using \(\boldsymbol\kappa\), we cancel the \(\omega / c\) from all four equations

\(\begin{align} \begin{aligned} \boldsymbol\kappa\cdot\mathbf{E} &= 0, &\mathbf{E} &= -\dfrac{\boldsymbol\kappa\times\mathbf{H}}{\varepsilon}, \\ \mathbf{H} &= \dfrac{\boldsymbol\kappa\times\mathbf{E}}{\mu}, &\boldsymbol\kappa\cdot\mathbf{H} &= 0. \end{aligned} \label{eqRedMax} \end{align}\)

This set of equations can be understood for both sides of eqs. \eqref{eqAmplitudes}, i.e. either for the \(\bf{E}\) and \(\bf{H}\) transients, or, after cancelling the factor \( e^{i(\mathbf{kr}-\omega t)} \), for the amplitudes \(\mathbf{E}_{\mathbf{k},\omega}\) and \(\mathbf{H}_{\mathbf{k},\omega}\). Later we will think in terms of the amplitudes, but we will drop the subscript \(\mathbf{k},\omega\).

For the simplest case of vacuum with \(\varepsilon = 1\) and \(\mu = 1 \) we can see that \(\boldsymbol\kappa , \bf{E} , \bf{H}\) form an orthogonal right-handed system.

Next, we insert the third equation of \eqref{eqRedMax} into second and move the constants on the left-hand side, obtaining:

\(\begin{align*}
-\varepsilon\mu\mathbf{E} = \boldsymbol\kappa\times ( \boldsymbol\kappa\times\mathbf{E} ).
\end{align*}\)

To the right-hand side we apply the “bac-cab” identity (see footnote):

\(\begin{align*}
-\varepsilon\mu\mathbf{E} = \boldsymbol\kappa\cdot ( \boldsymbol\kappa\cdot\mathbf{E} ) – \mathbf{E}\cdot ( \boldsymbol\kappa\cdot\boldsymbol\kappa ).
\end{align*}\)

Finally, we use the first eq. from \eqref{eqRedMax} to remove the first term on the right hand side. The equation must hold for any \(\bf{E}\), therefore

\(\begin{align} \varepsilon\mu = \boldsymbol\kappa\cdot\boldsymbol\kappa. \end{align}\)

This is the dispersion relation, only the physical dispersion has been removed from it. And that’s what we want, to see the geometry clearly.

Examples

In the vacuum case, \(\varepsilon = 1\) and \(\mu = 1 \), we can choose the direction of propagation along z-axis, \(\boldsymbol\kappa = (0,0,\kappa_{z}) \) and fix the electric field amplitude along x-axis, \( \mathbf{E} = (E_0,0,0) \). We get \( \kappa_z = 1 \) and the magnetic field amplitude is then \( \mathbf{H} = (0,E_0,0) \). The HLU field amplitudes in vacuum are the same.

If we replace the vacuum with simple transparent dielectric with \( \varepsilon > 1 \) and \(\mu = 1 \), we have \( \varepsilon = \kappa_z^{2} \), so \( \kappa_z = \sqrt{\varepsilon} \equiv n \), which means that the size of the geometrical wave vector is the (positive real) refractive index. The magnetic field amplitude is then \( \mathbf{H} = (0,n E_0,0) \), so the magnetic field is now greater than in vacuum, but it should be understood that for the same magnetic amplitude, the electric amplitude is smaller, as some of the electric field strength is spent polarizing the dielectric. For the same frequency, the wave vector is longer than in vacuum, so for the same period, the wavelength is shorter, means the phase of the wave propagates slower.

Other interesting case is the pure metallic response with \( \varepsilon < 0 \), where \( \kappa_z = \sqrt{\varepsilon} \equiv ik \), and \(k\) being the positive real absorption coefficient. Here we have non-propagating, pure evanescent wave in the positive z-direction. The magnetic field amplitude is \( \mathbf{H} = (0,i k E_0,0) \), so the magnetic field oscillation is \( \pi/2 \) shifted with respect to the electric field.

Eventually, for general dielectric response, \(\varepsilon = \varepsilon_1 + i\varepsilon_2 \), we can have any value of \( \varepsilon_1 \) while \( \varepsilon_2 \) is non-negative, so we have \( \kappa_z = \sqrt{\varepsilon} \equiv N = n+ik \). For wave propagating and diminishing along positive direction of z-axis both real and imaginary part of \( \kappa_z \) must be non-negative.

Then we have the curious case when we start to mix in the permeability \( \mu \), because we can reach a situation, when both the real parts of \( \varepsilon \) and \( \mu \) are negative (while the imaginary parts are non-negative). So looking at \( \varepsilon\mu = \kappa_z^{2} =(n+ik)^{2} \), the only way to get non-negative \( k \) is to have negative \( n \). Or, for purely real negative \( \varepsilon \) and \( \mu \), one can look at the fields from \eqref{eqRedMax} to realize that to achieve Poyinting vector, \( \mathbf{S} = c\mathbf{E}\times\mathbf{H}^\ast \) (energy flow, in HLU), oriented in positive z-direction, the \( \kappa_z \) must be negative.

---

We introduced the nabla operator:

\(
\newcommand{\parder}[2]{\frac{\partial {#1}}{\partial {#2}}}
\nabla \equiv (\parder{}{x},\parder{}{y},\parder{}{z}).\)

Used identity

\(
\mathbf{a} \times ( \mathbf{b} \times \mathbf{c} ) =
\mathbf{b} \cdot ( \mathbf{a} \cdot \mathbf{c} ) –
\mathbf{c} \cdot ( \mathbf{a} \cdot \mathbf{b} ).
\)