## Electromagnetic radiation and polarization (waves) ### General idea In (nonlinear) optics and material studies, when we say polarization we typically talk about polarization of material, namely the behavior of electrons. But the polarization of wave itself is important in many aspects. Here we do a brief recap for propagation and polarization of waves. Wave equation is presented in [[Maxwell equation recap#Wave equations and continuity of charges]]. - $\mathbf{E}$ (Electric field) and $\mathbf{B}$ (Magnetic field) are always perpendicular to each other and oscillate in phase. - $\mathbf{k}$ (Wave vector) represents the direction of wave propagation and perpendicular to the wavefronts. Both $\mathbf{E}$ and $\mathbf{B}$ are perpendicular to $\mathbf{k}$ in homogeneous materials and parallel to Poynting vector $\mathbf{S}$ in general. But for heterogeneous materials or metamaterials, things could be complicated. One example is [[Birefringence|birefringence]]. The extraordinary ray has the direction perpendicular to the wavefronts but not aligned with $\mathbf{S}$. This is called beam walk-off. - $\mathbf{S}$ ([[Poynting vector and energy|Poynting vector]]) represents the energy flow and points in the same direction as $\mathbf{k}$, defined by $\mathbf{S}=\mathbf{E}\times\mathbf{B}$. For a linear polarized wave, an illustration is shown in the `.gif` file below, ![[Electromagneticwave3D.gif]] Image from https://commons.wikimedia.org/wiki/File:Electromagneticwave3D.gif. And for a circular polarized wave, a representation is shown as the following image, ![[Circular.Polarization.Circularly.Polarized.Light_Right.Handed.Animation.305x190.255Colors.gif]] Image from https://commons.wikimedia.org/wiki/File:Circular.Polarization.Circularly.Polarized.Light_Right.Handed.Animation.305x190.255Colors.gif. Since $\mathbf{E}$ and $\mathbf{B}$ are both three dimensional vectors, they can be decomposed to three orthogonal directions. The direction of $\mathbf{E}$ is the direction of (wave) polarization, which could be described using Jones vector formalism. ### Jones calculus #### Introduction to Jones vector >[!Notice] >Jones vector formalism is used to tread **fully polarized** light. For partially or unpolarized light (like sunlight or scattered light), [Mueller calculus](https://en.wikipedia.org/wiki/Mueller_calculus) is applied. The degree of polarization is defined as >$\text{degree of polarization}= \frac{I_{pol}}{I_{pol}+I_{unpol}}$ >$I$ stands for intensity. Fully polarized light has this value being one. >In nonlinear optics studies, we almost have only fully polarized light. Consider a simple plane wave case. A beam of light propagating along $z$ direction. Assume it's monochromatic and consider frequency $\omega$ only. Then the electromagnetic wave can be written as $\mathbf{E}(\mathbf{r}, t) = \mathbf{E}_{0} e^{i(\omega t-\mathbf{k} \cdot \mathbf{r})}$ As explained above, this $\mathbf{E}_{0}$ is orthogonal to the propagating direction $z$. So if write the vector explicitly, we have $\mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ k \end{pmatrix}$ And with the direction defined, for this single beam, we can write the $\mathbf{E}_0 = \begin{pmatrix} a_{x} e^{i\phi_{x}} \\ a_{y}e^{i\phi_{y}} \\ 0 \end{pmatrix}$ Here $a_{x}$ and $a_{y}$ are amplitude in $x$ and $y$ direction. This $\mathbf{E}_0$ is not a function of position and time, and have the (initial) phase $\phi_{x}$ and $\phi_{y}$ inside. The wave would then be $\mathbf{E}(\mathbf{r}, t) = \begin{pmatrix} a_{x} e^{i(\omega t - kz + \phi_x)} \\ a_{y} e^{i(\omega t - kz + \phi_y)} \\ 0 \end{pmatrix}$ From $\mathbf{E}_{0}$ we already have clues on how it polarize. And the **Jones vector** is defined as $\mathbf{J} = \begin{pmatrix} a_x e^{i \phi_x} \\ a_y e^{i \phi_y} \end{pmatrix}= \begin{pmatrix} a_x \\ a_y e^{i \delta} \end{pmatrix}$ where $\delta=\phi_{y}-\phi_{x}$. These two expression are exactly the same, cause we maintained the same phase difference. The Jones vector completely specified the polarization state of the light field at a particular point in space, and the field on $x-$ and $y-$ direction are described by an amplitude and corresponding phase. In many cases, people refer the Jones vector as the normalized Jones vector, namely $\mathbf{J}_{\text{norm}} = \frac{1}{\sqrt{a_x^2 + a_y^2}} \begin{pmatrix} a_x \\ a_y e^{i \delta} \end{pmatrix}$ Two states of polarization are orthogonal if and only if we have their Jones vectors being orthogonal. This can be expressed as $\mathbf{J}_1 = \begin{pmatrix} a_{1x} e^{i \phi_{1x}} \\ a_{1y} e^{i \phi_{1y}} \end{pmatrix}, \quad \mathbf{J}_2 = \begin{pmatrix} a_{2x} e^{i \phi_{2x}} \\ a_{2y} e^{i \phi_{2y}} \end{pmatrix}$ and their complex inner product being zero, namely $\langle \mathbf{J}_1, \mathbf{J}_2\rangle = \mathbf{J}_1^* \mathbf{J}_2 = a_{1x} a_{2x} e^{i(\phi_{2x} - \phi_{1x})} + a_{1y} a_{2y} e^{i(\phi_{2y} - \phi_{1y})} = 0$ #### Types of polarizations ##### Linear polarization For linear polarized light, there is no relative phase difference, or $\delta =\phi_{y}-\phi_{x}=0\text{ or }\pi$. Two simplest cases, for linearly polarized light along $x-$ and $y-$ axis, we have $\mathbf{J}_x = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad \mathbf{J}_y = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ And for more general cases, with $\theta$ being the polarization angle (angle between $x$ axis and polarization direction), $\mathbf{J} = \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}$ ##### Circular polarization In this case amplitudes in both directions are the same. For left and right circular polarized light, we have $\mathbf{J}_{\sigma^{-}} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix}, \quad \mathbf{J}_{\sigma^{+}} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -i \end{pmatrix}$ >[!Notice] >The left- and right-handed polarization is illustrated below. >![[Drawing 2024-09-26 02.41.28.excalidraw.svg]] >For left-handed polarization, $\delta =\phi_{y}-\phi_{x}=\frac{\pi}{2}$, so $e^{i\pi/2}=i$. For right-handed polarization, $\delta=-\frac{\pi}{2}$, we have $-i$. >In general, when $\delta>0$, we have left-handed polarization, $\delta=0$ linearly polarization, otherwise right-handed. ##### Elliptical polarization If phase difference $\delta =\phi_{y}-\phi_{x}=\frac{\pi}{2}$, we have the unnormalized $\mathbf{J}$ being $\mathbf{J}_{\sigma^{-}} = \begin{pmatrix} a_x \\ a_y i \end{pmatrix}$ and $\delta=-\frac{\pi}{2}$, $\mathbf{J}_{\sigma^{+}} = \begin{pmatrix} a_x \\ -a_y i \end{pmatrix}$ For more general cases, one should keep the original definition, namely $\mathbf{J} = \begin{pmatrix} a_x e^{i \phi_x} \\ a_y e^{i \phi_y} \end{pmatrix}= \begin{pmatrix} a_x \\ a_y e^{i \delta} \end{pmatrix}$ since no simplification could be made. This polarization could be rotated to get arbitrary elliptical polarization. The rotation can be done by [[Polarizer, retarder and polarization rotator#Polarization rotator|rotator]]. The Poincaré sphere is a good tool for polarization visualization for such complex case. #### Manipulating polarization states What makes the Jones vector formalism more useful is that we may define a **Jones matrix** to describe how certain devices or optical components could change the polarization states, and easily relate to group operations like rotations (see [[Index (group theory)]] for more info). People use different notations to write this $2\times 2$ matrix, could be $\mathbf{T}$, $\mathbf{M}$ or $\mathcal{M}$ or other notations This depends on contexts or whether we have other matrixes like Mueller matrix. Here we will use $\mathbf{M}$. Specific devices are shown in [[Polarizer, retarder and polarization rotator]]. Assume $\mathbf{J}_{1}$ is the input Jones vector, and we have a device with Jones matrix $\mathbf{T}$, the output Jones vector would be $\mathbf{J}_{2}=\mathbf{M} \mathbf{J}_{1}$ here $\mathbf{M} = \begin{pmatrix} m_{11} & m_{12} \\ m_{21} & m_{22} \end{pmatrix}$. For some typical devices, we have the following results. ##### Linear polarizers For those in $x$ direction, the matrix should be $\mathbf{M}_{\text{LP}}^{0} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ in $y$ direction, the matrix should be $\mathbf{M}_{\text{LP}}^{90^\circ} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ For direction with $\theta$ being the polarization angle (angle between $x$ axis and polarization direction), we have $\mathbf{M}_{\text{LP}}^{\theta} = \begin{pmatrix} \cos^2\theta & \cos\theta \sin\theta \\ \cos\theta \sin\theta & \sin^2\theta \end{pmatrix}$ This result can be understand as rotating the linear polarizer from $x$ direction by angle $\theta$, namely $\mathbf{M}_{\text{LP}}^{\theta}=\mathbf{R}(\theta) \mathbf{M}_{\text{LP}}^{0}\mathbf{R}^{-1}(\theta)$ Here the rotation matrix, as typically defined, is $\mathbf{R}(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$, and $\mathbf{R}^{-1}(\theta) = \mathbf{R}(-\theta) = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$. So we have $\mathbf{M}_{\text{LP}}^{\theta}= \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}=\begin{pmatrix} \cos^2\theta & \cos\theta \sin\theta \\ \cos\theta \sin\theta & \sin^2\theta \end{pmatrix}$ >[!Example] >For example, if $\mathbf{J}_{1}=\begin{pmatrix} a \\ b \end{pmatrix}$, and we have a linear polarizer in $x$ direction. $\mathbf{M}_{\text{LP}}^{0} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ >$\mathbf{J}_{2}=\mathbf{M}_{\text{LP}}^{0} \mathbf{J}_{1} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} =\begin{pmatrix} a \\ 0 \end{pmatrix}$ > ##### Wave retarders Wave retarder could make the polarization (or say, phase) in one direction being delayed with respect to the orthogonal polarization. These devices are typically based on [[Birefringence|birefringence]]. $\mathbf{M}^{0}_{\Gamma} = \begin{pmatrix} 1 & 0 \\ 0 & e^{-i\Gamma} \end{pmatrix}$ This result assumes that the fast axis is $x$ direction, or say that $y$ direction is delayed. Here $\Gamma$ is the phase delayed. If it's birefringence based, then $\Gamma=\frac{2\pi d\Delta n}{\lambda}$ $d$ is the thickness, $\Delta n=n_{e}-n_{o}$, and $\lambda$ is the wavelength. For some special case, like half-wave plate, $\Gamma=\pi$, corresponding to $\frac{\lambda}{2}$, we write the phase with respect to wavelength, $\mathbf{M}_{\lambda/2}^{0} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and quarter-wave plate, with a phase delay being $\Gamma=\frac{\pi}{2}$, corresponding to $\frac{\lambda}{4}$, $\mathbf{M}_{\lambda/4}^{0} = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}$ We also have full-wave plate, $\mathbf{M}_{\lambda}^{0} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ This will not change the polarization states. But since the wave plate is designed for specific wavelength, such device could be used to filter out a specific wavelength of light if combined with linear polarizers. >[!Notice] >Half-wave plates, quarter-wave and full-wave plates are all designed for specific working wavelength. This is typically done by controlling the thickness of birefringence crystals. >To have a widely adjustable device suitable for more than one wavelength and phase retardation, one will need the [[Polarizer, retarder and polarization rotator#Babinet–Soleil compensator|Babinet–Soleil compensator]]. By rotating the wave plate, we could achieve some special effects. For example, apply a $45^\circ$ degree rotation (with respect to $x$ axis), $\mathbf{M}_{\lambda/2}^{45^\circ}=\mathbf{R}(45^\circ) \mathbf{M}_{\lambda/2}^{0}\mathbf{R}^{-1}(45^\circ)$ Insert $\mathbf{R}(45^\circ) = \begin{pmatrix} \cos 45^\circ & -\sin 45^\circ \\ \sin 45^\circ & \cos 45^\circ \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{pmatrix}$ into above expression, one can see that $\mathbf{M}_{\lambda/2}^{45^\circ} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ This may conversion a $x$ polarized light, $\mathbf{J}_x = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, becomes $y$ polarized, since $\mathbf{J}_{out}=\mathbf{M}_{\lambda/2}^{45^\circ} \mathbf{J}_x=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\1 \end{pmatrix}=\mathbf{J}_y$ Check [[Polarizer, retarder and polarization rotator]] for more information. ##### Polarization rotator For rotator, the Jones matrix is $\mathbf{M}_{\text{Rot}}^{\theta}=\mathbf{R}(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ The expression is the same as rotation matrix. It could rotate the polarization without having additional modifications on the polarization states. Applying a rotator is different from rotating the optical components. For example, rotate a half-wave plate by angle $\theta$ would be $\mathbf{R}(\theta) \mathbf{M}_{\lambda/2}^{0}\mathbf{R}^{-1}(\theta)$, while using a rotator will give $\mathbf{M}_{\text{Rot}}^{\theta} \mathbf{M}_{\lambda/2}^{0}=\mathbf{R}(\theta) \mathbf{M}_{\lambda/2}^{0}$. >[!Info] >Check *Polarisation States and Jones Vectors. (2022, September 16). https://phys.libretexts.org/@go/page/57095* for more contents. (PDF version [[Optics (Konijnenberg, Adam, and Urbach).pdf]])