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电子工程代写|光子简介代写Introduction to Photonics代考|Polarization Rotators

Optically active and magneto-optic materials (Sects. 2.4.1 and 2.4.2) have circularly polarized eigenstates with the propagation indices $n^{\pm}$; they act as circular retarders, inducing a phase shift of
$$\Delta \phi_{\mathrm{V}}=\left(n^{-}-n^{+}\right) k_{0} d$$
between the two circularly polarized states. In a circularly polarized base [Eq. (1.78)], the Jones matrix has the form
$$\boldsymbol{T}{\mathrm{c}}=\left[\begin{array}{cc} 1 & 0 \ 0 & \mathrm{e}^{-\mathrm{j} \Delta \phi{\mathrm{v}}} \end{array}\right]{\mathrm{c}},$$ which, as we shall see from an inspection of Eq. (1.124), corresponds to a polarization rotator that rotates an incoming state by an angle of $$\varphi=-\Delta \phi{\mathrm{V}} / 2=\left(n^{+}-n^{-}\right) k_{0} d / 2 .$$

Polarizers (also called polarization filters) are components that transmit one particular polarization state only; an incoming state is decomposed into the transmitted eigenstate and its orthogonal complement, which is absorbed or directed into a different direction; in other words, a polarizer projects the input state onto the transmitted eigenstate. The matrix of a polarizer for $x$-polarized light is therefore
$$\boldsymbol{T}=\left[\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right]$$

电子工程代写|光子简介代写Introduction to Photonics代考|Polarization Eigenstates

An eigenvector or eigenstate (or eigenmode) of a matrix $T$ is a vector that, if multiplied with $T$, remains unchanged apart from a (complex) factor, called eigenvalue. The eigenvectors of a Jones matrix are the polarization eigenstates of the corresponding optical element; to determine these states, we have to solve the equation
$$\mathbf{T} \mathbf{I}=\lambda_{T} \mathbf{I}$$

or
$$\left(\boldsymbol{T}-\lambda_{T} \mathbf{1}\right) \mathbf{J}=\mathbf{0},$$
where 1 is the unit matrix
$$\mathbf{1}:=\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}\right] ;$$
explicitly,
$$\left[\begin{array}{cc} T_{11}-\lambda_{T} & T_{12} \ T_{21} & T_{22}-\lambda_{T} \end{array}\right]\left[\begin{array}{l} J_{1} \ J_{2} \end{array}\right]=\mathbf{0}$$
For this system to have non-trivial (i.e., non-zero) solutions, the determinant $\operatorname{det}(T-$ $\lambda_{T} 1$ ) must be zero. Thus, the characteristic equation
$$\left(T_{11}-\lambda_{T}\right)\left(T_{22}-\lambda_{T}\right)-T_{21} T_{12}=0$$
has to be solved, yielding two eigenvalues $\lambda_{T}^{(1)}$ and $\lambda_{T}^{(2)}$. Corresponding eigenvectors $\mathbf{J}^{(1,2)}$ are found by inserting the values $\lambda_{T}$ into one of the equation of Eq. (1.99); the length of the eigenvectors is not defined, since any multiple $a \mathbf{J}$ of an eigenvector $\mathbf{J}$ is also an eigenvector. It is, however, convenient to normalize the eigenvectors to unit length.

Once the set of eigenvectors is given, any arbitrary state can be written as a linear combination of these eigenvectors,
$$\mathbf{J}=a_{1} \mathbf{J}^{(1)}+a_{2} \mathbf{J}^{(2)} ;$$
the output state of the optical element represented by $T$ is then
$$\boldsymbol{T} \mathbf{J}=a_{1} \lambda_{T}^{(1)} \mathbf{J}^{(1)}+a_{2} \lambda_{T}^{(2)} \mathbf{J}^{(2)}$$

光子简介代考

电子工程代写|光子简介代写Introduction to Photonics代考|Polarization Rotators

$$\Delta \phi_{\mathrm{V}}=\left(n^{-}-n^{+}\right) k_{0} d$$

$$\boldsymbol{T} \mathrm{c}=\left[\begin{array}{cccc} 1 & 0 & 0 & \mathrm{e}^{-\mathrm{j} \Delta \phi \mathrm{v}} \end{array}\right] \mathrm{c},$$

$$\varphi=-\Delta \phi \mathrm{V} / 2=\left(n^{+}-n^{-}\right) k_{0} d / 2 .$$

$$\boldsymbol{T}=\left[\begin{array}{llll} 1 & 0 & 0 & 0 \end{array}\right]$$

电子工程代写|光子简介代写Introduction to Photonics代考|Polarization Eigenstates

$$\mathbf{T I}=\lambda_{T} \mathbf{I}$$
$$\left(T-\lambda_{T} \mathbf{1}\right) \mathbf{J}=\mathbf{0},$$

$$1:=\left[\begin{array}{llll} 1 & 0 & 0 & 1 \end{array}\right] ;$$

$$\left[\begin{array}{lll} T_{11}-\lambda_{T} & T_{12} T_{21} & \left.T_{22}-\lambda_{T}\right]\left[J_{1} J_{2}\right]=\mathbf{0} \end{array}\right.$$

$$\left(T_{11}-\lambda_{T}\right)\left(T_{22}-\lambda_{T}\right)-T_{21} T_{12}=0$$

$$\mathbf{J}=a_{1} \mathbf{J}^{(1)}+a_{2} \mathbf{J}^{(2)} ;$$

$$\boldsymbol{T} \mathbf{J}=a_{1} \lambda_{T}^{(1)} \mathbf{J}^{(1)}+a_{2} \lambda_{T}^{(2)} \mathbf{J}^{(2)}$$

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