Voigt effect

Following the notation of Hubert,[2] the generalized dielectric cubic tensor ${\displaystyle \varepsilon _{r}}$ take the following form:

$${\displaystyle (1)\qquad \varepsilon _{r}=\epsilon {\begin{bmatrix}1&0&iQm_{y}\\0&1&-iQm_{x}\\-iQm_{y}&iQm_{x}&1\end{bmatrix}}+{\begin{bmatrix}B_{1}m_{x}^{2}&B_{2}m_{x}m_{y}&0\\B_{2}m_{x}m_{y}&B_{1}m_{y}^{2}&0\\0&0&B_{1}m_{z}^{2}\end{bmatrix}}}$$

where ${\displaystyle \varepsilon }$ is the material dielectric constant, ${\displaystyle Q}$ the Voigt parameter, ${\displaystyle B_{1}}$ and ${\displaystyle B_{2}}$ two cubic constants describing magneto-optical effect depending on ${\displaystyle m_{i}^{2}}$. ${\displaystyle m_{i}}$ is the reduce ${\displaystyle m_{i}=M_{i}/M_{s}}$. Calculation is made in the spherical approximation with ${\displaystyle B_{1}=B_{2}}$. At the present moment, there is no evidence that this approximation is not valid, as the observation of Voigt effect is rare because it is extremely small with respect to the Kerr effect.

To calculate the eigenvalues and eigenvectors, we consider the propagation equation derived from the Maxwell equations, with the convention ${\displaystyle {\vec {n}}={\vec {k}}c/\omega }$:

$${\displaystyle (2)\qquad n^{2}{\vec {E}}-{\vec {n}}({\vec {n}}\cdot {\vec {E}})=\varepsilon {\vec {E}}}$$

When the magnetization is perpendicular to the propagation wavevector, on the contrary to the Kerr effect, ${\displaystyle {\vec {E}}}$ may have all his three components equals to zero making calculations rather more complicated and making Fresnels equations no longer valid. A way to simplify the problem consists to use the electric field displacement vector ${\displaystyle {\vec {D}}=\varepsilon {\vec {E}}}$ . Since ${\displaystyle {\vec {\nabla }}\cdot {\vec {D}}=0}$ and ${\displaystyle {\vec {k}}\parallel {\vec {z}}}$ we have ${\displaystyle {\vec {D}}=(D_{x},D_{y},0)}$. The inconvenient is to deal with the inverse dielectric tensor which may be complicated to handle. Here calculation is made in the general case which is complicated mathematically to handle, however one can follow easily the demonstration by considering ${\displaystyle \phi =0}$.

Eigenvalues and eigenvectors are found by solving the propagation equation on ${\displaystyle {\vec {D}}}$ which gives the following system of equation:

$${\displaystyle (3)\quad {\begin{cases}(\varepsilon _{xx}^{-1}-{\frac {1}{n^{2}}})D_{x}+\epsilon _{xy}^{-1}D_{y}=0\\\\\varepsilon _{yx}^{-1}D_{x}+(\varepsilon _{yy}^{-1}-{\frac {1}{n^{2}}})D_{y}=0\end{cases}}}$$

where ${\displaystyle \varepsilon _{ij}^{-1}}$ represents the inverse ${\displaystyle ij}$ element of the dielectric tensor ${\displaystyle \varepsilon _{r}}$, and ${\displaystyle n^{2}=\varepsilon }$. After a straightforward calculation of the system's determinant, one has to make a developpement on 2nd order in ${\displaystyle Q}$ and first order of ${\displaystyle B_{1}}$. This led to the two eigenvalues corresponding the two refraction indices:

$${\displaystyle n_{\parallel }^{2}=\varepsilon +B_{1}}$$

$${\displaystyle n_{\perp }^{2}=\varepsilon (1-Q^{2})}$$

The corresponding eigenvectors for ${\displaystyle {\vec {D}}}$ and for ${\displaystyle {\vec {E}}}$ are:

$${\displaystyle (4)\qquad {\vec {D}}_{\parallel }={\begin{pmatrix}\cos(\phi )\\\sin(\phi )\\0\end{pmatrix}}\qquad {\vec {D}}_{\perp }={\begin{pmatrix}-\sin(\phi )\\\cos(\phi )\\0\end{pmatrix}}\qquad {\vec {E}}_{\parallel }=\varepsilon ^{-1}{\vec {D}}_{\parallel }={\begin{pmatrix}{\frac {\cos(\phi )}{B_{1}+\varepsilon }}\\{\frac {\sin(\phi )}{B_{1}+\epsilon }}\\0\end{pmatrix}}\qquad {\vec {E}}_{\perp }=\varepsilon ^{-1}{\vec {D}}_{\perp }={\begin{pmatrix}{\frac {\sin(\phi )}{(Q^{2}-1)\epsilon }}\\{\frac {\cos(\phi )}{(1-Q^{2})\epsilon }}\\{\frac {-iQ}{(1-Q^{2})\epsilon }}\end{pmatrix}}}$$

Knowing the eigenvectors and eigenvalues inside the material, one have to calculate ${\displaystyle {\vec {E_{r}}}=(E_{rx},E_{ry},0)}$ the reflected electromagnetic vector usually detected in experiments. We use the continuity equations for ${\displaystyle {\vec {E}}}$ and ${\displaystyle {\vec {H}}}$ where ${\displaystyle {\vec {H}}}$ is the induction defined from Maxwell's equations by ${\textstyle {\vec {\nabla }}\times {\vec {H}}={\frac {1}{c}}{\frac {\partial {\vec {D}}}{\partial t}}}$. Inside the medium, the electromagnetic field is decomposed on the derived eigenvectors ${\displaystyle {\vec {E}}_{t}=\alpha {\vec {E}}_{\parallel }+\beta {\vec {E}}_{\perp }}$. The system of equation to solve is:

$${\displaystyle (5)\quad {\begin{cases}\alpha {\Big (}{\frac {D_{y\parallel }}{n_{\parallel }}}{\Big )}+\beta {\Big (}{\frac {D_{y\perp }}{n_{\perp }}}{\Big )}+E_{ry}=E_{0y}\\\\\alpha {\Big (}{\frac {D_{x\parallel }}{n_{\parallel }}}{\Big )}+\beta {\Big (}{\frac {D_{x\perp }}{n_{\perp }}}{\Big )}+E_{rx}=E_{0x}\\\\\alpha E_{x\parallel }+\beta E_{x\perp }-E_{rx}=E_{0x}\\\\\alpha E_{y\parallel }+\beta E_{y\perp }-E_{ry}=E_{0y}\end{cases}}}$$

The solution of this system of equation are:

$${\displaystyle (6)\quad E_{rx}=E_{0}{\frac {(1-n_{\perp }n_{\parallel })\cos(\beta )+(n_{\perp }-n_{\parallel })\cos(\beta -2\phi )}{(1+n_{\parallel })(1+n_{\perp })}}}$$

$${\displaystyle (7)\quad E_{ry}=E_{0}{\frac {(1-n_{\perp }n_{\parallel })\sin(\beta )-(n_{\perp }-n_{\parallel })\sin(\beta -2\phi )}{(1+n_{\parallel })(1+n_{\perp })}}}$$

The rotation angle ${\displaystyle \delta \beta }$ and the ellipticity angle ${\displaystyle \psi }$ are defined from the ratio ${\displaystyle \chi =E_{ry}/E_{rx}}$ with the two following formulae:

$${\displaystyle (8)\quad \tan 2\delta \beta ={\frac {2\operatorname {Re} (\chi )}{1-|\chi |^{2}}}\qquad (9)\quad \sin(2\psi _{K})={\frac {2\operatorname {Im} (\chi )}{1-|\chi |^{2}}},}$$

where ${\displaystyle \operatorname {Re} (\chi )}$ and ${\displaystyle \operatorname {Im} (\chi )}$ represent the real and imaginary part of ${\displaystyle \chi }$. Using the two previously calculated components, one obtains:

$${\displaystyle (10)\qquad \chi ={\frac {(B_{1}+n_{0}^{2}Q^{2})}{2n_{0}(n_{0}^{2}-1)}}{\frac {\sin[2(\phi -\beta )]}{\cos(\beta )^{2}}}+\tan(\beta ).}$$

This gives for the Voigt rotation:

$${\displaystyle (11)\qquad \delta \beta =\operatorname {Re} \left[{\frac {B_{1}+n_{0}^{2}Q^{2}}{2n_{0}(n_{0}^{2}-1)}}\right]\sin[2(\phi -\beta )],}$$

which can also be rewritten in the case of ${\displaystyle B_{1}}$, ${\displaystyle n_{0}}$, and ${\displaystyle Q}$ real:

$${\displaystyle (12)\quad \delta \beta ={\frac {2\Delta n}{n_{0}^{2}-1}}\sin[2(\phi -\beta )],}$$

where ${\displaystyle \Delta n={\frac {n_{\parallel }-n_{\perp }}{2}}}$ is the difference of refraction indices. Consequently, one obtains something proportional to ${\displaystyle \Delta n}$ and which depends on the incident linear polarisation. For proper ${\displaystyle \phi -\beta }$ no Voigt rotation can be observed. ${\displaystyle \Delta n}$ is proportional to the square of the magnetization since ${\displaystyle B_{1}\propto M_{s}^{2}}$ and ${\displaystyle Q\propto M_{s}}$.

The calculation of the rotation of the Voigt effect in transmission is in principle equivalent to the one of the Faraday effect. In practice, this configuration is not used in general for ferromagnetic samples since the absorption length is weak in this kind of material. However, the use of transmission geometry is more common for paramagnetic liquid or cristal where the light can travel easily inside the material.

The calculation for a paramagnetic material is exactly the same with respect to a ferromagnetic one, except that the magnetization is replaced by a field ${\displaystyle \mu _{0}{\vec {H}}=\mu _{0}H_{0}(\cos \phi ,\,\sin \phi )}$ (${\displaystyle H_{0}}$ in ${\displaystyle \mathrm {A/m} }$ or ${\displaystyle \mathrm {G} }$). For convenience, the field will be added at the end of calculation in the magneto-optical parameters.

Consider the transmitted electromagnetic waves ${\displaystyle {\vec {E}}_{t}}$ propagating in a medium of length L. From equation (5), one obtains for ${\displaystyle \alpha }$ and ${\displaystyle \beta }$:

$${\displaystyle \alpha ={\frac {2E_{0}n_{\parallel }\cos(\theta -\phi )}{1+n_{\parallel }}}}$$

$${\displaystyle \beta ={\frac {2E_{0}n_{\perp }^{2}\sin(\theta -\phi )}{n_{\perp }+1}}}$$

At the position z = L, the expression of ${\displaystyle {\vec {E}}_{t}}$ is:

$${\displaystyle (13)\quad {\vec {E}}_{t}=e^{-i\omega [t+{\frac {(n_{\parallel }+n_{\perp })L}{2c}}]}{\Big [}\alpha {\vec {E}}_{\parallel }e^{i{\frac {\omega \Delta nL}{c}}}+\beta {\vec {E}}_{\perp }e^{-i{\frac {\omega \Delta nL}{c}}}{\Big ]}}$$

where ${\displaystyle {\vec {E}}_{\parallel }}$ and ${\displaystyle {\vec {E}}_{\perp }}$are the eigenvectors previously calculated, and ${\displaystyle \Delta n={\frac {n_{\parallel }-n_{\perp }}{2}}}$ is the difference for the two refraction indices. The rotation is then calculated from the ratio ${\displaystyle \chi ={\frac {E_{t_{y}}}{E_{t_{x}}}}}$, with development in first order in ${\displaystyle B_{1}}$ and second order in ${\displaystyle Q}$. This gives:

$${\displaystyle (14)\quad \chi ={\frac {c-i~\omega L(1+n_{0})(B_{1}+n_{0}Q^{2})\sin[2(\beta -\phi )]}{4cn_{0}(1+n_{0})\cos ^{2}(\beta )}}={\frac {c-i~\omega L(1+n_{0})\Delta n\sin[2(\beta -\phi )]}{c~(1+n_{0})\cos ^{2}(\beta )}}}$$

Again we obtain something proportional to ${\displaystyle \Delta n}$ and ${\displaystyle L}$, the light propagation length. Let us notice that ${\displaystyle \Delta n}$ is proportional to ${\displaystyle (\mu _{0}H_{0})^{2}}$ in the same way with respect to the geometry in reflexion for the magnetization. In order to extract the Voigt rotation, we consider ${\displaystyle n_{0}=\eta +i~\kappa }$, ${\displaystyle Q=Q_{r}+i~Q_{i}}$ and ${\displaystyle B_{1}}$ real. Then we have to calculate the real part of (14). The resulting expression is then inserted in (8). In the approximation of no absorption, one obtains for the Voigt rotation in transmission geometry:

$${\displaystyle (15)\quad \delta \beta =(\mu _{0}H)^{2}{\frac {B_{1}+n_{0}^{2}{\Big [}{\frac {2L\omega }{c}}(1+n_{0})Q_{i}Q_{r}+Q_{r}^{2}-Q_{i}^{2}{\Big ]}}{n_{0}(1+n_{0})}}}$$