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I understand that a permittivity tensor is symmetric if there is no absorption. On the other hand is it always Hermitian? Under what cases would that not be the case?

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  • $\begingroup$ The permittivity tensor is Hermitian when absorption losses can be neglected. If the tensor is non-real, then it corresponds to complex optical axes. The axes in this case correspond to general elliptically-polarised light, where left- and right-rotating polarisations travel at different speeds (This is "birefringence for elliptically-polarised light"). See, for example, the magneto-optic effect :) $\endgroup$ Commented May 30 at 13:22

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According to this text, the energy dissipated by a dielectric is given by

$$ Q = \frac{i \omega}{16 \pi} \left\{ \left( \varepsilon_{i k}^{*} - \varepsilon_{k i } \right) E_{i} E_{k}^{*} + \left( \mu_{i k}^{*} - \mu_{k i } \right) H_{i} H_{k}^{*} \right\} \tag{1} \label{1} $$

That means that if the matrix is non hermitian, it absorbs energy.

I'm not aware of what conditions would make it real in addition, which would be a symmetric matrix.

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