For a general permittivity tensor, why is the imaginary part of the off-diagonal component, e.g.:
$$\epsilon_{xy}$$
equal to:
$$\frac{\epsilon_{xy} + \epsilon_{yx}^*}{2i}$$
instead of:
$$(\epsilon_{xy} + \epsilon_{xy}^*)/2i$$
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$\begingroup$ Where did you get this relation? $\endgroup$– Er JioCommented Feb 15 at 18:38
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$\begingroup$ Because when you take complex conjugate of a matrix in addition of conjugating elements you must replace index of column with index of row I.e. $[\epsilon_{xy}]^\dagger=[\epsilon^*_{yx}]$ $\endgroup$– Sancol.Commented Feb 15 at 21:28
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$\begingroup$ I don't understand why we need to take the Hermitian conjugate of the epsilon tensor though? Could you explain the intuition behind this? Thanks! $\endgroup$– photonicaCommented Feb 16 at 1:03
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1 Answer
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We can split the permittivity $\epsilon$ into to part 1) refractive part $\epsilon_r$ and 2) absorptive part $\epsilon_a$ as follows: $$\epsilon = \epsilon_r + i\epsilon_a (1)$$ permittivity tensors $\epsilon_r$ and $\epsilon_a$ must be hermitian because both are physical quantities i.e. $$\epsilon_r^\dagger = \epsilon_r (2)$$ $$\epsilon_a^\dagger = \epsilon_a (3)$$ Therefore getting hermittian conjugate of equation (1) we have $$\epsilon^\dagger = \epsilon_r - i\epsilon_a (4)$$ Adding and subtracting (1) and (4) get $$\epsilon_r = \frac{\epsilon +\epsilon^\dagger}{2} (5)$$ $$\epsilon_r = \frac{\epsilon -\epsilon^\dagger}{2i} (6)$$
Be ware when you take complex conjugate of a matrix in addition of conjugating elements you must replace index of column with index of row i.e. $$[\epsilon_{xy}]^†=[\epsilon^*_{yx}]$$
