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Getting a diagonal susceptibility tensor:

Define the linear susceptibility tensor as $\chi_{ij}$: $P_i = \epsilon_0\chi_{ij}E_j$, using standard notation for the electric field and the polarization. If one assumes a toy model with atoms having a harmonic potential: $$\ddot{x} + 2\gamma\dot{x} + \omega_0^2x = -eE_x(t)/m$$ Now if we put $E_x(t)$ in terms of exponentials/sinusoid with a forcing frequency $\omega$, we can find the steady-state solution in the same manner as a forced harmonic oscillator. Then the $x$ can be used to write $P = -Nex$, with $N$ denoting number density. Finally, we can end up with an expression where $P \propto E$ and obtain the scalar susceptibility.

In principle, we can change the value of $\omega_0$ for $y$ and $z$ and write three different equations for the Cartesian axes to generate a diagonal susceptibility tensor, where not all non-zero terms are equal. Obviously, this tensor is symmetric and if we do a similarity transformation, it will stay symmetric.

Proving symmetry?

I'm not sure how to extend this basic idea to actually prove that the susceptibility tensor ought to be symmetric in general, which is pointed out here on page 2.

The matrix $\chi$ is known as the susceptibility tensor, and is a tensor of rank 2. This is the most general representation of the susceptibility of a linear and uniform dielectric medium. It may be shown that, for a lossless and non-optically active material, $\chi_{ij}=\chi_{ji}$.

While the optically-inactive part does seem intuitive, I'm not sure what is meant by lossless medium in this case. Does it mean $\gamma = 0$ in the toy model? As far as I understand, the $\gamma$ is introduced to account for retarding motions due to presence of other atoms in surroundings. Another question: how do you describe a material to be optically inactive, i.e., what equations can one write using it?

Also, I did manage to find a proof for the symmetry of dielectric tensor. There are actually two proofs, one using Onsager's theorem and one using Fluctuation-Dissipation theorem. However, I'm looking for a much simpler proof, hopefully in the terms of energy and momentum conservation only, and not having any Thermodynamic machinery.

Edit: The first link is from the PHYS 3003 Light and Matter course page by Tim Freegarde, School of Physics & Astronomy, University of Southampton, UK. The second link is titled "Symmetry of the Dielectric Tensor" by Curtis R. Menyuk, Computational Photonics Lab., UMBC.

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  • $\begingroup$ Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$ – Qmechanic Dec 3 '14 at 19:20
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I think a short argument would be that the energy is, up to a multiplicative factor, given by $$U\sim P_iE_i=\chi_{ij}\, E_i\, E_j$$ Therefore, $\chi$ is nothing but $$\chi_{ij}=\frac{\partial^2 U}{\partial E_i \partial E_j} $$ and therefore is symmetric.

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  • $\begingroup$ It does not work as it stands: If $\chi_{ij}$ had an anti-symmetric part, $\frac{\partial^2 U}{\partial E_i \partial E_j}$ would see the symmetric part only, as the symmetric one would be canceled out in $U = \chi_{ij} E_iE_j$ in view of the symmetry of $E_iE_j$. However I suspect that some energetic reasoning proves the wanted symmetry. $\endgroup$ – Valter Moretti Aug 21 '14 at 17:02
  • $\begingroup$ The wanted symmetry is equivalent to say that $P_i \sim \frac{\partial U(E)}{\partial E_i}$... $\endgroup$ – Valter Moretti Aug 21 '14 at 17:15
  • $\begingroup$ @ValterMoretti I agree with the math, but disagree with the interpretation. This whole $P\propto E$ relation is a linear response theory. It is derived the other way around: The energy, to leading order in $\vec E$, is given by $U\approx U_0 +\chi_{ij}E_iE_j$. The lack of a linear term is because $U$ is minimal when $\vec E=0$. Therefore, WLOG $\chi$ can be chosen to be symmetric. $P_i$ is defined as $\partial U/\partial E_i$. $\endgroup$ – yohBS Aug 21 '14 at 21:37
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    $\begingroup$ But isn't this line of reasoning misleading/incorrect because it concludes that $\chi_{ij}$ is symmetric for all materials? $\endgroup$ – theindigamer Aug 22 '14 at 1:45
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    $\begingroup$ I think it is. millersville.edu/~jdooley/macro/derive/elpol/alphasym/… $\endgroup$ – yohBS Aug 26 '14 at 9:58

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