# Is polarization matrix always diagonalizable?

In chapter 31 of Feynman lectures Vol 2, he covers polarization , polarization tensor and its diagonalisation, he proves that for a crystal, the tensor matrix is symmetric hermitian and hence diagonalizable, however if one reads carefully , one notices he talks clearly for crystals only and avoids generality, so is this diagonalizability valid only for crystals as they have proper symmetries or if i take any amorphous blob made of asymmetric organic molecules, it must be diagonalizable ?

I doubt this because there can be some hysteresis like phenomenon.

The polarization matrix is always symmetric. (I think this is what you are asking, I think you aren't asking about diagonalizability since even non-symmetric matrices can be diagonalized.) And, to clarify, when I use Feynman's term polarization matrix I am referring to what I think is more often and more accurately called the polarizability tensor.

Let's see why this must be true. Let's start by accepting Feynman's statement above Eq. (31.6) that the differential work $$dW$$ need to change the polarization by $$d\mathbf{P}$$ is given by $$dW=\mathbf{E}\cdot d\mathbf{P}.$$ Let's compare this to the equation for the work $$dW$$ needed to change the extension $$\mathbf{x}$$ of a isotropic spring by the amount of $$d\mathbf{x}$$ when the spring constnat is $$k$$. This energy $$dW$$ is given by $$dW = \mathbf{F} \cdot d\mathbf{x}.$$ Here we see that $$\mathbf{E}$$ is like $$\mathbf{F}$$ and the polarization $$\mathbf{P}$$ is like the displacement $$\mathbf{x}$$.

Next Feynman integrates the electric field from $$\mathbf{0}$$ along a linear path to some value $$\mathbf{E}$$ to find the total work $$W=\frac{1}{2} \mathbf{E}\cdot \boldsymbol{\alpha}\cdot\mathbf{E},$$ which is analogous to the formula $$W=\frac{1}{2} \mathbf{F}\cdot \frac{1}{k}\cdot\mathbf{F}.$$ Here we see that the polarizability $$\mathbf{\alpha}$$ is analogous to $$1/k$$, the inverse of the spring constant.

These two equations can be written in the form $$W=\frac{1}{2} \mathbf{P}\cdot \boldsymbol{\alpha}^{-T}\cdot\mathbf{P},$$ which is analogous to the formula $$W=\frac{1}{2} \mathbf{x}\cdot {k}\cdot\mathbf{x}.$$

Having gotten this far, the key insight Feynman states is that since the microscopic theory (where all charges are accounted for explicitly) is conservative, then so too must be the macroscopic theory where the bound charges represented by $$\mathbf{P}$$ are treated separately from the free charge density.

Therefore the expression $$W=\frac{1}{2} \mathbf{E}\cdot \boldsymbol{\alpha}\cdot\mathbf{E}$$ for the work done to establish the field $$\mathbf{E}$$ must be valid not only when the field is increased linearly from $$\mathbf{0}$$ to $$\mathbf{E}$$ but for any path in electric field space starting at $$\mathbf{0}$$ and ending at $$\mathbf{E}$$.

Taking the differential of $$W=\frac{1}{2} \mathbf{P}\cdot \boldsymbol{\alpha}^{-T}\cdot\mathbf{P}$$, we find $$dW = \frac{1}{2} \mathbf{P} \cdot \left(\boldsymbol{\alpha}^{-T} + \boldsymbol{\alpha}^{-1}\right)\cdot d\mathbf{P}.$$ Comparing this with $$dW=\mathbf{E}\cdot d\mathbf{P}$$, we find that $$\mathbf{E}=\frac{1}{2}\left(\boldsymbol{\alpha}^{-T} + \boldsymbol{\alpha}^{-1}\right) \cdot \mathbf{P},$$ and comparing this with $$\mathbf{E}=\boldsymbol{\alpha}^{-1} \mathbf{P},$$ we find that $$\boldsymbol{\alpha}^{-1} = \boldsymbol{\alpha}^{-T},$$ so that $$\boldsymbol{\alpha}$$ is symmetric.

The key point was that if $$\boldsymbol{\alpha}$$ is not symmetric, then the electric field can be manipulated so that $$\oint dW = \oint \mathbf{E} \cdot d \mathbf{P}$$ around a closed path in electric field space is nonzero, violating conservation of energy. This applies in a general linear dielectric, not just in a crystal.

Feynman says this is valid for crystals as if this is an experimental observation, but then goes on to show that given the assumptions about the polarization (that it is determined by the electric field at the same time and it is linear function of electric field vector - thus no hysteresis and no nonlinearity), the polarisation tensor has to be symmetric - because otherwise work done by external field during a specially devised cycle would not be zero.

So the proof does not require a crystal, but in practice crystals come closest to the assumptions needed.