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Recently I read this paper https://doi.org/10.1103/PhysRevA.30.1860 by Hillery and Mlodinow about the (canonical) quantization of electrodynamics in nonlinear dielectric media. They assume that the medium is lossless, nondispersive and uniform and that there are no free sources. As I understand it, by uniform, they mean that the susceptibilities don‘t depend on the spatial coordinates. However, after reading the paper, I still don‘t understand why the assumption of a uniform medium was necessary, as it seems to me that you could do the exact same derivation of the Hamiltonian for susceptibilities that depend on the spacial coordinates.

I also found this https://arxiv.org/abs/0901.3439 article by Hillery where he gives a pedagogical survey of nonlinear quantum optics. In chapter 8.1 he basically presents the aforementioned paper and elaborates on it. There he briefly mentions again that the formalism only works in uniform media and adds that if you do a quantization using the dual vector potential instead, you can drop the requirement of a uniform medium (There was also a paragraph about this different quantization scheme in the appendix of the original paper). Sadly though, he doesn‘t explicitly say why the method employing the dual vector potential works for inhomogeneous media in contrast to the other method using the vector potential. I don‘t see any difference between the two methods concerning inhomogeneities.

I am aware that there are also more general quantization schemes, e.g. presented in https://doi.org/10.1103/PhysRevA.42.6845 which can handle inhomogeneous and dispersive media. However, for reasons of simplicity, right now I just want to look at nondispersive media but I still would like to include inhomogeneities so that the model can handle spatially finite nonlinear crystals (which themselves can be homogenous) with vacuum around them. Also I prefer quantization via the vector potential.

To summarize, why does the quantization scheme presented by Hillery and Mlodinow only work for homogenous media? Also where lies the difference between this method and the method using the dual vector potential in that the latter apparently works for inhomogeneous media?

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  • $\begingroup$ Looking at Ref.2, is it related to the gauge condition Eq. 8.8? The authors don't explicitly solve this equation, but if you have an inhomogeneous medium then when substituting in the Taylor expansion in terms of $\mathbf{D}$ you would get spatial derivatives of the inverse susceptibilities that complicate things. $\endgroup$
    – fulis
    Commented Oct 12, 2023 at 21:18
  • $\begingroup$ @fulis First of all, thank you for your comment. As I understand it, we just need to be able to express $A_0$ in terms of $\mathbf{A}$ and $\mathbf{\Pi}$. This is possible with Eq. 8.8, Helmholtz's Theorem and the expansion given in Eq. 8.9, at least in principal. The actual calculation may be difficult for an inhomogeneous medium, but that shouldn't negate the fact that $A_0$ isn't an independent variable and can therefore be eliminated. Thus, the whole procedure should work for an inhomogeneous medium. Or am I overlooking something? $\endgroup$
    – WillHallas
    Commented Oct 17, 2023 at 17:58
  • $\begingroup$ I think you are correct. There is a book based on Ref. 2, 'The quantum theory of nonlinear optics'. In this book the authors write (sec 3.5) "In all these cases, it is essential to quantize an inhomogeneous dielectric. This is much simpler if we use the dual potential of the displacement field as the basic field variable.", suggesting that the standard treatment is not incompatible with inhomogeneous media, just more cumbersome. $\endgroup$
    – fulis
    Commented Oct 24, 2023 at 19:26
  • $\begingroup$ I think this might be related to the mode expansion of the field that they do, since the wave equation these modes satisfy (eq. 3.95 in the book, 8.47 in Ref. 2) is simpler for the dual potential, in that the susceptibility acts before the second curl. I didn't attempt to derive the mode functions using the standard potential but I could see it being harder. $\endgroup$
    – fulis
    Commented Oct 24, 2023 at 19:26

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