# The operator calculation of Helmholtz equation?

I am reading the beam propagation method (BPM) in optical imaging paper. I find a paper solve the Helmholtz equation in the inhomogeneous media. The paper is:

Light propagation in graded-index optical fibers, M. D. Feit and J. A. Fleck, Jr

They use the operator form to solve the equation, I never see this kind of operator calculation to solve the classical eletromagnetic proble before. I can intuitively understand how it is going on. But I want to learn more about this and acquire such method. However, i am not a mathmatical stduent. I don't want to learn a full book of the function analysis or some group theory to just acquire this. I mean is there some good,simple and not very long pages reading material to learn this kind of technique.

Besides, there are some places that i don't understand in the paper.

1. Question 1:

Like the equation (3) How to obtain this? Now the the expansion for $$(1+x)^{1/2}$$ is not hold for the Laplacian operator?

1. The second place i don't understand is: Why it is equivalent to the Helmholtz equation? Intuitively, I know that using the angular spectrum to solve the Helmholtz equation, we can get maybe the top solution. However, is there or mathmatical derivation of this? From top to expression to the equation (10).

• $(a+b)(a-b)=a^2-b^2$ for the former issue. The latter issue may have some meaning in the sense of semi groups, using the identity of the former issue you raised and factorising the left hand side of the last written equation. Jan 22 at 8:47
• Sorry I do not know books as you desire. A physicist usually learns how practically deal with this mathematical machinery when learning quantum mechanics… Jan 22 at 8:57
• Sorry, for the former issue can you say something more about that. Sorry for my stupid, i do not get your point. Jan 22 at 9:13
• For the second i also not get your point, pretty sorry. What do you mean of the factorising the left ahnd side of last equation. You mean factorising the Helmoltz equation or factorising the exp term? You say factorising mean thing like talyor expansion? Maybe, i am too stupid, i do not get your point. Jan 22 at 9:16
• For quantum mechanics, i learned some by myself. I know that it is filled with operator thing like momentum or Hamiltonian operator. But to be honesty, i do not have the ability to use such operator method to solve a general PDE like eletromagnetic problem. Do you have some book like some quantum mechanics book recommend, and which chapter i need focus on? @Valter Moretti Jan 22 at 9:21

Consider $$A := \sqrt{\nabla_\perp^2 + k^2I}$$ and $$B=kI$$ where $$k$$ is a real positive number. Since $$[A,B]=0$$, it holds $$(A-B)(A+B) = A^2 +AB-BA -B^2 = A^2 +BA-BA -B^2 = A^2-B^2\:.$$ Hence $$A = \frac{A^2-B^2}{A+B} + B\:.$$ That is your first identity.

Regarding the second issue, the equation you want to solve is $$\left(\nabla^2_\perp + k^2 + \frac{\partial^2}{\partial z^2}\right)E=0 \tag{1}\:.$$ You can factorize it as $$\left(\sqrt{\nabla^2_\perp + k^2} + i \frac{\partial }{\partial z}\right) \left(\sqrt{\nabla^2_\perp + k^2} - i \frac{\partial }{\partial z} \right) E=0\:.$$ In summary, if you are able do solve both equations $$\left(\sqrt{\nabla^2_\perp + k^2} \pm i \frac{\partial }{\partial z}\right) E_\pm=0$$ you are done by using twice the form of the solution. As the paper declares on its first page, the solutions of the equations above are $$E_\pm(x,y,z) = e^{\mp i z\sqrt{\nabla^2_\perp + k^2}} E_\pm(x,y,0)\:.$$ Using the initial identity, $$E_\pm(x,y,z) = e^{\mp i z\left[\frac{\nabla_\perp^2}{\sqrt{\nabla^2_\perp + k^2}+ k}+ k\right]} E_\pm(x,y,0) =e^{\mp i k z} e^{\mp i z\frac{\nabla_\perp^2}{\sqrt{\nabla^2_\perp + k^2}+k}} E_\pm(x,y,0)\:.$$ So, the information stored in $$e^{\mp i z\frac{\nabla_\perp^2}{\sqrt{\nabla^2_\perp + k^2}+k}} E_\pm(x,y,0)\:,$$ taking the trivial phase $$e^{\mp i k z}$$ into account, permits to solve (1). In this sense the said operation is equivalent'' to solve (1). I omitted a huge number of mathematical subtleties which do not affect the physical substance here.

COMMENT. Reading quite superficially the paper I realized that what I indicated by $$k$$ is actually a function of $$x,y$$ and not a constant. If it is really true my derivation does not hold, but also the your first identity declared in the paper is false. It is of crucial relevance that $$\nabla_\perp^2$$ commutes with $$k$$.

• Thank you very much. It solve my question for long time. It looks like a black magic to me. I want to know how you acquire such technique. Like by reading some quantum mechanics book and doing some exercises? My research is more about classical optics or electromagetic problem. I read some paper that use some of the quantum mechanics's mathmatical method to handle the classic electromagnetic problem. But I do not have such ability to do it by myself. Do you have some reading material like some quantum mechanics book recommend, and which chapter i need focus on? Thank you very much Jan 22 at 10:19
• Sorry I am basically a mathematician concerning this stuff, so I am not the right person to ask. What you should look for is a practically minded book of mathematical methods for engineering. Jan 22 at 10:30
• For you question, unfortunately the $\nabla^{2}$ is not commuting with k(x,y,z). You can check the book Introduction to Optical Waveguide Analysis: Solving Maxwell's Equation and the Schrodinger Equation, chapter 5, Beam Propagation Method. There is a prove of this. It is very simple, you definitely can prove it by yourself. However, i think you derivation still hold. Because this paper first assume that they are the commutator. Then the non commutator error can be estimated by formula called Baker-Campbell-Hausdorff formula, or maybe just the Taylor expansion. Jan 22 at 10:31
• The error depend on how you split the k(x,y,z) with the Lapacian operator term, error will be like $O(\delta z ^2)$ or $O(\delta z ^3)$ As soon as the $\delta z$ is small enough, it won't cause many problem. Jan 22 at 10:34
• Ok it makes sense. Jan 22 at 10:40

For the first part of the question

we pose : $$x=\Delta_{\perp}\;\;, y=\frac{\omega}{c}n$$

the second member of the equation becomes:

$$\frac{x^{2}}{\sqrt{x^{2}+y^{2}}+y}+y=\frac{x^{2}(\sqrt{x^{2}+y^{2}}-y)}{(\sqrt{x^{2}+y^{2}}+y)(\sqrt{x^{2}+y^{2}}-y)}+y=\frac{x^{2}(\sqrt{x^{2}+y^{2}}-y)}{x^{2}}+y=(x^{2}+y^{2})^{\frac{1}{2}}$$

• Thank you very much. Jan 22 at 11:00
• what is idea behind this change. Why people want to have such transformation? I do not get the idea behind it why people want to do so? what advantage this can bring to? Jan 22 at 12:17