A Can the Huygens-Fresnel principle be translated into path integrals?

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Wave optics, including diffraction, seems to be apt for path integral language. In fact, Feynman's double slit language is purely "diffraction". Also, the PDE for the wave equation results in a solution via Green's function, and the Green function is where "the path integral lives".

I have googled and there are a few references with somewhat blurry claims. I have not seen a explicit derivation.

Does anybody know if this program is possible, and if so, where is it derived?
 

vanhees71

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Sure, the standard way to derive diffraction is the Kirchhoff theory and various approximations to it (Fraunhofer and Fresnel observation). Exact electromagnetic diffraction theory is pretty complicated and has been first worked out by Sommerfeld (afaik in his habilitation thesis).

Of course, all this can be reformulated as path integrals, but the result is the same as it must be.
 
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And what's the resulting Lagrangian?
 

vanhees71

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I don't understand what you mean by "resulting Lagrangian"?
 
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For particles, Feynman showed that the path integral, with the classical Lagrangian for particles, "was" the solution to the Schrödinger equation (more precisely, the path integral is the Green function of the Schrödinger equation).

Classical waves are solutions of the wave equation. The papers:


show that for a series of partial differential equations (elliptic, parabolic and hyperbolic), their solutions can be recast into a path integral form.

For the Schrödinger equation, which is parabolic (if we exclude the complex number issue), this is just Feynman path integral.

But now I want to think about classical waves. As seen in the papers, the wave equation (a hyperbolic PDE) has a solution in terms of a path integral. The problem is the notation of the papers is not transparent at all (at least to me). What I would like to see if the path integral solution corresponding to the wave equation can be put in a way that formally resembles the Feynman path integral for particles, and see what the corresponding Lagrangian is (in the same way the Lagrangian for the Feynman path integrals for classical particles is the Lagrangian for classical particles). I assume that it should be the Lagrangian that gives the wave equation through the Euler-Lagrange equations, but I would like to be able to see it with my eyes.
 

DrDu

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I think the classical wave picture corresponds to the case where in the path integral solution only tree level graphs are considered.
 
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I think the classical wave picture corresponds to the case where in the path integral solution only tree level graphs are considered.
Thank you. Do you have a reference for that?
 

DrDu

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I had some discussion with Urs Schreiber here in the forum. Alas I don't find the thread any more.
 
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I continue without being able to fill in the details.

Let me recap what the OP asks for: be able to express the solution of the wave equation in terms of a path integral.

In theory, the process should be easy, since the Lagrangian of Geometrical Optics is simpler than the Lagrangian of Classical Mechanics, in the sense that "there is no potential" (there is the refraction index, but usually the refraction index is constant on big areas).

But on the contrary, the Lagrangian of Geometrical Optics has a square root. And in Classical Mechanics, it is clear to put the Lagrangian in quadratic terms, in order to quantize the theory.

Since the analogy of Geometrical Optics is a particle of mass zero, this cannot be done, at least trivially (there is a way to rephrase the Lagrangian with arbitrary mass, even zero, by adding a metric, though).

The process of quantization (or wavization, as it is sometimes called) is quite easy to do by promoting the energy and the momentum to the corresponding operators (well, then one has the Helmoltz equation). But I would like to see this in the Feynman quantization.

Any ideas of how I could see this in an explicit way?
 

vanhees71

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Have a look at WKB method or, for electrodynamics, eikonal method. Mathematicians call it singular perturbation theory. Formally it's the formal expansion in powers of ##\hbar## for quantum mechanics, starting with ##1/\hbar## (that's why it's singular, starting with ##\exp(\mathrm{i}/\hbar S_{\text{cl}}##. It can be derived from the path integral formally as saddlepoint approximation.

In classical electrodynamics/optics the expansion parameter is ##1/\lambda##. The approximation is valid if the refractive index varies slowly, i.e., over distances much larger than ##\lambda##.

Characteristically the method has trouble with sharp edges, i.e., in QM at the classical points of reflection on a potential or in electromagnetism at the boundary of the shadow of an obstacle.
 
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Yes, for example the WKB method for waves is heavily used in delightful's Adam's Rays, Waves and Scattering.

But there is no single discussion about the path integral corresponding to solutions of the hyperbolic wave equation.

Instead, for parabolic PDE (e.g. heat equation), the Feynman-Kac solution is widely known and used.

So, for sure there is some mathematical issue that makes "easy" to define path integrals for parabolic PDE but "hard" for hyperbolic ones.

But intuitively in physical terms, it seems clear that the solution of the wave equation should be a path integral: Huygens principle, the double slit interference pattern for waves ...

In addition to this, the quantization via "operators" is basically trivial.

Maybe it is very easy, and as trivial that nobody bothers to right it down, but I cannot find the solution by myself, yet.
 

vanhees71

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Hm, do you mean realtivistic QFTs? There of course you apply the path integral to fields rather than to paths in single-particle phase space as in non-relativistic physics. Mathematically the equivalent to the WKB method in QFT is the derivation of the loop expansion, which formally is an expansion in powers of ##\hbar##. See p. 119ff in

 

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