Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.
W3Counter Web Stats

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public β tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

206 submissions , 164 unreviewed
5,106 questions , 2,251 unanswered
5,412 answers , 23,072 comments
1,470 users with positive rep
822 active unimported users
More ...

  Theoretical penetration limit for evanescent waves

+ 9 like - 0 dislike
4596 views

Consider a problem in classical electrodynamics, when a monochromatic beam experiences total internal refraction when traveling from a medium with n>1 to a medium with refractive index 1 - see schematic below. Using Fresnel equations one gets the penetration depth d=1k2x+k2y(2πλ)2,

where kx and ky are the perpendicular components of the wave vector, and λ is the wavelength (in the vacuum).

At least in theory, it is possible to have an evanescent wave of an arbitrary penetration depth d. However, in such case one needs to use a plane wave, thus a wave of unbounded spatial size. For a beam with a finite variance x2 (and ky=0 to reduce the problem to two dimensions) there seems to be a relation that d/x2 is lesser than a constant.

The main questions: is there any strict bound in the form of a measure of penetration depthf(transversal beam size,n)

(perhaps in the style of Heisenberg uncertainty principle, or using other moments of x, y and d)?

Schematic of an evanescent wave size


This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user Piotr Migdal

asked Sep 18, 2011 in Theoretical Physics by Piotr Migdal (1,260 points) [ revision history ]
edited Jan 7, 2015 by Dilaton
Most voted comments show all comments
@daaxix Thanks for your answer, and for your effort. However, as of now, for me the hardest part is to make a complete proof. I already had a few "things than may result in a proof" or hand-waving arguments. The general procedure is in some sense "obvious" (well, for me Frourier transform is as natural as eating a beakfest :)).

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user Piotr Migdal
My answer is the correct way, classically. You could have some QM argument, but I suspect that a QM argument would end up being fairly unmanageable. You won't have an analytic solution because of the number of atoms involved, and without a periodic potential (as in an amorphous solid) you cannot use Bravais lattice types of approximations. I really like this question, I've never seen an actual answer before myself...

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user daaxix
@daaxix I don't understand your point. There is nothing QM here - pure Maxwell equations / wave physics (here we consider waves of much larger size than any molecules).

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user Piotr Migdal
Right, the only "more correct" approach would be a full QM treatment, which is unmanageable, my point was that my approach is as correct as you can manage without going to the QM level...

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user daaxix
@daaxix I don't claim that your approach is wrong. But I don't see result "a measure of penetration depth≤f(transversal beam size,n)" or proof that there isn't such.

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user Piotr Migdal
Most recent comments show all comments
@VladimirKalitvianski: For a planewave you get an exponentially decaying amplitude (with distance in the z-direction). For a general wave, you have a superposition of different exponentially decaying waves. The resulting penetration depth (e.g. when intensity drops to e2, or some weighted d=d|A(d)|2dd) won't be just max d.

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user Piotr Migdal
This paper seems to proove, that a local plane wave expression for the evanescent field E(x,z)=E(x,0)eκzei(kxxωt) is valid at least in the region z∼w, where w is the beam waist. Unfortunately, I failed to find this paper in open access.

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user straups

2 Answers

+ 1 like - 0 dislike

There always is the Heisenberg principle, which, applied to a nearly unidirectional motion, states that ΔxTΔpT2

for the transversal displacement and transversal component of momentum. Using the fact that for photons, momentum is directly proportional to wave vector through p=k, we can get rid of and obtain a completely classical relation valid for beams, ΔxTΔkT12.
Here ΔxT is the square root of x2.

Now let for total internal reflection, the light is incident at the boundary at an angle θ>θcrit to the normal. For the following, we will need the parallel component of the k vector ky=ksinθ

and its standard deviation Δky=ΔkTcosθ.

The mean penetration depth in the ensemble of the evanescent waves will still be expressed by the same formula, d=1k2y(2πλ)2,

but obtains a standard deviation Δd=|dky|Δky=kyΔkyd3=sinθcosθ(nsinθ)213(λ2π)2ΔkT.
One might rewrite this as an uncertainty relation multiplying by ΔxT: ΔdΔxTkyd3cosθ2.

This deviation would actually enlarge the depth that the evanescent waves can reach. Indeed, the beam is composed of plane waves at angles slightly larger and slightly smaller than θ. Those that reach θcrit induce an evanescent wave of infinite penetration depth; those that have even smaller incident angle continue as plane waves in the other medium and reach infinite distances trivially.

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user Vašek Potoček
answered Sep 14, 2012 by Vašek Potoček (45 points) [ no revision ]
A nice approach. However, I tried to bound it from above and only for evanescent wave (for non-evanescent obviously d=). So there is a problem to separate both parts and to get the opposite inequality (yes, also I tried Heisenberg uncertainty principle). But anyway, thx for the insight, maybe it can help me.

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user Piotr Migdal
I mention this near the end. Maybe my wording was wrong. The evanescent waves created by reflected beams sufficiently near the critical angle have penetration depths surpassing any upper bound, thus smoothly transiting to the non-evanescent regime. If we only separated evanescent waves from non-evanescent ones, the supremal d would always be +. Since this is not a very informational result, we might replace maximum by mean, which is independent of beam width, or perhaps mean plus 1 or 2 standard deviations.

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user Vašek Potoček
+ 1 like - 0 dislike

Because the math may be somewhat dense, I outline the procedure below:

  1. Assume we have an input field with a known spatio-temporal distribution.
  2. Fourier transform that input field in space and time to get the plane wave components.
  3. Apply Fresnel analysis to each component and obtain the plane wave spectrum for the evanescent side of the interface.
  4. Superpose all the evanescent components back together (if possible).
  5. The superposition should result in a single penetration depth, which may in fact be infinite if all components of the input field are not above the critical angle.
  6. Find the resulting exponential decay term and take 1e of that value to be the penetration depth.

Suppose that we have some incident electric field incident on an interface completely polarized in the p-plane, given by f(r,t)ˆp, where ˆp is the unit vector in the direction perpendicular to the Poynting vector for the propagating fields. We also assume that the incident field contains no evanescent wave components. Any vector field whose components are elements of the Schwartz class (of functions) can be represented as a superposition of plane waves \cite{Lax02}, so f(r,t) can represented as f(r,t)=F(k,ω)ei(krωt)d3kdω

where F(k,ω) is the 4-dimensional Fourier transform of f(r,t), (3 spatial dimensions and 1 temporal dimension). In general, k can be a complex valued vector, but here we assume that the k is propagating, (i.e., homogeneous) so it is real valued. Additionally, because of the symmetries of a planar interface, we can assume ky=0, and kz0, when we define the interface to lie in the xy-plane, with the Poynting vector of the incoming beam point somewhere in the +z direction, and when, for simplicity, we assume that the beam is only varying spatially in the xz-plane. For this case, the general Fresnel amplitude coefficient for the p-polarization is given by : τp(kx,ω)=2ϵa(ω)ϵb(ω)μb(ω)(ckxω)2ϵa(ω)ϵb(ω)μb(ω)(ckxω)2+ϵb(ω)ϵa(ω)μa(ω)(ckxω)2=2ϵa(ω)n2b(ω)(ckxω)2ϵa(ω)n2b(ω)(ckxω)2+ϵb(ω)n2a(ω)(ckxω)2
where n(ω)=ϵ(ω)μ(ω) is the complex index of refraction, ϵ(ω) is the dielectric permittivity, μ(ω) is the magnetic permeability, and ω is angular frequency. Is this case we assume that we are at optical frequencies, i.e., that μ(ω)1, then n2(ω)ϵ(ω)
which implies that the Fresnel coefficient becomes: τp(kx,ω)=2n2a(ω)n2b(ω)(ckxω)2n2a(ω)n2b(ω)(ckxω)2+n2b(ω)n2a(ω)(ckxω)2
From Maxwell's equations we have the following relationship for each plane wave component: kk=(ωc)2ϵ(ω)μ(ω)=(ωn(ω)c)2
where c is the speed of light. In our case this results in: k2x+k2z=(ωn(ω)c)2
We now need to answer the question, what is the resultant k on the other side of the interface? From the generalized Snell's law we know that, for our case, kx=kx. So we must compute kz using the generalized Snell's law: na(ω)sin(θa)=nb(ω)sin(θb)na(ω)nb(ω)sin(θa)=sin(θb)
where θ is the angle of the ks with respect to the z-axis, and the sine functions can be complex valued. If the left hand side of the above is real and na(ω)>nb(ω) the we have the following equation where C is real: eiθbeiθb2i=Ceiθbeiθb=2Ci
In this case we have that (θb)=π/2, and the imaginary part varies, to meet the above equation. Note that in whatever latex interpreter StackExchange is using () is the real part and () is the imaginary part. Then ie(θb)+ie(θb)=2Cie(θb)+e(θb)=2Ccosh((θb))=na(ω)nb(ω)sin(θa)(θb)=cosh1(na(ω)nb(ω)sin(θa))
since eiπ/2=i,eiπ/2=1. Now since θa is defined to the z-axis, we have the following definition for k: k=ωna(ω)c(sin(θa)0cos(θa)).
Note, that we have: kk=krekim+2ikrekim=(ωc)2n2b(ω)
for the complex valued k=kre+kim. Since we are assuming that nb(ω) is real valued, then the imaginary part of the dot product must be zero, and hence the real and imaginary parts of k are perpendicular. Therefore, since kx=kx and kx is real, kz is either real or purely imaginary. When kz is purely imaginary then we have the definition of an evanescent wave. Those components in the input beam which have kz purely real will not be evanescent. Now we just need to compute kz. We know that k=ωnb(ω)c(sin(θb)0cos(θb))=ωnb(ω)c(kx0cos(θb)).
It is easy to show that for θb=π/2+(θb) and the above solution for (θb) that cos(θb)=isinh[(θb)]=isinh[cosh1(na(ω)nb(ω)sin(θa))]=isinh[cosh1(cωnb(ω)kx)]
Now, we have each k and associated k, and the τp. The field on the nb side of the interface is then τp(kx,ω)F(k,ω)ei(krωt)
for each k and each ω. We must superpose these to obtain the field. First, we shall rewrite the above in terms of kx.

k=ωnb(ω)c(kx0isinh[cosh1(cωnb(ω)kx)])

Note that the above is only for the evanescent components! We have the following relations (a restatement of above): kx=kxky=ky=0kz=isinh[cosh1(cωnb(ω)kx)]

The superposition then becomes: Eτp(kx,ω)F(kx,kz,ω)esinh[cosh1(cωnb(ω)kx)]zei(kxxωt)dkxdkzdt+[propagating components]
where E is the domain for the evanescent components. Then by analyzing the inner integral (wrt kx) we get rect((kxkx,0)D)τp(kx,ω)F(kx,kz,ω)esinh[cosh1(cωnb(ω)kx)]zeikxxdkx=F1{rect((kxkx,0)D)τp(kx,ω)F(kx,kz,ω)esinh[cosh1(cωnb(ω)kx)]z},
i.e., the inverse Fourier transform of a somewhat nasty function for the evanescent part where rect((kxkx,0)D) is the frequency window for the evanescent components. Evaluating the above inverse Fourier transform and computing the 1e point of it would yield the depth (there is an integral of an exponentially decaying term here), I'm still working on this part...I'll update as it is completed.

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user daaxix
answered Dec 15, 2012 by daaxix (10 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol in the following word:
ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...