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Theoretical penetration limit for evanescent waves

+ 9 like - 0 dislike

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 $$d = \frac{1}{\sqrt{k_x^2+k_y^2-(\tfrac{2\pi}{\lambda})^2}},$$ where $k_x$ and $k_y$ are the perpendicular components of the wave vector, and $\lambda$ 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 $\langle x^2\rangle$ (and $k_y=0$ to reduce the problem to two dimensions) there seems to be a relation that $\langle d\rangle/\sqrt{\langle x^2\rangle}$ is lesser than a constant.

The main questions: is there any strict bound in the form of $$\text{a measure of penetration depth}\leq f(\text{transversal beam size},n)$$ (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
@VladimirKalitvianski: The penetration length is highly dependent on the incident angle. No, I am not talking about the Goos-Hanchen displacement (while it is always present in the presented setup. I'm not that interested into it).

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user Piotr Migdal
I have a feeling that the penetration length is independent. There is a phenomenon of the beam displacement along the media interface, but the penetration length is determined with $n$ and $\lambda$. A finite size beam is a superposition of plane waves for each of which everything is clear.

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user Vladimir Kalitvianski
I understand but there is no notion of a penetration length for a couple of beams, for example. Each angle gives its own length and the resulting "penetration length" is the longest by definition. One cannot make a "weighted" length, can one?

This post imported from StackExchange Physics at 2015-01-07 19:25 (UTC), posted by SE-user Vladimir Kalitvianski
@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 $e^{-2}$, or some weighted $\langle d \rangle=\int d |A(d)|^2 dd $) 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 $\mathcal{E}(x,z)=\mathcal{E}(x,0)e^{−\kappa z}e^{i\left(k_xx−ωt\right)}$ 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
Most recent comments show all comments
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

2 Answers

+ 1 like - 0 dislike

There always is the Heisenberg principle, which, applied to a nearly unidirectional motion, states that $$\Delta x_T \Delta p_T \ge \frac\hbar2$$ for the transversal displacement and transversal component of momentum. Using the fact that for photons, momentum is directly proportional to wave vector through $p = \hbar k$, we can get rid of $\hbar$ and obtain a completely classical relation valid for beams, $$\Delta x_T \Delta k_T \ge \frac12.$$ Here $\Delta x_T$ is the square root of $\langle x^2 \rangle$.

Now let for total internal reflection, the light is incident at the boundary at an angle $\theta > \theta_{\it crit}$ to the normal. For the following, we will need the parallel component of the $k$ vector $$k_y = k \sin \theta$$ and its standard deviation $$\Delta k_y = \Delta k_T \cos \theta.$$

The mean penetration depth in the ensemble of the evanescent waves will still be expressed by the same formula, $$d = \frac1{\sqrt{k_y^2-\left(\frac{2\pi}{\lambda}\right)^2}},$$ but obtains a standard deviation $$\Delta d = \left| \frac{\partial d}{\partial k_y} \right| \Delta k_y = k_y \Delta k_y d^3 = \frac{\sin\theta\cos\theta}{\sqrt{(n\sin\theta)^2-1}^3}\left(\frac{\lambda}{2\pi}\right)^2 \Delta k_T.$$ One might rewrite this as an uncertainty relation multiplying by $\Delta x_T$: $$\Delta d \Delta x_T \ge \frac{k_y d^3 \cos\theta}2.$$

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 $\theta$. Those that reach $\theta_{\it crit}$ 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 $d=\infty$). 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 $+\infty$. 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 $\frac{1}{e}$ 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 $f({\bf{r}},t){\hat{\bf{p}}}$, where ${\hat{\bf{p}}}$ 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 $f({\bf{r}},t)$ can represented as \begin{align} f({\bf{r}},t)=\int_{-\infty}^\infty F({\bf{k}},\omega)e^{i({\bf{k}}\cdot{\bf{r}}-\omega t)} d^3kd\omega \end{align} where $F({\bf{k}},\omega)$ is the 4-dimensional Fourier transform of $f({\bf{r}},t)$, (3 spatial dimensions and 1 temporal dimension). In general, ${\bf{k}}$ can be a complex valued vector, but here we assume that the ${\bf{k}}$ is propagating, (i.e., homogeneous) so it is real valued. Additionally, because of the symmetries of a planar interface, we can assume $k_y=0$, and $k_z\geq 0$, 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 : \begin{align} \tau_{\text{p}}(k_x,\omega) = &2\frac{\epsilon_a(\omega)\sqrt{\epsilon_b(\omega)\mu_b(\omega) - \left(\frac{ck_x}{\omega}\right)^2}}{\epsilon_a(\omega)\sqrt{\epsilon_b(\omega)\mu_b(\omega)-\left(\frac{ck_x}{\omega}\right)^2}+\epsilon_b(\omega)\sqrt{\epsilon_a(\omega)\mu_a(\omega)-\left(\frac{ck_x}{\omega}\right)^2}}\\ =&2\frac{\epsilon_a(\omega)\sqrt{n_b^2(\omega) - \left(\frac{ck_x}{\omega}\right)^2}}{\epsilon_a(\omega)\sqrt{n_b^2(\omega)-\left(\frac{ck_x}{\omega}\right)^2}+\epsilon_b(\omega)\sqrt{n_a^2(\omega)-\left(\frac{ck_x}{\omega}\right)^2}}\\ \end{align} where $n(\omega)=\sqrt{\epsilon(\omega)\mu(\omega)}$ is the complex index of refraction, $\epsilon(\omega)$ is the dielectric permittivity, $\mu(\omega)$ is the magnetic permeability, and $\omega$ is angular frequency. Is this case we assume that we are at optical frequencies, i.e., that $\mu(\omega) \approx 1$, then \begin{align} n^2(\omega) \to \epsilon(\omega) \end{align} which implies that the Fresnel coefficient becomes: \begin{align} \tau_{\text{p}}(k_x,\omega) = 2\frac{n_a^2(\omega)\sqrt{n_b^2(\omega) - \left(\frac{ck_x}{\omega}\right)^2}}{n_a^2(\omega)\sqrt{n_b^2(\omega)-\left(\frac{ck_x}{\omega}\right)^2}+n_b^2(\omega)\sqrt{n_a^2(\omega)-\left(\frac{ck_x}{\omega}\right)^2}}\\ \end{align} From Maxwell's equations we have the following relationship for each plane wave component: \begin{align} {\bf{k}}\cdot{\bf{k}}= &\left(\frac{\omega}{c}\right)^2\epsilon(\omega)\mu(\omega)\\ = &\left(\frac{\omega\cdot n(\omega)}{c}\right)^2 \end{align} where $c$ is the speed of light. In our case this results in: \begin{align} k_x^2 + k_z^2= &\left(\frac{\omega \cdot n(\omega)}{c}\right)^2 \end{align} We now need to answer the question, what is the resultant ${\bf{k}}'$ on the other side of the interface? From the generalized Snell's law we know that, for our case, $k_x = k'_x$. So we must compute $k'_z$ using the generalized Snell's law: \begin{align} n_a(\omega)\sin(\theta_a)=&n_b(\omega)\sin(\theta_b)\\ \implies \frac{n_a(\omega)}{n_b(\omega)}\sin(\theta_a)=&\sin(\theta_b) \end{align} where $\theta$ is the angle of the ${\bf{k}}$s 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 $n_a(\omega) > n_b(\omega)$ the we have the following equation where $C$ is real: \begin{align} &\frac{e^{i\theta_b}-e^{-i\theta_b}}{2i}=C\\ \implies &e^{i\theta_b}-e^{-i\theta_b}=2Ci\\ \end{align} In this case we have that $\Re(\theta_b) = \pi /2$, and the imaginary part varies, to meet the above equation. Note that in whatever latex interpreter StackExchange is using $\Re(\cdot)$ is the real part and $\Im(\cdot)$ is the imaginary part. Then \begin{align} &ie^{-\Im(\theta_b)}+ie^{\Im(\theta_b)}=2Ci\\ \implies &e^{-\Im(\theta_b)}+e^{\Im(\theta_b)}=2C\\ \implies &\cosh(\Im(\theta_b))=\frac{n_a(\omega)}{n_b(\omega)}\sin(\theta_a)\\ \implies &\Im(\theta_b)=\cosh^{-1}\left(\frac{n_a(\omega)}{n_b(\omega)}\sin(\theta_a)\right) \end{align} since $e^{i\pi/2} = i, e^{-i\pi/2}=-1$. Now since $\theta_a$ is defined to the $z$-axis, we have the following definition for ${\bf{k}}$: \begin{align} {\bf{k}}=\frac{\omega \cdot n_a(\omega)}{c}\begin{pmatrix} \sin(\theta_a)\\ 0\\ \cos(\theta_a) \end{pmatrix}. \end{align} Note, that we have: \begin{align} {\bf{k}}'\cdot {\bf{k}}' = \|{\bf{k}}'_{\text{re}}\| - \|{\bf{k}}'_{\text{im}}\| + 2i{\bf{k}}'_{\text{re}}\cdot {\bf{k}}'_{\text{im}} = \left(\frac{\omega}{c}\right)^2n^2_b(\omega) \end{align} for the complex valued ${\bf{k}}'= {\bf{k}}'_{\text{re}} + {\bf{k}}'_{\text{im}}$. Since we are assuming that $n_b(\omega)$ is real valued, then the imaginary part of the dot product must be zero, and hence the real and imaginary parts of ${\bf{k}}'$ are perpendicular. Therefore, since $k'_x = k_x$ and $k_x$ is real, $k'_z$ is either real or purely imaginary. When $k'_z$ is purely imaginary then we have the definition of an evanescent wave. Those components in the input beam which have $k'_z$ purely real will not be evanescent. Now we just need to compute $k'_z$. We know that \begin{align} {\bf{k}}'=\frac{\omega \cdot n_b(\omega)}{c}\begin{pmatrix} \sin(\theta_b)\\ 0\\ \cos(\theta_b) \end{pmatrix} =\frac{\omega \cdot n_b(\omega)}{c}\begin{pmatrix} k_x\\ 0\\ \cos(\theta_b) \end{pmatrix}. \end{align} It is easy to show that for $\theta_b = \pi/2 + \Im(\theta_b)$ and the above solution for $ \Im(\theta_b)$ that \begin{align} \cos(\theta_b) = &i\sinh[\Im(\theta_b)]\\ = &i\sinh\left[\cosh^{-1}\left(\frac{n_a(\omega)}{n_b(\omega)}\sin(\theta_a)\right)\right]\\ = &i\sinh\left[\cosh^{-1}\left(\frac{c}{\omega n_b(\omega)}k_x\right)\right] \end{align} Now, we have each ${\bf{k}}$ and associated ${\bf{k}}'$, and the $\tau_p$. The field on the $n_b$ side of the interface is then \begin{align} \tau_{\text{p}}(k_x,\omega)F({\bf{k}},\omega)e^{i({\bf{k}}'\cdot{\bf{r}}-\omega t)} \end{align} for each ${\bf{k}}$ and each $\omega$. We must superpose these to obtain the field. First, we shall rewrite the above in terms of $k_x$.

\begin{align} {\bf{k}}'=\frac{\omega \cdot n_b(\omega)}{c}\begin{pmatrix} k_x\\ 0\\ i\sinh\left[\cosh^{-1}\left(\frac{c}{\omega n_b(\omega)}k_x\right)\right] \end{pmatrix} \end{align}

Note that the above is only for the evanescent components! We have the following relations (a restatement of above): \begin{align} k_x'&=k_x\\ k_y'&=k_y=0\\ k_z'&=i\sinh\left[\cosh^{-1}\left(\frac{c}{\omega n_b(\omega)}k_x\right)\right] \end{align} The superposition then becomes: \begin{align} \int_{-\infty}^\infty\int_{\bf{E}}\tau_{\text{p}}(k_x,\omega)F(k_x,k_z,\omega)e^{-\sinh\left[\cosh^{-1}\left(\frac{c}{\omega n_b(\omega)}k_x\right)\right]z}e^{i(k_x x-\omega t)}dk_x dk_z dt + \big[\text{propagating components}\big] \end{align} where $\bf{E}$ is the domain for the evanescent components. Then by analyzing the inner integral (wrt $k_x$) we get \begin{align} &\int_{-\infty}^\infty\mathrm{rect}\left(\frac{(k_x-k_{x,0})}{D}\right)\tau_{\text{p}}(k_x,\omega)F(k_x,k_z,\omega)e^{-\sinh\left[\cosh^{-1}\left(\frac{c}{\omega n_b(\omega)}k_x\right)\right]z}e^{ik_x x}dk_x\\ =&\mbox{$\mathscr{F}$}^{-1}\left\{\mathrm{rect}\left(\frac{(k_x-k_{x,0})}{D}\right)\tau_{\text{p}}(k_x,\omega)F(k_x,k_z,\omega)e^{-\sinh\left[\cosh^{-1}\left(\frac{c}{\omega n_b(\omega)}k_x\right)\right]z}\right\}, \end{align} i.e., the inverse Fourier transform of a somewhat nasty function for the evanescent part where $\mathrm{rect}\left(\frac{(k_x-k_{x,0})}{D}\right)$ is the frequency window for the evanescent components. Evaluating the above inverse Fourier transform and computing the $\frac{1}{e}$ 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 ]

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