Currently I am reading about super lens and came across these two waves, propagating and evanescent. If a negative index material is used as a lens then both propagating and evanescent can be passed comparing to a conventional lens in which only propagating can be passed. What is the difference between these two waves on image formation?
[Physics] Difference between propagating and evanescent waves
opticswaves
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I believe there is an acoustic analogy to evanescent waves, therefore a spring and mass lattice model must approximate evanescent waves in the same way that a finite difference model of the continuous medium example below must simulate evanescent waves.
In homogeneous mediums scalar acoustic fields fulfil the Helmholtz equation:
$$\left(\nabla^2 + \frac{\omega^2}{c_a^2}\right) \psi = 0$$
where $\omega$ is the waves angular frequency, $c_a = \sqrt{\frac{K}{\rho}}$ the speed of sound in the medium, $\rho$ the medium's mass density and $K$ its bulk modulus.
Therefore, if you have a plane wave travelling in a medium of relatively low speed and it is incident at a glancing angle on a plane interface with a medium of relatively high speed and if the angle of incidence $\theta$ is greater than the critical angle $\theta_c$ where:
$$\sin\theta_c = \frac{c_{a,1}}{c_{a,2}} = \sqrt{\frac{K_1\,\rho_2}{K_2\,\rho_1}}$$
where $c_{a,1}$ is the speed of sound in the first (slower) medium and $c_{a,2}$ that of the second (faster) medium, then there will be total internal reflexion. Therefore, there will be an evanescent wave tunnelling into the second (faster) medium, with a Goos-Hänchen phase shift. Scalar wave theory (i.e. waves fulfilling the Helmholtz equation) is all that is needed to account for this phenomenon in optics, so there is a precise mathematical analogy between this acoustic example and my answer here where I derive the evanescent wave and the Goos-Hänchen phase shift.
So now, if you have a discretised model of such a system: i.e. two "mediums" made of a spring and mass lattices with a plane interface between them, you can have total internal reflexion and the continuum, Helmholtz-equation-fulfilling waves as long as $\lambda = 2\pi c_a/\omega$ is much longer than the lattice spacing in both mediums. Therefore, you must get approximately evanescent waves on the high velocity medium side.
It is likely even possible for discretised exactly exponentially decaying waves to be derived mathematically in the lattice without thinking of them as approximation to the continuum Helmholtz equation modes: the lattice will support discretised sine waves. You could probably derive the theory with $z$-Transformed discretised propagation equations instead of Laplace / Fourier transformed Helmholtz equations.
I have added the acoustics tag to your question because there is definitely an acoustic element to this question. Hopefully an acoustics expert can confirm the above and describe mechanical evanescent waves more directly.
Further answer after new comments
Great, total internal reflection is what I was now going to try. In fact the system I'm looking for needn't be discrete - I'd be happy with a continuous model. I tested on discrete one because I could make band gap there. What bothers me in total internal reflection is that it seems to be fundamentally multidimensional. Do you have any ideas on getting this effect in 1 dimension?
This is a very good point. I agree with you that you need two or more dimensions for total internal reflexion and this is a fundamental limitation because TIR is a glancing phenomenon - the result of a component of the wavevector projected onto an interface such that the wavevector component $k_x^\prime$ is higher than the wavevector magnitude $k^\prime$ in the new medium so that the new axial component $k_z^\prime$, which fulfils the Fourier transformed Helmholtz equation ${k_z^\prime}^2 + {k_x^\prime}^2 = k^2$ so that ${k_z^\prime}^2 < 0$.
If you are interested in one dimension, then I believe that truly there are no truly one dimensional evanescent plane waves.
For general background, you may wish to take a look at the following document:
They deal with evanescent waves in §3.3 and one dimensional systems in §4. An interesting piece of trivia is that the acoustic coupling between the loudspeaker and the listener's ear in a Sony Walkman (the authors are perhaps letting slip their age here - presumably this comment includes things like modern earpieces for iPods and the like) is through near-field, evanescent wave transmission. This of course has the advantage of being not bothersome to other nearby people because the radiative component of the acoustic field is minimised. Evanescent waves come up again in §7.2, where they look at cut-off modes in ducts. However, all these examples are two dimensional: a circular duct supports cylindrical Bessel modes whose pressure varies like $J_n(U_{n,m} r/r_0) \cos(n\,\phi+\delta)\,e{i\,k_z\,z}$ where $r$ is the radial co-ordinate, $\phi$ the azimuthal co-ordinate, $z$ the axial co-ordinate, $r_0$ the duct radius and $U_{n,m}$ the $m^{th}$ zero of the first derivative of $n^{th}$ order first kind Bessel function $J_n$. The transverse component of the wavevector has magnitude $U_{n,m}/r_0$ and the substitution of the solution into Helmholtz's equation yields:
$$k_z^2 + \frac{U_{n,m}^2}{r_0^2} = k^2$$
thus leading to "cut-off" modes (i.e. evanescent ones with imaginary axial wavenumber) when $U_{n,m} > k\,r_0$.
In one dimension, something that might be construed as an evanescent wave is the transitional acoustic field in a tuned acoustic grating. In §4.4.1 of the Rienstra &Hirschberg reference above, they derive the transmission $T$ and reflection $R$ coefficients across a plane interface and they are precisely analogous to these electromagnetic case for transverse electromagnetic waves on two-conductor, translationally invariant transmission-lines, namely:
$$R = \frac{Z_1-Z_2}{Z_1+Z_2};\quad T = \frac{2\,Z_2}{Z_1+Z_2}$$
where $Z_1, Z_2$ are the acoustic impedances of the two mediums in question. Thus if we assemble a stack of alternating, uniform media, we can build lossless acoustic reflectors, and the wave penetrating into the reflector dwindles exponentially with depth as the grating losslessly converts the forward running into the reflected field. If:
$$\psi_j=\left(\begin{array}{c}a_j\\b_j\end{array}\right)$$
represents the field at the $j^{th}$ interface in the stack, then the transfer matrix across the interface from medium 1 into medium 2 is:
$$K_j=\frac{1}{1+R}\left(\begin{array}{cc}-R&1\\1&-R\end{array}\right)$$
so that the transfer matrix through a layer of thickness $t_1$ of medium 1 followed by a layer of thickness $t_2$ of medium 2 is:
$$P=\frac{1}{1-R^2}\left(\begin{array}{cc}e^{i\,k_1\,t_1}&0\\0&e^{-i\,k_1\,t_1}\end{array}\right)\left(\begin{array}{cc}-R&1\\1&-R\end{array}\right)\left(\begin{array}{cc}e^{i\,k_2\,t_2}&0\\0&e^{-i\,k_2\,t_2}\end{array}\right)\left(\begin{array}{cc}R&1\\1&R\end{array}\right)$$
which has eigenvalues:
$$\lambda = a\pm\sqrt{a^2-b^2}$$
where:
$$a = \cos (k_1 t_1-k_2 t_2)-R^2\,\cos (k_1 t_1+k_2 t_2));\qquad b = 1-R^2$$
By checking stationary points of this expression we can show that the eigenvalues both lie inside the closed unit circle and that there are choices for $k_1\,t_1\,k_2 \,t_2$ for any $R$ such that both eigenvalues lie strictly within the unit circle. Therefore, because $\psi_{2\,j} = P^j \psi_0$, we see that this means $|\psi_j|^2$ decreases exponentially with grating cycle number $j$ and is of the order of $|\lambda_{max}|^{2\,j} |\psi_0|^2$, where $\lambda_{max}$ is the maximum magnitude eigenvalue, which we can choose to have either exactly unit magnitude, or strictly less than unit magnitude. So the grating steadily converts forwards running waves into backwards running waves as the wave penetrates deeper into the reflector. This, I believe, is a pretty good analogue for the penetration and gradual conversion of wave direction that happens beyond a totally internally reflecting optical interface.
We can even chirp the grating to get essentially any frequency response of the reflexion that we like. This is what is done in one-dimensional gratings in one-moded fibre optic waveguides. Leon Paladian at the University of Sydney came up with a general method to design the band structure of the photonic crystal (grating) to achieve a given frequency response in the Mid 1990s. Check his personal page (and here) for publications in this field. I should declare a friendship with Leon, but was not involved in this particular work.
I think you are probably getting a bit too worried about words and their meaning and are possibly trying to ascribe more precision to natural English words than they can give you without further precise description in mathematical language.
As you witness, the mathematics describing photon tunnelling and evanescent waves is exactly the same as that describing electron tunnelling into classically forbidden regions. Maxwell's equations are both classical equations and can be interpreted as the propagation equation for a one-photon state. So both phenomena - photon and electron tunnelling - are equally quantum mechanics and the mechanics of waves. The two aren't mutually exclusive: the propagation equations in quantum mechanics naturally lead to D'Alembert's and like equations. If you interpret Maxwell's equations for a tunnelling wave into a system of dielectric layers as the propagation equation for a lone photon, the energy density as a function of position for the properly normalized solution is interpreted as the probability to destructively detect the photon at the point in question with a detector when one photon states propagate into the layers separately and the Poynting vector becomes the flux of this probability.
Question from OP
I fully agree with what you say. Maxwell <-> photon tunnelling and Shrodinger <-> electron tunnelling are the "same" thing. But Shrodinger equation leads to quantised states (example hydrogen atom), hence it is part of quantum mechanics. Can Maxwell equations lead to quantised states as Shrodinger does? What is puzzling to me, I think, is the following: Why is Shrodinger description of particles (wave nature) considered quantum mechanics and Maxwell description is not?
Certainly Maxwell's equations lead to bound states. Look at the bound states of an optical fiber, which are shift-invariant eigenfunctions of the form $\Psi(x,\,y)\,e^{i\,\beta\,z}$, where the z direction is along the optical axis of the fiber, and where the propagation constant $\beta$ lies between the core and cladding wavenumbers. This is the discrete spectrum of the relevant Sturm-Liuoville system. As a quantum mechanical description when there is one photon in the mode system at a time, the propagation equation is actually the propagation for a pseudo particle called various things - polariton is probably the most apposite to an optical fiber propagation. The pseudo particle is a quantum superposition of pure EM field one-photon states and excited matter states in the fiber's material.
You ask why this kind of thing isn't called "quantum mechanics". Well it most certainly is part of quantum mechanics and the reason it isn't often referred to as such is probably historical. There is no nonrelativistic description of the photon - Maxwell's Equations are fully Lorentz-covariant - in contrast with the atomic electron Schrödinger equation which describes a nonrelativistic approximation. Such approximations admit position co-ordinates where the wavefunction can be loosely interpreted as definining, through its magnitude, the probability of "finding" an electron at a given position. This kind of thing isn't possible for the relativistic photon, or, for that matter, the relativistic electron described by the Dirac equation (note that Maxwell's equations can indeed be written as a Schrödinger equation and also that Maxwell's equations are equivalent to the Dirac equation for a massless particle). See my answer here and also here and here for further details. The question of photonic wave functions is addressed in detail in the works of Iwo Bialynicki-Birula, for example, cited in my answers.
Best Answer
The difference between propagating and evanescent is most obvious when looking at the intensity functions of the wave. You are probably familiar with a propagating wave, as its intensity is oscillating like a sine or cosine wave; this oscillation is what causes them to propagate or travel. An evanescent wave has an exponential wave function, typically an exponentially decaying curve if referring to intensity of light. This causes the wave to not really be able to travel through space and through a conventional lens.
I'm not really familiar with the super lenses you speak of, but I believe that the lenses are able to make evanescent waves visible due to the complex component of a evanescent wave function.
This comes into effect with reflection and refraction of light, particularly total internal reflection. If you were to have a block of glass and shine light into it which undergoes total internal reflection, no light would be visible where the light would have come out of the material if it would have went straight through; however, if another block of glass were put at this location with an extremely small distance between the two blocks (on the order of picometers, I believe), the light would come out of the second block of glass where a complex wave was created from the original electromagnetic wave in the space between the blocks, then was able to be changed back to a real wave in the second block of glass.