On wikipedia they say that the evanescent wave has a magnitude of: $$E=E_0 e^{\alpha y-j\beta x}$$ Where $j=\sqrt{-1}$, but what direction does this vector point in and why? (let us say the boundary lies in the zy-plane)


I am not sure if my original question was clear. The electric field is a vector and hence can be written as: $$\vec E= E_0 e^{\alpha y-j\beta x} \hat a $$ where $\hat a$ is some unit vector. My question is what direction is $\hat a$ pointing in?

  • $\begingroup$ By 'point', so you mean the direction of propagation or polarization? $\endgroup$ – Emilio Pisanty Sep 6 '15 at 17:35
  • $\begingroup$ @EmilioPisanty both, although I am uncertain if it has a direction of propagation since it is not really a propagating wave?! $\endgroup$ – Quantum spaghettification Sep 6 '15 at 17:37
  • $\begingroup$ I notice that they use the traditional $i$ for $\sqrt{-1}$. Out of curiosity, is there any reaction why you changed it? $\endgroup$ – HDE 226868 Sep 6 '15 at 17:46
  • 2
    $\begingroup$ @HDE226868 I started using Microsoft edge on windows 10 and it always corrects a lower case 'I' to an upper case 'I' so to save the hassle I just used j. $\endgroup$ – Quantum spaghettification Sep 6 '15 at 17:49
  • $\begingroup$ Good question Joseph, nice answer lemontwist. By the by, see see for example Evanescent Waves in Quantum Electrodynamics with Unquantized Sources dating from 1973: "The identity of these evanescent waves with virtual photons is established". Also see near and far field on Wikipedia $\endgroup$ – John Duffield Sep 7 '15 at 19:38

To summarise: it depends on the ratio of refractive indices, the angle of incidence and the polarisation state of the incident wave.

Details: The polarisation of the electric field will be perpendicular to the wavevector. Why? Because where there are no free charges then $\nabla \cdot \vec{E} = 0$. If we represent the wave as $\vec{E} = \vec{E_0} f(\omega t - \vec{k}\cdot \vec{r})$, where $f$ can be any function (including a sinusoid), then $$ \nabla \cdot \vec{E} = \vec{k}\cdot \vec{E_0} f^{\prime}( \omega t - \vec{k}\cdot \vec{r})$$ For this to be true for all $\vec{r}, t$, then $\vec{k}\cdot \vec{E_0}=0$. This does not uniquely define the E-field vector.

If the plane of incidence is the $x,y$ plane, then in general the wavevector of the evanescent wave is given by $$ \vec{k}_e = i\alpha \hat{y} + \beta \hat{x}.$$ In order to ensure continuity of the E-field parallel to the interface plane (a consequence of Faraday's law) then the polarisation plane of the evanescent wave must be the same as that of the wave incident at the interface. For the simpler case of s-polarisation (incident E-field perpendicular to the plane of incidence and parallel to the plane of the interface), then the evanescent wave is also polarised perpendicular to the plane of incidence. In the example here, this means if the incident E-field is polarised in the z-direction, then so is the evanescent wave. $$\vec{E} = E_0 \exp i (\omega t - i\alpha y - \beta x)\ \hat{z} = E_0 \exp(\alpha y) \exp i(\omega t -\beta x)\ \hat{z},$$ which is just a plane wave propagating in the positive x direction, parallel to the interface, with an exponentially decaying amplitude in the y direction (where the form of the wavevector in the wikipedia article implies that if $\alpha$ is real and positive, $y$ becomes more negative as you move away from the interface).

However, there is another possibility. The incident light could be polarised in the plane of incidence (p-polarisation). In which case the E-field of the reflected and evanescent waves will also be in the plane of incidence, but will have components $E_x$ and $E_y$, such that $\vec{k} \cdot \vec{E} =0$. The relative magnitudes of $E_x$ and $E_y$ will depend on the ratio of the refractive indices and the angle of incidence. This is calculated by noting that as the waves on either side of the boundary (where $y=0$) are of the form $f(\omega t - k_{ix} x)$ and $f(\omega t - k_{ex} x)$, where $k_{ix}$ and $k_{ex}$ are the x-components of the wavevectors either side of the interface, then $k_{ix}=k_{ex} = 2\pi \sin \phi /\lambda$, where $\phi$ is the incidence angle, to ensure a constant phase relationship along the interface. It can then be shown that $$\frac{E_x}{E_y} = i \frac{k_y}{k_x} = i \frac{(N^2 \sin^2 \phi -1)^{1/2}}{N\sin \phi},$$ where $N$ is the high/low ratio of the refractive indices (see Azzam 2011). Because $\vec{k}$ is complex this means that there is a 90 degree phase difference between $E_x$ and $E_y$ which means that the electric field is elliptically polarised in general and that the wave propagating in the x-direction is no longer purely transverse, but has a longitudinal component.

This animation may make things clearer.

For an arbitrary incident polarisation you have to represent it as the sum of a s-polarised and a p-polarised component, then use the Fresnel equations to work out what the transmission coefficients are and calculate the three components of the evanescent wave (note that a form of these are given in the link above, but that these are the intensities rather than the amplitudes).

  • $\begingroup$ To help prevent link rot, the linked page is now in the Wayback Machine :). The animation, in any case, is here. $\endgroup$ – Emilio Pisanty Oct 4 '15 at 18:43
  • $\begingroup$ It is not the direction of propagation that I am after, it is the direction of the electric field. I think you have shown it in one case but what about in general? $\endgroup$ – Quantum spaghettification Oct 4 '15 at 18:46
  • $\begingroup$ @Joseph Done, and also linked to a better animation/applet. $\endgroup$ – Rob Jeffries Oct 5 '15 at 6:17

An evanescant wave decays (generally into a metal), which is why $e^{\alpha y}$ doesn't have a prefix of $j$. The value of the EM field in the metal is highest at the interface ($y = 0$), and then decays (with decay length $\frac{1}{\alpha}$) into the metal.

So if you have $y$ as your vertical axis, you have the half-space where $y > 0$ (dielectric material) and the half-space where $y < 0$ (metal). Where $y > 0$ you will have a standing wave. Where $y < 0$ you will have a decaying wave.

At times you can have a propagating wave (surface plasmon polariton) that propagates along the interface (in your case it would be $x$ as the horizontal axis, where $y = 0$), but there must be particular frequency requirements met in that circumstance.

The clues for "direction" come from your exponents. Which term has an imaginary exponent? In your case, the x-direction does, therefore it will have a propagating wave. Which term has a real exponent? In your example, it is the y-direction, therefore you will have an evanescent wave in y.

  • $\begingroup$ I wasn't going to comment or downvote since I assumed this was an answer to the original (less clear) question. However, since it has now been upvoted after the edited question appeared I will point out that this answer says nothing about the electric field vector direction, which is what the (edited) question asks for. $\endgroup$ – Rob Jeffries Oct 6 '15 at 15:58
  • $\begingroup$ Yes, this was an answer to the original question, which did not ask about a unit vector. $\endgroup$ – lemontwist Oct 11 '15 at 20:21

A general answer is that the field vector is independent from the decay direction, and hence the field vector can be pointing to any direction. However, on the decay direction (that is the $-y$ direction in your formula), all field components decrease. The only constrain for the field vector is that all field components must satisfy the boundary condition at the interface which may result in some given direction depending on the specific case.

Different from some of the other answers, I think the field can have the longitudinal component which is parallel to the propagation direction. In fact, the group velocity of a propagating wave is related to the longitudinal component due to the dielectric or some other medium interface. See, for example, arXiv:1509.02625 the appendix A for all of the evanescent field components outside of an optical fiber.

  • $\begingroup$ My answer specifically gives a longitudinal component, but only if there is a p polarised component in the incident wave. I will edit to make that clear. $\endgroup$ – Rob Jeffries Oct 6 '15 at 6:14
  • $\begingroup$ So your answer should really be "it depends"? $\endgroup$ – Xiaodong Qi Oct 6 '15 at 6:20
  • $\begingroup$ That is what my answer says - it depends on the ratio of refractive indices, the angle of incidence and the polarisation state of the incident wave. I will add that as a summary statement, thanks. $\endgroup$ – Rob Jeffries Oct 6 '15 at 6:30
  • $\begingroup$ Glad you realized that. Thanks for your efforts to formulate the answer clearly in a good style :) $\endgroup$ – Xiaodong Qi Oct 6 '15 at 6:35

This isn't a general answer, but one famous example of an evanescent wave is a surface plasmon polariton. (If the surface is the x-y plane, then it is evanescent in the z direction on both sides of the surface.) I put an animation of the E-field of an SPP on wikipedia ... here it is:

surface plasmon polariton animation

Again, this is just one random example of a lossless SPP. But maybe food for thought.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.