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I was working out the evanescent magnetic field of a perpendicularly polarized incident wave. The plane of incidence is the $xz$ plane. For the evanescent field the wave vector $\mathbf{k}_T$ is give as, $$\mathbf{k}_T=k_T\left[\frac{n_1}{n_2}\sin\theta_I\hat{\mathbf{x}}+i\left(\sqrt{\frac{n_1^2}{n_2^2}\sin^2\theta_I-1}\right)\hat{\mathbf{z}}\right]\tag{1}$$ Where $\theta_I$ is the angle of incidence and subscript $T$ denotes transmission. This can we written as, $$\mathbf{k}=k\hat{\mathbf{x}}+i\kappa\hat{\mathbf{z}}\tag{2}$$ This is derived from, $$\mathbf{k}_T=k_T(\sin\theta_T\hat{\mathbf{x}}+\cos\theta_T\hat{\mathbf{z}})\tag{3}$$ via substitution using Snell's law.

Now for an evanescent field, the incident angle has passed the critical angle so total internal reflection is occurring. For this the Snell's law equivalent value in place of $\sin\theta_T$ will be greater than $1$. In fact $\theta_T$ cannot be considered as an angle. What I would like to know here is the magnitude of $\mathbf{k}_T$?

The magnitude should have been $k_T$ from $(2)$ (if $\theta_T$ was an angle). But that's not the case here I suppose because the vector has complex components now so talking the magnitude should look something like, $$|\mathbf{k}_T|^2=k^2+\kappa^2 \tag{4}$$ Which is not $k_T$. And that can be shown easily, consider a vector, $$\mathbf{a}=a\hat{\mathbf{x}}+\sqrt{1-a^2}\hat{\mathbf{y}} $$ Taking the norm gives us, $$\|\mathbf{a}\|=1$$ This is incorrect if $a$ is larger than 1, but this is exactly what we are doing when we are considering the norm of $(3)$ is $k_T$. According to the convention of keeping positive definiteness, $$\mathbf{a}=a\hat{\mathbf{x}}+i\sqrt{a^2-1}\hat{\mathbf{y}} $$ Taking the norm gives us, $$\|\mathbf{a}\|=\sqrt{2a^2-1}$$ Where the convention says, $$\mathbf{a}\cdot\mathbf{a}=\sum_{i=1}^{n}a_i\bar{a}_i$$ Final remarks on the magnetic field: It has the form, $$\mathbf{B}=\frac{E_0}{\omega}e^{-\kappa z}[\kappa\sin(kx-\omega t)\hat{\mathbf{x}}+k\cos(kx-\omega t)\hat{\mathbf{z}}]\tag{5}$$ Which I found can only be possible if the magnitude of the wave vector is $k_T$. This is taken from Griffith. Again this is derived from, $$\mathbf{B}=\frac{1}{v_2}\hat{\mathbf{k}}\times\mathbf{E}\tag{6}$$ where $v_2$ is the velocity of electromagnetic wave in the transmitted medium.

I know $(4), (5)$ and $(6)$ cannot all be correct simultaneously. One of them has to be wrong for this case. I want to know which one and why?

In my understanding $(4)$ perhaps is wrong. The convention of keeping the property of positive definiteness does not apply in this case because the wave vector of the evanescent wave is not a geometric entity. But I am not entirely satisfied with my own argument. I guess my question boils down to the conventions of taking a dot product. In what way should I take the dot product?

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  • $\begingroup$ "magnitude" is an extremely ambiguous term, and ultimately quite meaningless on its own, for a complex vector. What do you mean by the term, and what do you want to use it for? In particular, there are two different meanings (the sum of complex squares of the components, and the sum of squares of their moduli), and they both do different things. $\endgroup$ Sep 20, 2020 at 11:23
  • $\begingroup$ What I want is the value of $\hat{\mathbf{k}}$. I want to calculate $(5)$ from $(6)$. I may have some misconception here but essentially what I want to know is how to get the unit vector of $(2)$. $\endgroup$
    – Nothingham
    Sep 20, 2020 at 12:21
  • $\begingroup$ Unit vector by what norm? Why that norm? What do you want to use it for? Those are absolutely essential questions - without clear answers, your query is not answerable. $\endgroup$ Sep 20, 2020 at 13:27

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Snell’s Law and the Fresnel Equations are all valid for complex media and any angle of incidence. You just need to be open to $\theta$ being complex as well. If you have an evanescent wavevector, then it’s imaginary, and so is $\theta$. Depending on what you want, you could take the imaginary part, and that will give you the decay length of the evanescent field into the medium beyond the interface. That would also be the amplitude of the imaginary wavevector.

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