What is the phase shift of a wave that tunnels through a barrier, meaning the difference in phase between the incoming (in front of the barrier) and the outgoing (behind the barrier) waves?

For example, in the situations of a wave function tunneling through a classically forbidden region, where the wave is no longer oscillating, but decaying exponentially within the barrier. Or of a beam of light incident at the critical angle of an interface in a frustrated total internal reflection setup.


There is no short answer to this question aside from: you must analyse in detail the boundary value problem wherein the tunnelling in question arises and calculate the answer.

For a reasonably simple and not too messy, example, consider the tunnelling of light through a layer of thickness $w$ as in my drawing below:

Definitions for Tunnelling Wave

and we imaging $s$-polarised light (electric field straight out of the page) of amplitude $a_1$ incident on a layer of thickness $w$. Instead of writing down the wavevector components $(k_x,\,k_y,\,k_z)$, we instead use complex propagation constants $(\gamma_x,\,\gamma_y,\,\gamma_z)$ to more readily handle tunnelling and evanescence, a plane wave's variation is then of the form $\exp(\gamma_x\,x+\gamma_y\,y+\gamma_z\,z)$. The layer has a refractive index $1/n$ times that of the other mediums and the wavenumber in this medium (number 2) is $k$, so that $\gamma_x^2+\gamma_y^2+\gamma_z^2 = -k^2$ in medium 2, and $\gamma_x^2+\gamma_y^2+\gamma_z^2 = -k^2 \,n^2$ in the other two mediums (note the minus sign).

So, assuming that the electric field is straight out of the page, in a medium with index $n$ the electric ($\vec{E}$) and magnetic ($\vec{H}$) field components take the form:

$$\begin{array}{lcl} E_y &=& \exp\left(\gamma_x\,x\right)\,\left(a\,\exp\left(\gamma_z\,z\right) + b\,\exp\left(-\gamma_z\,z\right)\right)\\ H_x &=& -\frac{i}{\omega\,\mu}\,\gamma_z\,\exp\left(\gamma_x\,x\right)\left(-a\,\exp\left(\gamma_z\,z\right) + b\,\exp\left(-\gamma_z\,z\right)\right)\\ H_z &=& -i\,\frac{i}{\omega\,\mu}\,\gamma_x\,\exp\left(\gamma_x\,x\right)\left(a\,\exp\left(\gamma_z\,z\right) + b\,\exp\left(-\gamma_z\,z\right)\right)\\ \end{array} $$

So now we down these equations for each of the three mediums and equate field components on the boundaries $z = \pm w/2$; we set $a_3=0$ given that the wave is incident from the drawing's left. Firstly, we take heed that the forms of $E$ and $H_z$ are simply scaled versions of one another; this can only be so if:


and, moreover, that this condition must also be fulfilled if the variation $\exp(\gamma_x\,x)$ in the $x$-direction is to match at each interface, i.e. each medium's transverse propagation constant $\gamma_{x,\,j}$ is the same for all the layers. This of course is the generalised version of Snell's law, and it holds whether we have propagating or evanescent waves. Note too in this multilayer problem we must include both "left-to-right" AND "right-to-left" running waves in the layer whether or not we have propagation or evanescence (tunnelling).

On writing down all these equations, insetting $\gamma_{x,1}=\gamma_{x,2}=\gamma_{x,3} = \gamma_x$ and eliminating $a_2,\,b_2$ we get:

$$\frac{b_3}{a_1} = 4\,\frac{\gamma_{z,1} \cosh(\gamma_{z,2}\,w)\,\cosh\left((\gamma_{z,1}+\gamma_{z,2})\frac{w}{2}\right) - \gamma_{z,2}\sinh(\gamma_{z,2}\,w)\,\sinh\left((\gamma_{z,1}+\gamma_{z,2})\frac{w}{2}\right)}{2\,\gamma_{z,1}\,\exp\left(-(\gamma_{z,1}+\gamma_{z,2})\frac{w}{2}\right) + (\gamma_{z,2}-\gamma_{z,1}) \exp(\gamma_{z,2}\,w) - (\gamma_{z,2}+\gamma_{z,1}) \exp(-\gamma_{z,2}\,w)}$$

wherein you need to set:

$$\gamma_{z,2} = i\,\sqrt{k^2 + \gamma_{x}^2} = -\sqrt{n^2\,k^2\,\sin\theta^2-k^2}$$ $$\gamma_{z,1} = \gamma_{z,3} = i\,\sqrt{n^2\,k^2 + \gamma_{x}^2} = i\,n\,k\,\cos\theta$$

and $\theta$ is the angle of incidence on the left. These equations will word for all values of $\theta$, I have arranged the last two lines so that $\gamma_{z,2}$ is real when total internal reflexion is happenning. The expression for $b_3/a_1$ gives you the transmitted amplitude, i.e. the amplitude of the wave that propagates again after tunnelling (this dwindles swiftly to nought as $w$ increases) and its phase lets you compute the phase of the transmitted wave relative to the incident one. You can also calculate the reflected complex amplitude:

$$\frac{b_1}{a_1} = \frac{-2\, \exp(-\gamma_{z,1}\,w)\,\left(\gamma_{z,1} \left(\cosh(\gamma_{z,2}\,w) + \exp\left((\gamma_{z,1}+\gamma_{z,2})\frac{w}{2}\right)\right) + \gamma_{z,2}\sinh(\gamma_{z,2}\,w) \right)}{2\,\gamma_{z,1}\,\exp\left(-(\gamma_{z,1}+\gamma_{z,2})\frac{w}{2}\right) + (\gamma_{z,2}-\gamma_{z,1}) \exp(\gamma_{z,2}\,w) - (\gamma_{z,2}+\gamma_{z,1}) \exp(-\gamma_{z,2}\,w)}$$

and when $w\to\infty$ and total internal reflexion is happening (i.e. $\gamma_{z,2}$ is real and negative) the last expression approaches:

$$\frac{b_1}{a_1}\to \exp(-\gamma_{z,1}\,w)$$

which has unity magnitude, because $\gamma_{z,1} = i\,k\,n\,\cos\theta$, thus showing that all the light is reflected from the first interface if the layer is so thick that tunnels through it.


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