Tunnelling in Total Internal Reflexion and Evanescent Fields

When an electromagnetic wave is totally internally reflected, it takes on a phase shift. Inside the optically less dense medium, where ray optics tells us there is no field, there is rather an evanescent field. This field is wholly analogous to quantum mechanical tunnelling, which is the penetration of a particle’s wavefunction into regions it is classically forbidden to go into.

The phase shift, and indeed sometimes the whole phenomenon of evanescent wave penetration, is called the Goos-Hänchen effect.

The best way to understand this phase shift is to solve and study solutions of the Helmholtz equation near the boundary between two dielectric mediums. You don’t quite have to solve the full Maxwell equations: the assumption that the light field can be modelled by one scalar field (approximately equal to one transverse component of the electric field) rather than the $(\vec{E},\vec{H})$ vectors is known as scalar diffraction theory and justified in chapters 1 and 8 of Born and Wolf, “Principles of Optics”.

First an intuitive explanation. When total internal reflexion happens, the field isn’t abruptly turned around by the interface, it actually penetrates some distance beyond the interface as an evanescent field. The phenomenon is actually wholly analogous to quantum tunnelling by a first quantised particle field described by e.g. the Schrödinger or Dirac equation into regions which are, through their being at a higher potential than the particle’s total energy, classically “forbidden” to the particle. Indeed, if you have a sandwich of lower refractive index material between two higher index materials such that an incoming wave is “totally internally reflected” from the first high-index to lower-index interface, then some of the light tunnels through the sandwich and again propagates freely (i.e. non evanescently) when it gets through the low refractive index layer. The power transmitted through the layer decreases exponentially with layer thickness, as with analogous quantum tunnelling through high but thin potential barrier problems.

So, given that the field penetrates some distance into the lower refractive index medium, the “effective” interface actually lies a small distance into the lower refractive index medium. The Goos-Hänchen phase shift is the phase delay arising from this short journey into and out of the lower index medium.

Now for some details. Let:

$$\psi_i = \exp(i\,n_i \vec{k} \cdot \vec{r}) = \exp(i\,n_i(k_x x + k_y y))$$

be the incident field with the plane of polarisation in the $x-y$ plane with the $x$-axis being the interface and $n_i$ the refractive index for $y>0$. The point is now that the boundary sees a scalar field variation of $\exp(i\,n_i\,k_x\, x)$ such that $n_i\,k_x > n_t\,k$ where $k = 2\pi/\lambda = \sqrt{k_x^2+k_y^2}$ is the field’s freespace wavenumber. So, to ensure continuity of the scalar field across the boundary, the $x$-component of the wavevector on the lower side of this boundary must also be $n_i\,k_x$. So what is the $y$-component of the wavevector in the lower medium. It has to be $k_y^\prime$ where $\sqrt{{k_y^\prime}^2 + (n_i k_x/n_t)^2} = k^2$ so as to fulfill the Helmholtz equation $(\nabla^2 +k^2 n_t^2)\psi = 0$ in the lower medium. Therefore, $k_y^\prime = \pm \sqrt{k^2-(n_i k_x/n_t)^2}$ which is imaginary by dint of the condition for total internal reflexion $n_i\, k_x > n_t\,k$. Now the solution $k_y = – \sqrt{k^2-(n_i k_x/n_t)^2}$ is unphsyical as it would have the field magnitude rising exponentially with penetration depth into the lower medium. So, in the lower medium, there is a field of the form:

$$\psi_{t,1}(x,y) = \exp(-\sqrt{(n_i\, k_x)^2 – (n_t\,k)^2}\,y + i\,n_i\,k_x\,x)$$

Fields that dwindle exponentially with distance into a medium like $\psi_{t,1}$ are known as evanescent fields (evanescere is classical Latin for “vanish”). In a fuller vector field analysis done by fully solving Maxwell’s equations, one can work out the Poynting vector and show that such fields do not bear optical power with them. Instead, they are very like inductive and capacitive energy stores; they of course have an energy density but it shuttles back and forth between neighbouring regions in the medium and so the nett power flux through any surface over a whole period is nought.

There is also the reflected field:

$$\psi_r(x,y) = \gamma_r \exp(i\,n_i \vec{k}_r \cdot \vec{r}) = \gamma_r\,\exp(i\,n_i(k_x x – k_y y))$$

where $ \gamma_r$ is a yet-to-be found reflexion co-efficient. To uphold continuity of the scalar field at the boundary, this field also has a corresponding evanescent field $\psi_{t,2}(x,y) = \gamma_r \psi_{t,1}(x,y)$. How do we find $\gamma_r$; in scalar field theory it is chosen to make the normal derivative to the interface of the scalar field continuous across the interface. So we have:

$$(1+\gamma_r) \left.\partial_y \psi_t(x,y)\right|_{y=0} = \left.\partial_y\left(\gamma_r\,\exp(i\,n_i(k_x x – k_y y)) + \exp(i\,n_i(k_x x + k_y y))\right)\right|_{y=0}$$

so that:

$$-\frac{1+\gamma_r}{1-\gamma_r} = -i\,g$$

where:

$$g=\frac{n_i\,k_y}{\sqrt{(n_i k_x)^2 – (n_t k)^2}}$$

and $\gamma_r$ is complex; on inverting the billinear relationship between $\gamma_r$ and $g$ we find that $\gamma_r$ is the unity magnitude complex number:

$$\gamma_r = -\frac{1 + i\,g}{1-i\,g}$$

i.e. to a phase delay of:

$$\phi_{GH}=-2\arctan\left(\frac{1}{g}\right) = -2\arctan\left(\frac{\sqrt{(n_i k_x)^2 – (n_t k)^2}}{n_i\,k_y}\right)$$

so its phase represents the Goos-Hänchen shift, a kind of “mean” radian penetration into the lower index medium by the field.

What Happens for a Thin Layer

If the classically totally internally interface is one side of a thin sandwich, rather than a half-infinite region, then something very different can happen. The fields are again evanescent in the less optically dense medium, but when they reach the other side of the layer, they begin propagating again. A thin classically totally internally reflecting layer will thus transmit some light and not actually reflect all the incident light.

Consider the tunnelling of light through a layer of thickness $w$ as in my drawing below:

Tunnelling Wave

Figure 1: Definitions for the 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:

$$\gamma_x=const$$

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 we 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 work 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.