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Up: PHY531-PHY532 Classical Electrodynamics

Displacements and Time Delays by Wave Packet Analysis

Kevin Schmidt
Department of Physics and Astronomy
Arizona State University
Tempe, AZ

1 Wave Packet Basics

If a plane wave in vacuum is incident from negative z on material bounded by the planes z = z1 and z = z2 with permitivity and permeability functions only of z the solution outside the material is given, for the two polarizations, as a reflected wave for z < - z1 and a transmitted wave for z > z2 . The incident plane wave can be specified by the value of the tangential wave vector $\mbox{\boldmath$\space {k} $}$t and the frequency $\omega$ . The electric field for z < z1 and z > z2 is therefore given by defining

 
$\displaystyle\mbox{\boldmath$ {\tilde E_\perp} $}$($\displaystyle\mbox{\boldmath$ {k} $}$t,$\displaystyle\omega$) = $\displaystyle\mbox{\boldmath$ {\hat {k}} $}$ x $\displaystyle\mbox{\boldmath$ {\hat {z}} $}$$\displaystyle\left \{
\begin{array}
{ll}
e^{i k_z z} + r_\perp(\mbox{\boldmath$...
 ...mbox{\boldmath$ {k} $}_t,\omega) e^{i k_z z} & z \gt z_2\\ \end{array}\right.\,$. (1)

 
$\displaystyle\mbox{\boldmath$ {\tilde E_\parallel} $}$($\displaystyle\mbox{\boldmath$ {k} $}$t,$\displaystyle\omega$) = $\displaystyle\mbox{\boldmath$ {\hat {k}} $}$ x ($\displaystyle\mbox{\boldmath$ {\hat {k}} $}$ x $\displaystyle\mbox{\boldmath$ {\hat {z}} $}$)$\displaystyle\left \{
\begin{array}
{ll}
e^{i k_z z} + r_\parallel(\mbox{\boldm...
 ...mbox{\boldmath$ {k} $}_t,\omega) e^{i k_z z} & z \gt z_2\\ \end{array}\right.\,$. (2)

and writing

$\displaystyle\mbox{\boldmath$ {E} $}$($\displaystyle\mbox{\boldmath$ {r} $}$,t) = $\displaystyle\int$$\displaystyle{\frac{d^2k_t}{(2\pi)^2}}$$\displaystyle\int$$\displaystyle{\frac{d\omega}{2\pi}}$e i$\scriptstyle\mbox{\boldmath$ {k} $}$t $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$t - i$\scriptstyle\omega$t$\displaystyle\left [ A_\perp(\mbox{\boldmath$ {k} $}_t,\omega) \mbox{\boldmath$...
 ...ox{\boldmath$ {\tilde E_\parallel} $}(\mbox{\boldmath$ {k} $}_t,\omega)
\right]$ (3)

where $\mbox{\boldmath$\space {k} $}$ = $\mbox{\boldmath$\space {k} $}$t + $\mbox{\boldmath$\space {\hat {z}} $}$kz , and

kz = $\displaystyle\left \{
\begin{array}
{ll}
\sqrt{\frac{\omega^2}{c^2}-k_t^2} & \o...
 ...c\\ i \sqrt{k_t^2 - \frac{\omega^2}{c^2}} & \omega < k_t c\\ \end{array}\right.$ (4)

and the positive square root is used.

Our source is a pulse starting far enough away from the material that it can be localized in the region z < z1 , and to have only Fourier components with real positive kz so that they propagate toward the interface. In that case the reflected wave components above do not contribute when calculating the amplitudes A$\scriptstyle\perp$ and A || with the result

A$\scriptstyle\perp$($\displaystyle\mbox{\boldmath$ {k} $}$t,$\displaystyle\omega$) = $\displaystyle\int$d 2rt$\displaystyle\int$dt$\displaystyle\left [ ({\mbox{\boldmath$ {\hat {k}} $}} \times {\mbox{\boldmath$...
 ...t {z}} $}}) \cdot \mbox{\boldmath$ {E} $}_0(\mbox{\boldmath$ {r} $},t) \right]e$- i$\scriptstyle\mbox{\boldmath$ {k} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$ + i$\scriptstyle\omega$t (5)

and

A || ($\displaystyle\mbox{\boldmath$ {k} $}$t,$\displaystyle\omega$) = $\displaystyle\int$d 2rt$\displaystyle\int$dt$\displaystyle\left [ ({\mbox{\boldmath$ {\hat {k}} $}} \times {\mbox{\boldmath$...
 ...t {z}} $}}) \cdot \mbox{\boldmath$ {E} $}_0(\mbox{\boldmath$ {r} $},t) \right]e$- i$\scriptstyle\mbox{\boldmath$ {k} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$ + i$\scriptstyle\omega$t (6)

where $\mbox{\boldmath$\space {E} $}$0($\mbox{\boldmath$\space {r} $}$,t) is the electric field of the incident pulse which is completely confined to z < z1 .

The algorithm is therefore, given the incident field, calculate the amplitudes A$\scriptstyle\perp$ and A || . For each value of $\mbox{\boldmath$\space {k} $}$t and $\omega$ that occurs, solve Maxwell's equations for the transmission t and reflection r coefficients for a plane wave incident. Plug all these in to Eqs. 1 and 2 and integrate to get the reflected and transmitted fields. To get the fields in the material, the solutions in the material region need to be included in Eqs. 1 and 2.

Often the solution for a pulse that is a beam of nearly monochromatic radiation at nearly a single angle of incidence is desired. In addition, often the exiting pulses are the same shape as the incoming one just time delayed and/or shifted along the transverse direction. Let's see if we can see how this can come about. If the pulse is strongly peaked around a particular $\mbox{\boldmath$\space {k} $}$t and $\omega$ , we can expand the r and t functions around these values. As we will see, it is more convenient to write them as expansions in the exponent.

Initially, let's ignore the vector character of the waves. An incoming pulse can be written as

F0($\displaystyle\mbox{\boldmath$ {r} $}$,t) = $\displaystyle\int$$\displaystyle{\frac{d^2k_t}{(2\pi)^2}}$$\displaystyle\int$$\displaystyle{\frac{d\omega}{2\pi}}$e i$\scriptstyle\mbox{\boldmath$ {k} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$ - i$\scriptstyle\omega$tA($\displaystyle\mbox{\boldmath$ {k} $}$t,$\displaystyle\omega$) (7)

The transmitted wave will be
FT($\displaystyle\mbox{\boldmath$ {r} $}$,t) = $\displaystyle\int$$\displaystyle{\frac{d^2k_t}{(2\pi)^2}}$$\displaystyle\int$$\displaystyle{\frac{d\omega}{2\pi}}$t($\displaystyle\mbox{\boldmath$ {k} $}$t,$\displaystyle\omega$)e i$\scriptstyle\mbox{\boldmath$ {k} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$ - i$\scriptstyle\omega$tA($\displaystyle\mbox{\boldmath$ {k} $}$t,$\displaystyle\omega$) (8)
where t($\mbox{\boldmath$\space {k} $}$,$\omega$) is the transmission coefficient. Writing the transmission coefficient as e i$\scriptstyle\alpha$($\scriptstyle\mbox{\boldmath$\space {k} $}$t,$\scriptstyle\omega$), where $\alpha$ is complex, and expanding $\alpha$ around the peak of A($\mbox{\boldmath$\space {k} $}$t,$\omega$) at $\mbox{\boldmath$\space {k} $}$t(0) and $\omega^{(0)}_{}$ this becomes
FT($\displaystyle\mbox{\boldmath$ {r} $}$,t) = $\displaystyle\int$$\displaystyle{\frac{d^2k_t}{(2\pi)^2}}$$\displaystyle\int$$\displaystyle{\frac{d\omega}{2\pi}}$e i$\scriptstyle\alpha$($\scriptstyle\mbox{\boldmath$ {k} $}$t,$\scriptstyle\omega$)e i$\scriptstyle\mbox{\boldmath$ {k} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$ - i$\scriptstyle\omega$tA($\displaystyle\mbox{\boldmath$ {k} $}$t,$\displaystyle\omega$)   
  = $\displaystyle\int$$\displaystyle{\frac{d^2k_t}{(2\pi)^2}}$$\displaystyle\int$$\displaystyle{\frac{d\omega}{2\pi}}$e i$\scriptstyle\alpha$($\scriptstyle\mbox{\boldmath$ {k} $}$t(0),$\scriptstyle\omega^{(0)}$) + i$\scriptstyle\left . \mbox{\boldmath$ {\nabla} $}_{k_t} \alpha(\mbox{\boldmath$ {k} $}_t,\omega) \right\vert _$0 $\scriptstyle\cdot$ ($\scriptstyle\mbox{\boldmath$ {k} $}$t - $\scriptstyle\mbox{\boldmath$ {k} $}$t(0)) + i$\scriptstyle\left . \frac{\partial \alpha(\mbox{\boldmath$ {k} $}_t ,\omega)}{\partial \omega}
\right\vert _$0($\scriptstyle\omega$ - $\scriptstyle\omega^{(0)}$) + ...e i$\scriptstyle\mbox{\boldmath$ {k} $}$ $\scriptstyle\cdot$ $\scriptstyle\mbox{\boldmath$ {r} $}$ - i$\scriptstyle\omega$tA($\displaystyle\mbox{\boldmath$ {k} $}$t,$\displaystyle\omega$) .   

For now we neglect the quadratic terms in the exponent, but if they are large this will be a poor approximation. We will come back to this point later. Notice that the lowest order term in the expansion e i$\scriptstyle\alpha$($\scriptstyle\mbox{\boldmath$\space {k} $}$t(0),$\scriptstyle\omega^{(0)}$) is an overall constant that simply multiplies the wave packet. Similarly, the terms multiplied by $\mbox{\boldmath$\space {k} $}$t(0) and $\omega^{(0)}_{}$ are also constants. If we take the case where the first derivatives of $\alpha$ are real, we can approximate the transmitted amplitude by
 

FT($\displaystyle\mbox{\boldmath$ {r} $}$,t) $\textstyle\simeq$ e i$\scriptstyle\phi$t($\displaystyle\mbox{\boldmath$ {k} $}$t(0),$\displaystyle\omega^{(0)}_{}$)$\displaystyle\int$$\displaystyle{\frac{d^2k_t}{(2\pi)^2}}$$\displaystyle\int$$\displaystyle{\frac{d\omega}{2\pi}}$e i$\scriptstyle\mbox{\boldmath$ {k} $}$ $\scriptstyle\cdot$ $\scriptstyle\left (\mbox{\boldmath$ {r} $} +
\left . \mbox{\boldmath$ {\nabla} $}_{k_t} \alpha(\mbox{\boldmath$ {k} $}_t,\omega) \right \vert _0 \right)-$i$\scriptstyle\omega$$\scriptstyle\left (t-
\left . \frac{\partial \alpha(\mbox{\boldmath$ {k} $}_t ,\omega)}{\partial \omega}
\right \vert _0 \right)$A($\displaystyle\mbox{\boldmath$ {k} $}$t,$\displaystyle\omega$)   
  = e i$\scriptstyle\phi$t($\displaystyle\mbox{\boldmath$ {k} $}$t(0),$\displaystyle\omega^{(0)}_{}$)F0$\displaystyle\left(\mbox{\boldmath$ {r} $}+
\left . \mbox{\boldmath$ {\nabla} $...
 ...ha(\mbox{\boldmath$ {k} $}_t ,\omega)}{\partial \omega}
\right \vert _0 \right)$ (9)
where $\phi$ comes from the remaining parts of the derivative. Notice that the result to this order is that the transmitted packet is shifted transversely by a distance given by the kt gradient of $\alpha$ and time delayed by the omega derivative of $\alpha$ .

Repeating the calculation for the reflected wave gives in the same approximation and writing r($\mbox{\boldmath$\space {k} $}$,$\omega$) = e i$\scriptstyle\beta$($\scriptstyle\mbox{\boldmath$\space {k} $}$t,$\scriptstyle\omega$)

 
FR($\displaystyle\mbox{\boldmath$ {r} $}$,t) = e i$\scriptstyle\psi$t($\displaystyle\mbox{\boldmath$ {k} $}$t(0),$\displaystyle\omega^{(0)}_{}$)F0$\displaystyle\left(\mbox{\boldmath$ {r} $}+
\left . \mbox{\boldmath$ {\nabla} $...
 ...ha(\mbox{\boldmath$ {k} $}_t ,\omega)}{\partial \omega}
\right \vert _0 \right)$ (10)

so that the result travels in the negative z direction and the packet is shifted transversely by a distance given by the kt gradient of $\beta$ and time delayed by the omega derivative of $\beta$ .

Before looking at these results in more detail lets examine the neglected terms. If we calculate the width of the packet in kt or $\omega$ , the second derivative terms of $\alpha$ or $\beta$ will change this. However the magnitude of these terms is of order the second derivative times factors of $\Delta$kt(0) or $\Delta$w (0) (there are also cross terms from the mixed derivatives). The requirement that these can be neglected is therefore that these are small. That is the change in the first derivatives is small over the range of frequencies and wave vectors contained in the packet. If these terms are included but higher order terms are neglected, and the packet also expanded in this way, we would have a quadratic form in the exponent. Diagonalizing this quadratic form would give terms that would contribute both to a new width of the packets and to a shifted position from the linear terms and from completing the square. The other neglected terms are the real parts of the derivatives of $\alpha$ and $\beta$ . Including them would again change the shape of the packet. However, in this case, the real part of the zeroth order term dominates.

To reiterate, we expand the transmission and reflection amplitudes. The key point is that a large overall phase does not change the energy in the wavep packet therefore for the imaginary part of the exponent, we need to keep the gradient terms since they are the lowest order components that change the energy in the wave packet. For the real parts, the constant term dominates the gradient and gives the lowest order component. Therefore, for a wave packet with a sufficiently narrow frequency range, so that only the lowest order physical terms need to be kept, Eqs. 10 and 11 with $\alpha$ and $\beta$ given by the imaginary parts of the exponent give the transmitted and reflected packets.

At this point, we can go back to the vector equations. Notice that the polarization factors are real. Therefore, within the same approximation as the expansions above, we can keep just the lowest order terms and use the polarizations with $\mbox{\boldmath$\space {\hat {k}} $}$0

2 Examples

Let's take a material has an index of refraction n($\omega$) which is independent of z except near z1 and z2 goes smoothly from 1 to n($\omega$) $\geq$ 1 so that reflections at the boundaries can be ignored. In this case, each component will smoothly change index of refraction through the material, and if z2 - z1 = L is the thickness, the transmission coefficient will be

t($\displaystyle\mbox{\boldmath$ {k} $}$t,$\displaystyle\omega$) = e i$\scriptstyle\sqrt{\frac{n^2(\omega)\omega^2}{c^2}-k_t^2}$L - i$\scriptstyle\sqrt{\frac{\omega^2}{c^2}-k_t^2}$L . (11)

The transverse shift of the beam will be

dt = - $\displaystyle\left . \frac{\mbox{\boldmath$ {k} $}_t L}{\sqrt{\frac{n^2(\omega) \omega^2}{c^2} -k_t^2}}
\right\vert _$0$\displaystyle\left .
+\frac{\mbox{\boldmath$ {k} $}_t L}{\sqrt{\frac{\omega^2}{c^2} -k_t^2}}
\right\vert _$ 0 (12)

Writing |k (0)t| = ${\frac{\omega}{c}}$sin $\theta_{i}^{}$ , or as |k (0)t| = ${\frac{n(\omega) \omega}{c}}$sin $\theta_{m}^{}$ , where $\theta_{m}^{}$ is the angle made by the packet through the material, ( n($\omega$)sin($\theta_{m}^{}$) = sin $\theta_{i}^{}$ ) we have,

dt = L$\displaystyle\left [ \tan \theta_i - \tan \theta_m
\right]\,$, (13)

which agrees with a calculation using ray optics.

Next look at the time delay of the packet through the medium,

$\displaystyle\tau$ = $\displaystyle{\frac{L}{c \cos \theta_m}}$$\displaystyle{\frac{\partial n(\omega) \omega}{\partial \omega }}$ - $\displaystyle{\frac{L}{c \cos \theta_i}}$ (14)

and since the L/cos $\theta_{i}^{}$ is the distance traveled by the original packet while L/cos $\theta_{m}^{}$ is the distance traveled by the packet through the material, this shows that the velocity of the packet through the material, the group velocity, is

vg = c$\displaystyle\left . \left [ \frac{\partial n(\omega) \omega}{\partial \omega}
\right ]^{-1} \right\vert _$ 0 (15)

Let's look at the Goos-Hänchen effect. Here we have a beam undergoing total internal reflection. Placing the interface at z = 0 , the reflection coefficient has, for a real index of refraction, unit magnitude and is a phase. The reflection coefficient for perpendicular polarization in terms of the incident angle $\theta_{i}^{}$ and the critical angle sin $\theta_{c}^{}$ is

$\displaystyle\mbox{\boldmath$ {r} $}$$\scriptstyle\perp$ = $\displaystyle{\frac{\cos \theta_i - i \sqrt{\sin^2 \theta_i - \sin^2 \theta_c}}{\cos \theta_i + i \sqrt{\sin^2 \theta_i - \sin^2 \theta_c}}}$ (16)

so that

$\displaystyle\beta$ = - 2$\displaystyle\tan^{-1}_{}$$\displaystyle\left (\frac{\cos \theta_i}{
\sqrt{\sin^2 \theta_i - \sin^2 \theta_c}} \right)\,$. (17)

With |kt|c = ni$\omega$sin $\theta_{i}^{}$ , the magnitude of the gradient with respect to kt is

|$\displaystyle\mbox{\boldmath$ {\nabla} $}$kt$\displaystyle\beta$| = $\displaystyle{\frac{\partial \beta}{\partial k_t}}$ = $\displaystyle{\frac{c}{n_i \omega \cos \theta_i}}$$\displaystyle{\frac{\partial \beta}{\partial \theta_i}}$ (18)

since the reflection coefficient only depends on the magnitude of kt . The lateral displacement is then readily evaluated

d$\scriptstyle\perp$ = |$\displaystyle\mbox{\boldmath$ {\nabla} $}$kt$\displaystyle\beta$| = $\displaystyle{\frac{2 c}{n_i \omega }}$$\displaystyle{\frac{\tan\theta_i}{\sqrt{\sin^2 \theta_i - \sin^2 \theta_c}}}$ . (19)

Jackson uses the perpendicular distance D between the beams rather than the lateral displacement d we have used. They are related by D = dcos $\theta_{i}^{}$ , and substituting $\lambda$ = c/(2$\pi$ni$\omega$) gives Jackson's result

D$\scriptstyle\perp$ = $\displaystyle{\frac{\lambda}{\pi}}$$\displaystyle{\frac{\sin \theta_i}{\sqrt{\sin^2 \theta_i - \sin^2 \theta_c}}}$ , (20)

which can be compared to Jackson Eq. 7.48. The parallel polarization result goes through similarly.


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Up: PHY531-PHY532 Classical Electrodynamics