A complement to the scalar wave theory of light

Hamilton's principle of stationary action is introduced to the students in analytical mechanics courses. Action S of a dynamical system is defined as the time integral of Lagrangian density $\mathcal{L}=\mathcal{L}(q,\dot{q},t)$:

$$\begin{equation}S={\int }_{{t}_{1}}^{{t}_{2}}\mathcal{L}(q,\dot{q},t)\ \mathrm{d}t.\end{equation} \tag{ 1 }$$

Sticking to the usual convention, we shall call $\mathcal{L}$ the Lagrangian for its subsequent appearances. For the fixed time instants t₁ and t₂, the principle says that the system will dynamically evolve along a path, for which the action will be stationary to the first order. That is, the action will not change due to a slight variation of the path. This statement is mathematically expressed as:

$$\begin{equation}\delta S=\delta {\int }_{{t}_{1}}^{{t}_{2}}\mathcal{L}(q,\dot{q},t)\ \mathrm{d}t=0.\end{equation} \tag{ 2 }$$

The above principle leads to the Euler–Lagrange equation that governs the dynamical evolution of the system:

$$\begin{equation}\frac{\partial \mathcal{L}}{\partial q}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right).\end{equation} \tag{ 3 }$$

In case of geometrical optics, the equivalent physical principle is the Fermat's principle which states that for fixed end points a ray of light will propagate along a trajectory along which the optical path length $\mathbb{L}=\int n\,\mathrm{d}s$ is stationary. That means the variation in optical path length will not change to the first order:

$$\begin{equation}\delta \mathbb{L}=\delta {\int }_{{P}_{1}}^{{P}_{2}}n\,\mathrm{d}s=0.\end{equation} \tag{ 4 }$$

Euler–Lagrange variation of the principle (equation (4)) leads to the differential ray equation governing the spatial evolution of the light rays:

$$\begin{equation}\frac{\partial n}{\partial \mathbf{r}}=\frac{\mathrm{d}}{\mathrm{d}s}\left(n\,\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}s}\right).\end{equation} \tag{ 5 }$$

This equation is non-linear and is difficult to solve when the refractive index n varies spatially.

Evans and Rosenquist [2] showed that if Fermat's principle is expressed in terms of an independent variable a, defined by $n=\left\vert \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}a}\right\vert$, where n denotes the refractive index and r denotes the position vector, then equation (5) can be expressed as:

$$\begin{equation}\frac{{\mathrm{d}}^{2}\mathbf{r}}{\mathrm{d}{a}^{2}}=-{\nabla }^{2}\left(\frac{{n}^{2}}{2}\right),\end{equation} \tag{ 6 }$$

where a is a parameter. This equation has the well-known form of Newton's second law of motion, in terms of the parameter a. In this formulation, the quantity that is equivalent to force is proportional to the gradient of the square of the refractive index. This specialized coordinate is applicable to geometrical optics, in which mass, velocity, kinetic and potential energies do not have their standard units. This correspondence is shown below in table 1. From table 1, we find that the optical equivalent of potential energy is a quadratic function of refractive index and the equivalent of total energy (Hamiltonian) is 0.

Table 1. Mechanical view of optics.

	Definitions in classical mechanics	Equivalent quantities in geometrical optics
Position	r(t)	r(a)
Time	t	a
Velocity	$\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\equiv \dot{\mathbf{r}}$	$\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}a}\equiv {\mathbf{r}}^{\prime }$
Potential energy	U(r)	$-\frac{{n}^{2}(\mathbf{r})}{2}$
Mass	m	1
Kinetic energy	$T=\frac{m}{2}{\left\vert \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\right\vert }^{2}$	$\frac{1}{2}{\left\vert \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}a}\right\vert }^{2}$
Total energy	$\frac{m}{2}{\left\vert \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\right\vert }^{2}+U(\mathbf{r})$	$\frac{1}{2}{\left\vert \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}a}\right\vert }^{2}-\frac{{n}^{2}}{2}=0$

A similar formulation of the problem is possible in terms of the optical Hamiltonian, expressed as a function of (x, y, z), and their conjugate momenta p_x , p_y and p_z . Here we take b as a stepping parameter along the ray. Then, the expression for optical Lagrangian becomes:

$$\begin{equation}L(x,y,z,{x}^{\prime },{y}^{\prime },{z}^{\prime },b)=n(x,y,z)\sqrt{{x}^{\prime 2}+{y}^{\prime 2}+{z}^{\prime 2}},\end{equation} \tag{ 7 }$$

where ${x}^{\prime }=\frac{\mathrm{d}x}{\mathrm{d}b}$ etc. The conjugate momenta are defined as:

$$\begin{align}\hfill {p}_{x}=\frac{\partial L}{\partial {x}^{\prime }}=n\frac{{x}^{\prime }}{\sqrt{{x}^{\prime 2}+{y}^{\prime 2}+{z}^{\prime 2}}}=n\,\frac{\mathrm{d}x}{\mathrm{d}s}={n}_{x}\\ \hfill {p}_{y}=\frac{\partial L}{\partial {y}^{\prime }}=n\frac{{y}^{\prime }}{\sqrt{{x}^{\prime 2}+{y}^{\prime 2}+{z}^{\prime 2}}}=n\,\frac{\mathrm{d}y}{\mathrm{d}s}={n}_{y}\\ \hfill {p}_{z}=\frac{\partial L}{\partial {z}^{\prime }}=n\frac{{z}^{\prime }}{\sqrt{{x}^{\prime 2}+{y}^{\prime 2}+{z}^{\prime 2}}}=n\,\frac{\mathrm{d}z}{\mathrm{d}s}={n}_{z},\end{align} \tag{ 8 }$$

where (n_x , n_y , n_z ) denote the components of refractive index vector n = (n_x , n_y , n_z ). The Hamiltonian, constructed by Legendre transformation, can be evaluated to be equal to zero ¹ :

$$\begin{equation}H(x,y,z,{p}_{x},{p}_{y},{p}_{z},b)={x}^{\prime }{p}_{x}+{y}^{\prime }{p}_{y}+{z}^{\prime }{p}_{z}-L=0.\end{equation} \tag{ 9 }$$

2.4.1. Quantum description of light rays

From the ray view of light, one can make the transition to the quantum description of light by treating momenta by appropriate linear differential operators: ${p}_{x}\to {\hat{p}}_{x}\equiv -i\frac{\lambda }{2\pi }\frac{\partial }{\partial x}$ etc. These act upon wavefunctions ψ that represent light rays via the eigenvalue equations:

$$\begin{equation}{\hat{p}}_{x}\psi ={n}_{x}\psi \hspace{28.45274pt}\,{\Rightarrow}\,\hspace{28.45274pt}{\hat{p}}_{x}^{2}\psi ={n}_{x}^{2}\psi .\end{equation} \tag{ 10 }$$

Repeating this exercise for all three components, and summing up the squared equations, we find the reduced wave equation:

$$\begin{equation}{-\hspace{-7.5pt}\lambda}^{2}{\nabla }^{2}\psi +{n}^{2}\psi =0.\end{equation} \tag{ 11 }$$

Quantum effect manifests in the limit ƛ↛0, and geometrical optics emerges in the limit when λ → 0. Gloge and Marcuse [8] used z as the stepping parameter to arrive at this equation. On the other hand, we chose b as the stepping parameter to show the internal consistency with the formulation presented in 'F = ma' optics.

We note that the so-called 'quantum' description of light rays is equivalent to the scalar wave description of light, usually taught at the undergraduate level as wave optics.

The electromagnetic theory asserts that light is a travelling transverse electromagnetic wave in which the electric field $\vec{E}(\mathbf{r},t)$ and the magnetic field $\vec{B}(\mathbf{r},t)$ are coupled in the form of a pulse (see section 3.2 of [10]). The spatio-temporal variation of the electric field in a region devoid of source electric charge and current can be expressed by the vector wave equation:

$$\begin{equation}{\nabla }^{2}\vec{E}-\frac{1}{{v}^{2}}\frac{{\partial }^{2}\vec{E}}{\partial {t}^{2}}=0,\end{equation} \tag{ 12 }$$

where v denotes the speed of light in the medium of propagation. Each component of the electric field vector E_i ≡ (E_x , E_y , E_z ) denotes a pulse that satisfies the scalar wave equation:

$$\begin{equation}{\nabla }^{2}{E}_{i}-\frac{1}{{v}^{2}}\frac{{\partial }^{2}{E}_{i}}{\partial {t}^{2}}=0.\end{equation} \tag{ 13 }$$

If we assume that the electric pulse E_i (r, t) can be expressed as a product of spatial part u(r) and temporal part T(t), then using the separation of the variables technique, we can deduce the differential equation for the spatial part of the light wave:

$$\begin{equation}{\nabla }^{2}u(\mathbf{r})+{k}^{2}u(\mathbf{r})=0,\end{equation} \tag{ 14 }$$

where k is a constant representing the wavenumber. This equation is called the reduced wave equation or the Helmholtz equation. One can also express this equation in the form:

$$\begin{equation}{-\hspace{-7.5pt}\lambda}^{2}{\nabla }^{2}u(\mathbf{r})+{n}^{2}u(\mathbf{r})=0,\end{equation} \tag{ 15 }$$

where n represents the refractive index of the medium and $-\hspace{-7.5pt}\lambda=\frac{\lambda }{2\pi }$ is the reduced wavelength of light waves in this medium. Clearly, equations (11) and (15) are identical, and thus ψ-representing a ray of light, and u-representing a pulse of the electric field, must be intimately related.

The elementary eigensolutions ³ of equation (11) in a homogeneous medium are given by [5] (a) plane waves $\left({\text{e}}^{\text{i}\mathbf{k}\cdot \mathbf{r}}\right)$, and (b) spherical waves $\left(\frac{\lambda }{r}{\text{e}}^{\text{i}\mathbf{k}\cdot \mathbf{r}}\right)$, for r > λ. In a typical wave optics experiment taught in the undergraduate physics curriculum, one shines an aperture (or a slit) of dimension a with monochromatic light of wavelength λ, and a screen is placed on the opposite side, at a distance L from the aperture, to observe possible diffraction patterns. Assuming that the monochromatic light incident on the aperture can be assumed to be plane waves, the Fresnel number of the optical setup can be defined as [10]:

$$\begin{equation}\mathcal{F}=\frac{1}{L}\left(\frac{{a}^{2}}{\lambda }\right).\end{equation} \tag{ 17 }$$

If the observation screen is close enough to the aperture, in a way such that $\mathcal{F}\;\gtrsim \;1$, then one is working in the near field limit. In this case, it is convenient to use the spherical wave basis to describe the near field diffraction. However, if it is far away, such that $\mathcal{F}\ll 1$, then one is working in the far field limit. In this limit, one can use the plane wave solution, as locally spherical wavefronts behave as plane waves (figure 1).

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These wavefronts may indeed be considered as the envelopes of Huygen's wavelets, generated from the wavefront at a previous step of light propagation. In the far field limit, these wavelets are just the plane waves and can be expressed as: ψ = Ae^ik⋅r where A is a complex constant (the initial phase for the monochromatic wave is assumed to have been absorbed within A). As such, there is no problem with the complex wavefunction, but this cannot represent a real-valued pulse of the electric field. The electric field pulse can be written as $u=\mathrm{R}\mathrm{e}\left[A{\text{e}}^{\text{i}\mathbf{k}\cdot \mathbf{r}}\right]$. These pulses are referred to as phasors, due to the phase contained in the exponent (note that k ⋅ r is a phase). This observation can also be taken as a justification for taking the electric field pulse u of equation (15) as a complex quantity, commonly practised to simplify calculations. The phasor addition of the waves is utilised in the derivation of the intensity patterns of many interference and diffraction experiments [12].

The discussion in the previous section 3 throws light on the scalar wave nature of the optical field. Specifically, we found that the zero energy wavefunctions of light rays satisfy Schrödinger's equation. Therefore, it might be possible to derive Snell's law of geometrical optics. To check this, we begin with the problem of a particle or a ray of light of energy E, in the region with potential V₀ at x < 0, incident obliquely on a potential barrier V₁ at x ⩾ 0. ⁴ We represent this particle (or ray of light) by the wavefunction ψ. In the context of geometrical optics, such a potential barrier arises due to the difference of the refractive indices of the two media, as shown in the following figure 2.

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Due to the presence of the potential V₀ at x < 0, the momentum of the particle (or a ray of light) represented by the wavefunction ψ is: ${k}_{0}=\sqrt{\frac{2m(E-{V}_{0})}{{\hslash }^{2}}}$. At x > 0, the momentum of the transmitted particle (or the light ray) is: ${k}_{1}=\sqrt{\frac{2m(E-{V}_{1})}{{\hslash }^{2}}}$. The reflected component of the original wavefunction continues to be in the region of potential V₀, and has a momentum eigenvalue k₀ (with the sign of the x component is reversed).

Since $\vert {{k}_{0}}_{x}\vert$ and ${{k}_{0}}_{y}$ for the incident and reflected wavefunctions are the same, the angle of incidence must be equal to the angle of reflection. The translation invariance of the problem in the Y direction demands:

$$\begin{equation}{{k}_{0}}_{y}={{k}_{1}}_{y}\,{\Rightarrow}\,{k}_{0}\,\mathrm{sin}\,{\theta }_{0}={k}_{1}\,\mathrm{sin}\,{\theta }_{1}.\end{equation} \tag{ 18 }$$

Using the above expressions of momenta, we find:

$$\begin{equation}\sqrt{E-{V}_{0}}\,\mathrm{sin}\,{\theta }_{0}=\sqrt{E-{V}_{1}}\,\mathrm{sin}\,{\theta }_{1}.\end{equation} \tag{ 19 }$$

At this point, we exploit the identifications made in the context of 'F = ma optics' and equation (16), that the total 'energy' of the light rays is 0, and the 'potential energy' of light rays in a medium with refractive index n is $-\frac{{n}^{2}}{2}$. Replacing these in equation (19), we find that it reduces to

$$\begin{equation}{n}_{0}\,\mathrm{sin}\,{\theta }_{0}={n}_{1}\,\mathrm{sin}\,{\theta }_{1},\end{equation} \tag{ 20 }$$

which is Snell's law, established from the scalar wave description of light in a purely algebraic manner. Students who are unfamiliar with Hamiltonian optics, but have a basic idea of Schrödinger's equation, can fathom this derivation.

From figure 2, the total wavefunction at x < 0 is given as:

where denotes the amplitude of reflection back into the region of space with potential V₀. On the other hand, the wavefunction at x > 0 can be written as:

where denotes the transmission amplitude and ${{k}_{0}}_{y}={{k}_{1}}_{y}$. The boundary conditions at x = 0 are given by:

• (a)  
  ψ₀(x = 0) = ψ₁(x = 0) and
• (b)  
  ${\left(\frac{\partial {\psi }_{0}}{\partial x}\right)}_{x=0}={\left(\frac{\partial {\psi }_{1}}{\partial x}\right)}_{x=0}$.

The first condition yields:

The second condition implies:

From equations (23) and (24), we can show that the reflection coefficient can be expressed as:

$$\begin{equation}R={\left\vert \left\vert \frac{{{k}_{0}}_{x}-{{k}_{1}}_{x}}{{{k}_{0}}_{x}+{{k}_{1}}_{x}}\right\vert \right\vert }^{2}.\end{equation} \tag{ 25 }$$

In equation (25), ${{k}_{0}}_{x}={k}_{0}\,\mathrm{cos}\,{\theta }_{0}=\sqrt{E-V}\,\mathrm{cos}\,{\theta }_{0}$ etc. Hence,

$$\begin{equation}R={\left\vert \left\vert \frac{{k}_{0}\,\mathrm{cos}\,{\theta }_{0}-{k}_{1}\,\mathrm{cos}\,{\theta }_{1}}{{k}_{0}\,\mathrm{cos}\,{\theta }_{0}+{k}_{1}\,\mathrm{cos}\,{\theta }_{1}}\right\vert \right\vert }^{2}={\left\vert \left\vert \frac{\sqrt{E-{V}_{0}}\,\mathrm{cos}\,{\theta }_{0}-\sqrt{E-{V}_{1}}\,\mathrm{cos}\,{\theta }_{1}}{\sqrt{E-{V}_{0}}\,\mathrm{cos}\,{\theta }_{0}+\sqrt{E-{V}_{1}}\,\mathrm{cos}\,{\theta }_{1}}\right\vert \right\vert }^{2}.\end{equation} \tag{ 26 }$$

Let us now see the implication of this discussion in the context of geometrical optics. If we take E = 0 and ${V}_{j}=-\frac{{n}_{j}^{2}}{2}$ in accordance with our understanding of refraction, the quantity R can be written as:

$$\begin{equation}R={\left\vert \left\vert \frac{{n}_{0}\,\mathrm{cos}\,{\theta }_{0}-{n}_{1}\,\mathrm{cos}\,{\theta }_{1}}{{n}_{0}\,\mathrm{cos}\,{\theta }_{0}+{n}_{1}\,\mathrm{cos}\,{\theta }_{1}}\right\vert \right\vert }^{2}={\left\vert \left\vert \frac{{n}_{0}\,\mathrm{cos}\,{\theta }_{0}-\sqrt{{n}_{1}^{2}-{n}_{0}^{2}\,{\mathrm{sin}}^{2}\,{\theta }_{0}}}{{n}_{0}\,\mathrm{cos}{\,\theta }_{0}+\sqrt{{n}_{1}^{2}-{n}_{0}^{2}\,{\mathrm{sin}}^{2}\,{\theta }_{0}}}\right\vert \right\vert }^{2}.\end{equation} \tag{ 27 }$$

As expected, R → 1, when the incident angle θ₀ → 90°, or if the condition for the total internal reflection n₀ sin θ₀ = n₁ is satisfied. We notice that equation (27) is the same as Fresnel's equation of the coefficient of reflection for s-polarized light. The polarization of light in terms of electromagnetic theory is depicted in the following figure 3.

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This is expected, of course; but a curious student may wonder if there is a way to find the corresponding expression for p-polarized light. In fact, this formulation does not seem to lead to that expression. This must be the result of an incomplete description of the polarization of the light that is inherent in the ray picture. Gloge and Marcuse [8] write 'we cannot expect that the total content of Maxwell's equations can be restored by our quantization concept, since the ray picture does not contain any information about the photon spin'.