Light waves can't have a wavelength

Question

The wave nature of light comes from Maxwell's equations. More precisely, the two wave equations that come from them:

$$\Delta\vec{E}=\mu\varepsilon \frac{\partial^2\vec{E}}{\partial t^2}\\ \Delta\vec{H}=\mu\varepsilon \frac{\partial^2\vec{H}}{\partial t^2}$$

Looking only at the electric field part, the solution of that equation is any function:

$$\vec{E}=\vec{F}(\vec{s}\cdot\vec{r}-vt)$$

That is what we call a wave - we make a disturbance in one point and we can measure the same disturbance elsewhere after a short period of time.

Now my question is why do we always talk about light in terms of wavelengths? It's not necessarily periodic - the wave equation doesn't say that it has to be. One possible explanation that I thought about is the possibility to dissolve any disturbance in Fourier series, but that would mean we would have to add light with infinitely small wavelengths. Since the series is infinite, that would definitely make problems thinking about QM. Why do we say light has wavelength and phase when it's not necessarily a sinusoidal wave or not even periodic at all?

Answer 1

You're right in saying that you can't always attribute a unique wavelength to an electromagnetic field. However, it's a simple consequence of Fourier analysis that you can write any field as a superposition of different frequencies, or equivalently wavelengths. Basically, you can look at sinusoidal fields as a basis in which you can decompose any given field. Since a lot of electromagnetic systems are linear, it's usually convenient to work with the basis functions (sinusoids) instead of a general functional form.

To see this explicitly, just take a (temporal) Fourier transform on both sides of the wave equation $\Delta\mathbf{E}(\mathbf{x},t) = \mu \epsilon ~\partial^2 \mathbf{E}(\mathbf{x},t)/\partial t^2$:

$$\Delta \tilde{\mathbf{E}}(\mathbf{x},f) = - \mu \epsilon (2\pi f)^2\tilde{\mathbf{E}}(\mathbf{x},f), $$

where my convention for the Fourier transform is $\tilde{\mathbf{E}}(\mathbf{x},f) = \int_{-\infty}^\infty dt~\mathbf{E}(\mathbf{x},t)e^{-i2\pi f t}$. The above is typically called the Helmholtz equation, which is just the frequency-domain representation of the wave equation. Whatever the solution is, you can convert it to the original (time-domain) field via an inverse Fourier transform

$$\mathbf{E}(\mathbf{x},t) = \int_{-\infty}^\infty df~\tilde{\mathbf{E}}(\mathbf{x},f)~e^{i 2\pi f t},$$ which explicitly tells you that the field is superposition of sinusoids with varying frequencies ($f$). If you prefer, you can also write this as a superposition of wavelengths with a simple change of variables $\lambda = c/f$: $$\mathbf{E}(\mathbf{x},t) = -c\int_{-\infty}^\infty d\lambda~\frac{\tilde{\mathbf{E}}(\mathbf{x},c/\lambda)}{\lambda^2}~e^{i 2\pi c t/\lambda}.$$

Answer 2

1. Something does not need to be sinusoidal in order to have a wavelength. It just needs to be periodic or repeat after a certain distance. More generally, if $\lambda$ is the smallest non-negative real number such that $f(x+\lambda) = f(x)$ is true for all $x$ then we would say that the function $f$ has a wavelength $\lambda.$
2. Working with sinusoids is mathematically convenient for a few different reason. One is that they are just a simple example of a periodic function. But more importantly the mathematical convenience comes from two facts: sinusoidal functions form a basis, and they are eigenvectors of the differentiation operator. The wave function in Fourier space then just takes the simple form of multiplication by the eigenvalue. See @Sahand's answer for more details. Note that many simple periodic functions can also be approximated with a single sinusoid function.
3. Finally, a lot of physical processes acutally do happen near fixed sinusoidal functions with definite wavelengths. For example, emissions spectrums are a useful scientific tool because atoms/molecules radiate/absorb light, and this has a dependence with the frequency/wavelength of the light. Fundamentally, this is because a particle of light has energy related to its wavelength. A photon at a fixed energy $E$ in free space would actually be a sinusoidal function with corresponding wavelength $E=h\nu$. (Setting aside the non-normalizability of such a function).

Answer 3

Sahand gave a really great answer but I'd like to offer a different perspective.

Edit:

It was pointed out in the comments that energy conservation isn't enough to give periodic solutions. Which is correct. I went back to a couple of my E&M books and the answer is actually a bit unsatisfying. As far as I can tell everyone decided to assume that the solution should be separable into time and space components. If you do that you get (I believe) only oscillatory solutions and a defined wavelength. To be honest the assumption was probably motivated by the experimental evidence that light was mostly an oscillatory wave and previous experience solving differential equations of that form in ways that described oscillatory waves.

So we guessed a functional form (for EM waves in free space) and that resulted in only oscillatory solutions which explained experimental results.

Old Answer:

If I recall correctly those wave equations are valid only in free space with no charges and no currents. In terms of pure mathematics the most common solutions would be exponentials and sinusoidal functions (which are really exponentials in disguise). In the absence of any other conditions on the nature of the allowed solutions both families of solutions are allowed.

However, there is another condition. In the absence of charges and currents the amount of energy in the disturbance described by these equations must be constant in time i.e. conservation of energy. That definitely rules out pure exponentials and I can't really think of any nonperiodic function that would satisfy conservation of energy. Except maybe the trivial solution of constant electric and/or magnetic fields. Basically, we must choose oscillatory solutions because of conservation of energy. I believe similar reasoning holds for an undamped mass/spring system.

Of course once you put charges and currents back in it is possible to have solutions that are not oscillatory (or at least not oscillatory forever) but still conserve energy. Negative exponentials show up in processes that involve absorption and positive exponentials show up when you're driving an electromagnetic oscillation especially near or at resonance.

Answer 4

Time domain Maxwell's equation never implies periodicity. If time harmonic representation of Maxwell's equation is done than by default idea of periodicity automatically arise in equations.

Answer 5

Panagopoulos, D. J. (2018). Man-made electromagnetic radiation is not quantized. Horizons in world physics, 296, 1-57.