Confusion about EM spectrum and the Fourier transform

Question

Since courses on signal analysis and electromagnetism I have become confused about what the spectrum of electromagnetic radiation really means. I know light is when electric and magnetic fields become coupled and start propagating through space with a certain frequency, according to Maxwell's equations, and this is usually depicted as a simple linearly polarized sine wave (or plane wave). Now a pure sine wave is infinitely long in space and needs infinite time to have a single frequency (frequency and wavelength are coupled by the speed of light), this is due to the Fourier transform: truncated signals get spread in the frequency domain and vice versa.

So when scientists and engineers talk about the frequency of light, how can they speak of just a frequency, and not a bandwidth of frequencies, because the light wave is not simply a pure sine wave it is actually a pure sine wave multiplied with a block function/window function. In fact if the light was an intricate signal with varying electric and magnetic fields, such as a radio plane wave, which is clearly not sinusoidal, we would obtain a continuous spectrum not just a Delta function with a slight bandwidth. Even crazier, how is it that spectral lines are said to be a single frequency, even though we need a slight bandwidth around each spectral line. Furthermore, when they speak of a spectrum, isn't this simply a Fourier transform of an arbitrary signal of electric and magnetic fields (here for simplicity I assume the fields propagate in the same direction and are linearly polarized, see sketch below), which is then decomposed into pure sinusoids (technically complex exponentials, but for real signals both are fine to use). Isn't this what a prism or diffraction grating does, splitting the light wave (which could be an arbitrary signal) into pure frequencies of sinusoids and the spectrum we observe is a physical Fourier spectrum created by the prism or diffraction grating.

Also, I am aware that the Fourier transform is a mathematical construct that simplifies the analysis of many problems, so in theory we can apply it to any signal or function. The reason I believe a certain EM spectrum is a physical Fourier spectrum is because Maxwell's equations have plane waves as fundamental solution to the electromagnetic wave equation and these can be superimposed to create any arbitrary forward traveling wave; by Fourier analysis.

Below is a sketch of what I imagine to be an arbitrary light wave or signal (linearly polarized) that is split into it's constituent frequency components, I've only drawn a few sine waves to illustrate the prism's effect, in reality it would be a continuum or a few spectral lines (with some spread). The exact refraction and effect on polarization by the prism is not important here.

Thank you for your time and effort.

With kind regards,

Jelle

(Undergraduate physics student)

Answer 1

Yes. A real light beam always has a finite bandwidth.

In quantum optics, for example, a single photon wavepacket may have a length of perhaps 25$\mu$ and so a duration of about a 100 fs. Consquently the experimentalists photon will be a linear superposition of a continuum of theorist's photons with corresponding bandwidth.

Answer 2

I wouldn't say plane waves are fundamental solutions to the EM wave-equation, I would: they are eigenfunction of eigenfunction of certain operators, which make them useful.

If we take a plane wave:

$$ \psi(x, t) = e^{i(\omega t - kx)} $$

and apply a time translation:

$$ t \rightarrow t'= t + \Delta t $$

to it:

$$ \psi(x, t') = e^{i(\omega (t+\Delta t) - kx)} = e^{i\omega\Delta t}\psi(x, t) $$

so the eigenvalue is a phase rotation that is linear in $\omega$. Likewise for space translations.

Since many systems can have their behavior parameterize by frequency of wavenumber, this is a very useful approximation.

As you pointed out, real signals have finite time duration, and hence finite bandwidth. You need finite bandwidth to transport information, of course.

Also: get used to it. When you learn the hydrogen atom, you start with a Coulomb potential:

$$ V(r) = \frac 1 {4\pi\epsilon_0} \frac e r $$

and solve for eigenstates $\psi_{nlm}$.

All these state evolve through time just like a plane wave: multiplication by a phase factor.

They are non-physical, though: as energy eigenstates, they are stationary and stable. There is no mechanism to transition.

The physical orbitals are changed by fine-structure and so on, and the addition of a background EM field allows transitions (and then bandwidth and lifetime have an inverse relationship, just like finite bandwidth time-domain signals).

Nevertheless, we (as not-an-atomic physicists) talk about the in-theory states and transition rates and what not.

So I'm tempted to say "nothing is exact", but then we come to rotations.

In the plane wave, we have eigenvalues based on spatial translation. Likewise, the $\psi_{nlm}$ for atoms, or electromagnetic waves in the multipole expansion, are Eigen-functions of rotations. In both of those cases, two of the spatial dimensions are captured by spherical harmonics:

$$ Y_{lm}(\theta, \phi) $$

which are eigenfunctions of $z$-rotations with eigenvalues that are, wait for it, phase rotations:

$$ \hat R_z(Y_{lm}(\theta, \phi)) = e^{im\theta}Y_{lm}(\theta, \phi) $$

In this case, I'm pretty sure it's exact, and we can proceed to solve problems using angular momentum.

The point is, we usually work with approximate eigenstates who's behavior under translations/rotations is a phase rotation; moreover in both the classical EM and quantum cases, the existence of these eigenfunction means there's a conserved quantity related to the rate of phase-change of the eigenvalue:

Time translation -> Energy

Space translation -> Momentum

Rotations -> Angular Momentum