Does a square (or any non-sinusoidal) wave a definite wavelength?

I'm currently reading/studying the FLP and I have a question regarding waves. In a chapter about QM, Feynman says that any short wave train doesn't have a definite wavelength. I understand this, because the wave train is composed of a number of frequencies of sine waves (revealed by a Fourier transform). The only [sinusoidal] waves that have a definite wavelength are infinite perfect sines (disregarding any phase difference).

My question is: what happens if the waveform is not a sinusoid? If the waveform is a [infinite] square wave, for example, it can be analyzed via Fourier to reveal that it's composed of several sinusoids of different wavelengths, but on the other hand, one can actually look at the pattern and measure a "definite wavelength".

In the image above, the wavelength I'm speaking about is $$v * 10ms$$, where v is the velocity of the wave [For example, assuming 1000 m/s, then $$\lambda = 10 meters$$]. Does make sense to speak of a definite wavelength in this case?

• It likely depends on the context that you are talking about wavelengths. Most people should at least understand what is meant as long as it's clear the wave isn't a sinusoid; but is periodic.
– JMac
Jun 27 '17 at 15:23

If you get from a Fourier transform the spectra of the square wave you get many wavelengths (and of course frequencies), and it has a bandwidth. Freq and wavelength is defined wrt to the basis waves, which are sinusoids. See below about other basis waves.

If you use it differently it'll be easy to have people confused, or just say you are wrong. The spectrum of the square wave is the delta function convolved with a sync function. In radio and EM transmission and reception they try to filter out the sidelobes (and any freqs outside what you want), and usually the square wave is rounded by some smoothing function.

In QM it's even more critical because the sinusoids are the eigenfunctions of the Hamiltonian (or some physical and important operator that you use with eigenfunctions as the basis for expanding any arbitrary solution. The eigenvalues or for the Hamiltonian the energy, are then we'll defined. A square wave or a limited wave train will not have a definite energy.

This is also reflected in the uncertainty principle in QM, and it is equivalently true for any wave. If your wave last infinitely long your uncertainty in the arrival time is infinite. Then the energy uncertainty has to be zero. In classical waves you need a long train to measure it's frequency accurately, equivalently to QM.

Now, you can choose other basis waves, or in QM, eigenfunctions of any operator (with some constraints) and expand on the basis of a those. The eigenvalues, or values you could actually measure and get a number, will just not be energy. Similarly for classical waves, you can choose mathematically any basis you want. In some signal processing work people use so called wavelets, and other things.

For a periodic waveform, the definition of wavelength of the waveform is the wavelength of the lowest nonzero component of the Fourier transform. This fits the intuitive definition and is also mathematically rigorous.

Using Fourier, a DC signal plus an infinite non-periodic wavelet, by using a factor of ka will definitely produce an infinite square waveform. I hope that answers your question.