# Understanding de Broglie Wavelengths

I understand the derivation and calculation of de Broglie wavelengths, but what exactly does the wave of a particle represent? What does the wavelength of a particle mean? I'm assuming it's not the wavelength of its probability wave because then there would be no uncertainty in momentum, unless it represents the average wavelength.

The idea of wave-particle duality for EMR makes sense to me in that it can be seen as oscillating changes in energy (a wave), and as photons from the photoelectric effect. But with particles, I don't understand what quantity would be oscillating in their wave.

Thanks!

## 1 Answer

The "wave" part of the wave-particle duality for particles such as electrons and protons (as opposed to EM radiation) is called their wavefunction. It does not have any classical analogue and any attempt at understanding it using classical intuition can only be a crude analogy. However, if you are happy with the concept of a probability wave then it is exactly that.

Why isn't this a problem with the uncertainty principle, then? Well, there is a corresponding uncertainty principle between the wavelength of any wave (or more precisely its wavenumber $k=2\pi/\lambda$) and its position in space. The wavelength of a wave is only precisely definable, to arbitrary precision, if you have an infinite wave; otherwise, you can only measure a finite number of periods and divide, and that will yield an imprecise measurement of $\lambda$. (Even worse, the amplitude will taper out near the edges, so it will be hard to tell where each peak or trough is.)

To have a better-defined wavelength, then you need a bigger wavepacket, but this means that the position of the wavepacket in space, which only makes sense to a precision of the wavepacket size, has bigger uncertainty. This trade-off game can be expressed as $$\Delta k\Delta x \gtrsim1,$$ and can be made precise using Fourier analysis of waves.

• Thanks! One question, however: since you said it's possible to know the exact wavelength (hypothetically), wouldn't the uncertainty principle than be broken because $\sigma_p$ would be 0? Or does this not break uncertainty because it is only possible in theory to calculate the wavelength exactly? – nsanger Apr 8 '13 at 23:31
• No. To know the wavelength exactly you need an infinite wavepacket, so $\sigma_x=\infty$, which is consistent with $\sigma_p\sigma_x\gtrsim\text{const}$. (In real situations, of course, you have $\sigma_p$ very small and $\sigma_x$ very large, subject to that condition.) – Emilio Pisanty Apr 9 '13 at 11:44