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I'm wondering what exactly is meant by the wavelength in De Broglie formula $p=\frac{h}{\lambda}$, where $p$ is the momentum of a particle and $\lambda$ is the wavelength. I know that a wave function might very well be messy without a defined wavelength. Can someone clear my confusion?

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You might have heard about the wave-particle duality: depending on how you look at it, matter can behave as being made of particles, or it can behave as being made of waves.

That all depends on the kind of experiment you are performing: roughly speaking, if the typical lengths of your experiment are large enough you will likely see particles; if the typical lengths of your experiment are small enough you will likely see waves.

So, when matter behaves as a wave, what is the associated wavelength? Answer: the de Broglie wavelength.

For instance, light passing through a double slit experiment will cast a diffraction pattern on the screen. We can predict the maxima and minima of intensity by using well know formulas (see, for example, here and here).

If the slits are very close apart, like under a nanometer, then a beam of electrons will also cast a diffraction pattern on the screen (the Davisson-Germer experiment), and the peaks in the diffraction patterns are predicted using very much the same formulas for light, except that you use the de Broglie wavelength for those electrons in the experiment.

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De Broglie wavelength of a particle is the associated wavelength of the particle when the particle behaves as a wave. Simply put, in wave-particle duality, De Broglie wavelength of a particle is the wavelength that the particle would have if it were a wave exhibiting the particle's wave-like properties.

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  • $\begingroup$ Are you assuming the particle is in a momentum eigenstate? $\endgroup$
    – WillO
    Feb 1, 2022 at 3:49

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