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The de Broglie wave equation states:

$$\lambda = \frac{h}{p},$$

where $\lambda$ is the wavelength of the “particle”, $h$ is Plank's constant, and $p$ is the momentum of the particle.

Momentum is usually written $\,p=mv$, where $m$ is the mass and $v$ is the velocity of the particle. But presumably $v$ is the relative velocity between the observer and the particle.

So does this mean that the wavelength of a particle depends upon the relative motion between the particle and the observer?

Or, perhaps more accurately, when a particle is incoming to another particle, in as much as an interaction between the particles depends on their relative speed, or the energy of impact, it thus also has something to do with their relative wavelengths.

Is that a conclusion, or simply a restatement of the premise, using different words that mean the same thing?

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Yes, the de Broglie wavelength of a particle depends on the relative velocity between the particle and an observer. I find it easier to think about the de Broglie frequency instead of the wavelength. They are related by $v=f\lambda$ or $f=v/\lambda,$ where $v=p/m.$

If the particle is moving towards the observer, the frequency appears higher, and if the particle is moving away from the observer, the frequency appears lower. This is just the Doppler effect. Therefore, if the particle is moving towards the observer, the wavelength appears shorter, and if the particle is moving away from the observer, the wavelength appears to be longer. As you said, this will have implications for interactions because the energy is related to the wavelength.

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  • $\begingroup$ Can this be readily observed e.g. with an electron beam an oscillating double slits? $\endgroup$ – Hagen von Eitzen Oct 29 '19 at 7:34
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So does this mean that the wavelength of a particle depends upon the relative motion between the particle and the observer?

Yes.

The second statement is more or less equivalent to the first.

Side note: The strange thing about this relationship is that it means that, e.g., wave train seen by one observer as consisting of one wavelength can be seen by another as containing some other number of wavelengths, say 3. This would be impossible if the wave was a real number whose phase was directly were observable, because you could, for example, count nodes.

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Your understanding is correct: the de Broeglie wavelength of a particle as measured by an observer depends on the relative motion of the observer and particle. It doesn't make a lot of sense to state that an interaction depends on the relative wavelengths of two particles, because an observer on either particle will perceive the wavelength of his particle as infinite.

However, an external observer watching the two particles will see each as having its own wavelength, and from the wavelengths (and masses) can deduce their momenta and kinetic energies as measured from any moving frame. The kinetic energy in the rest frame of the center of mass of the two particles will be the same as calculated from any other frame.

Note that the above is not quantum mechanically precise.

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  • $\begingroup$ I'm not sure how an external observer watches the particles. Doesn't the external observer have to interact with the particles in order to "watch them," for example, by hitting the particle with an electron and measuring the response? $\endgroup$ – vy32 Oct 28 '19 at 20:54
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    $\begingroup$ Sure. "...the above is not quantum mechanically precise." It would require a large number of identical experiments to get a clear measurement of the particles' wavelengths. $\endgroup$ – S. McGrew Oct 28 '19 at 21:41
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Yes, the observed wavelength of a particle depends upon the relative motion between the observer and the particle. This is called the Doppler Effect.

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