1. The de Broglie formula states that: My question is:
"If an object slows down and eventually stops, does the wavelength of the matter wave become infinite?"  2.

3. Originally Posted by Cogito Ergo Sum The de Broglie formula states that: λ = h/(m*v)

My question is:
"If an object slows down and eventually stops, does the wavelength of the matter wave become infinite?"
Yes, although with a caveat. It isn't possible for a particle to have an exact momentum, unless you have no idea where it is, due to the Uncertainty Principle that the product of the uncertainty in the momentum and the uncertainty in the position must be greater than h-bar. Therefore, if a particle had a momentum of exactly zero, the uncertainty in its position would be infinite. The chance of the particle being within observational range would be too small to be worth pursuing. Any observable, "at rest" particle must have a nonzero probability of nonzero momentum.  4. Originally Posted by Cogito Ergo Sum The de Broglie formula states that: λ = h/(m*v)

My question is:
"If an object slows down and eventually stops, does the wavelength of the matter wave become infinite?"
Interesting, I have had a rather lively debate with someone about this very topic not long ago. But mvb beat me to it, I wouldn't have anything to add to his answer   5. Originally Posted by mvb Therefore, if a particle had a momentum of exactly zero, the uncertainty in its position would be infinite.
Isn't this equivalent to saying the de Broglie wavelength is infinite? I'm not sure what the "meaning" of the de Broglie wave is, but it almost looks like just another way of viewing the Heisenberg relationship.  6. Originally Posted by Strange  Originally Posted by mvb Therefore, if a particle had a momentum of exactly zero, the uncertainty in its position would be infinite.
Isn't this equivalent to saying the de Broglie wavelength is infinite? I'm not sure what the "meaning" of the de Broglie wave is, but it almost looks like just another way of viewing the Heisenberg relationship.
Strange I'm not sure it is quite the same. My understanding of the Uncertainty Principle has always been that to know the position to any degree of accuracy you need to have a "wave-packet", involving what is effectively a Fourier series of superimposed frequencies that interfere constructively to produce an envelope of probability in only one localised region of space. This superposition of wavelengths, via de Broglie, implies a lack of knowledge of momentum. The "monochromatic" case, i.e. with exact momentum, would imply a wavefunction distributed evenly throughout space, i.e. total uncertainty about position. But this has nothing to do with the magnitude of the momentum, i.e. the wavelength of the monochromatic wave. Or so it seems to me.  de broglie 