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I've seen some examples/problems where the de Broglie wavelength of large objects (like a tennis ball) is calculated and this doesn't really make sense to me. So the baseball consists of many smaller particles like electrons, which all have a wave function of their own. So what I'm wondering:

  1. Wouldn't interactions between the electrons/atoms be like measurements? Like they would collapse each other's wave functions all the time. For sure if I would be looking at the tennis ball there will be some measurements at the edge at least (though maybe the problem's are supposed to be in a vacuum, if that would completely change the situation..?). I can't really see how this collapsing of the individual particles would relate to the wave of the whole tennis ball.
  2. If there's a wave for the whole thing, how would the individual constituents know that they are all one tennis ball together? Could you also talk about the wave of two tennis balls traveling together (maybe attached with a string to make them into one object, though it also wouldn't make sense if that would make a difference right?)?
  3. Could you say if the wave is centered at some location, like in the middle of the tennis ball?

I could ask the same questions for large molecules, of which I know that they have also been successfully used to make interference patterns in slit experiments.

I hope the questions make sense, I think it mostly comes from errors in how I look at the problem, but if someone could clear that up that would be great!

friso
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  • I think the wavefunction of bound particles doesn't come into play here. You are correct that the deBroglie wavelength of smallish molecules has been verified. – Carl Witthoft Aug 20 '21 at 14:32

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It is more accurate to say there is one wavefunction that describes all the particles together. Let's say you have two particles. A general wavefunction would be $$\Psi(\vec r_1,\vec r_2)$$ where $|\Psi(\vec r_1,\vec r_2)|^2$ is the probability density of finding the first particle at position $r_1$ and the second one at $\vec r_2$. This is easily extended to an arbitrary number of particles: $$\Psi(\vec r_1,\vec r_2,\dots, \vec r_N).$$ So to get the wavefunction of a macroscopic object you could just write down the wavefunction of all its constituents (which gets hard very quickly).

As a sidenote quantum mechanical particles are indistinguishable so if you swap two particles, the wavefunction has to stay the same up to a sign: $$\Psi(\vec r_1,\vec r_2)=\pm\Psi(\vec r_2,\vec r_2)$$ This means you can't just write down any wave function you want as long as you want to describe a physical system. It has to obey this exchange rule for all of its arguments.

AccidentalTaylorExpansion
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