Does the Wave Function Represent the Spatial Distribution of a Particle?

Does the wave function represent the spatial distribution of a particle? If so, why use probability to determine the particles position? Why not say that the particle occupies the entire region of space given by the wave function?
- Noel Delgado (age 41)
OroValley, AZ, US

Hi Noel,

A particle's wavefunction encodes its spatial distribution, but doesn't give it to you directly. It turns out that if you carefully measure the position of a particle, you will find it to be at a narrow location (the more careful your measurement, the more narrow the outcome) somewhere within the spatial extent of its wavefunction. The likelihood it has of being in a little volume around a given position is equal to the magnitude squared of its wavefunction's value at that position times the little volume.

It should be emphasized that this measurement of position tells you where the particle is after the measurement, but it doesn't really tell you where the was beforehand. To know where the particle was beforehand, the best you can do is know the wavefunction and the facts given above.

I suppose you could say that "a particle occupies the entire region of space given by the wave function." However, all realistic wavefunctions have tiny tiny but non-zero "tails" for all possible positions in the universe. It isn't quite sensible to claim that every particle "occupies the entire universe," so the safest thing to remember is the full probabilistic description.

Cheers,

David Schmid

(published on 01/17/2014)

I understand that the wave function does not give a particle’s position directly. In regards to a particle occupying the entire universe, don’t we say the same thing about a particle’s field (electric, magnetic, gravitational)? By imposing probabilities we instead make the more in-sensible assertion that the particle has a tiny tiny but non-zero probability of appearing anywhere in the universe. It seems to me that if we simply took the electron cloud at face value we might somehow be able to avoid the weirdness of the "measurement postulate".
- Noel Delgado (age 41)
Oro valley

That's a great question. In some modern interpretations of quantum mechanics, we don't use probability to describe the quantum state itself. It just is what it is, spread out, just like classical fields, exactly as you point out. In the end, however, that elegant description must make some contact with what we have actually observed.

Whenever the quantum state should give a spread of different large-scale events we only see one version, not the whole output predicted by the basic quantum time-dependence. Geiger counters click at specific times, and those times look completely random. Detector screens flash at specific little regions within the area of some particle's state. and the pattern of regions looks completely random. And so forth. It's as if the quantum state collapses to some little fraction of its former self, a little fraction consistent with just one macro outcome. We can correctly calculate the probability of an outcome happening by asking how much (in squared units) of the quantum state would have led to that outcome.  

We can avoid saying that the quantum state really does collapse in a measurement, but at the cost of saying that all parts of it keep going. Since those parts include more than what we observe, that would require that their be multiple versions of us coming from each quantum "measurement". That's called the Many Worlds interpretation of quantum mechanics. It may be right, but you'll probably agree it also has some weirdness to it.

Mike W.

(published on 03/29/2014)