High School Why the square of the wave function equals probability?

[Prem1998]

If the problem is just to avoid negative probabilities, then why isn't the modulus of the value of wave function equal to the probability of finding the particle? I mean, is it proved by mathematics that the integration of the square of wave function value over a particular region is equal to the probability that the particle will be found in that region?

 

[Nugatory]

I mean, is it proved by mathematics that the integration of the square of wave function value over a particular region is equal to the probability that the particle will be found in that region?

It's not proved by mathematics, it's proved by experiment. There are many interesting mathematical constructs out there. One of them - superimpose amplitudes and then square - yields results that match the way the universe is observed to behave, so that's the one we use.

 

[Prem1998]

It's not proved by mathematics, it's proved by experiment. There are many interesting mathematical constructs out there. One of them - superimpose amplitudes and then square - yields results that match the way the universe is observed to behave, so that's the one we use.

What kind of experiment can possibly prove this?

 

[vanhees71]

http://arxiv.org/abs/0811.2068
http://arxiv.org/abs/1007.4193

 

[RockyMarciano]

What kind of experiment can possibly prove this?

Let's leave aside for a moment the well known fact that strictly speaking "proving" something by experiment is not usually considered possible , only disproving is so let's center on this. The Born rule is just the most reasonable probabilistic interpretation of the mathematical formulation of QM, in that sense there is nothing to disprove about the rule itself as long as the formalism works within its own premises and definitions of states and superpositions, transitions and observables. If you identify observables with operators you have to take into account the cross-terms and that must be reflected in the way you calculate probabillities, this obviously cannot be disproved with experiments anymore than the arithmetic operations can be.

For instance what the papers linked in the previous post show is that once one identifies observables with operators the experiments confirm quantum interference as the best approximation.

 

[vanhees71]

Another theoretical argument is Gleason's theorem

https://en.wikipedia.org/wiki/Gleason's_theorem

 

[Vanadium 50]

then why isn't the modulus of the value of wave function equal to the probability of finding the particle?

Because then you get the wrong answer.

 

[mikeyork]

This is not a proof, but an argument about consistency: First, physicists like analytic functions. The modulus of the probability amplitude is not analytic at 0. The squared modulus is. Second probabilities are purely relative until normalized so that they sum to unity. The "length" of a vector is the sum of the squares of its components. In QM state vectors are normalized to unity. When written as a superposition this means the sum of the square moduli of its projections should add to unity. So it is natural to treat the squared moduli of the projection of the state vector onto an eigenvector as being the probability of finding that eigenstate. However, it is important to remember that it is the unsquared projections that contain the physics. The relative phases of projections are physically significant (as in interference for instance) so the probabilities obtained by squared moduli do not completely describe all the physics. And when they do describe the physics they do so only in the form of an approximation to relative frequencies of mutiple measured outcomes and converge to exact multiple frequencies only for an infinite number of samples. The projections {probability "amplitudes") however describe a single system exactly.