# What is the physical basis of Born's interpretations?

Did anyone has any idea how Born came up with the probabilistic interpretation of quantum mechanics. It is by all means very bizarre. And then it leads to the idea of copenhagen interpretation. Also no one emphasis on any kind of meaning attached to $\psi$ the wavefunction rather it's norm squared. One last thing why we work just with complex number why not quaternion if we are happy doing abstract thing without give any physical foundational meaning to complex number?

• I think it is easier to see what the QM equations means in $\mathbb{R}^2$ instead of $\mathbb{C}$. – Emil Apr 9 '18 at 6:19
• he explains a bit in his Nobel lecture nobelprize.org/nobel_prizes/physics/laureates/1954/… – physicopath Apr 9 '18 at 8:06
• If you’re interested in a possible interpretational justification of the Born rule, that would be a strength of the transactional interpretation of quantum mechanics. ayuba.fr/mach_effect/cramer1986.pdf – Gilbert Apr 9 '18 at 15:22

It is interesting to point out that in the original rapid communication Born intepreted $\Psi$ itself as a probability density. It's only in a footnote, probably added after the paper was finished, where he points out that the correct formula involves square. This was a rather obvious guess as the wavefunction is not only nonpositive, it's also complex in the collision processes that Born's paper was devoted to. This guess was supported then by calculations of scattering crossections compared to the experiments.
The physical meaning of the wavefunction is that it is a probability amplitude. To properly understand this you should realize that $P(x)=|\psi(x)|^2$ is not the only probability distribution that appears in the experiments! As a matter of fact quite often it is not the particle position that is measured. E.g. you can measure the particle momentum and then you will have to use, $$P(p)=|\phi(p)|^2,\quad \phi=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}dx\,e^{-ipx}\psi(x)$$ You can't predict $P(p)$ knowing only $P(x)$! To get it you have to also know the phases of $\psi(x)$. The constant phase doesn't matter but relative phases in different points do. So measuring sufficient number of observables you can actually reconstruct those phases up to the constant. These phases also govern the dynamics of the quantum state - different relative phases mean different dynamics.