stevendaryl said:
For a while, maybe. But if no satisfactory alternative theory was found, many of them would shift to pretending that the evidence doesn't REALLY show what it seems to show. In other words, many theorist would just live in denial about it. And that's exactly what we do have.
The reason there is eternal fighting about it is because 1 & 2 are contradictory, and there is no agreement about how to fix it. On the other hand, there is a "rule of thumb" that allows us to get past the contradiction, which is that we have heuristics for when to treat a system as a measurement device obeying 1, and when to treat it as a quantum mechanical system. You can't consistently do both.
Let me sketch a scenario showing that rule 1 is contradictory with rule 2.
Suppose that you have a Stern-Gerlach type situation in which an electron that is spin-up in the z-direction is deflected left, and makes a black spot on the left side of a photographic plate. An electron that is spin-down makes a black spot on the right side.
Now suppose that we set up initial conditions that are precisely left-right symmetric. We send an electron through the Stern-Gerlach device that is spin-up in the x-direction.
According to rule 1, eventually the system evolves into a final state that is not left-right symmetric. Either the electron goes left and makes a spot on the left, or goes right and makes a spot on the right. So the final state does not satisfy the left-right symmetry of the initial state.
That is not possible, if everything obeys the laws of quantum mechanics. If the initial state is left-right symmetric, and the Hamilton is similarly left-right symmetric, then the final state will be left-right symmetric.
Obviously I'm to stupid to understand, why there's a contradiction between 1 and 2. Particularly your example of an apparent paradox is completely incomprehensible to me, because it contradicts the standard interpretation of QT. Of course, as Bohr has already analyzed, you cannot perform the SG expeirment with free electrons in practice. So let me put Ag atom instead of electron (because that was the atom Stern and Gerlach used at the time)
If you send an Ag atom with indetermined spin-z component through a Stern-Gerlach apparatus it is deflected with the probabilities given by the corresponding (pure or mixed) state either to the left or the right. That's the point, why Born introduced the probability interpretation of the quantum state in the first place: You cannot split a single Ag atom into pieces, because you never find a smeared classical-field like entity that's smeared according to the wave-function squared, but you always find a single spot on the screen after it run through the SG-magnet. It ends either "to the left" or "to the right" and thus either, through the entanglement between position and spin-z component through the running through the magnet we conclude that the spin component is either up or down, depending on where the electron landed. You can even prepare pure spin-z eigenstates by just looking at the corresponding partial beam (through filtering away all electrons running in the other region of space). I.e., here you have a paradigmatic example for a von Neumann filter measurement. Of course, for the single electron there's no way to predict, which value will be found. You can only say that with the probabilities given by the state the Ag atom is prepared in
In the case of the original experiment it was a beam of Ag atoms from an oven running through a little opening, so it's some mixed state of roughly given by (written as product of the spatial/momentum part and the spin part)
$$\hat{\rho} \propto \int_{\text{aperture}} \mathrm{d}^3 p \exp[-\vec{p}^2/(2m k T)] |\vec{p} \rangle \langle \vec{p}| \otimes \hat{1}_{\text{spin}}/2.$$
Of all Ag atoms the probability for either spin-z component up or down is 1/2, i.e., the symmetric situation you assume. Of course, each single atom will go either the one or the other direction, and for the single atom the situation is not symmetric. Only the probability distribution is symmetric, and that's what's also predicted by QM. This can be even calculated in good approximation analytically (I've still not found the time to write this up completely, but it's really not too complicated).
The final state of the Ag atom is a quantum state again and only describes probabilities, and this distribution obeys the left-right symmetry you rightly assume.
In experimental terms: What's symmetric is the distribution of many Ag atoms, all prepared in the same initial state, running through the setup. Each single atom "breaks" the symmetry of course, landing either "left" or "right" from the symmetry plane.
It's the same with any random experiment. Although a perfectly fair die is symmetric, any outcome is showing 1 of the six edges and thus breaks the cubic symmetry. Only "statistically", i.e., the "average" over many outcomes is symmetric, i.e., the probability for each outcome is the same, 1/6.
