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I collect here info from another thread, to have a more focussed discussion.

Nonrelativistic particles have no different interpretation than relativistic ones.

Particle detectors respond to the momentum of a particle, not to its position.

Scattering experiments are interpreted in the momentum picture. Nobody is interested in the position of particle tracks, only in their momentum (which tells about masses).

covering the size of an uranium atom (N=92), say.

This is ridiculous - such measurement devices are impossible!

Whereas the form in which I stated the probability interpretation assumes nothing. it makes claims only for those projectors that are actually realizable. it is therefore much more realistic.

''this particular assumption'' refers to the assumption that |psi(x_1,...,x_n|^2 is the probability density of observing simultaneously particle k at position x_k (k=1:N).The probability interpretation says that _if_ you can set up an experiment that measures a self-adjoint operator for a system in state psi then the probability of observing the k-th eigenvalue is psi^* P_k psi, where P_k is the projector to the k-th eigenspace. It says _nothing_ at all about which particular operators are observable in this sense.

Everything beyond that is interpretation, and hence (at the current state of affairs) a matter of philosophy. In particular, which operators can be measured is not part of the probability interpretation but a matter of theoretical and experimental developments.

Regarding what is arbitrarily _precisely_ measurable, there is a no go theorem by Wigner (I can give references if you want to check that) that states that _only_ quantities commuting with all additive conserved quantities are precisely measurable. The position operator is not among these.

Nobody comparing QM with experiments is making use of this particular assumption.

It is stated in the beginning as an interpretation aid without proof, and never taken up again in the context of real measurements where the claim would have to be justified. It is very common to make this sort of idealized assumption to get started; but once the formalism is established, this assumption is never used again.

For example, Landau & Lifgarbagez begin in Section 2 of their Vol. 3 with such a statement, but immediately replace it in (2.1) and (3.10) by the more correct version about the interpretation of the expectation value <K> = Psi^* K Psi, where K is an arbitrary observable (linear integral operator) depending on the form and values of the measurement. From then on, only the latter interpretation is used; never the fictitious, idealized introductory remark.

And it cannot be different, since quantum mechanics is used in many situations where the state vectors used in the formalism have no interpretation as a function of position - the whole of quantum information theory and the whole of quantum optics belonging to this category.

Please show me a comparison with experiment that does this.For nonrelativistic particles, absolutely everyone comparing QM with experimentsdoesmake use of the wave-mechanics interpretation of [itex]|\psi(x)|^2[/itex] as a probability density.

Nonrelativistic particles have no different interpretation than relativistic ones.

Particle detectors respond to the momentum of a particle, not to its position.

Scattering experiments are interpreted in the momentum picture. Nobody is interested in the position of particle tracks, only in their momentum (which tells about masses).

It could be this _only_ if you can prepare an experiment that realizes such a P_k. But this is a matter of experimental technique and not one about the interpretation of quantum mechanics. But there are no such operators since the spectrum of position is continuous.Yes, this is true. QM does not talk about the specifics of observations and measuring devices. For example, P_k can be a projection on the k-th eigenvalue of the position operator.

No. This unrealistic assumption is needed _only_when one wants to insists on a probability density interpretation of |psi(x)|^2. And for the position representation of an N-particle state, one would need an even more ideal precise measuring device that can measure the simultaneous presence of N particles in N different, arbitrarily small regionsYes, this is true. QM does not talk about the specifics of observations and measuring devices. For example, P_k can be a projection on the k-th eigenvalue of the position operator. Then psi^* P_k psi is the probability (density) for measuring position k in the state described by psi. QM tacitly assumes that some ideal precise measuring device can be constructed, which does exactly that

covering the size of an uranium atom (N=92), say.

This is ridiculous - such measurement devices are impossible!

Whereas the form in which I stated the probability interpretation assumes nothing. it makes claims only for those projectors that are actually realizable. it is therefore much more realistic.

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