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Answers to questions like this discuss how real measurements retain uncertainty due to limitations of our instruments.

Is this uncertainty quantum or classical?

If it's classical, i.e., the wave function truly collapses into a single eigenstate but due to our instrumentation we cannot detect exactly which one and need to use statistics/ mixed states to model our classical uncertainty, as suggested here, this seems highly unphysical as it suggests, e.g., after observing a particle's momentum, even with measurement uncertainty about the particular momentum eigenstate, it is truly (ignoring boundary conditions) in a single momentum eigenstate and delocalized with equal probability across all space.

This is, of course, not what we experience in day-to-day life. I can observe the momentum of particles and, even with tremendous measurement uncertainty, not see them immediately delocalized across all space as would be expected if my observation truly collapsed the wave function to a single momentum eigenstate.

If this uncertainty is quantum, and observations in reality only sharpen wavefunctions around a particular eigenstate while retaining contributions of "nearby" eigenstates, it would seem we need new operators corresponding to this sharpening phenomenon. Projecting with our existing operators, which collapse wave functions to single eigenstates, don't give us the sharpening we're looking for. How do we fix our operators so that the Born rule and other postulates of QM still apply, while making projection consistent with our sharpened but still multi-eigenstate states?

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Is this uncertainty quantum or classical?

The measurement uncertainty is classical, estimated with classical statistical mechanics etc.

If it's classical, i.e., the wave function truly collapses into a single eigenstate but due to our instrumentation we cannot detect exactly which

The "collapse"is a sloppy way, in my opinion, to say that new boundary conditions because of measurement will need a new wavefunction solution.

If your are talking of eigenstates of the momentum operator, the plane waves are defined up to infinity and the probability of measuring the momentum would be practically zero.

You are missing the concept of wave packets. If we want to model a real particle as it is going through in the bubble chamber, we would have to use a wavepacket to describe it. Thus it is no longer one "eigen" function , the momentum has a spread.

Here is a bubble chamber photo , where the location of the particles leaving the basic vertex, which is the quantum mechanical interaction under study, are seen like classical particles, the small ionization bubbles their charge leaves in the chamber, the footprint.

bubblec

We do not need to use the wavepacket description when measuring the momentum, because the measurement errors in space are orders of magnitude larger than the space probability distribution of the wavepacket describing the particle quantum mechanically (think of the Heisenberg uncertainty). So we treat the charged particles with classical electrodynamics equations and the energy losses classically, and measure the momenta from the curvature in the imposed magnetic field. We accumulate events like the one above in large statistical numbers to study the quantum mechanical crossections of interactions and decays.

The usual eigen-function solutions belong to simple solutions of quantum mechanical equations of bound systems, and their wavefunctions cannot be measured, only the spectra which have a width of the quantum mechanical energy uncertainty of the eigen levels.

Wave functions are unmeasurable, only the probability distributions given by the complex conjugate squared is measurable.

I can observe the momentum of particles and, even with tremendous measurement uncertainty, not see them immediately delocalized across all space as would be expected if my observation truly collapsed the wave function to a single momentum eigenstate.

This is a misunderstanding in the case of the momentum operator eigen-function, as described above. Unique states can be defined in the solutions of quantum mechanical equations. The wavefunctions of the hydrogen atom which give the unique spectra, are not plane waves. Eigen functions can have cvarious shapes.

Actually plane wave solutions of the quantum mechanical differential equations are used to construct the Quantum Field Theory , which allows us to calculate crossections and decays for particle interactions and on which the standard model of particle physics is built.

QFT needs graduate studies.

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  • $\begingroup$ The last sentence with a link to MIT seems a bit unnecessary. $\endgroup$ – MannyC Mar 6 at 9:29
  • $\begingroup$ @MannyC it is an open course $\endgroup$ – anna v Mar 6 at 10:27

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