1
$\begingroup$

First of all, I'm an undergrad. student of engineering physics, so I must explicit my lack of formal knowledge on the subject and total confusion with its implications. I also understand it may be impossible to explain in words, so feel free to do the math.

I've read and had some classes introducing the subject of wavefunction collapse, eigenstates and similar subjects, and understand that an "observation" is a casual term for "measurement", which can be said to be a synonym of "interaction". I also understand collapse is an allegory of the mathematical implications we can derive from formulas involved and does not represent reality necessarily.

The problem arises here and I can't find a satisfactory answer on the internet. What EXACTLY constitutes an "interaction"? Don't all particles in the universe interact with fields and particles all the time (unninterrupted)? If so, wouldn't all particles have its eigenstates determined all the time by endless interactions (if not true, what are examples of non-collapsing interactions?)? And, if the previous question is true, does that mean every particle "experience" different collapses of wavefunctions of other particles compared to us (therefore, meaning we could have a "relativistic" collapse)?

$\endgroup$
  • 2
    $\begingroup$ You are wading into issues of quantum interpretations, but you should not think of every interaction (for example, between two electrons) as a measurement. Think of measurement as an interaction of a small quantum system with a large classical measuring device. Particles are interacting with other particles all the time. Measurement is when they interact with something like a Geiger counter. $\endgroup$ – G. Smith Dec 12 '19 at 23:50
  • $\begingroup$ Asking what is different about interacting with large classical objects gets into decoherence theory, or at least I think it does. I myself am content to “shut up and calculate” using the standard Copenhagen interpretation. $\endgroup$ – G. Smith Dec 12 '19 at 23:57
  • 1
    $\begingroup$ A simple example of particles “interacting all the time” without the wave function collapsing is a hydrogen atom. The proton and electron constantly feel each other’s electrostatic attraction. But an isolated hydrogen atom can exist (for awhile, at least) in a superposition of energy eigenstates without the wavefunction collapsing into one energy eigenstate. $\endgroup$ – G. Smith Dec 13 '19 at 0:06
0
$\begingroup$

Nikolas,

Any measurement requires an interaction between the quantum system and an appropriate instrument. However, not any interaction represents a measurement.

An interaction between an electron and a fluorescent screen allows you to determine the electron's position. So, the interaction involved here represents a measurement. The electron also interacts with Earth's gravitational field, but this interaction does not allow you to determine the position of the electron, so it does not represent a measurement of the electron's position.

"I also understand collapse is an allegory of the mathematical implications we can derive from formulas involved and does not represent reality necessarily."

This is true. It is possible to give the wavefunction a realistic interpretation (as some sort of field) but this requires a rejection of locality, which is a pretty well established physical principle. This is because the collapse is instantaneous. So, it is more reasonable to understand the wavefunction as an abstract methematical object that represents our incomplete information about the system. Each measurement results in an increase of information about the system so the wavefunction will "collapse".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.