0
$\begingroup$

In quantum physics, the configuration of a particle is fully defined by its wave function. When a measurement of a particular observable ( e.g., position, angular momentum, etc.) is made on the particle, its wave function collapses to one of its eigenstates (of the operator used to represent the observable) whose corresponding eigenvalue gives the result of the measurement. What happens if the measurement of two observables is made simultaneously? Does the wave function collapse into a linear combination of two eigenstates (one for each of the two operators corresponding to the observable)? How do we get the result of the measurements, then?

Note: I know that if the operators do not have a common eigenfunction, the corresponding observables cannot be measured simultaneously with accuracy ( Uncertainty principle), but I would like to know the scenario in terms of collapsing wave functions.

$\endgroup$
3
  • $\begingroup$ To be clear, are you asking about the simultaneous measurement of compatible observables or of incompatible ones? If the latter, then you're asking, to the letter, "I know this is a mathematical impossibility, but how do you describe it mathematically?". $\endgroup$ Mar 14, 2018 at 12:51
  • $\begingroup$ I am asking about both the cases along with calculations for each $\endgroup$
    – D.K.
    Mar 14, 2018 at 13:52
  • $\begingroup$ Measurement is irreversible energy, momentum and angular momentum transfer. The reason why we can't make two incompatible measurements at once is because the required energy, momentum and angular momentum transfer for one measurement necessarily changes the energy, momentum and angular momentum of the quantum system required to make the other. $\endgroup$ Oct 17, 2022 at 19:18

1 Answer 1

3
$\begingroup$
  • If the observables are compatible then you just project on their shared eigenfunctions.

  • If the observables are incompatible then they are incompatible, period. It's not a question of whether you can "observe them simultaneously with accuracy" or not: if the observables are incompatible then there isn't a shared eigenprojector and the very notion of simultaneous measurement is meaningless. You can't say what "happens" to the wavefunction because the scenario is nonsensical to begin with.

$\endgroup$
4
  • $\begingroup$ But we can measure two incompatible quantities with some uncertainty, right? How is measurement possible if the the wave function does not collapse? $\endgroup$
    – D.K.
    Mar 14, 2018 at 15:11
  • 2
    $\begingroup$ @D.K. No, you can't perform simultaneous measurements of incompatible quantities. You can perform partial measurements of a given quantity and then subsequently perform a measurement of other (possibly incompatible) quantities, but that is not what your question's phrasing entails. $\endgroup$ Mar 14, 2018 at 16:06
  • $\begingroup$ can you elaborate on what you mean by partial measurement? $\endgroup$
    – D.K.
    Mar 14, 2018 at 17:21
  • $\begingroup$ @D.K. If you want to perform an actual location measurement, then you have to restrict the system with an aperture, which will change the momentum of the system by an unknown amount, so your "second" momentum measurement will have a different result than one that was not disturbed by the location measurement. In case of compatible variables we can arrange the necessary measurement interactions in such a way that a single detection gives us the "correct" information about both. The only "surprise" is how simple the mathematics for all of this is (commuting operators). $\endgroup$ Oct 17, 2022 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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