When a wave function is said to "collapse" to a single point during a measurement, is there uncertainty about the point's position or is it known infinitely precisely?
Asked
Active
Viewed 148 times
0
-
It is known precisely, although in practice one cannot construct experiments that measure to arbitrarily precision. – Charlie May 28 '21 at 16:04
1 Answers
3
"Wavefunction collapse" doesn't mean collapsing to a single point. The wavefunction is always defined at all points in space.
A better understanding is that if you were to make a position measurement of your system, then the wavefunction "collapses" to a state with definite position. This wavefunction is $0$ everywhere except at the location where the particle was observed to be at.
Now, the above is an idealization. States with definite position are not physical. So yes, there will be some inherent "uncertainty", and the "collapsed wavefunction" will be a spike of finite width centered at the location the particle was observed to be at. As before, the wavefunction is still defined at all points in space.
More generally, measurement of any observable (not just position) causes the system's state vector to "collapse" to an eigenstate associated with the measured value of the observable. So once you go more general, "collapse" doesn't refer to any sort of physical (position) space at all.
BioPhysicist
- 56,248
-
So in actual physicality , the measured wave function will still be a combination of multiple eigenstates but heavily weighted at some point, am I correct? – Sidarth May 28 '21 at 20:27
-