Suppose we have a wavefunction of a single particle in a potential. We measure it's position. After collapse the wavefunction collapses to a single eigenstate of position. This means that the wavefunction takes the form of a $\delta$-function in position space. But we all know that $\delta$-functions are not continuous, thus failing the law of quantum mechanics that wavefunction needs to be continuous and differentiable everywhere. Is something wrong with the concept of observation and collapse. Or there are special cases where the law may be violated?
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4$\begingroup$ Is a measurement of position ever made with perfect precision? $\endgroup$Buzz– Buzz ♦2024-10-24 20:07:12 +00:00Commented Oct 24, 2024 at 20:07
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$\begingroup$ @Buzz position of a particle is the value that we get after measuring position with an apparatus. We measure and get one and only one value . Isn't that the position of the particle? If yes then why is it not the precise position . And in quantum mechanics we don't consider that the instrument must be inaccurate. $\endgroup$Reader– Reader2024-10-25 04:11:40 +00:00Commented Oct 25, 2024 at 4:11
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$\begingroup$ Do you mean the apparatus gives the position value in a range of coordinates (depending on how many pixels the apparatus is made of )? Then also the wavefunction collapses to some discontinuous function . $\endgroup$Reader– Reader2024-10-25 04:22:16 +00:00Commented Oct 25, 2024 at 4:22
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2$\begingroup$ @Reader we are taught in QM classes that a measurement collapses the wavefunction to an eigenstate, but this is an idealisation that never happens in the real world. Measurement always involves some exchange of energy and momentum and for a position measurement to collapse the target to a position eigenstate would require infinite energy, so it never happens. Any real position measurement always produces a superposition of position states, but hopefully a tight superposition is $\Delta x$ is small. The position eigenstate is unphysical, just as the momentum state is unphysical. $\endgroup$John Rennie– John Rennie2024-10-25 05:47:10 +00:00Commented Oct 25, 2024 at 5:47
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1$\begingroup$ @Reader Correct! $\endgroup$John Rennie– John Rennie2024-10-25 07:19:45 +00:00Commented Oct 25, 2024 at 7:19
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2 Answers
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The problem is that the observable position $x$ is quite complicated. For example, the eigenvectors $|x\rangle$ are not normalizable and it representation in position gives a delta $$ \langle x^{'}|x\rangle = \delta(x-x^{'}). $$
This is one of the reasons why the free particle is not a simple case to study in quantum mechanics, but one of the weirdest. To circumvent it, usually people consider a discretized space, where $x$ can only assume values in a countable set. We have similar problem with momentum operator and momentum basis, where the procedure to get a discretized momentum basis is to impose periodic boundary conditions.
But all of this are not a flaw in quantum theory. This is just a consequence of our imposition about a well defined position $|x\rangle$ or well defined momentum $|p\rangle$ states. Any of these are physically meaningful; by Heisenberg uncertainty relation, the first should have infinite momentum uncertainty and the second infinite position uncertainty.
In practice, any of these states would be a result of a measuring process. What we get in a position or momentum measuring process is something physically more realistic and mathematically less problematic.
For all other quantum problems we solve, we consider interactions and bounded states, with gives us discrete eigenvectors and no such problem occurs.
Is something wrong with the concept of observation and collapse. Or there are special cases where the law may be violated?
The concept of observation and collapse is a matter of controversy for some (but not much) physicists. It is a non-continuous and non-deterministic process, where we have irreversible loss of information. Look that the time evolution of a quantum system, described by the Schrodinger equation, is continuous and deterministic, and this is why we impose that the wave function should be continuous. The continuity of wave function is a criterion of mathematical consistence in this case. But it could not apply for the measurement postulate.
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We measure it's position. After collapse the wavefunction collapses to a single eigenstate of position. This means that the wavefunction takes the form of a $\delta$-function in position space. But we all know that $\delta$-functions are not continuous, thus failing the law of quantum mechanics that wavefunction needs to be continuous and differentiable everywhere.
What leads to a contradiction here is the implicit assumption that we can have a perfect/ideal measurement process - which determines position with infinite precision. Such measurement process implies a non-integrable probability density - even if it were classical.
In practice measuring position means localizing a particle in a very narrow potential well - the width of which is the resolution of our measurement instrument (more precisely, the width of the eigenfunction of this state, a real physical potential well cannot be infinitely deep.)
Although QM and relativity can be viewed as theories about how we measure things, they do not encompass all the nuances of physical reality vs. mathematical description.
