Given the fact that every wavefunction exists everywhere in space, shouldn't a measurement at any location cause all wave functions to collapse since a measurement at any point measures all wavefunctions?
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Qmechanic
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Chris Laforet
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What does "measurement at any point" mean? What is the nature of that measurement? – Lourenco Entrudo Nov 10 '23 at 02:22
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For instance, you might fire photons at an atom to locate the electron in the atom, but wouldn't that photon measure the wavefunction of that electron wherever the photon is (because the electron wavefuction is everywhere, so isn't not finding the electron at any location in space the same as finding it somewhere specific)? – Chris Laforet Nov 10 '23 at 02:32
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Wavefunction collapse is, so far as we can guess, a function of some macroscopic measurement apparatus significant perturbing the system to the point that it crashes down probabilistically into one of the eigenstates. But if we do not sufficiently perturb the rest of the universe, why should something similar happen to the rest of it? – Matt Hanson Nov 10 '23 at 02:43
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The photon-electron system will become an superposition of entangled "photon finds electron here" and "photon fails to find electron here", over all possible "here"s. It will thus look as if the "measuring" has "collapsed" the wavefunction, each in its own internally consistent way, over many different ways. – naturallyInconsistent Nov 10 '23 at 03:52
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This may be of interest - Parallel Worlds Probably Exist. Here’s Why – mmesser314 Nov 10 '23 at 04:04
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There is nothing wrong with Laforet's question, @hft answered it more or less but not completely, see SenorO's comment. Maybe the right answer is that nobody really knows and it is unanswerable, I sure cannot but that should not be the reason to close it. – hyportnex Nov 10 '23 at 08:35
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https://physics.stackexchange.com/questions/73867/state-collapse-in-the-heisenberg-picture/786980#786980 – alanf Nov 10 '23 at 08:44
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To the extent the question can be made sense of, the answer is no. To the extent the question can not be made sense of, the answer does not matter--this is why the simplest answer is just "no." Some ostensible "questions" are just unanswerable muddles. When I say a "muddle" I mean something like a grouping of words and punctuation that looks syntactically correct, but is semantically disturbed/deranged/absurd. – hft Nov 10 '23 at 16:51
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I feel like these muddles usually arise when a questioner dearly desires for some abstract concept to be directly demonstrable--but not all concepts are directly demonstrable. Something similar is happening here, I think, where OP seems to want the wavefunction to be some physical thing such as the physical system being studied. But the wavefunction is a tool that describes the physics of the physical system, it is not "the physical system itself." – hft Nov 10 '23 at 16:57
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Given the fact that every wavefunction exists everywhere in space, shouldn't a measurement at any location cause all wave functions to collapse since a measurement at any point measures all wavefunctions?
No. Or, alternatively, this question has no answer since it is just a muddle.
Wavefunctions are used to describe microscopic systems, such as atoms or molecules. Wavefunctions provide us with as much information as we could know about the given microscopic system (provided we can solve the governing equations), but wavefunctions are not the microscopic system.
Also, wavefunction "collapse" is not a physical collapse. It describes a change of our knowledge of the given microscopic systems (due to the measurement result being obtained) that we force into our equations "by hand."
For example, a wavefunction $\psi(\vec r)$ might describe the probability amplitude to measure the position of an electron in a hydrogen atom. If you measure the position of that electron in that hydrogen atom, you could measure any value, say, $\vec r_0$. (Any sets of three real numbers are the eigenvalues of the position operator). If this $\vec r_0$ is actually measured (and the probability density that this would happen is $|\psi(\vec r_0)|^2$), then the wavefunction "collapses" to the state $|\vec r_0\rangle$. This just means that, if we would like to further describe the dynamics of the microscopic system, we need to evolve the state $|\vec r_0\rangle$, not the state $\int d^3r \psi(\vec r)|\vec r\rangle$.
So, that specific wavefunction "collapsed." All this means is that our new description of the system will use the prior information. We measured $\vec r_0$ and so we know the state right now is $|\vec r_0\rangle$. And we know after an additional amount of time $t$ (if we don't measure anymore) the state will be $e^{-i\hat H t}|\vec r_0\rangle$.
hft
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This is all a good correction of OPs understanding of wavefunctions, but it doesn't really answer their question. Their question essentially boils down to (leaving errors in), "if a non-interaction of a particle through a non-zero $\Psi(\vec{r})$ collapses a wave function, then doesn't a non-interaction collapse every wave function since every wave function is non-zero everywhere?" – Señor O Nov 10 '23 at 05:52
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muddle or not, your answer is not quite the same as that of @benrg who appears to believe that the "collapse" is real if the answer is "found here", so which is the right one and why? As a failed engineer, I have no opinion just am curious... – hyportnex Nov 10 '23 at 19:56
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@hyportnex We are both answering "no." I don't think benrg is writing that the wavefunction is "real" any more than I am. I mean, the wavefunction is "real" in the sense that I write it down and I use it to calculate. It is "real" in the same sense that the glyph "2" is "real." But the glyph "2" is not the same as two birds in a bush. I could use the glyph "2" when I write down how many physical birds there are in the bush. When one bird flies out of the bush I can write the glyph "1" to indicate how many birds remain. This doesn't have any effect on any other birds anywhere else in the world. – hft Nov 10 '23 at 20:04
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I mean, there might be a non-absurd question buried somewhere inside OP's stated question. But for the most part I think that OP's question has been addressed as much as it can be addressed. – hft Nov 10 '23 at 20:06
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In a comment you clarified your question:
(because the electron wavefuction is everywhere, so isn't not finding the electron at any location in space the same as finding it somewhere specific)?
No, it isn't the same. Realistic position measurements (unlike those in undergraduate QM) have only a small number of possible outcomes, perhaps even as small as two: "found here" and "not found here". The wavefunction collapse is a projection to the eigenspace associated with the measurement outcome. The eigenspace for "not found here" is very large since it includes everywhere else the particle might be, so the projection has correspondingly little effect.
The collapse after a "found here" result will turn the wave function into a narrow spike at the location of the measurement device, but the collapse after "not found here" will just force the wave function to zero at the location of the measurement device, without localizing it elsewhere.
benrg
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What does "collapse" mean? Depending on your formalism, yes, any measurement of anything will "collapse" the wave function of the universe into an eigenspace of that measurement. But surely that eigenspace is considerably more than one-dimensional.
WillO
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