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If you perform a measurement the wavefunction collapses to a certain position. Does this mean there is a very high probability current to that position? Comparable to an strong electrical current towards a central charge? With probability current I mean the following general definition: $$J(x) = \frac{\hbar}{2mi}\left( \psi^*(x)\frac{\partial\psi(x)}{\partial x} -\psi(x)\frac{\partial\psi^*(x)}{\partial x} \right) \tag{1}$$

Note this is not a homework question, I placed the equation and the image as examples.

enter image description here

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    $\begingroup$ Have you tried to do it yourself? Take for example a "sharp" (normalized) Gaussian, compute the current and plot it if needed. $\endgroup$
    – Tobias Fünke
    Feb 24 at 9:41
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    $\begingroup$ How on Earth is this getting close votes as a "homework-like" question? $\endgroup$
    – Michael Seifert
    Feb 24 at 12:43
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    $\begingroup$ @MichaelSeifert Well, the question, as in my understanding asks: "Given a specific form of $\psi$, what is the form of $J$?" - and as I pointed out in my comment above, this is just a calculation. There is nothing conceptual asked here, as far as I can see. $\endgroup$
    – Tobias Fünke
    Feb 24 at 14:06
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    $\begingroup$ @TobiasFünke: My reading is not that the OP is asking about what the probability current is directly before or after the collapse (which, as you correctly point out, can be easily calculated.) They're asking about whether there is a probability current associated with the moment of collapse. There's a conceptual question about whether this concept is well-defined in the first place, and whether we should expect such a sudden probability current. ... $\endgroup$
    – Michael Seifert
    Feb 24 at 14:22
  • $\begingroup$ Off the top of my head, I suspect that having $\nabla \cdot \vec{J} + \partial \rho/\partial t = 0$ could rely on the time-evolution operator being unitary, and since collapse is a non-unitary process we might not expect any version of the continuity equation to hold at all. $\endgroup$
    – Michael Seifert
    Feb 24 at 14:23

2 Answers 2

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If you perform a measurement the wavefunction collapse to a certain position.

The conceptual meaning of wave function collapse is that quantum phenomena are not accessible to direct observation - we can only observe behavior of "classical" objects, resulting from their interaction with quantum ones - so-called measurement. Collapse is an ad-hoc prescription for describing what happens when such interaction takes place (one might even call it a philosophical concept.) Such a prescription allows to test quantum behavior experimentally - making it a physical theory, rather than just an unproven mathematical conjecture.

Of course, any "classical" object is really a quantum object, usually a macroscopic one, that behaves classically for practical purposes. Thus, we can model this interaction from first principles, describing the "classical" object as a quantum one - then we do not need to invoke a notion of collapse at all... but it also implies that we have accepted correctness of quantum mechanics. In other words performing measurement of the wavefunction collapse does not make sense - we cannot measure a theoretical/philosophical concept.

Remark:

  • The discussions of wave faction collapse always pose a risk of circular reasoning: the collapse is needed to justify quantum mechanics, but once we believe QM to be true, we do not need the collapse... but without it the QM is not justified.
  • Knowledgeable people would point out that there are formulations of QM that avoid using the notion of collapse. This however confirms my main point: collapse is not a physical description of a process, but a theoretical ansatz, that should not be modeled. Moreover, any quantum theory still faces the same conceptual question of relating unobservable phenomena to the classical world accessible to our senses.
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  • $\begingroup$ Nature of wavefunction collapse is a thing of many debates, with the resolution depending on considered interpretations of QM. Some interpretations have it as a real process, some not. It is ok to ask what happens during measurements, and yes, it is good to note that what we suppose depends on interpretations of QM. $\endgroup$
    – M.S.
    Mar 25 at 14:52
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Collapse isn't part of quantum theory since it isn't compatible with quantum mechanical equations of motion such as the Schrodinger equation. If collapse happens then it will require a modification of quantum theory such as the GRW theory

https://arxiv.org/abs/0710.0885

What happens instead is that quantum systems evolve by local equations of motion so the state (and the probability current) changes continuously. A measurement is just a process that copies information from one system $S$ to a measurement device $M$. In general there are multiple versions of $S$ one for each possible measurement result and there is a corresponding version of $M$ for each version of $S$. The multiple versions of $M$ undergo a process decoherence that prevents interference between then and each version evolves approximately according to classical equations of motion:

https://arxiv.org/abs/quant-ph/0306072

So $M$ has a particular state relative to each version of $S$, which is called the relative state:

https://arxiv.org/abs/2008.02328

This way of looking at quantum theory is commonly called the many world interpretation, but since collapse would require a modification of quantum theory it should just be called quantum theory:

https://arxiv.org/abs/2205.00568

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    $\begingroup$ Wrong, this answer confuses it in a way common at /. Starting with 0710.0885 that is about spontaneous collapse, i.e. outside measurements. Ending with many-world interpretation that has severe issues. Notice that wavefunction collapse (at measurements) is a regular part of some interpretations, and it does not need a modification of QM. Only having the collapse outside measurements needs a modification. $\endgroup$
    – M.S.
    Mar 25 at 15:04
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    $\begingroup$ Wave function collapse isn't compatible with quantum mechanical equations of motion. So anyone who advocates collapse while using those equations of motion is engaged in making up an ad hoc theory on the fly in every calculation they do. $\endgroup$
    – alanf
    Mar 25 at 17:03
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    $\begingroup$ If a collapse were not spontaneous, it would be predicted by the initial conditions and therefore not referred to as collapse. If there were an issue with MWI which was more severe than an aesthetic one, this would imply the same for the Schroedinger equation. Because that's all MWI uses. $\endgroup$
    – Connor Behan
    Mar 25 at 17:19
  • $\begingroup$ @ConnorBehan There simply is no "collapse". You won't even find a proper physical definition of that word in the literature for all I know. There is a misunderstanding of the standard theory, though: it predicts the outcomes of measurements AFTER an infinite amount of time. That is what "measurement" means: a lasting, irreversible record of an interaction between a quantum system and the rest of the world. QM simply does not have a notion of "classical point-like events". Every "measurement" in quantum mechanics is, at least, a worldline with a starting point that extends to infinity. $\endgroup$
    – FlatterMann
    Mar 25 at 21:16

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