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If the only observer for the Schrödinger's cat experiment was a camera filming the box from the outside while the box was opened automatically without direct human intervention, and the only observation performed was through watching the recorded video ten years later, will the wave function collapse into one of the two states when watching the video for the first time or at the moment of filming it? Also following the many-worlds interpretation, will the universe “branch out” at the moment of watching the video for the first time?

The question is not limited to the Schrödinger's cat experiment, but to any other experiment where a function wave is supposed to collapse, e.g. double-slit experiment.

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    $\begingroup$ See also the "Wigner's friend" thought experiment. $\endgroup$
    – Kevin
    Jun 18 at 18:09
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    $\begingroup$ Considering that the light travels at a finite speed, you ALWAYS watch somewhat later. $\endgroup$
    – fraxinus
    Jun 19 at 12:55
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    $\begingroup$ If you don't like (or can't understand) a particular interpretation of QM, just ignore it. The basic feature of all "interpretations" is that there is no possible way to know which one of them is "correct". $\endgroup$
    – alephzero
    Jun 20 at 14:40
  • $\begingroup$ @alephzero "Shut up and calculate" has been a time-honored tradition. $\endgroup$ Jun 21 at 13:37
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The collapse of the wave function happens whenever the quantum system initially described by the wave function becomes entangled with environment — the part of the Universe that wasn't tracked by the wave function. This may be a human, but this could just as easily be a video camera. If the initial wave function described the system being watched and the camera, then the collapse happens whenever the state of both becomes entangled with something else, whose being a living thing is once again irrelevant.

Technically, the collapse just means that the initial subsystem no longer can be described by a wave function, because the subsystem has additional correlations with the environment.

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  • $\begingroup$ Is there a way to model these additional correlations in a simple system (ideally of few particles)? $\endgroup$
    – alessandro
    Jun 18 at 18:24
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    $\begingroup$ This answer is not wrong, but it happily dodges the interpretational issues that the question inevitably brings up. The first paragraph might make it seem like there is a well-understood mechanism that explains how entanglement with the environment causes collapse (there isn't), and the second paragraph is a low-key defense of MWI. $\endgroup$
    – Pedro
    Jun 18 at 22:08
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    $\begingroup$ I would simply emphasize that collapse is not a well-defined concept so it's not possible to give a precise answer to the question. Presumably collapse happens after decoherence i.e. entanglement with the environment, but a) it's unclear what exactly that means, and b) someone has yet to come up with a reasonable physical theory that explains how that happens. $\endgroup$
    – Pedro
    Jun 19 at 7:56
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    $\begingroup$ This is actually a deeply insightful answer; it shows how the formalism abstracts the system under consideration. There's no fundamental difference between opening the box (thus performing a measurement and revealing the outcome), and watching a video (thus performing a measurement and revealing the outcome). It's just a box within a box. The only difference is what's included in the system; and because there's an important aspect of the system that's QM in nature, the whole system may be described as a QM system, even though, internally, aspects of it can be described classically. $\endgroup$ Jun 19 at 11:15
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    $\begingroup$ @ChappoHasn'tForgottenMonica - In some sense I suppose that's true, but I'm not saying that the "box" is some feature of quantum mechanics, but that, when describing a phenomenon involving an element that has to be treated as a QM object (e.g., Schrödinger's cat involves a radioactive atom), a physicists can choose what belongs under the term "system" (can choose what to "put in the box"), and successfully treat the whole thing using QM, even though some aspects of that system can be understood without it from an inside perspective. $\endgroup$ Jun 26 at 21:24
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I agree with the answers given by Ruslan and Xcheckr. I would like, however, to caution against a common mistake of confusing of what an observer means in physics and philosophy:

  • in philosophy it means a human, since only a human can consciously process the observed information
  • in physics it means a certain physical effect that the observer has on the measured object. E.g., the Landau & Livshitz textbook specifies that any macroscopic object can be an observer.

Remarks

  • One may argue about what a macroscopic object means in this context or demand a more precise term - however the important distinction here is between the physical object and the mental process. It also often gets circular when one tries to describe precisely the nature of measurement in quantum terms... however, such a description is grounded in accepting QM and its measurement postulate.
  • This problem is known more generally under If a tree falls in a forest and no one is around to hear it, does it make a sound?
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    $\begingroup$ This is the best answer. "Observe" in this context means any interaction between the system and the external macroscopic world (resulting in the loss of quantum coherence). Nothing at all to do with human observing. +1 $\endgroup$
    – joseph h
    Jun 18 at 9:44
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    $\begingroup$ Note that even an individual particle "observer" is enough to destroy the inference pattern in the double slit experiment. It only requires 1 qbit to decohere. $\endgroup$
    – PyRulez
    Jun 19 at 20:07
  • $\begingroup$ @PyRulez this is not how I interpret it: if a particle is not observed, the interference is not destroyed. $\endgroup$ Jun 19 at 20:32
  • $\begingroup$ @RogerVadim you actually need to erase the quantum information of the particle, and even then the inference pattern is conditioned on some information generated by the erasure. See delayed choice quantum eraser. $\endgroup$
    – PyRulez
    Jun 19 at 21:09
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    $\begingroup$ @philipxy isn't this how theories are actually made, even though they look neat and shiny in textbooks half a century later? ;) $\endgroup$ Jun 20 at 13:55
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Collapse is not a physical phenomenon! It may be real in a metaphysical sense, but this is a physics website.

Collapse is a numerical tool. Here is how we use it:

We divide the "universe" into the system (the tiny thing we care about) and the environment (what we model classically). Collapse happens every time the system interacts with the environment or shoots a particle out into the environment (both of these are called "measurements"). The nature of said interaction determines how we collapse the wave function.

We have an accuracy-speed tradeoff: If we make the system too big, we waste time, but we still get the right answer. If we make the system too small, we risk getting the wrong answer: Our math assumes that whatever the system interacts with is not previously entangled with the system. Interactions between things generate entanglement, breaking this assumption. Fortunately, most environments will dilute this into insignificance for the "obvious" choice of what belongs in the "environment".

What about the cat? Let’s make the "system" anything inside the box. For any real box, the interactions with the environment through the walls (do you hear it scratching?) are so pervasive that the cat is either alive or dead in our model. It is, for all practical purposes impossible, to create or isolate a superposition of kilogram masses in drastically different macroscopic states.

But now let’s make the "system" the entire universe. The cat stays in a superposition of alive or dead forever, even well after the box is opened. The owner ends up in a superposition of happiness and grief over their kidnapped cat. This sounds contrary to our intuition indeed! But we currently have no better model!

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    $\begingroup$ What about the measurement postulate? Isn't collapse a part of making QM a testable theory? $\endgroup$ Jun 19 at 14:27
  • $\begingroup$ @Roger Vadim: QM is testable without collapse. Let your "system" include the entire machinery to make the particles for N experiments, send them through i.e. some Stern-Gerlach device, and count the fraction "up" and "down". You never use collapse, so you will end up with a superposition of states. But in the limit of large N the amplitudes will be concentrated in the "yes quantum mechanics is consistent with the fraction up" region. $\endgroup$ Jun 19 at 18:35
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    $\begingroup$ Interesting... however, what does one measure in your experiment? Besides, large N is essentially macroscopic limit - does one still observe quantum properties in this case? $\endgroup$ Jun 19 at 19:12
  • $\begingroup$ in this conceptualization is the camera part of the system or part of the environment? Typically we use the camera to track the system, but we cannot say it belongs to things we ignore, as we need and want a model of how the system interacts with the camera $\endgroup$
    – lurscher
    Jun 19 at 19:29
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    $\begingroup$ To track a quantum state one needs to accept quantum mechanics, together with the measurement postulate. Quantum behavior needs to have classical manifestation. $\endgroup$ Jun 20 at 5:59
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"Also following the many-worlds interpretation, will the universe “branch out” at the moment of watching the video for the first time?"

No, not quite. The Everett interpretation, which was somewhat misleadingly called the many-worlds interpretation by Wheeler when he popularised it, asserts that the quantum mechanics applicable at the microscopic level also applies without modification at the macroscopic level, and that quantum mechanics predicts that as quantum-mechanical observers we will observe what appears to be classical physics. A particle in a quantum superposition interacts with another quantum system, and the interaction causes the two to become correlated, so that the observer system is in a superposition of an observer seeing state 1 and an observer seeing state 2. The components of the superposition are orthogonal, and don't interact with each other, they can't see each other, and thus from the point of view of the observers it is as if each outcome happens in a separate universe. But this is no more a 'branching of universes' than an electron passing through two slits at once. The electron passing through one slit 'cannot see' the electron passing through the other (so for example they do not electrostatically repel). From the electron's point of view, the orthogonal components of the superposition it is in do not interact with one another; it is as if they are in separate worlds, or a sum over alternative histories.

When you see two sets of ripples cross over on a pond, they seem to pass through each other as if the other wasn't there. To the ripples, it is as if there were two ponds, with a different set of ripples on each. But there is only actually one pond. Both waves are in the same universe, in linear superposition.

The reason the components are orthogonal is related to a topic in classical physics called 'normal modes of vibration'. There is an interesting physics experiment where you hang two pendulums on the same piece of string stretched between two support posts. Start one pendulum swinging, and it gradually comes to a stop while the other starts swinging, and then the cycle reverses. This happens when oscillators are weakly coupled, their oscillations become synchronised. If the system of differential equations governing their state is written as a matrix differential equation, you can separate the coupled multidimensional joint states into a sum of independent one-dimensional oscillators by finding the eigenvectors of the matrix, which are orthogonal to one another. In each orthogonal state, the motion of one pendulum is correlated in some way to the motion of the other. We say that one pendulum 'observes' the other. The particular breakdown into orthogonal states is governed by the nature of the interaction - the coupling terms.

I'm going to digress here to explain what I mean about the normal modes, as it's often misunderstood. But it's not needed to understand the overall answer to the question. Feel free to skip. (Or delete, if you feel it doesn't help clarify.)

Two uncoupled 1D simple harmonic oscillators look like this:

$\left( {\begin{array}{c} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{array} } \right) = \left( {\begin{array}{cc} -K_1 & 0 \\ 0 & -K_2 \\ \end{array} } \right) \left( {\begin{array}{c} x_1 \\ x_2 \\ \end{array} } \right)$

We introduce interaction between them by putting values in the off-diagonal entries.

$\left( {\begin{array}{c} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{array} } \right) = \left( {\begin{array}{cc} -K_1 & p \\ q & -K_2 \\ \end{array} } \right) \left( {\begin{array}{c} x_1 \\ x_2 \\ \end{array} } \right)$

We can usually diagonalise the matrix $M = U^{-1}DU$ where $U$ is a unitary matrix of eigenvectors, and $D$ is a diagonal matrix of eigenvalues.

$\left( {\begin{array}{c} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{array} } \right) = U^{-1} \left( {\begin{array}{cc} -D_1 & 0 \\ 0 & -D_2 \\ \end{array} } \right) U \left( {\begin{array}{c} x_1 \\ x_2 \\ \end{array} } \right)$

Move $U^{-1}$ over to the other side:

$U \left( {\begin{array}{c} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{array} } \right) = \left( {\begin{array}{cc} -D_1 & 0 \\ 0 & -D_2 \\ \end{array} } \right) U \left( {\begin{array}{c} x_1 \\ x_2 \\ \end{array} } \right)$

And we substitute variables to find a superposition of $x$ states that behaves as a pair of uncoupled oscillators.

$\left( {\begin{array}{c} y_1 \\ y_2 \\ \end{array} } \right) = U \left( {\begin{array}{c} x_1 \\ x_2 \\ \end{array} } \right)$

so

$\left( {\begin{array}{c} \ddot{y}_1 \\ \ddot{y}_2 \\ \end{array} } \right) = \left( {\begin{array}{cc} -D_1 & 0 \\ 0 & -D_2 \\ \end{array} } \right) \left( {\begin{array}{c} y_1 \\ y_2 \\ \end{array} } \right)$

The $y$ states oscillate independently, each as if the other didn't exist, but each represents an $x$ state where $x_1$ is correlated with $x_2$. The $y$ states are called 'normal modes of vibration', and this sort of thing happens whenever linear wave phenomena interact. The quantum version is similar in principle, but more complicated, with a block-diagonal matrix to represent more complicated systems, but this is essentially drawing an analogy between simple harmonic motion and the Klein–Gordon equation.

End of digression.

So the radioactive particle is in a superposition of states, the cat in the box becomes correlated with it, to become a superposition of a dead cat and a live one, the camera becomes correlated with the cat, entering a superposition of a film of the cat dying and a film of an angry cat fighting to get out. If you can keep the camera sufficiently isolated from interactions, the observer viewing the film years later becomes correlated with the film only when it is viewed, becoming a superposition of someone seeing a film of a dead cat, and someone seeing a film of a live one. That's very difficult to do with film - atoms are constantly bumping into one another and interaction spreads, even if the observer system isn't aware of it. But it is easy enough to do with photons flying through empty space. Transmit the film as a TV broadcast from Alpha Centauri, and watch it four years later. The film in transit in space is to physics no different to the film stored in a box in the back of a cupboard. What matters is when the chain of interactions first reaches the observer. If you can arrange for that to be when the observer first sees the film, then that's when the observer splits. It doesn't matter if the cat interacts with the environment on Alpha Centauri, because that can't interact with the observer faster than light.

The Everett interpretation has no wave function collapse, and no actual splitting of universes. It just applies the ordinary rules of quantum mechanics that everyone accepts apply at the microscopic level and says the same rules apply everywhere, at every scale. It's local, deterministic, and realist. It makes no distinction between 'observers' and any other physical system, and it doesn't rely on consciousness, intelligence, or other vitalist nonsense to trigger unexplained and unobservable 'collapse' effects. But because each orthogonal component of a superposition does not see any of the others, it implies that most of the universe is forever unobservable to us, and people have philosophical objections to that!

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    $\begingroup$ This is amazing writing! $\endgroup$ Jun 20 at 14:10
  • $\begingroup$ +1. There are so many elegant explanations here! You turn the seemingly mysterious concept of quantum entanglement (well, for this physics amateur at least) into a relatively straightforward cascade of related superimpositions. :-) $\endgroup$ Jun 26 at 5:47
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The answer is definitely at the time of filming. In the modern quantum information approach to such questions, as long as the information about an observable is "out there" in the universe somewhere, it has already "collapsed". There need not be an external human agent that has to watch anything. Inanimate objects are enough. A lot of modern quantum information thus consists of questions about how to quantify information and how it leads to the observed "collapse".

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    $\begingroup$ Completely unrelated maybe, but isn't everything "out there" then? Does this not mean that any wave is at any time collapsed? $\endgroup$
    – arynaq
    Jun 18 at 8:47
  • $\begingroup$ @arynaq If the particles (or whatever) are interacting just with each other (in a quantum way), then an individual part of it might be considered collapsed (if you're being weird with definitions) but the system as a whole wouldn't be. If, however, it was an open system, with information about the system's state leaking out, then it would be said to have collapsed when the information leaked, because quantum processes no longer behave as though it's an isolated quantum system. $\endgroup$
    – wizzwizz4
    Jun 18 at 16:28
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    $\begingroup$ @arynaq To unpack slightly differently -- a wave function is a description of a system of particles. When the system has interacted with anything else to the degree that the wave function can no longer describe the system, then it has collapsed. If you were to write two wave functions, A describing the whole system and B describing just half of it, then B can be collapsed without A being collapsed. $\endgroup$ Jun 18 at 16:42
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The various answers here don't really do this question its fullest justice. There are two different notions of "wave function collapse" in "standard textbook quantum theory":

  1. One of these is that when a system whose quantum state is initially pure, becomes entangled with a larger environment, its state must now be described as mixed, if one wants to exclude the environment. That means instead of a single quantum vector wave function, we must use a density operator, mathematically. In terms of conceptualization, pure states are extremal points; mixed states are in the middle.

  2. The second of these is the von Neumann-Lueders "collapse postulate", which introduces a new primitive undefined term, "measurement", or "observation", into the quantum theory, in which a random replacement of one wave function with another occurs, representing a single observational outcome.

Both of these exist together in "standard textbook quantum theory". The question is what, if any, relationship there is between these two things, and what is the significance of the undefined term "measurement" or "observation". Your concerns relate specifically to the second, in fact. Whether it can be dispensed with is subject to a lot of debate. My own, though hardly unique, viewpoint is that it can't. In the classical limit of quantum theory, the von Neumann-Lueders collapse looks an awful lot like the "gain of information" in Bayesian probability, and moreover becomes indispensible to make sense of what we're seeing as "really being classical mechanics", and thus I think it makes sense that this interpretation should be retained in the non-classical regime as well, because the structure of the mathematical formalisms are identical; the only difference is whether $\hat{x}$ and $\hat{p}$ commute or not or, equivalently, if $\hbar$ is or isn't zero.

So then to answer your question about the camera. The question is whether and where a von Neumann-Lueders collapse should be instituted here. If we take the viewpoint that it can be dispensed with, then of course the answer is "no", we shouldn't have such a thing, but if we take the "knowledge acquisition" view, then we have to further elaborate that the wave function represents the knowledge or information held by a particular agent, and so we must specify what agent that is. There is nothing that says that agents must be conscious: quantum agents need only be able to possess and acquire information, and thus we can perfectly as well regard the camera as an agent. If we do, the wave function - which represents the camera's "knowledge" or available information - "collapses" every time the camera takes a frame and records it. But if we regard the agent as a human looking at the camera, then the wave function - or better the density operator, as we cannot necessarily say the human's knowledge is extremal - should collapse when the human gets information from that camera, i.e. looks at the video.

That said, if we are talking about the first sense of "collapse", this "collapse" is actually not a discrete event: it's something that can happen gradually because the evolution from the non-entangled to entangled configuration is fully continuous.

Where things get interesting is if we try to model a system/agent pair as a single quantum system. Then we find that as the agent goes through its knowledge-acquisition process, which we should model by the physics of how the agent operates, a "collapse" of the form of type 1 occurs throughout that, and we also know from its subjective viewpoint that it should see a vN-L collapse with a single outcome. This suggests there is some relation between the two things. Yet, the full state will still be a pure superposition involving multiple outcomes. This is the famous "Wigner's friend" problem.

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TL;DR


Whenever the box is opened. - after $10$yr if the camera was inside the box, immediately if not.

Innocent viewer


...was a camera filming the box from the outside while the box was opened

If the camera is assumed outside the box, the poor$^1$ cat died (or lived) the moment the box was opened. That's even before the camera could click a picture. That's because the whole point of the box was to hide any possible determinism of the state of cat i.e. allow for quantum superposition. In other words, the box demarcates the system which is undergoing quantum mechanical evolution. Everything outside is the surroundings.

It doesn't matter that "the only observation performed was through watching the recorded video ten years later". When the box is opened to the surroundings, the system has been observed - the cat aka the system must take a stance. If we consider the camera alone to constitute surroundings, it's act of recording the state of the box is an implicit act of observation. The collapse occurs then. It doesn't matter that the footage hasn't been viewed - the camera is the surroundings and it has observed.

The unwarned viewer who would stumble upon the possibly sad footage a decade later, is blameless.

This also highlights that the observer need not be a living being - any observation by any observer$^2$, regardless of their nature will cause wavefunction collapse. Since all such observations involve opening the box , it is the opening of the box to any surroundings and much less it being recorded that makes the collapse inevitable.

To be precise


the collapse of the wavefunction - the observation of an eigenstate- must be assumed the moment the Hamiltonian describing the system is no longer valid - this happened the moment the system was exposed to the surroundings. The earlier Hamiltonian must now include an interaction term with the environment and can no longer tread its earlier evolution trajectory - this sudden discontinuity is the collapse.

Spirit of the question


If instead I go with the spirit of your question and let the camera be considered part of the system (say the original box and the camera both lie enclosed in a bigger room), the wavefunction$^3$ collapse will occur the moment the footage became available for observation to outsiders. If it involved opening the (stale) room a decade later, the collapse occurs then. If it involved a near instantaneous live stream, the collapse occurred the moment the livestream began. If the transmission was pre-scheduled to occur a decade later, the collapse occurs at the start of transmission.

Interestingly, in this case the camera itself would exhibit quantum superposition. Unlike the previous case, the camera must simultaneously posses the recording of both fates of the cat - only one of which is seen during a viewing of the footage in an ensemble of rooms.


$^1$ it's been put in radioactive environments for almost a century now.

$^2$ this includes any matter or radiation that can interact with the contents of the box and whose influence wasn't already included in the Hamiltonian, as discussed later

$^3$ this wavefunction is different from earlier as it involves the camera-box coupling. The collapse of this entire wavefunction is what is being talked about now.

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