I know this is probably a common point of confusion, but I have a specific question about measurements in Quantum Mechanics. I read an explanation on this, but still have a point of confusion.

The explanation of why a measurement somehow affects what occurs in a quantum experiment, for example the double slit experiment, seems to be that by measuring, even just by watching, we interact with the system and cause the "wave function to collapse". But it seems to me, with such a broad definition of "measurement", all particles, everywhere in the universe, would in some small way be measured at any given instance.

This answer explains that light is a form of measurement, but light, after all, is not the only way to perform measurements, as we very often measure things purely with gravity, and would all particles in the universe not be subject to (and the source of) some trace amount of gravity? Or have some interaction with some other particle, in some way shape or form? It seems like the answer would be yes. So it seems like we would never be able to observe an experiment without that collapse.

With "measurements" in quantum mechanics, interactions per se, why are they not always occurring?

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    $\begingroup$ I find it astounding that no answer so far (except those in the link I proposed as a duplicate question) even refers to the measurement problem. The simple answer to your question is: no-one knows, it is a very famous, and long-standing, open question. $\endgroup$ – Stéphane Rollandin Oct 9 '18 at 8:30
  • $\begingroup$ measurement is still a cutting-edge research topic. measurement is not interaction, some ppl think measurement may have sth to do with entanglement. $\endgroup$ – Shing Oct 9 '18 at 8:37
  • $\begingroup$ This feels like a semantical argument. When you say "all particles, everywhere in the universe, would in some small way be measured", you're inherently limited to particles that are observed in any capacity, at which point your conclusion becomes a logical inevitability. If, instead, you include particles that aren't observed; then how would you know that the wave function collapses or not, since these particles are not actually observed? $\endgroup$ – Flater Oct 9 '18 at 10:14
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    $\begingroup$ @StéphaneRollandin Decoherence (which is not controversial: it can be derived from the pure quantum evolution of the joint system+environment and has been observed in the lab) goes a long way toward describing how the "collapse" occurs. Yes, there remain subtle interpretational issues, so I wouldn't say that decoherence fully solves the measurement pb. But I don't think these issues are relevant to the much more practical question of how can we possibly design experiments which avoid collapse inbetween intentional measurements. $\endgroup$ – Luzanne Oct 9 '18 at 13:30
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    $\begingroup$ I disagree with the current duplicate tagging: the answers there do not address the question of how uncontrolled collapse is avoided in real world qu experiments, nor do the answers here address the broader interpretational implications of the proposed duplicate. Not all qm questions that use the M-word should be deferred as "unanswerable w/o a full solution of the measurement pb". If it really has to be marked as duplicate, then maybe of this one: Why doesn't gravity mess up the double slit experiment? $\endgroup$ – Luzanne Oct 12 '18 at 18:21

What you describe is the process known as decoherence: any interaction of a quantum system with its environment (e.g. with photons or other particles passing by, and, yes, most likely interacting through gravity, although we don't have a theory to fully describe this yet) has the potential to destroy its genuinely quantum nature, turning quantum superpositions into mere classical statistical ones. This process is indeed the first half of a measurement, the second half being the reading out of the result which resolves the remaining statistical superposition into a single result.

But decoherence is not an all or nothing, instantaneous process: it is progressive in time, and the weaker the interaction between a system and its environment, the slower it will decohere. When we actually do a measurement we deliberately arrange for the interaction to be strong enough and we wait long enough for full decoherence to occur, so that a result can be obtained. But in between deliberate measurements, we can arrange for decoherence to be so weak as to be negligible, at least for the duration of the experiment, so that the evolution is (almost) truly quantum. It's relatively easy for, say, single atoms at very low temperature, but it becomes harder and harder the bigger the system is (it is for example a well-known and very real hurdle to design quantum computers with enough qubits). In practice, gravity is not usually the limiting factor here, because it is such a weak interaction.

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    $\begingroup$ What do you mean by gravity being a weak interaction in this context? Assuming gravitons exist, the probability of any particle interacting with them is very high, because we are in a sea of gravitons. In an accelerator, before a particle hits the target, it hits a gazillion gravitons on its way. If there were only one graviton, it would collapse the wave function of the particle. It is the collective nature that makes a difference, not the "strength". How would your argument change for the Higgs interactions? They are not nearly as weak as gravitational and also are happening constantly. $\endgroup$ – safesphere Oct 8 '18 at 20:41
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    $\begingroup$ @safesphere I am not an experimentalist, but my understanding is that in practice people doing this kind of experiments (say quantum computing) have much more severe sources of decoherence to worry about before worrying about gravity (the coupling constant of which is extremely small compared to say electromagnetism) or Higgs processes (which have very low probability at low energy, due to the large mass of the Higgs). $\endgroup$ – Luzanne Oct 8 '18 at 20:54
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    $\begingroup$ @safesphere The fact that processes happen continuously say nothing about their strength. And, at the risk of repeating myself, the strength of the coupling does matter, because decoherence is not an all-or-nothing, instantaneous event: it is a process than happens over time, slower or faster depending of the strength of the coupling with the environment. see also physics.stackexchange.com/a/228766/132157 $\endgroup$ – Luzanne Oct 8 '18 at 21:59
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    $\begingroup$ @safesphere: Regarding "the probability of any particle interacting with them is very high, because we are in a sea of gravitons.": You are currently in a sea of neutrinos, roughly 10^11 neutrinos pass through your thumbnail per second. However, the strength of neutrino interactions is so low, the probability of a single lifetime interaction anywhere in your body is about 25%. The number of particles isn't enough to overcome their individual irrelevance. $\endgroup$ – Eric Towers Oct 9 '18 at 15:00
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    $\begingroup$ @EricTowers I am not sure what is so hard to understand here. If you are near a supernova, you'd be vaporized by neutrinos in an instant. Everything is relative. The important part is the physical effect. Solar neutrinos have none, why should this be surprising? Is the magnitude of the $10^{11}$ number supposed to be impressive somehow? This number is simply insufficient for neutrinos to have a tangible effect. In contrast, the Higgs does cause mass and gravitons do cause gravity (assuming they exist). So their number vs. strength is sufficient. So what exactly is your objection? $\endgroup$ – safesphere Oct 9 '18 at 17:59

This topic gave me trouble as well. The fundamental basis for answering it is to look at decoherence.

Basically, any interaction in quantum mechanics yields the expected coherent result that comes from two particles interacting. Often this leads to an entanglement of their states. If we had constructed the particles with a known previous state, we can make statements about the state of the particles (such as probabilistic statements about momentum or spin).

However, what if we don't know any information about one of the particles? What if it came in from the outside environment? As such, we have no knowledge of the state. The best we can do is talk about its state as a random variable and apply statistics. The result is a density function showing the probability that our particle under test is in any given state.

Do this enough times with particles whose state is Independent and Identically Distibuted (IID), and the "quantumness" of the particle starts to go away. As the number of interactions goes up, the central limit theorem starts to apply, and the variance in the predicted resulting state diminishes. Eventually, when the variance is low enough, we start to say the particle is "measured" and that it has a state that matches the expected value.

This is, of course, a relatively new viewpoint. The original use of measurement was in explaining how the unusual quantum world could interact with the "classical" world, and in particular with classical beings such as us human beings. This has lead to the famous interpretations of Quantum Mechanics. Decoherence is another way to explain this effect. Instead of offering the philosophically perfect measurement of one of the interpretations, it offers a statistical process whose limit is the same as the predicted results of the other interpretations.

Of interest may be the concept of weak measurements. Weak measurements are designed to provide some measurement while retaining most of the quantum coherence.


Not every interaction is measurement or collapses the wave function. When light reflects off a mirror, the phase information is preserved. As each single photon hits the mirror and scatters on an electron, the photon doesn't hit the mirror in just one point or interacts with just one electron. Instead each single photon hits the entire mirror and interacts with all electrons in the mirror. In other words, due to the uncertainty principle, the interaction is a superposition of interactions with every electron in the mirror. This uncertainty preserves the wave function of the photon from collapsing.

The same concept applies to other collective processes, including the photon's travel through space, whether flat or curved by gravity. If the photon is allowed to take any trajectory, then the photon takes all of them simultaneously with different probabilities and therefore acts like a wave. In this case the photon's trajectory through space is a superposition of all possible trajectories. Therefore gravity does not collapse the photon's wave function (at least while away from black holes).

Furthermore, certain particles have a low probability of interaction, e.g. neutrinos that can fly through the universe as through a virtually empty space. Also, the hypothetical particles of dark matter may not nteract at all other than via gravity while the gravity interactions almost always would be a collective process described above that would not collapse the wave function.

Science is about predicting practical results. Your question however seems rather hypothetical. Whether the answer is yes or no, there seems to be no practical difference either way. Finally, quantum mechanics alone does not describe the universe as a whole. This requires quantum gravity to view spacetime as a function rather than a set of independent variables and effectively make this world a projection. Thus your question cannot be fully answered until quantum gravity has been developed.

  • $\begingroup$ Can you clarify "due to the uncertainty principle.." This principle is normal taught within the context of measurement. What meaning are you giving to it here? thanks. $\endgroup$ – Bruce Greetham Oct 8 '18 at 19:56
  • $\begingroup$ @BruceGreetham The uncertainty principle is a reflection of the wave properties of matter. For example, in a double slit experiment, it is uncertain, which slit the particle passes through. In fact, if you measure, you would kill the uncertainty along with the wave effects of the interference. Here it is the same concept. With the slits, a photon passes through all of them (no matter how many) simultaneously. With the mirror, a photon hits it in every point. You can view this as uncertainty of position per the uncertainty principle. $\endgroup$ – safesphere Oct 8 '18 at 20:42
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    $\begingroup$ @safesphere I am not convinced by your theory that collective processes cause no decoherence. Do you have any references to back up your claim? The textbook example of decoherence, namely a system interacting with a thermal bath of harmonic oscillators, is certainly a "collective" process, yet decoherence does occur in this case. $\endgroup$ – Luzanne Oct 8 '18 at 20:51
  • $\begingroup$ @Luzanne so are you saying here that the photon wavepacket does a collective interaction with the uniform mirror or the double slit (or empty space) : these are all examples of collective processes which don't decohere. But then the photon hits the photoelectric plate and does another collective interaction with all the electrons in the plate. But this time there is decoherence so now the photon gets measured. $\endgroup$ – Bruce Greetham Oct 8 '18 at 21:17
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    $\begingroup$ @Luzanne I did not say that any collective process preserved the wave function. Only those processes do, in which the collection (e.g. the mirror) does not change its state. A very thin mirror affected by light would not reflect it the same way. As a reference please see this answer: physics.stackexchange.com/questions/368333/… $\endgroup$ – safesphere Oct 8 '18 at 21:55

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