2
$\begingroup$

I think this must be a very basic question but I couldn't find the answers anywhere. I was starting reading about Quantum Mechanics and these questions came in mind:

As I understand the quantum universe is considered stochastic, and the act of measurement is what makes the wavefunctions collapse into a single state, what is called wavefunction collapse, making the system randomly assume only one of its possible states.

Copenhagen says that the system transforms from a superposition of states to a single unique state.

The problem is shouldn't gravity make all particles keep measuring themselves and act as observers all the time (and thus making the wavefunctions keep collapsing and denying an uncollapsed state)? In the sense that all particles are in constant interaction with each other's gravitational fields.

E.g. The schrodinger cat for instance, say if he is dead he is emitting gravity from a lay down position and if he is alive from a straight up position. The problem I'm seeing is that if gravity carries information and it's a wave or force, then how a superposition of states is possible to exist?

Is gravity being a particle a requirement for superposition being true?

I hope my question was clear enough now, thank you!

$\endgroup$
  • 1
    $\begingroup$ Maybe this will help: physics.stackexchange.com/q/8295 $\endgroup$ – Nick Sep 22 '13 at 2:46
  • 1
    $\begingroup$ I honestly really don't understand the question at all. I don't understand paragraph #2, and my reaction to #3 is no, I don't see why it should, and you haven't given any reason why you think it should. $\endgroup$ – Ben Crowell Sep 22 '13 at 4:32
  • $\begingroup$ A related concept to your question is gravitational decoherence, arxiv.org/abs/gr-qc/0306084 $\endgroup$ – Tarek Sep 22 '13 at 9:49
  • 1
    $\begingroup$ Comment to the question (v4): It would be good if OP (or anybody else?) could motivate and explain the question better via an edit; possibly throwing in a reference, to ensure that OP and the readers are on the same page. E.g., what prestory lead OP to ask the questions Why doesn't gravity act as a measurement? and Shouldn't gravity make all particles keep measuring themselves? in the first place? $\endgroup$ – Qmechanic Sep 22 '13 at 16:17
  • $\begingroup$ I've tried my best to make it understandable. It's just that I'm used to debate a subject in order do understand it. Even if anything does not make sense I don't see why the question has to be closed instead of answered. $\endgroup$ – eJunior Oct 6 '13 at 13:03
5
$\begingroup$

No, gravity doesn't "cause" a measurement in any greater way then the electric force or another force. In fact, a difference is that between two particles, gravity is about $10^{40}$ times weaker than the electric force, so its effects – and your hypothetical "auto-measurement" effects – are negligible relatively to the electric force. None of them exists.

The idea that gravity makes the wave functions "collapse" was vaguely proposed by Roger Penrose and one may easily show that it is completely wrong. If such an effect existed, it would lead to the loss of coherence and interference in experiments we know to remain coherent and in the violations of the equivalence principle (all forms of energy must accelerate in the same way in the gravitational field).

The fact that the results of a measurement are "sharp" boils down to the fact that the wave function isn't a classical "objectively existing" wave but just a probability amplitude wave and the actual phenomena that exist, the results of measurements, are the sharp outcomes we know; the wave function is just a tool to predict the probabilities. The "sharp" results of the measurements have nothing to do with gravity or any other particular "mechanism" one could think of. There is no mechanism; the probabilistic character of the wave function is a fundamental postulate of quantum mechanics.

When you throw a die, the probability of each outcome is 1/6: $$(1/6,1/6,1/6,1/6,1/6,1/6)$$ When a particular outcome, like 4, is obtained, the probability distribution "shrinks" to $$(0,0,0,1,0,0)$$ But the shrinking isn't due to any "force". It's due to our learning about the result. In classical physics, one may imagine that the number 4 was already "guaranteed" before the die landed, due to determinism. But in quantum mechanics, it's not possible: the results are really random. But the "shrinking" of the wave function is still just about the change of the knowledge, not about the physical shrinking of any classical wave.

Another, related way to falsify your thinking is to point out that the wave function, because it is not a classical wave or field, doesn't self-attract. In fact, any such self-attraction would mean that the dynamics is nonlinear in the wave function. But linearity of the evolution operators as a function of wave functions is a principle of quantum mechanics, linearity, and it holds completely generally, including cases with gravity.

$\endgroup$
  • $\begingroup$ I see that the collapse has to do with knowledge being obtained rather than a force being applied. But if someone was capable of measuring the gravitational waves departing from a particle it would just as easily gather knowledge, right? The fact that it is a weaker force (please correct me if I'm wrong) would only mean that there is too much 'noise' involved to accurately measure a particle by its gravitational waves, is that right? But it wouldn't be rather a technological impairment than a physical one? [Sorry if what I'm saying doesn't make real sense, please enlight what is wrong] $\endgroup$ – eJunior Sep 22 '13 at 12:31
  • $\begingroup$ Hi @ejunior, the fact that the gravitational wave is so weak means (because of the rules of quantum mechanics) that normal bodies only emit a few gravitons per very long periods of time (gravitational waves come in gravitons $E=hf$ just like photons), that's on the source side, and it also means that it's very unlikely that a graviton will be detected and/or interacting with the apparatus. To summarize, it's very unlikely that a "measurement" will take place. Of course if you measure sufficient information about gravitational waves encoding the position, it will "collapse" but you likely won't $\endgroup$ – Luboš Motl Sep 23 '13 at 13:02
  • $\begingroup$ I see. Only an add: A particle doesn't necessarily emit fewer Gravitons because the energy of a graviton isn't actually measured, so no one can know for sure if 'fewer gravitons are emitted'. But it is a plausible explanation, if you consider gravity as a particle instead of a force or a dimension then there is 'time' between measures what opens space for uncollapsed states. Personally I dont like gravitons tough, but its a possible explanation for the problem. $\endgroup$ – eJunior Sep 24 '13 at 22:59
  • $\begingroup$ I'm curious about this from a more philosophical point of view and I've been bugged by it for a long time: If you say the only things that actually exist are the measured results, then does this mean that before the measurement was made utterly nothing existed at all? Does this also mean the universe has only existed as long as there's been humans around to observe it, so talk of universe as lasting "billions and billions of years" pre-dating humans is meaningless nonsense and thus whole other fields of science are wrong or hogwash? Or is it that the "measurement" we've made of everything $\endgroup$ – The_Sympathizer Jan 12 '18 at 9:31
  • $\begingroup$ up to this point by our observations has actualized it, and what we have can be described in terms of such a history, but it would still be meaningless to talk of that history as something which had actually took place before the measurement, something actually taking place in the absence of creatures capable of obtaining knowledge for that by definition requires an independent-of-knowledge existence at that point in time, and thus the universe again, only actually has existed for as long as humans have or at least suitably conscious enough organsims? $\endgroup$ – The_Sympathizer Jan 12 '18 at 9:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.