2
$\begingroup$

Consider a particle in a box system. Assume its state to be a superposition of the ground and the first excited energy states. Consider two observers A and B (rest of the world). A made the measurement of the energy of the system and got energy corresponding to one of the states. Consider two scenarios:

  1. A made the measurement and B is not aware of it at all. For B would the state still be the superimposed state?

  2. A made the measurement and B know(s) but is unaware of the result. Would it be ok to say that for B the state of the system is still the same as it was initially?

Are these two scenarios equivalent?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ In written English, spaces are used after the period at the end of each sentence. I have edited the question for you to adhere to this convention. $\endgroup$
    – DanielSank
    Commented Mar 26, 2015 at 5:11
  • $\begingroup$ This question is related to the question of Wigner's friend. Wigner's conclusion from this thought experiment was, it must be said, rather unorthodox and isn't generally accepted as a resolution to the question. $\endgroup$
    – Michael Seifert
    Commented Jul 8, 2015 at 17:59

4 Answers 4

Reset to default
1
$\begingroup$

It sounds to me like you're asking about "Quantum Bayesianism", or "QBism" as it's called by David Mermin, one of its principal proponents, e.g., http://arxiv.org/abs/1311.5253 Or you can google those two terms in quotes for lots more references. Is that what you're talking about? And do their interpretations answer your questions?

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ This is more like a comment. Please add this below the question, not as an answer. $\endgroup$
    – rmhleo
    Commented Aug 14, 2015 at 5:23
  • $\begingroup$ Sorry, I'm new here and not familiar with stackexchange's (n)etiquette. I'd also mention that the op's two "scenarios" seem to implicitly be trying to distinguish two cases: 1) observer B isn't even aware a measurement was made $\equiv$ ontological superposition, and 2) B knows a measurement was made but doesn't know its outcome $\equiv$ epistemic mixture (aka density matrix). There's no sensible way to "span" all the stuff implicitly and explicitly brought up by the op in the space of a stackexchange answer. I think relevant references are maybe the best kind of "answer" here. $\endgroup$
    – user89220
    Commented Aug 16, 2015 at 7:03
  • $\begingroup$ @John_Forkosh Well there is no limit in space, but it is true that a very long answer is not too convenient. However you can shorten your answer by giving references, as you do, but still address the more relevant parts of the question. Your comment is a good way of making your answer more complete. Please, edit your answer and add it, feel free to add any other element like pictures, equations or more links. $\endgroup$
    – rmhleo
    Commented Aug 16, 2015 at 9:04
0
$\begingroup$

A measurement is a form of interaction. A's interaction with the state is information that B is not unaware of. Which is equivalent to say that A's interaction is information that B cannot use. Thus, your two statements are equivalent in that the best B can do is use the information about the state that he does have to make a prediction. A practical example of this makes the situation more clear.

Consider an experiment where A measured the spin of an unpolarized particle to be +1/2 in the z-direction. B is unaware of A's measurement. B subsequently measures the z-direction spin and finds it to be +1/2. This is a reasonable from a prediction using the superimposed state (unpolarized) state with 50% probability of being +1/2 and 50% probability of being -1.2

Now consider the same situation but with B being aware of A's measurement but unaware of the result. It remains that the best B can do is still make a probabilistic prediction.

Now consider the same experiment repeated over and over again. On B's measurement after A's measurement, B will see that 50% of the time the particle is +1/2 and 50% of the time the particle is -1/2. Thus, to B, the state is a superposition as expected. However, if A were to join B on the subsequent measurement and tell B of his early results, then B could use the collapsed state and predict the correct spin 100% of the time (since A already measured it). But, both predictions accurately conform to the results of the experiment, however one prediction (A's prediction) contains more knowledge about the system.

Improve this answer
$\endgroup$
10
  • $\begingroup$ but does this not make the system an statistical ensemble of systems among which hafl are in up state and the other half in down state. and morever because there is an information which B has no knowledge of does it not make the system possess a 'hidden-variable' just out of reach of B? $\endgroup$
    – user35122
    Commented Feb 20, 2014 at 16:25
  • $\begingroup$ I don't understand your first sentence. As for a "hidden-variable"... That concept is in regards to the probablistic nature of quantum mechanics. Just because an observer lacks information in making a prediction does not mean there is some fundamental hidden variable in your theory. For example, if we tried to predict the orbit of Mercury yet didn't account for the small deviation due to Venus, this does not mean that the fundamental laws of gravity have a hidden variable. It just means we didn't account for all the interactions. $\endgroup$
    – mcFreid
    Commented Feb 20, 2014 at 20:37
  • $\begingroup$ by statistical ensemble i mean a large collection of systems given by superposition of up spin and down spin becomes in this case a collection of large no of systems among which half of them are in up-spin state and the rest in down spin state! $\endgroup$
    – user35122
    Commented Feb 21, 2014 at 14:39
  • $\begingroup$ Yes, I know what a statistical ensemble is - I just don't see how that changes what I wrote. $\endgroup$
    – mcFreid
    Commented Feb 21, 2014 at 14:41
  • $\begingroup$ so each of the system would be in one particular state, not the superposition.I am just wondering if these two pictures are the same! $\endgroup$
    – user35122
    Commented Feb 21, 2014 at 15:57
0
$\begingroup$

The relevant issue is not what B knows, but rather whether the content of the measurement result has affected B. If the content of the result has affected B then B will not be able to arrange interference between the different results of that measurement. For example, suppose that the information about the result has been displayed on a computer screen and the light from the screen has interacted with B's skin while his back is turned, then he will not be able to produce interference between the results.

See papers about decoherence such as C. Jess Riedel, Wojciech H. Zurek, Quantum Darwinism in an Everyday Environment: Huge Redundancy in Scattered Photons, Phys. Rev. Lett. 105, 020404 (2010) http://arxiv.org/abs/1001.3419.

There is a slight complication. When two systems are entangled the correlations between them are explained by decoherent systems carrying quantum information in observables in such a way that the content of the information doesn't change the expectation values of those observables. See David Deutsch, Patrick Hayden, 'Information Flow in Entangled Quantum Systems', Proc. R. Soc. Lond. A 456(1999):1759-1774. available at http://arxiv.org/abs/quant-ph/9906007. And also David Deutsch, 'Vindication of quantum locality', Proc. R. Soc. A 468(2012), 531-544. available at http://arxiv.org/abs/1109.6223.

Improve this answer
$\endgroup$
-1
$\begingroup$

It shouldn't make a difference because in both cases the particle is measured.

This kind of question can be rather muddy - physicists try to stay away from using the words 'observer' and 'measurement' in quantum mechanics because they raise more questions than they answer.

At any rate, most physicists prefer 'interaction with a macroscopic system'.

Improve this answer
$\endgroup$
3
  • $\begingroup$ and what about the state according to B now? $\endgroup$
    – user35122
    Commented Feb 16, 2014 at 16:21
  • $\begingroup$ In both cases A's measurement collapsed the wave function, it's just that B either doesn't know it collapsed or what it collapsed to. It won't matter for A or the particle what B knows. $\endgroup$
    – Kvothe
    Commented Feb 16, 2014 at 16:26
  • $\begingroup$ so if B has to do some calculations with this system what state should he use? $\endgroup$
    – user35122
    Commented Feb 16, 2014 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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