Somebody could help me to clarify how its possible that different choices of the observable to measure can lead to different outcomes of the observed system state?
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2Can you be more specific? – Hugo V Oct 31 '18 at 21:07
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I often read that quantum mechanics postulates that the act of the observation defines the system proprieties and that it really depends on what observables the experimenter choose to measure..so that different choices of the observer leads to different system’s state..but sometimes it seems trivial to me because I find obvious that different measured quantities lead to different result..I just want to clarify this – Andrea Scaglioni Oct 31 '18 at 21:15
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Possible duplicates: https://physics.stackexchange.com/q/284724/2451 and links therein. – Qmechanic Dec 17 '18 at 15:47
1 Answers
While it is trivially true that measuring different things gives us different kinds of information, there is a more interesting sense in which the state of the system is affected by what we choose to measure. In the real world, the assumption that a measurable property exists whether or not we measure it is inconsistent with the experimental facts.
Here's a relatively simple example. Suppose we have four observables, $A,B,C,D$, each of which has two possible outcomes. (For example, these could be single-photon polarization observables.) For mathematical convenience, label the two possible outcomes $+1$ and $-1$, for each of the four observables. Suppose for a moment that the act of measurement merely reveals properties that would exist anyway even if they were not measured. If this were true, then any given state of the system would have definite values $a,b,c,d$ of the observables $A,B,C,D$. Each of the four values $a,b,c,d$ could be either $+1$ or $-1$, so there would be $2^4=16$ different possible combinations of these values. Any given state would have one of these 16 possible combinations.
Now consider the two quantities $a+c$ and $c-a$. The fact that $a$ and $c$ both have magnitude $1$ implies that one of these two quantities must be zero, and then the other one must be either $+2$ or $-2$. This, in turn, implies that the quantity $$ (a+c)b+(c-a)d $$ is either $+2$ or $-2$. This is true for every one of the 16 possible combinations of values for $a,b,c,d$, so if we prepare many states, then the average value of this quantity must be somewhere between $+2$ and $-2$. In particular, the average cannot be any larger than $+2$. This gives the CHSH inequality $$ \langle{AB}\rangle +\langle{CB}\rangle +\langle{CD}\rangle -\langle{AD}\rangle\leq 2, $$ where $\langle{AB}\rangle$ denotes the average of the product of the values of $a$ and $b$ over all of the prepared states.
In the real world, the CHSH inequality can be violated, and quantum theory correctly predicts the observed violations. The quantity $\langle{AB}\rangle +\langle{CB}\rangle +\langle{CD}\rangle -\langle{AD}\rangle$ can be as large as $2\sqrt{2}$. Here are a few papers describing experiments that verify this:
Kwiat et al, 1995. “New high-intensity source of polarization-entangled photon pairs,” http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.75.4337
Kwiat et al, 1998. “Ultra-bright source of polarization-entangled photons,” http://arxiv.org/abs/quant-ph/9810003
Weihs et al, 1998. “Violation of Bell’s inequality under strict Einstein locality conditions,” http://arxiv.org/abs/quant-ph/9810080
The fact that the CHSH inequality is violated in the real world implies that the premise from which it was derived cannot be correct. The CHSH inequality was derived above by assuming that the act of measurement merely reveals properties that would exist anyway even if they were not measured. The inequality is violated in the real world, so this assumption must be wrong in the real world. Measurement plays a more active role.

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1Nice answer (+1). I would not say however that the real world is contextual. The real world is quantum instead! Contextuality is a way out to preserve realism in a hidden-variable theory compatible with violation of BCHSH inequality when locality is not assumed (e.g. dealing with intra particle entanglement) or compatible with the no-go result of Kocken-Specker's theorem....I would say. – Valter Moretti Nov 01 '18 at 11:47
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1Quantum theory is non-contextual instead. The outcomes of a measurement of an observable do not depend on the other observables simultaneously measured. Conversely it violates realism conditions: however you fix a quantum state there are necessarily observables which do not have definite values as a consequence of Gleason theorem as first observed by Bell il 1966. – Valter Moretti Nov 01 '18 at 11:53
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@ValterMoretti Those are good comments. "Contextual" isn't a word that I've used much in conversations with other physicists, so I may be using it incorrectly. I will look into this and re-orient my understanding of how it should be used. I reworded that sentence to avoid using the word "contextual". Thank you for calling this to my attention! – Chiral Anomaly Nov 01 '18 at 12:40
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@DanYand first of all i want to say that your answer is a really nice one cause it 's helping me a lot to understand this facts from a matemathical point of view, even if im not even close to having your knowledge about these issues. So if we want to oversimplify things a bit it is also correct to say that different choices about the observables to measure would lead to different system's future state?potentially different behaviours of the system?i know that once the masurement is done the evolution of the system follows the shrödinger equation but i would like to know if its correct to say – Andrea Scaglioni Nov 01 '18 at 13:35
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@DanYand that the observer has a causal role in the evolution of the system because he define its initial state – Andrea Scaglioni Nov 01 '18 at 13:37
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1@AndreaScaglioni Yes, I think it is correct to say that different choices about the observables to measure can lead to different future states. Even without considering quantum physics, one of the defining features of measurement (as a physical process using equipment made of molecules) is that it lets the thing being measured influence the rest of the world in a practically irreversible way that depends on the outcome. Think of observing the location of an object based on the light that it scatters. In quantum physics, this aspect of measurement is even more essential. – Chiral Anomaly Nov 02 '18 at 02:36
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@DanYand You're right ,its true. But in Quantum Mechanics its role is so fundamental that it amaze me. Anyway i want to thank you for your time and especially for your first explanation in the main answer.It helped me a lot to figure out these things from a more technical viewpoint. – Andrea Scaglioni Nov 02 '18 at 08:07
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1Could you help me see where is the locality assumption in this derivation? I assume it must be somewhere but I cannot figure it out – Nov 11 '18 at 21:48
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1@Wolphramjonny If I'm using the words correctly (see Valter Moretti's comments above), the derivation I showed here doesn't use the locality assumption. It uses the noncontextuality assumption instead. According to page 1 in http://arxiv.org/abs/0808.2456, "Local hidden variable theories are a special type of noncontextual hidden variable (NCHV) theories..." I don't remember the distinction clearly enough to review it here, but I think that author (Cabello) has several papers that discuss the distinction between the assumptions of "local realism" and noncontextuality. – Chiral Anomaly Nov 11 '18 at 22:07