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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?


marked as duplicate by Qmechanic quantum-mechanics Dec 17 '18 at 15:46

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    $\begingroup$ Can you be more specific? $\endgroup$ – Hugo V Oct 31 '18 at 21:07
  • $\begingroup$ 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 $\endgroup$ – Andrea Scaglioni Oct 31 '18 at 21:15
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/284724/2451 and links therein. $\endgroup$ – Qmechanic Dec 17 '18 at 15:47

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:

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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    $\begingroup$ Nice 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. $\endgroup$ – Valter Moretti Nov 1 '18 at 11:47
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    $\begingroup$ Quantum 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. $\endgroup$ – Valter Moretti Nov 1 '18 at 11:53
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    $\begingroup$ @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. $\endgroup$ – Chiral Anomaly Nov 2 '18 at 2:36
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    $\begingroup$ Could you help me see where is the locality assumption in this derivation? I assume it must be somewhere but I cannot figure it out $\endgroup$ – Wolphram jonny Nov 11 '18 at 21:48
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    $\begingroup$ @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 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. $\endgroup$ – Chiral Anomaly Nov 11 '18 at 22:07

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