18
$\begingroup$

This question already has an answer here:

I have never found experimental evidence that measuring one entangled particle causes the state of the other entangled particle to change, rather than just being revealed.

Using the spin up spin down example we know that one of the particles will be spin up and the other spin down, so when we measure one and find it is spin up we know the other is spin down. Is there any situation after creation that the particle with spin up will change to spin down?

$\endgroup$

marked as duplicate by Norbert Schuch, Jon Custer, Chair, ahemmetter, rob Dec 15 '18 at 16:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 4
    $\begingroup$ What you describe cannot happen - a change of the entangled particle cannot be observed. If you knew it was already spin up then you already measured it, so you already collapsed the state of the first one to spin down, which is what you will then measure. $\endgroup$ – OrangeDog Dec 13 '18 at 12:11
  • $\begingroup$ I think you need to really understand non-commuting observables in order to understand both why entanglement seems weird and why it isn't really. $\endgroup$ – Harry Johnston Dec 14 '18 at 5:32
  • $\begingroup$ I find this video to have the easiest to understand explanation of why hidden variables cannot explain quantum effects. $\endgroup$ – kasperd Dec 14 '18 at 10:47
47
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Dec 13 '18 at 22:12
2
$\begingroup$

Just for a moment, if one (you) imagine that the entangled particles are determinated in their states from the beginning of their entanglement, would it change something in the outcome of the experiment? In the pair production of two photons from nonlinear optical crystals the result of the experiments shows a statistical dependency; the orientation of the electric field components of the photons is orthogonal to each over.

Producing such a pair of photons we are not able to control the direction of the polarisation, we only able to design the process so that the polarisations of two particles are orthogonal on each over, the orientation by itself is randomly distributed around 360°. The uncertainty is in the phenomenon, that using a polariser to find the orientation of the particles, we do not get a result in 50% of our measurements (with the best designed polarizer and for the right wavelength).

Let the polarizer be orientated to the vertical (0°). Getting a result, we will know that the photon has had an orientation from -45° to 45° and from 135° to 225°. Not getting a result, we are allowed to say, that the photon has an orientation in the opposite to the above mentioned angles. To count for the statistics these not measured photons we are able only after getting the information from the second measurement device, that a photon in their setup was measured. In reality simply the number of measured incidents is compared.

What is the proof of entanglement

Firstly a lot of experiments for a given process, in this case for photons from nonlinear optical crystals in a well definded setup. And later the knowledge, that in this process are produced entangled particles.

Is there any situation after creation that the particle with spin up will change to spin down.

This happens all the time due to “impurities” in the experimental setup. The produced particles colliding with other particles and are under the influence of external fields. The longer the distance and the time between production and measurement, the weaker the measurable coincidence. But the swap of a spin by itself is impossible.

$\endgroup$
2
$\begingroup$

A pretty simple explanation that should be mostly fine:

Suppose you have a "magic" black box that produces entangled particles with spin up and spin down. One is always up and the other is always down.

If you measure one with up and down base states, you will get up in 50% of cases and down in the other 50%, while the other particle will be always just opposite. You can easily argue that here, you merely didn't know what came out and when you measured one you just figured out what both were. Everything works just fine, no need for any QM yet.

But suppose you measure this setup in left/right base instead. Then, obviously, the first spin you measure will have 50% to be left, and 50% to be right, doesn't matter if it was up or down before. Now, predictions for the second spin are different. In classical physics, you are measuring left/right with 50% for the other particle too. This happens because you have 50% it was up * 50% it gets measured right + 50% it was down * 50% it gets measured right (same for left) But QM entanglement says that nope, we know with 100% certainty that the second spin will be right if the first was left and vice versa.

To summarize: measure in up/down axis: both give 50% for each value of the first particle and 100%/0% for the second particle. Measure in left/right axis: both give 50% for each value of the first particle. Classical physics gives 50%/50% for the second one, while QM gives 100%/0%.

This can be tested, has been tested extensively in various configurations and QM entanglement was found to match reality, while the classical physics doesn't.

You STILL have several possibilities:

  1. Typical QM interpretation is right that stuff doesn't have state until measured.
  2. When you measured one particle, you changed both particles instantly.
  3. Everything including the measurement outcomes had been predefined before you even thought to make the box to measure this.
  4. Maybe something else.

But you can't have good old classical physics where stuff behaves as you would expect and want it to.

$\endgroup$
  • $\begingroup$ I think "axis" is a more appropriate word than "base". And you don't really explain why classical physics would predict 50% left/right for the second particle. $\endgroup$ – Acccumulation Dec 13 '18 at 16:20
  • $\begingroup$ Edited a bit with that classical physics addition. I used "base" as for example with light you might have LCP/RCP, where "axis" doesn't really seem right, while "base" still works. $\endgroup$ – Zizy Archer Dec 14 '18 at 7:19
2
$\begingroup$

What you describe is essentially a variation a local hidden-variable theory which try to explain features of quantum mechanics (like observable effects of entanglement) via deterministic but not yet observable properties of objects (hidden variables), ruling out faster-than-light interaction between distant events.

Local hidden-variable theories are disproved by experiments featuring a Bell's inequality violation, which cannot be explained without accounting for entanglement. It's not a proof of entanglement of course, rather, it's a proof that reality cannot be explained by local hidden-variable theories alone.

$\endgroup$
1
$\begingroup$

In a comment Nagora says "It's a bad question: we have no "proof" of any scientific theory, we only have proof that discarded ones are wrong." But even that isn't correct (or is a bit of a non sequitur). "Entanglement exists" isn't a theory, it's a particular claim that exists within a wider theoretical framework. We don't prove or disprove individual claims, we evaluate theories as a whole with respect to the evidence. You can't evaluate, say, whether neutrinos travel faster than light except within a particular theoretical framework: however you're measuring the speed of the neutrinos depends on assumptions about how your measuring devices interact with neutrinos.

We have a particular theory of quantum mechanics that predicts entanglement, and it comports with the experimental evidence. If you are willing to accept radically different assumptions, you could come up with a theoretical framework without the concept of entanglement.

$\endgroup$
  • $\begingroup$ Anytime you disprove a theory, you also prove its exact logical inverse. Usually, that's not very interesting as it may contain virtually anything. But if you define "entanglement exists" to mean that a measurement can actually have an effect on another measurement that's not within the same light-cone, the logical inverse is the theory of local realism. This theory can be disproven, and has been disproven. Consequently, it has been proven that entanglement exists. Whether that's what QM describes is another question, but we know that some kind of "spooky action at a distance" exists. $\endgroup$ – cmaster Dec 13 '18 at 22:39
  • $\begingroup$ Local realism hasn't been disproven, it's been proven to be inconsistent with other "self-evident" premises. And the inverse is not local realism, it's just locality. Realism is a separate premise. $\endgroup$ – Acccumulation Dec 13 '18 at 22:45
  • $\begingroup$ The last time I checked, the direct evidence was still somewhat inconclusive anyway. Have there been significant experimental improvements recently? $\endgroup$ – Harry Johnston Dec 14 '18 at 5:30
1
$\begingroup$

To my knowledge, there exists no evidence that there is a causal interaction between two entangled but separated particles when the other one is measured. The preparation of such experiment produces correlated states, and if the system remains sufficiently isolated, the correlation survives through the time evolution and is revealed at the time of the measurement.

CHSH and Bell's inequality prove that classical probability with uniformly random distributions cannot be used for calculating averages for such correlated systems - instead one should use the usual quantum mechanical approach. However, without further assumptions concerning the time evolution of the states, this argument is neutral regarding the causality vs correlation question.

$\endgroup$

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