5 ms·

The gap I still have in my understanding: One hypothesis is that there is no spooky action. One particle was 'always' going to resolve one way, likewise with

by mrkeen 3y ago

The gap I still have in my understanding: One hypothesis is that there is no spooky action. One particle was 'always' going to resolve one way, likewise with the other. Like inspecting 'heads' on one side of a coin 'forces' 'tails' onto the other side. I accept that this coin-hypothesis has been disproved by people who actually know what they're talking about. But to me, this implies that you should build your spooky FTL message by transmitting "Measured" rather than heads/tails: Have an array of 8 particles at both locations. They represent measured/unmeasured rather than heads/tails. You got yourself a one-way single-use FTL byte. For this to not work (which I'm sure it doesn't) you'd need to be unable to distinguish between a measured/unmeasured particle. To me this is equivalent of being unable to prove that there is anything spooky going on. So how can you have it both ways? How can you theoretically know that something was sent when measurement itself would destroy any evidence of something being sent?

bowsamic 3y ago
> I accept that this coin-hypothesis has been disproved by people who actually knows what they're talking about. The classical idea is more like as follows: you have a guy Charles who makes two letters each with a card in that has written on it either number 0 or 1, and gives one letter to Alice and another to Bob. Now, "entanglement" here is that if Charles writes 0 in Alice's letter he writes the number 1 in Bob's, and vice versa. Alice and Bob are also aware of this. This means that even at a distance, if Alice opens her letters and sees a 0, she knows that Bob's has 1 in. For a long time, Einstein and co. through the "EPR paradox" (Einstein-Podolsky-Rosen) thought that QM would be something like this, but we just didn't know how to access the actual letter "in transit" yet, which they called an "element of reality". They were wrong. It turns out that Bell's theorem basically says that in this situation the correlation between the two letters would be less strong than quantum theory predicts for entangled states, so this cannot be the full explanation. Bell's theorem says you either need to give up the idea that there is no "spooky action", or give up determinism (no true "element of reality" that really holds what's going on), or you can retain both if you let go of another concept called statistical independence. > Have an array of 8 particles at both locations. They represent measured/unmeasured rather than heads/tails. You got yourself a one-way single-use FTL byte. You can't have a quantum state where something is the superposition of measured and not measured, this violates a fundamental postulate of quantum mechanics: that measurement "chooses" one of the states in the measurement basis. I think you touch on that later, in that you can't tell via entanglement if a measurement took place elsewhere, which is partly why we can't experimentally demonstrate a non-local theory yet. Basically, it isn't that some signal is transmitted, it's that if they are entangled the states of each system cannot be independently known. That is, like in the letters example I gave: it can be either 0 for Alice and 1 for Bob, or 0 for Alice and 1 for Bob. Now, the reason why information is not transmitted is the same: Bob didn't know what was in his letter before he measured it, so it's randomly symmetric between 0 and 1, there is no information transferred. It turns out that this kind of symmetry between the two states tells you nothing that can be used to transmit information.
- vjerancrnjak 3y ago
  Yes, Bell's theorem is such a strong result. 1. locality 2. (local) reality 3. statistical independence Only 2 can hold at once and there's consensus to have 1. and 3., with that we lose the hidden variables. Instead, having 1. and 2. hold would be so interesting.
  - tsimionescu 3y ago
    In some sense, MWI also gets rid of 3, though it's not commonly seen like that. Essentially in MWI any experiment has all possible results, so the concept of statistical independence jsut doesn't make sense. Either way, MWI is by far the most popular interpretation that is both local and realistic.
- jemmyw 3y ago
  > Bell's theorem says you either need to give up the idea that there is no "spooky action", or give up determinism Can you not also say that the determinism includes the observer, aka superdeterminism (which is just determinism because that would surely be global).
  - bowsamic 3y ago
    This quote from my post > or you can retain both if you let go of another concept called statistical independence. is superdeterminism
- ngcc_hk 3y ago
  The key difference, unlike the usual sock example or your letter example, it is not about knowing. The superimposed quantum states have not been fixed. It is both colour or or letters or states at the same time all the time unlike it is measured. It is not the knowledge about a pre-existed, the individual state does not exist until measurement. That is the crazy part.
  - bowsamic 3y ago
    Yes they have, it's just that in quantum mechanics you can expand the state in different bases. The state still "exists" before it is measured, just in a non-separable superposition state (i.e. an entangled state)
vlovic 3y ago
The answer is Bell's Theorem. Scott Aaronson has some good explanations.
- gwill 3y ago
  i won't have time to read this until later, but here's a post by him on it: https://scottaaronson.blog/?cat=33 https://scottaaronson.blog/?cat=33
honeyed_coffee 3y ago
The spooky part occurs when two parties who share this entangled state know what measurement to perform. If the measurement choice aligns for both parties, their outcomes can be correlated precisely. If the measurement choices are not aligned, the outcomes are also random
- magicalhippo 3y ago
  > If the measurement choices are not aligned, the outcomes are also random Just to be a bit pedantic, as it can otherwise lead to some confusion: the measurement outcomes are always random. If the particles are entangled, then it is the correlations that are not random.
  - taeric 3y ago
    This is still a part that annoys me. I have asked why you can't use the correlations to facilitate communication, and people always seem to think I'm asking why you can't do this per particle. I get that the the individual measures are basically useless on their own. Question is if the correlations can be confirmed so well, why can't that be used?
    - deleted 3y ago
      [deleted]
    - drwiggly 3y ago
      From what I'm gathering. Alice measures at angle X, gets value V1 Calls Bob on the phone, okay I measured angle X. Bob measures at angle X, also gets value V1 Bob measures at angle Y, gets value V2. Bob calls Alice back says, okay I measured angle Y. Alice measures angle Y, also gets V2. The correlation here is nobody can do other measurements while the other party is in the process of measuring. Each party can't know the other party is done until traditional communication has happened. If each party acted independently they would randomly change the state on the other side and each party would get what appears to be random values.
elcritch 3y ago
> One hypothesis is that there is no spooky action. One particle was 'always' going to resolve one way, likewise with the other. Like inspecting 'heads' on one side of a coin 'forces' 'tails' onto the other side. > I accept that this coin-hypothesis has been disproved by people who actually know what they're talking about. The quantum erasure experiment is to my mind the best example of the spooky action at a distance. The outcome of the particle pairs is measured after the entanglement occurs implying not just "spooky action at a distance" but also in time. https://www.youtube.com/watch?v=8ORLN_KwAgs https://www.youtube.com/watch?v=8ORLN_KwAgs
- yetihehe 3y ago
  Calling it "action" may be a mistake. Action implies something changes (there is a state called "before" and state called "after"). But nothing changes "over there", we just now know what actually happened (there is no state called "before" or it's value is "unknown", only state called "after").
  - mrkeen 3y ago
    This is the simplest explanation, but it's also the null hypothesis. I believe that "nothing changes over there" implies not only no spooky action, but also no entanglement.
    - yetihehe 3y ago
      This is about word used for description of that thing. Just using entanglement is ok, but using "action" suggests that something changes. Words have some meaning and incorrect usage causes misunderstandings.
      - elcritch 3y ago
        > Action implies something changes (there is a state called "before" and state called "after"). But nothing changes "over there", we just now know what actually happened (there is no state called "before" or it's value is "unknown", only state called "after"). > using "action" suggests that something changes By my understanding, the delayed part is important. It implies something did change as the state of the earlier entanglement will be different based on a measurement which occurs after the entangled state has been created. If this is correct, then there has been an "action" at a distance. Though an alternative understanding could be that the state is set based on a future event which hasn't occurred yet. That would avoid needing an "action" to change the state after the entanglement event, but implies other things.
dist-epoch 3y ago
> you'd need to be unable to distinguish between a measured/unmeasured particle This is true. There is no way to tell if a measurement you do happened on an entangled particle or on a "free" one. You can only tell in retrospect, when you correlate results, which requires classical communication.
- yetihehe 3y ago
  So it can't be called "action"? Could be also described as correlation is not causation and action implies causation.
sebzim4500 3y ago
Disclaimer: I am not an expert, but I understood this stuff pretty well 5 years ago. If you have a single particle there is no local experiment that you can do to determine whether that is entangled to a particle somewhere else. You need to use both particles in the experiment in order to prove it one way or the other. Mathematically, this is due to the fact that the density function for one half of an entangled pair is the same as the maximally mixed state for one particle. See CHSH[1] for an example of an experiment you can do if you have access to both particles and you want to show they are entangled. [1] https://en.wikipedia.org/wiki/CHSH_inequality https://en.wikipedia.org/wiki/CHSH_inequality
Workaccount2 3y ago
You cannot tell if a particle has been measured or not without...measuring it. The key to understanding this is that _any_ interaction with anything counts as a "measurement" from the particles perspective. I like to think that our simulation has a bandwidth compression algorithm that only decodes the state of a particle when it needs to calculate an interaction. Just like how your dungeon crawler doesn't spawn mobs until they are within interaction distance. But that's just fun head canon.
- mrkeen 3y ago
  I don't dispute this, but it's indistinguishable from the 'no spooky action' hypothesis. I can invent a story that the coin in my pocket has two sides, each of which is in a superposition of heads and tails. If you disprove that by repeatedly looking at both sides and noting that they always end up different, then I can invent the story that the observed side tells the unobserved side what to become. If you disprove that by separating the sides by a great distance before observing them both, I can invent the story that they coordinate instantaneously, ignoring the speed of light.
  - slfnflctd 3y ago
    I've never quite grasped this stuff to my satisfaction, and I strongly suspect part of why is because of some really bad popular examples (metaphors, allegories, thought experiments, whatever) which contain factually or logically incorrect language. This comment chain is a prime example of the ways quantum mechanics are (mis)understood and how many of us struggle to fully reason out the implications of each piece of it. I hope we can all learn something.
    - Jensson 3y ago
      Fact is that the position isn't "unknown", it is in both positions at once. The double-slit experiment is strong evidence for this, we can see that the particle went through both slits at once by measuring a lot of particles and seeing where they end up. If you block one slit and then shoot particles, then block the other slit and shoot particles, you end up with a completely different pattern than if you have both open at once. This isn't difficult to understand, the only difficult part is people not wanting to believe it is true and trying to come up with classical explanations for quantum mechanics, even though we have hard proof that classical physics can't explain this. The difficult part is that even though the particle went through both slits, it only ever ends up in a single spot at the end, that is the "wave function collapse", it was a wave that went through both slits, and then it collapses to a single point when we measure it at the end. That is still not well understood by anyone today. https://en.wikipedia.org/wiki/Double-slit_experiment https://en.wikipedia.org/wiki/Double-slit_experiment
stephc_int13 3y ago
At the quantum scale, you can't observe without interaction. And the interaction is always destructive (of information, not of the particle).
gaze 3y ago
You can’t tell if a particle has been measured or not.
rssoconnor 3y ago
While you don't get action at a distance, you do get something like "correlation without causation". This ability does let you "cheat" at particular games of chance, which is spooky in its own right. https://r6.ca/blog/20150816T185621Z.html https://r6.ca/blog/20150816T185621Z.html
cwmma 3y ago
You can not tell whether a particle was measured or not by looking at its entangled pair.
d_tr 3y ago
From the perspective of the MWI there is no spooky action. If you measure one of the particles and look at the result, you then know what term of the quantum state you ended up in, so you know the probability distribution of the other particle's results. Nature does not need to work according to what our brains consider reasonable, but the MWI getting rid of several very weird notions makes it very appealing, I won't lie...
- Jensson 3y ago
  > From the perspective of the MWI there is no spooky action. There is a spooky world split that separates the entire universe instead of just having a spooky thing having to the particle. Some prefers having a small spooky thing over a large spooky thing.
  - a_gnostic 3y ago
    The universe is not split, though. You just perceive it that way. Think of it as a wave function, where some of the time, some waves cancel each other out, and are not observed, and some of the time they are amplified and easily observable.
Sai_ 3y ago
> You got yourself a one-way single-use FTL byte. But you don’t. If I have the entangled version of the MEASURED byte and measure, only I know I’ve measured it. The corresponding MEASURED particles on your side still appear fuzzy to you because you haven’t measured them. If you measure them, they will reveal their state which correlates to mine BUT that doesn’t tell you if I have already measured my side of byte or not. In effect, the heads/tail coin on your side is still spinning even after I have grabbed my side of the tail/head coin. You may get HEAD as your result but that only tells you that I will get TAIL when I measure on my side, not that I have measured it already. If you and I agree to measure at the same time, then I will know your state by determining mine but this is the same as knowing your state because I know how a two sided coin works, not FTL.
huytersd 3y ago
Yeah you can’t know whether something has been measured or not. You measure it, you get a value. You don’t know if it has been measured/collapsed on the other end.

by mrkeen 3y ago

bowsamic 3y ago

> I accept that this coin-hypothesis has been disproved by people who actually knows what they're talking about. The classical idea is more like as follows: you have a guy Charles who makes two letters each with a card in that has written on it either number 0 or 1, and gives one letter to Alice and another to Bob. Now, "entanglement" here is that if Charles writes 0 in Alice's letter he writes the number 1 in Bob's, and vice versa. Alice and Bob are also aware of this. This means that even at a distance, if Alice opens her letters and sees a 0, she knows that Bob's has 1 in. For a long time, Einstein and co. through the "EPR paradox" (Einstein-Podolsky-Rosen) thought that QM would be something like this, but we just didn't know how to access the actual letter "in transit" yet, which they called an "element of reality". They were wrong. It turns out that Bell's theorem basically says that in this situation the correlation between the two letters would be less strong than quantum theory predicts for entangled states, so this cannot be the full explanation. Bell's theorem says you either need to give up the idea that there is no "spooky action", or give up determinism (no true "element of reality" that really holds what's going on), or you can retain both if you let go of another concept called statistical independence. > Have an array of 8 particles at both locations. They represent measured/unmeasured rather than heads/tails. You got yourself a one-way single-use FTL byte. You can't have a quantum state where something is the superposition of measured and not measured, this violates a fundamental postulate of quantum mechanics: that measurement "chooses" one of the states in the measurement basis. I think you touch on that later, in that you can't tell via entanglement if a measurement took place elsewhere, which is partly why we can't experimentally demonstrate a non-local theory yet. Basically, it isn't that some signal is transmitted, it's that if they are entangled the states of each system cannot be independently known. That is, like in the letters example I gave: it can be either 0 for Alice and 1 for Bob, or 0 for Alice and 1 for Bob. Now, the reason why information is not transmitted is the same: Bob didn't know what was in his letter before he measured it, so it's randomly symmetric between 0 and 1, there is no information transferred. It turns out that this kind of symmetry between the two states tells you nothing that can be used to transmit information.

vjerancrnjak 3y ago
Yes, Bell's theorem is such a strong result. 1. locality 2. (local) reality 3. statistical independence Only 2 can hold at once and there's consensus to have 1. and 3., with that we lose the hidden variables. Instead, having 1. and 2. hold would be so interesting.
- tsimionescu 3y ago
  In some sense, MWI also gets rid of 3, though it's not commonly seen like that. Essentially in MWI any experiment has all possible results, so the concept of statistical independence jsut doesn't make sense. Either way, MWI is by far the most popular interpretation that is both local and realistic.
jemmyw 3y ago
> Bell's theorem says you either need to give up the idea that there is no "spooky action", or give up determinism Can you not also say that the determinism includes the observer, aka superdeterminism (which is just determinism because that would surely be global).
- bowsamic 3y ago
  This quote from my post > or you can retain both if you let go of another concept called statistical independence. is superdeterminism
ngcc_hk 3y ago
The key difference, unlike the usual sock example or your letter example, it is not about knowing. The superimposed quantum states have not been fixed. It is both colour or or letters or states at the same time all the time unlike it is measured. It is not the knowledge about a pre-existed, the individual state does not exist until measurement. That is the crazy part.
- bowsamic 3y ago
  Yes they have, it's just that in quantum mechanics you can expand the state in different bases. The state still "exists" before it is measured, just in a non-separable superposition state (i.e. an entangled state)

vlovic 3y ago

The answer is Bell's Theorem. Scott Aaronson has some good explanations.

gwill 3y ago
i won't have time to read this until later, but here's a post by him on it: https://scottaaronson.blog/?cat=33 https://scottaaronson.blog/?cat=33

honeyed_coffee 3y ago

The spooky part occurs when two parties who share this entangled state know what measurement to perform. If the measurement choice aligns for both parties, their outcomes can be correlated precisely. If the measurement choices are not aligned, the outcomes are also random

magicalhippo 3y ago
> If the measurement choices are not aligned, the outcomes are also random Just to be a bit pedantic, as it can otherwise lead to some confusion: the measurement outcomes are always random. If the particles are entangled, then it is the correlations that are not random.
- taeric 3y ago
  This is still a part that annoys me. I have asked why you can't use the correlations to facilitate communication, and people always seem to think I'm asking why you can't do this per particle. I get that the the individual measures are basically useless on their own. Question is if the correlations can be confirmed so well, why can't that be used?
  - deleted 3y ago
    [deleted]
  - drwiggly 3y ago
    From what I'm gathering. Alice measures at angle X, gets value V1 Calls Bob on the phone, okay I measured angle X. Bob measures at angle X, also gets value V1 Bob measures at angle Y, gets value V2. Bob calls Alice back says, okay I measured angle Y. Alice measures angle Y, also gets V2. The correlation here is nobody can do other measurements while the other party is in the process of measuring. Each party can't know the other party is done until traditional communication has happened. If each party acted independently they would randomly change the state on the other side and each party would get what appears to be random values.

elcritch 3y ago

> One hypothesis is that there is no spooky action. One particle was 'always' going to resolve one way, likewise with the other. Like inspecting 'heads' on one side of a coin 'forces' 'tails' onto the other side. > I accept that this coin-hypothesis has been disproved by people who actually know what they're talking about. The quantum erasure experiment is to my mind the best example of the spooky action at a distance. The outcome of the particle pairs is measured after the entanglement occurs implying not just "spooky action at a distance" but also in time. https://www.youtube.com/watch?v=8ORLN_KwAgs https://www.youtube.com/watch?v=8ORLN_KwAgs

yetihehe 3y ago
Calling it "action" may be a mistake. Action implies something changes (there is a state called "before" and state called "after"). But nothing changes "over there", we just now know what actually happened (there is no state called "before" or it's value is "unknown", only state called "after").
- mrkeen 3y ago
  This is the simplest explanation, but it's also the null hypothesis. I believe that "nothing changes over there" implies not only no spooky action, but also no entanglement.
  - yetihehe 3y ago
    This is about word used for description of that thing. Just using entanglement is ok, but using "action" suggests that something changes. Words have some meaning and incorrect usage causes misunderstandings.
    - elcritch 3y ago
      > Action implies something changes (there is a state called "before" and state called "after"). But nothing changes "over there", we just now know what actually happened (there is no state called "before" or it's value is "unknown", only state called "after"). > using "action" suggests that something changes By my understanding, the delayed part is important. It implies something did change as the state of the earlier entanglement will be different based on a measurement which occurs after the entangled state has been created. If this is correct, then there has been an "action" at a distance. Though an alternative understanding could be that the state is set based on a future event which hasn't occurred yet. That would avoid needing an "action" to change the state after the entanglement event, but implies other things.

dist-epoch 3y ago

> you'd need to be unable to distinguish between a measured/unmeasured particle This is true. There is no way to tell if a measurement you do happened on an entangled particle or on a "free" one. You can only tell in retrospect, when you correlate results, which requires classical communication.

yetihehe 3y ago
So it can't be called "action"? Could be also described as correlation is not causation and action implies causation.

sebzim4500 3y ago

Disclaimer: I am not an expert, but I understood this stuff pretty well 5 years ago. If you have a single particle there is no local experiment that you can do to determine whether that is entangled to a particle somewhere else. You need to use both particles in the experiment in order to prove it one way or the other. Mathematically, this is due to the fact that the density function for one half of an entangled pair is the same as the maximally mixed state for one particle. See CHSH[1] for an example of an experiment you can do if you have access to both particles and you want to show they are entangled. [1] https://en.wikipedia.org/wiki/CHSH_inequality https://en.wikipedia.org/wiki/CHSH_inequality

Workaccount2 3y ago

You cannot tell if a particle has been measured or not without...measuring it. The key to understanding this is that _any_ interaction with anything counts as a "measurement" from the particles perspective. I like to think that our simulation has a bandwidth compression algorithm that only decodes the state of a particle when it needs to calculate an interaction. Just like how your dungeon crawler doesn't spawn mobs until they are within interaction distance. But that's just fun head canon.

mrkeen 3y ago
I don't dispute this, but it's indistinguishable from the 'no spooky action' hypothesis. I can invent a story that the coin in my pocket has two sides, each of which is in a superposition of heads and tails. If you disprove that by repeatedly looking at both sides and noting that they always end up different, then I can invent the story that the observed side tells the unobserved side what to become. If you disprove that by separating the sides by a great distance before observing them both, I can invent the story that they coordinate instantaneously, ignoring the speed of light.
- slfnflctd 3y ago
  I've never quite grasped this stuff to my satisfaction, and I strongly suspect part of why is because of some really bad popular examples (metaphors, allegories, thought experiments, whatever) which contain factually or logically incorrect language. This comment chain is a prime example of the ways quantum mechanics are (mis)understood and how many of us struggle to fully reason out the implications of each piece of it. I hope we can all learn something.
  - Jensson 3y ago
    Fact is that the position isn't "unknown", it is in both positions at once. The double-slit experiment is strong evidence for this, we can see that the particle went through both slits at once by measuring a lot of particles and seeing where they end up. If you block one slit and then shoot particles, then block the other slit and shoot particles, you end up with a completely different pattern than if you have both open at once. This isn't difficult to understand, the only difficult part is people not wanting to believe it is true and trying to come up with classical explanations for quantum mechanics, even though we have hard proof that classical physics can't explain this. The difficult part is that even though the particle went through both slits, it only ever ends up in a single spot at the end, that is the "wave function collapse", it was a wave that went through both slits, and then it collapses to a single point when we measure it at the end. That is still not well understood by anyone today. https://en.wikipedia.org/wiki/Double-slit_experiment https://en.wikipedia.org/wiki/Double-slit_experiment

stephc_int13 3y ago

At the quantum scale, you can't observe without interaction. And the interaction is always destructive (of information, not of the particle).

gaze 3y ago

You can’t tell if a particle has been measured or not.

rssoconnor 3y ago

While you don't get action at a distance, you do get something like "correlation without causation". This ability does let you "cheat" at particular games of chance, which is spooky in its own right. https://r6.ca/blog/20150816T185621Z.html https://r6.ca/blog/20150816T185621Z.html

cwmma 3y ago

You can not tell whether a particle was measured or not by looking at its entangled pair.

d_tr 3y ago

From the perspective of the MWI there is no spooky action. If you measure one of the particles and look at the result, you then know what term of the quantum state you ended up in, so you know the probability distribution of the other particle's results. Nature does not need to work according to what our brains consider reasonable, but the MWI getting rid of several very weird notions makes it very appealing, I won't lie...

Jensson 3y ago
> From the perspective of the MWI there is no spooky action. There is a spooky world split that separates the entire universe instead of just having a spooky thing having to the particle. Some prefers having a small spooky thing over a large spooky thing.
- a_gnostic 3y ago
  The universe is not split, though. You just perceive it that way. Think of it as a wave function, where some of the time, some waves cancel each other out, and are not observed, and some of the time they are amplified and easily observable.

Sai_ 3y ago

> You got yourself a one-way single-use FTL byte. But you don’t. If I have the entangled version of the MEASURED byte and measure, only I know I’ve measured it. The corresponding MEASURED particles on your side still appear fuzzy to you because you haven’t measured them. If you measure them, they will reveal their state which correlates to mine BUT that doesn’t tell you if I have already measured my side of byte or not. In effect, the heads/tail coin on your side is still spinning even after I have grabbed my side of the tail/head coin. You may get HEAD as your result but that only tells you that I will get TAIL when I measure on my side, not that I have measured it already. If you and I agree to measure at the same time, then I will know your state by determining mine but this is the same as knowing your state because I know how a two sided coin works, not FTL.

huytersd 3y ago

Yeah you can’t know whether something has been measured or not. You measure it, you get a value. You don’t know if it has been measured/collapsed on the other end.