# Exactly why FTLC is impossible with entangled photons?

• B
• Karagoz
In summary, it is written that almost all physics books and courses state that entangled photons cannot be used for faster than light communication. The question of used entangled photons to communicate faster than light has been discussed before, but I was not able to understand why it is impossible. My goal is not to prove that many physicists and scientists are wrong about this, but to provide some superficial knowledge on quantum physics. According to my understanding, the probability for a photon to pass through a filter depends on the angle Φ between the photon and the filter polarization axis. Imagine some photons that are entangled and sent far away with Jack and Joe. They have agreed beforehand that they’ll send message to each other every 24 hours (or every 12 pm).
Karagoz
It’s written almost all physics books and courses that entangled photons can’t be used for faster than light communication. The question about used entangled photons to communicate faster than light has been discussed before, but I didn’t get exactly why faster than light communication with entangled photons is impossible.

My goal is not to prove so many physicists and scientists that they’re wrong about entangled photons can’t be used to faster than light communication, nor to show that I have “invented” a way to entangled photons for faster than light communication.

From my superficial knowledge on quantum physics:

The probability P for a photon to pass through the filter depends on the angle Φ between the photon and the filter polarization axis:P = cos^2(Φ)

Imagine some photons that are entangled and sent far away with Jack and Joe.

They have agreed beforehand that they’ll send message to each other every 24 hours (or every 12 pm)

Jack has some photons, and Joe has other photons. The photons are entangled:

Jack will give a signal to Joe, 0 or 1.

The signal is 0 when he passes his photon through a filter like this, and changes its axis:

The likelihood of the photon passing through is %50, because Φ = 45 degrees.
If Jack’s photon passes through the filter and changes its axis, then Joe’s photon will change to this axis:

The signal is 1 is when he passes his photon through a filter like this, and changes its axis:

The likelihood of the photon passing through is %50, because Φ = 45 degrees.

If Jack’s photon passes through the filter and changes its axis, then Joe’s photon will change to this axis:

After 12 pm Jack sends his photon through a filter. It’s going to pass through it by chance of %50 (since Φ = 45 degrees). He use polarizing beam splitter to test same photon again and again until it passes through the filter. So if the photon doesn’t pass in the first time, Jack tries the same photon again and again until it passes through it.When Jack’s photons pass through the filter, the axis of the photon changes. When axis of Jack’s photon changes, so does Joe’s photon change its axis too.

After 12 pm Joe knows that his photon has changed its axis because Jack sent him a message. But Joe doesn’t know the axis of his photon.Joe passes his photon through a filter like this:

If his photon is changed to this axis:

Then it’s %100 likely his photon will pass through that filter.

If his photon is changed to this axis:

Then it’s %100 likely his photon won’t pass through the filter.So by trying to get the phtoton pass the filter, Joe will know what message is sent from Jack (whether it’s 0 or 1).Now Jack’s photons is one of this axis:

or

He can pass his photon through another filter and change reset the photons axis again. He can use polarizing beam splitter to try same photon again and again until it passes through the filter and changes its axis:

Jack can repeat this process again next day, and send a message (0 or 1) next day.

By increasing the number of photons, Jack can increase the number of 0s and 1s, and he can send codes that can be interpreted as full messages.

Or is there something I’m missing or taking it wrong?

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Karagoz said:
After 12 pm Jack sends his photon through a filter. It’s going to pass through it by chance of %50 (since Φ = 45 degrees). He use polarizing beam splitter to test same photon again and again until it passes through the filter. So if the photon doesn’t pass in the first time, Jack tries the same photon again and again until it passes through it.

Which it will never do. The polarizations at both output ports of the PBS are well defined. A photon that leaves via one exit port has 0 probability to leave through the other exit port, when you return it to the PBS.

zonde
There are a number of things about your premises that are incorrect.

1. "He can pass his photon through another filter and change reset the photons axis again. ..."
Entanglement is broken once the photon's polarization is first measured. There is no such thing as "again and again" for the entangled partner. It's one and done.

2. "If Jack’s photon passes through the filter and changes its axis, then Joe’s photon will change to this axis..."
There is no such thing as saying "Jack's observation changes Joe's results" any more than "Joe's observation changes Jack's results". The ordering of observations is irrelevant to the observed results. That is because time of measurement is not a factor (in the sense there is no observable difference).

Demystifier
Karagoz said:
The likelihood of the photon passing through is %50, because Φ = 45 degrees.
If Jack’s photon passes through the filter and changes its axis, then Joe’s photon will change to this axis:
View attachment 223759
No, nothing happens to Joe's photon if Jack's photon passes through the filter and changes it axis.

Karagoz said:
If Jack’s photon passes through the filter and changes its axis, then Joe’s photon will change to this axis:

View attachment 223761
See above.

DrChinese said:
There are a number of things about your premises that are incorrect.

1. "He can pass his photon through another filter and change reset the photons axis again. ..."
Entanglement is broken once the photon's polarization is first measured. There is no such thing as "again and again" for the entangled partner. It's one and done.

2. "If Jack’s photon passes through the filter and changes its axis, then Joe’s photon will change to this axis..."
There is no such thing as saying "Jack's observation changes Joe's results" any more than "Joe's observation changes Jack's results". The ordering of observations is irrelevant to the observed results. That is because time of measurement is not a factor (in the sense there is no observable difference).

I didn't get your 2nd point.

From what I know:

If Φ = 45 in the both entangled photons, then the probability of the first one passing through the filter is 50%.

But if the first photon passes through the filter, then it's 100% certain that the other entangled photon too passes through filter.

And it's because when the first photon changes its position, the other photon will change its position too instantly if they are entangled photons.

So if Jack's photon passes through a filter and changes its axis, won't the Joe's photon change its axis too?

Karagoz said:
So if Jack's photon passes through a filter and changes its axis, won't the Joe's photon change its axis too?
No.

StevieTNZ said:
No.
DrChinese said:
There are a number of things about your premises that are incorrect.

1. "He can pass his photon through another filter and change reset the photons axis again. ..."
Entanglement is broken once the photon's polarization is first measured. There is no such thing as "again and again" for the entangled partner. It's one and done.

2. "If Jack’s photon passes through the filter and changes its axis, then Joe’s photon will change to this axis..."
There is no such thing as saying "Jack's observation changes Joe's results" any more than "Joe's observation changes Jack's results". The ordering of observations is irrelevant to the observed results. That is because time of measurement is not a factor (in the sense there is no observable difference).

But this is what's written in Norwegian high-school physics study book:
Let's assume that the entangled photons move in opposite directions and have the same frequency and same polarization. We place a polarization filter in the way of each of the photons so that the filters have parallel polarization axes. What do you think will happen? You probably suggest that if the photons are polarized perpendicular to the filters (V), both photons are blocked and if they are polarized parallel to the filters (P), then both pass through. There you are right. But what if the photons are in a state of superposition of P and V? Then you might think that each photon will pass through with a probability of P(passing through) = cos 2 φ, and that the two photons pass through independently. For example, if the angle was 60°, one would think that the probability that each photon would pass through was 1/4 and that the likelihood that both photons would pass through was therefore (1/4)^2 = 116. But it is not. If one photon is passed through, then it's 100% certain that the other photon will pass through too! The same happens if we place the filters so that one photon passes its filter before the other passes. Attempts show that the distance between the filters does not matter. At the same moment as the first photon passes a filter, both photons change their state to P.

If the first photon pass through a filter, it's 100% certain the second pass through that filter too. Won't it mean that if the first one changes its state, the second one instantly changes its state too?

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Maybe I'm mistaken.
I've always been under the impression that if we start with the entangled state, for example, |H>|V> - |V>|H>, photon 1 goes through a 45 degree filter (or doesn't), the state is now |45>|V> - |135>|V - |45>|H> - |135>|H>.

Perhaps further clarification is needed.

Karagoz said:
My goal is not to prove so many physicists and scientists that they’re wrong about entangled photons can’t be used to faster than light communication

Karagoz said:
But this is what's written in Norwegian high-school physics study book:

It kind of sounds like you are.

The problem isn't that you don't have the answer. You got some really excellent responses. The problem is you don't believe the answer. Not sure what can be done here.

bhobba
It kind of sounds like you are.

The problem isn't that you don't have the answer. You got some really excellent responses. The problem is you don't believe the answer. Not sure what can be done here.

I don't get why you try to read my intention, and why you think I'm trying to prove all scientists wrong.

Because you got some really excellent responses, which have gone nowhere.

bhobba
Karagoz said:
But this is what's written in Norwegian high-school physics study book:
"Let's assume that the entangled photons move in opposite directions and have the same frequency and same polarization."

The book is in error. With any of the usual entanglement states, neither photon has its own polarization state.
Furthermore, whatever axis Joe chooses he will see ±1 with equal probability (the eigenvalues of the measurement operator). The same goes for Jack. Later they can check for correlations at their leisure, that will depend on the original entanglement state and their respective axis choices.
For example with
StevieTNZ said:
|H>|V> - |V>|H>,
and both choosing the same axis, they will get opposite values (correlation = -1), no communication.

One more thing, in the future use Alice and Bob or you will be kicked out of the quantum club.

Karagoz said:
Or is there something I’m missing or taking it wrong?
There are number of things in your model that are not aligned with actual experiments. Couple of them:
1) Polarization of photons is described in 0-180 deg range i.e. 0 and 180 deg is the same polarization. So your arrows in pictures are misleading.
2) You can not have polarization entangled photons with known polarization. You might describe one side as mixture of H and V polarized photons (actually QM formalism describes it as superposition of H and V modes rather than mixture).
And of course the objections noted by Cthugha and DrChinese p.1.

To avoid different errors that take you away from actual experimental situation I would recommend you to read this quite accesible description of entanglement experiment:
Entangled photons, nonlocality and Bell inequalities in the undergraduate laboratory

Concerning the statement that "At the same moment as the first photon passes a filter, both photons change their state to P.", you can assume this but it is an interpretation. There are others who would say that this is the wrong interpretation and the right one says something different. But the questions concerning the "right" interpretation are not sorted out so the best option is to keep an open mind about it.

Karagoz
Karagoz said:
It’s written almost all physics books and courses that entangled photons can’t be used for faster than light communication. The question about used entangled photons to communicate faster than light has been discussed before, but I didn’t get exactly why faster than light communication with entangled photons is impossible.

Or is there something I’m missing or taking it wrong?

What about a simpler, but related problem - effectively hidden variables. You send out a pair of socks in sealed containers. One container has a right sock and the other a left sock. The two containers are sent to two recipients.

Each recipient, upon opeing the container, not only sees what sock they have, but knows immediately what sock the other recipient has.

How do you use this for FTL communication between the two recipients?

What is the difference between this (hidden variables) and quantum entanglement in terms of the capability for FTL communication?

Karagoz
StevieTNZ said:
I've always been under the impression that if we start with the entangled state, for example, |H>|V> - |V>|H>, photon 1 goes through a 45 degree filter (or doesn't), the state is now |45>|V> - |135>|V - |45>|H> - |135>|H>.

Is this correct? I understand only by detecting the photon after the filter will determine whether it passed or not.

PeroK said:
What about a simpler, but related problem - effectively hidden variables. You send out a pair of socks in sealed containers. One container has a right sock and the other a left sock. The two containers are sent to two recipients.

Each recipient, upon opeing the container, not only sees what sock they have, but knows immediately what sock the other recipient has.

How do you use this for FTL communication between the two recipients?

What is the difference between this (hidden variables) and quantum entanglement in terms of the capability for FTL communication?

But can't one person change the state of his sock from left to righ? (Change his photon from one state to another?)

Won't it cause the other person's photon change its state instantly?

This is what's written in the same physics book:
"Entangled photons can affect each other immediately over arbitrary distances. However, it is unfortunately impossible to use this to send information faster than the light.
Consider entangled photons that are sent through two polarization filters. A person changes the axis of filter A back and forth in an attempt to communicate with a person at filter B. The person at B observes that his photon sometimes passes the filter and sometimes it's blocked. The results are closely related to the axis of filter A at all times. The problem is only that the person at B has no possibility of interpreting the observations, thus reading the message from A if he does not know the axis of filter A in each case.
"

I didn't get why person B can't interpret the observations and why he can't know the direction of filter A?

If his photon is blocked, won't he know that it's blocked because
1) his photon changed its state, because
2) the person by filter A changed state of his photon, because
3) the person by filter A changed the axis of filter A and made his photon pass through it

Why it doesn't work that way?

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I might be wrong but as I understand it you can only examine the sock and discover if it's left or right (which it is will be random). If it's a right sock then the other one will be a left sock. I don't think you can force your sock to be left or right.

Karagoz said:
But can't one person change the state of his sock from left to righ? (Change his photon from one state to another?)

No, of course not. That's the point. You measure the sock and you get L/R. You cannot compel your sock to one or the other. If you take your sock out and it's L, then you pull it to pieces and resew it as a R sock, that doesn't change the other sock to L.

Karagoz
I would recommend to have a look at the outcomes of the actual experiment.

If Alice does nothing to her photon and Bob performs a measurement on his photon, he has a probability of 50% to get either H or V as outcome regardless of the axis which he chooses.
If Alice performs a measurement, she can predict what Bob will measure. But Alice, too, has a probability of 50% to get either H or V regardless of the axis which she choses.
How is Bob supposed to determine whether he got his result from "his probability" or from "Alice's probability"?

Let's imagine we repeat the measurement ten times for both possibilities which I mentioned above and invent some data:
1) Alice does nothing. For each measurement, Bob chooses a certain axis. He gets outcomes H,V,H,H,V,H,V,V,H,V.
2) Alice performs measurements. For each measurement, she chooses a certain axis. For each of his measurements, Bob, too, chooses a certain axis. He gets outcomes V,V,H,V,H,H,V,H,H,V.

If you think that Alice can communicate with Bob via her measurements, you should be able to reconstruct a message from the second sequence but not from the first.

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Karagoz
Karagoz said:
[]

I didn't get why person B can't interpret the observations and why he can't know the direction of filter A?

If his photon is blocked, won't he know that it's blocked because
1) his photon changed its state, because
2) the person by filter A changed state of his photon, because
3) the person by filter A changed the axis of filter A and made his photon pass through it
Even if you could do this the communication between A and B is no faster than the photons.

Karagoz said:
"Entangled photons can affect each other immediately over arbitrary distances. However, it is unfortunately impossible to use this to send information faster than the light.

This is extremely misleading. It appears to assume that the photons have separate states before a measurement. They do not. Only the two-photon system has a state. This is where photons are different from socks.

Before a measurement there is only the two-photon system. After the measurement, there are two photons. Neither has communicated with the other.

The simple reason you cannot send a message is that nothing you do changes the initial two-photon state. It simply reveals something about that state. If you could choose the result - force the system to change to something you want - hence change the probabilities of what is observed, then you could send a message.

But, all you can do is measure the system and that changes nothing about the original state.

All the complicated arguments about filters and the like are just a smokescreen. Nothing changes the probabilities associated with that first measurement.

Karagoz
PS here's the nub. Suppose we have an entangled two-particle system and observer B decides to measure something (on one particle) and the probability is ##p## of getting a certain result.

1) B does that on a large number of experiments and a proportion ##p## have the said result.

Now, they rerun the experiment, where (by agreement) A measures something before B.

2) B gets the said result with the same proportion ##p##.

Then then rerun the experiment and A measures something else before B.

3) B gets the said result with the same proportion ##p##

Nothing A does can possiblity change the proportions of B's measurements.

What A could do is make his/her measurement, then send a message to B telling them what the result of the measurement was. B could wait for that message and then might know certainly what the result of his/her measurement will be in that particular case. The same proportion of results ##p## would build up over time, but now B does know in advance the results in each case.

But, unless B knows the result of A's measurement, nothing has changed as far as he/she is concerned. B is simply unable to tell whether A has made any measurement, or not, and what result they got. B's measurements are always in proportional agreement with the original state of the two-particle system.

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PeterDonis and Karagoz
Karagoz said:
but I didn’t get exactly why faster than light communication with entangled photons is impossible.
We have a bunch of older threads on this subject, but if you don't want to go digging for them, here's an explanation that may work for you. It's actually just what everyone else has been saying in this thread, but worded a bit differently

First, you understand from what has already been said that you only get one measurement on each photon? After that the entanglement is broken.

Second (and this is the bit whose implications you are missing), there is no way of creating either member of the pair in a particular state. So let's say that Alice measures first, and sets her polarizer to vertical. 50% of the time Alice's photon will pass through and 50% of the time it will be absorbed - that 50-50 split is because we cannot create the entangled pair in a state in which Alice is more likely to be vertical than horizontal, or vice versa.

This means that Bob receives a stream of photons half of which are vertical (the partners of the ones that didn't pass Alice's filter) and half of which are horizontal (the partners of the ones that did pass Alice's filter). If Bob's filter is oriented vertically, Bob will find that 50% of the photons pass his filter and 50% don't.

Now suppose that Alice changes her angle to something else, which we'll call ##\theta##. The stream of photons reaching Bob is made up 50% of photons polarized along the angle ##\theta## and 50% polarized along the angle ##\theta+\pi/2##. What fraction of these pass Bob's filter, which is still oriented vertically? The ones that are oriented at ##\theta## will pass with probability ##\cos^2\theta##; the ones that are oriented at ##\theta+\pi/2## will pass with probabiilty ##\cos^2(\theta+\pi/2)=\sin^2\theta##; the total number passing will be ##1/2\cos^2\theta+1/2\sin^2\theta=1/2(cos^2\theta+\sin^2\theta)=1/2## - and this is exactly what Bob observed before Alice changed the angle of her polarizer, so Bob cannot tell that Alice made any change and Alice cannot use the change to send a signal to Bob.

Play with the algebra a bit more and you should be able to satisfy yourself that no matter what either of them does with their polarizers, both will always see the same thing - half the incoming photons pass and half don't. In fact, they can't even tell that they're dealing with entangled photons. They aren't seeing anything different from if they were receiving two random streams of random photons from two unrelated sources, and they only find out about the entanglement if they get together afterwards and compare notes to see exactly what happened with each pair.

It would be a different story if we could arrange for all the pairs to be such that all will pass Alice's polarizer when it is vertical; then Bob would be able to infer the setting of Alice's polarizer by looking at the fraction that pass his polarizer - but we can't do this. The reason is in some math that goes beyond a B-level thread (OK for an I-level thread though) so you'll have to take my word for it that there is no entangled state in which we also know with certainty the initial state of Alice's photon.

Derek P, bhobba, zonde and 1 other person
That was a great explanation. I understood the most part. But this part:

Nugatory said:
Second (and this is the bit whose implications you are missing), there is no way of creating either member of the pair in a particular state. So let's say that Alice measures first, and sets her polarizer to vertical. 50% of the time Alice's photon will pass through and 50% of the time it will be absorbed - that 50-50 split is because we cannot create the entangled pair in a state in which Alice is more likely to be vertical than horizontal, or vice versa.

This means that Bob receives a stream of photons half of which are vertical (the partners of the ones that didn't pass Alice's filter) and half of which are horizontal (the partners of the ones that did pass Alice's filter). If Bob's filter is oriented vertically, Bob will find that 50% of the photons pass his filter and 50% don't.

Can you explain why is it that way?

Why 50% of the photons pass through if the polarizer is vertical?

And why the "partners" of photons that didn't pass through and the vertical polarizer and got absorbed become vertical themselves?

Isn't it 100% certain vertical photons pass through vertical polarizers?

Karagoz said:
Can you explain why is it that way?
Why 50% of the photons pass through if the polarizer is vertical?
It's that way for any angle, not just vertical; if we're working with entangled pairs of photons, 50% pass through Alice's polarizer no matter what the angle. As for why it's that way? That's part of the math of how entanglement works. The high school text you quoted earlier left all of that out because there's no easy way to explain it using only high school math... but if you're willing to work at it, Giancarlo Ghirardi's book "Sneaking a look at God's cards" explains it well without pushing beyond the limits of high-school math and a B-level thread.

And why the "partners" of photons that didn't pass through and the vertical polarizer and got absorbed become vertical themselves?
When a photon encounters a polarizer at any angle, there are two possibilities. Either the photon becomes polarized parallel to that axis and passes through the filter, or it become polarized perpendicular to that axis and is absorbed because it can't pass through the filter. Either way, you've determined the polarization of the photon on that axis and the way that high school text explains it, that also determines the polarization of the other photon on that same angle. In the most common case of entanglement, if one of the photons is polarized parallel to that axis so will pass through its filter, the other one will be polarized perpendicular to that axis so won't pass a filter set to the same angle. That's the case I was assuming, so I said that if one passes the other won't: if one is vertical the other will be horizontal.
But it's possible to create your entangled pairs in such a way that they both end up the same way (both parallel or both perpendicular when encountering filters set to the same angle), so you'll also see some examples worked out that way. Either way though, if you work through the algebra (which comes down to ##\sin^2+\cos^2=1##), you'll come to same conclusion: given that 50% of the photons that reach Alice pass her filter, then 50% of the photons that reach Bob will pass his filter - no matter what angles they choose and no matter how they change the angles.
Isn't it 100% certain vertical photons pass through vertical polarizers?
If a photon passes a vertical filter, then it is 100% certain that it will pass another vertical filter. But it won't be entangled after it's encountered the first filter - that's the "you only get one measurement" rule.

Karagoz said:
Why 50% of the photons pass through if the polarizer is vertical?
Maybe it will help if you look at experimental setup that generates polarization entangled photons:
1. Let's take vertically (V) polarized photon beam and split the photons using specific crystal into two horizontally polarized photons (H1 and H2). These two photons will be classically correlated. No entanglement what's so ever.
2. Let's take H polarized photon beam and split the photons using specific crystal into two vertically polarized photons (V1 and V2). Again these two photons will be classically correlated.
3. Now arrange the two crystals and source beam so that half of the source photons are converted into H1 and H2 and other half into V1 and V2 in such a way that H1 and V1 end up in one beam and H2 and V2 end up in another beam. Now the photons become polarization entangled.

As you can see for the photons to be polarization entangled there should be H and V polarized photons in the same beam. Obviously analyzing this beam with polarization beam splitter will give you half of the photons with one polarization and other half with orthogonal polarization no matter at what angle you place polarization beam splitter.

Karagoz said:
"Entangled photons can affect each other immediately over arbitrary distances. However, it is unfortunately impossible to use this to send information faster than the light.

That physics book is a load of bollocks.

Nothing is affected instantaneously.

Put a red slip of paper in an envelope and green in another. Take one envelope to the other side of the galaxy, universe - its doesn't matter - just some very large distance. Open one envelope and you instantly know the other slip. Did anything affect the envelope? Of course not - they are simply what's called correlated.

Actually QM is not our most fundamental theory - its Quantum Field Theory (QFT) and action at a distance is based on the so called cluster decomposition property:

Note - correlated systems are specifically precluded. The concept of instantaneous action at a distance doesn't even apply to them - they are precluded by definition. There is no mystery - the electrons or whatever are simply correlated.

What the issue is, is this - in QM correlations have different statistical properties to classical probability. This is hardly surprising because QM is simply an extension of normal probability theory:
https://www.scottaaronson.com/democritus/lec9.html

So what is all this stuff written about these correlations - why is it so 'captivating'. Well to those like me nothing is particularly worrisome - a bit interesting that's all. But to others they say - this is weird - we want it the same as ordinary probability. That's when things start to, IMHO, get weird - the phenomena itself not weird - but if you want to interpret it classically then you get issues. Bell showed - yes you can have it like ordinary probability theory - but only if you allow instantaneous communication between objects. That's all it is - people who like QM more 'common-sense' are led to this instantaneous communication idea.

To each their own - personally I just say - correlations are just that - correlations - they are not even part of the foundations of locality in QFT.

Thanks
Bill

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StevieTNZ said:
I've always been under the impression that if we start with the entangled state, for example, |H>|V> - |V>|H>, photon 1 goes through a 45 degree filter (or doesn't), the state is now |45>|V> - |135>|V - |45>|H> - |135>|H>.
StevieTNZ said:
Is this correct?
Seems fine. If you take projection of the first photon onto 45 deg angle the second photon will have a state |V>-|H>, and for projection onto 135 deg angle the second photon will have a state |V>+|H>. So the other photon has certain polarization in 45/135 deg basis.
StevieTNZ said:
I understand only by detecting the photon after the filter will determine whether it passed or not.
This depends on interpretation.

bhobba said:
That physics book is a load of bollocks.

Nothing is affected instantaneously.
Strong words on quite controversial topic. Ok, you have a viewpoint.

bhobba said:
Put a red slip of paper in an envelope and green in another. Take one envelope to the other side of the galaxy, universe - its doesn't matter - just some very large distance. Open one envelope and you instantly know the other slip. Did anything affect the envelope? Of course not - they are simply what's called correlated.
Thanks to Bell we know that this is not relevant for entanglement.

bhobba said:
Actually QM is not our most fundamental theory - its Quantum Field Theory (QFT) and action at a distance is based on the so called cluster decomposition property:

Note - correlated systems are specifically precluded. The concept of instantaneous action at a distance doesn't even apply to them - they are precluded by definition. There is no mystery - the electrons or whatever are simply correlated.
So Cluster decomposition principle does not apply to correlated systems. We observe correlated systems in experiments. So we know that CDP is false in some cases (it has limited domain of applicability). So exactly why CDP is relevant for the topic?

bhobba said:
What the issue is, is this - in QM correlations have different statistical properties to classical probability. This is hardly surprising because QM is simply an extension of normal probability theory:
https://www.scottaaronson.com/democritus/lec9.html

So what is call this stuff written about these correlations - why is it so 'captivating'. Well to those like me nothing is particularly worrisome - a bit interesting that's all. But to others they say - this is weird - we want it the same as ordinary probability. That's when things start to, IMHO, get weird - the phenomena itself not weird - but if you want to interpret it classically then you get issues. Bell showed - yes you can have it like ordinary probability theory - but only if you allow instantaneous communication between objects. That's all it is - people who do not like ordinary QM and want it more 'common-sense' are led to this instantaneous communication idea.
The very first proof of Bell's theorem had something to do with probabilities. But the world has moved on since then. There is Eberhard's proof (as part of this paper https://journals.aps.org/pra/abstract/10.1103/PhysRevA.47.R747) or if you have no access to Eberhard's paper there is layman level Nick Herbert's proof https://www.physicsforums.com/threads/a-simple-proof-of-bells-theorem.417173/#post-2817138. There is no need for probabilities in these proofs.

zonde said:
Strong words on quite controversial topic. Ok, you have a viewpoint.

Its simply the truth. Those that think otherwise misunderstand Bell. There is no controversy about it at all. Bell says nothing about particles instantaneously affecting others unless you want to view the world in a certain way some find more like common sense - but of course you do not have to. In fact QM says nothing about particles even existing apart from observation so even saying observing one particle affects the other is making assumptions the theory says nothing about. It's this very strange quirk of QM that is most likely responsible for the different statistical properties it has that classical probability like the slips of paper I mention does not. But regardless its still the same - just a correlation.

People that think otherwise, including the author of the physics quote from a Textbook, should read carefully the theorem and think a bit. If its not obvious even then I will spell out the detail in full as I have many times before.

To be fair its a common misconception, but misconception it certainly is.

Here is a simple explanation of Bell:
http://www.drchinese.com/Bells_Theorem.htm

If anyone hasn't read it do so and THINK.

I will spell it out if required - but really its pretty obvious.

Thanks
Bill

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zonde said:

Well others don't seem to think so
https://arxiv.org/pdf/0709.3909.pdf

But if you want to discuss that one I suggest you start a thread posting the paper that supports the claim - the link you gave is not to a generally downloadable paper.

Regarding Nicks supposed simple proof it has been discussed before, as per the link you posted, many thinking as I do he is a bit of a crackpot (I repeat the link for reference):

For example he says: Most sources of light in nature are unpolarized (SPOT produces a 50/50 random sequence of "0"s and "1"s). However sunlight scattered in the atmosphere becomes partially polarized with a polarization angle and degree of polarization that varies across the sky. SPOT would be an good tool to measure the polarization pattern of scattered sunlight in the heavens but we will use SPOT instead to probe the nature of quantum reality here on Earth

Sounds like stats to me. But we have professors of probability that post here - I would defer to their view.

Thanks
Bill

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bhobba said:
Its simply the truth. Those that think otherwise misunderstand Bell. There is no controversy about it at all. Bell says nothing about particles instantaneously affecting others unless you want to view the world in a certain way some find more like common sense - but of course you do not have to. In fact QM says nothing about particles even existing apart from observation so even saying observing one particle affects the other is making assumptions the theory says nothing about. It's this very strange quirk of QM that is most likely responsible for the different statistical properties it has that classical probability like the slips of paper I mention does not. But regardless its still the same - just a correlation.
It seems you are arguing that there is no instantaneous effects in QM model.
So as I understand I should read your phrase "Nothing is affected instantaneously." as "Nothing is affected instantaneously in QM."
But then it does not seem that the book speaks about theory. After all individual photons are not elements of QM as you noted.

And btw Bell (I mean the author himself) speaks about remote operations (not exactly particles however) affecting measurement : "It is the requirement of locality, or more precisely that the result of a measurement on one system be unaffected by operations on a distant system with which it has interacted in the past, that creates the essential difficulty."

bhobba said:
Well others don't seem to think so
https://arxiv.org/pdf/0709.3909.pdf
Khrennikov in his paper is not quoting Eberhard's paper to which I gave the link. And Khrennikov in his paper seems to be arguing about seriousness of "fair sampling loophole" in rather obscure and sophisticated way. But "fair sampling loophole" is closed in recent experiments. So it does not seem relevant IMO.

bhobba said:
Regarding Nicks supposed simple proof it has been discussed before, as per the link you posted, many thinking as I do he is a bit of a crackpot
This is Ad hominem fallacy. The argument is simple enough that you should not relay on credibility of the author. I came up with very similar proof before I learned about Nicks proof. And I saw this type of proof discussed in one of the recent threads and it came from different source (book).

zonde said:
I came up with very similar proof before I learned about Nicks proof. And I saw this type of proof discussed in one of the recent threads and it came from different source (book).

You miss the point - his proof, right or wrong, regardless of what you think of him, used probability.

If you want to discuss Eberhard' proof post the full proof and in another thread at least at the I level then it can be discussed.

Thanks
Bill

bhobba said:
You miss the point - his proof, right or wrong, regardless of what you think of him, used probability.

If you want to discuss Eberhard' proof post the full proof and in another thread at least at the I level then it can be discussed.

Thanks
Bill
zonde said:
Khrennikov in his paper is not quoting Eberhard's paper to which I gave the link. And Khrennikov in his paper seems to be arguing about seriousness of "fair sampling loophole" in rather obscure and sophisticated way. But "fair sampling loophole" is closed in recent experiments. So it does not seem relevant IMO.This is Ad hominem fallacy. The argument is simple enough that you should not relay on credibility of the author. I came up with very similar proof before I learned about Nicks proof. And I saw this type of proof discussed in one of the recent threads and it came from different source (book).
I don't understand why you two are arguing when you both must believe that a separable conditional probability ##P(xy|\alpha\beta)=P(x|\alpha)P(y|\beta)## cannot reproduce the predictions of QT ? The least that is required is ##P(xy|\alpha\beta)=P(x|\alpha)P(y|\alpha\beta)## meaning there must be common up-to-date information sharing. This could be correlation or communication. Both look the same and we can't use the model to decide - or can we ?

bhobba

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