High School Does the EPR experiment imply QM is incomplete?

[DrChinese]

that's why the simulation works fine.

It doesn't work too bad, but there is the one problem where it produces the wrong values for about half the photon angle settings - at least the ones I tried.

0 and 0: 100%
0 and 30: about 75%
0 and 60: about 75% OOPS should be 25%
0 and 90: 100% OOPS should be 0%.

Just saying...

 

[Boing3000]

The fact is, you seem to be ignoring Bell. I know you have read the material, but you are skipping the essential elements - and that is what PeterDonis is referencing.

I ignore it so much that I have actually translated the logic of its proof in functional code.

You see, there is no progress to be made if you keep thinking that Bell's proof is wrong in anyway, or that I am trying to prove it wrong.

 

[Boing3000]

It doesn't work too bad, but there is the one problem where it produces the wrong values for about half the photon angle settings - at least the ones I tried.

0 and 0: 100%
0 and 30: about 75%
0 and 60: about 75% OOPS should be 25%
0 and 90: 100% OOPS should be 0%.

Just saying...

The correlation is 100% at 90°.
I think you are mixing up correlation with identity. In other word cos(delta) with 1/2 + (cos(delta))^2

Edit: Ho, and I just see your last line. So much for the tone...

 

[DrChinese]

That's incorrect. There is no local hidden variable. And that's the whole point of Bell.
There is a set(of one element) of hidden variable that can easily reproduce QM prediction. But those variable must be non-local.
...
There isn't complete independence, because V accomplish that. The dependence just can NOT be guessed by A nor B. (unless they compare their result)

No one is arguing that non-local hidden variables won't work as an algorithm. The issue is that the measurement setting of either A or B... and the outcome at that setting... must be part of the algorithm. You are in denial that must occur.

1. Assume A and B are randomly given a "+" or a "-" initial value (same for both). To be specific, it is "+".
2. A is measured first at 138 and yields value of "+", its initial value, and needs nothing else.
3. But B, which is also "+", needs to know about the 138. That value is transmitted to B via FTL means.
4. B can now give answers for any measurement setting simply by applying the Cos^2(theta) function.

All good. There is FTL communication in this model.

 

[DrChinese]

The correlation is 100% at 90°.

Photons don't produce those statistics, my friend. It's COS^2(theta).

If you're not sure, try taking a couple of polarized lenses and crossing them. 0% light transmitted.

 

[DrChinese]

The correlation is 100% at 90°.
I think you are mixing up correlation with identity. In other word cos(delta) with 1/2 + (cos(delta))^2

True, I use match % and not proper correlation %. However, on that basis, the proper correlation at 90 degree is -1. At 45 degrees it is 0. At theta=0 degrees, it is 1. You don't produce those either with your simulation.

 

[PeterDonis]

The wave function picks up new information after the preparation and before measurement ?

No. The wave function for a system of multiple particles is a function of the position of all the particles (if we are working in the position basis), not just the position of the particle you are interested in. So the wave function for the probability of a particular measurement result for the particle you are interested in does not depend only on quantities associated with the particle you are interested in.

Have you ever heard of "hidden variable", "non-locality", "entanglement" and "correlation" ?

Of course. And what you are saying indicates to me that you do not understand how all of these things actually function in the proof of Bell's Theorem or in discussions of the EPR experiment. In fact, per my comment above, it's not even clear to me that you understand what a wave function is for a system containing more than one particle.

I strongly advise you to take a step back at this point and carefully consider all the responses you have already received before posting further. You are getting very close to a thread ban, since you are not the OP of this thread and the things you are bringing up don't seem to me to be making any positive contribution to discussion of the question the OP asked.

 

[kurt101]

Further, entanglement can exist between particles that have never existed in a common light cone.

Does anyone have doubts that the collapse algorithm would model this kind of entanglement correctly? I have not tried simulating it, but just reading through the experiment (the one with photons) it seems like it would.

 

[kurt101]

That's incorrect. There is no local hidden variable. And that's the whole point of Bell.
There is a set(of one element) of hidden variable that can easily reproduce QM prediction. But those variable must be non-local.
That's not my diagram. My diagram is $A{\rightarrow}V{\leftarrow}B$. A cannot reach B in any way.

I strongly suspect you are agreeing on substance, but somehow disagreeing on language. I also have trouble when you say "A cannot reach B in any way". I think this is just a difference in what we want to call it. Because I would rather say A gives B its state when it interacts with the polarizer. I believe it is the same algorithm as what you are saying, but you would rather just make another variable called V that is shared between A and B until right after A interacts with its polarizer and then A no longer is associated with V, but B still uses V until it interacts.

Hopefully were just saying the same thing.

 

[kurt101]

No. The wave function for a system of multiple particles is a function of the position of all the particles (if we are working in the position basis), not just the position of the particle you are interested in. So the wave function for the probability of a particular measurement result for the particle you are interested in does not depend only on quantities associated with the particle you are interested in.

For calculating the probability of an outcome you square the sum of the probability amplitudes. What is interesting to me about this math as it relates to the collapse algorithm is that the QM math result only gives you terms that include 2 outcomes. For example if you have outcomes with probability amplitudes a, b, c; then you would get a probability of aa + ab + ac + bb + ba + bc + cc + ca + cb. You don't ever see a term like abc or aab for example. Does this imply that you really only have a superposition of 2 outcomes when you make a single measurement? (Even when you have the possibility for many outcomes) And bringing this back to the collapse algorithm, the algorithm only considers entanglement for exactly 2 particles at a time, until the interaction takes place and then presumably the entanglement is between a different set of 2 particles.

 

[DrChinese]

I strongly suspect you are agreeing on substance, but somehow disagreeing on language. I also have trouble when you say "A cannot reach B in any way". I think this is just a difference in what we want to call it. Because I would rather say A gives B its state when it interacts with the polarizer. I believe it is the same algorithm as what you are saying, but you would rather just make another variable called V that is shared between A and B until right after A interacts with its polarizer and then A no longer is associated with V, but B still uses V until it interacts.

I don't know about agreeing on substance or not. The issue is whether a non-local influence is occurring. Clearly, that is a viable possibility per Bell. And, for example, Bohmian Mechanics postulates a manner in which that can occur. However, there is mutual influence between A and B in that model.

So that is different than what Boing was saying. And his simulation did not feature mutual influence, it was based on either A influencing B or vice versa. But he was also denying that influence, which kinda misses the "out" that a non-local influence provides.

 

[PeterDonis]

For calculating the probability of an outcome you square the sum of the probability amplitudes.

The sum of the probability amplitudes for that outcome. You appear to be misunderstanding what that means.

Suppose there are three possible outcomes; call them A, B, and C. For outcome A, suppose there are two ways that can happen, with amplitudes $a_1$ and $a_2$. For outcome B, suppose there are three ways it can happen, with amplitudes $b_1$, $b_2$, and $b_3$. And for outcome C, suppose there is only one way it can happen, with amplitude $c$.

Then the probabilities are: for A, $\left( a_1 + a_2 \right)^2$; for B, $\left( b_1 + b_2 + b_3 \right)^2$; and for C, $c^2$. There are no "cross terms" between outcomes.

You should re-think things in the light of the above.

 

[Boing3000]

No one is arguing that non-local hidden variables won't work as an algorithm.

That's good, we are making progress.

The issue is that the measurement setting of either A or B... and the outcome at that setting... must be part of the algorithm.

It is. That has been explained over and over again.

You are in denial that must occur.

No, I am not. You are just enable to understand the sharing of a variable and the precise direction of the flow of information.There is no function that take A and B as input. There are two function that take A(thus V) and the other take B(thus V). This is the same V. Not only the same value, it is the same placeholder.
There is no way for function A to get any information about B. None. Yet the sharing of the variable allow for the correlation to hold

1. Assume A and B are randomly given a "+" or a "-" initial value (same for both). To be specific, it is "+".
2. A is measured first at 138 and yields value of "+", its initial value, and needs nothing else.
3. But B, which is also "+", needs to know about the 138. That value is transmitted to B via FTL means.
4. B can now give answers for any measurement setting simply by applying the Cos^2(theta) function.

All good. There is FTL communication in this model.

I have no idea why you, among all, would bring such trivial model. I have argue against it because step 3 is not possible. Not by virtue of the theory of relativity, nor even the lack of formula to be put in an algorithm (making it wishful thinking).
It is not possible because B may have transmitted a message to A, before B felt any need(??) to know anything. The absence of commutativity at every step makes it impossible.

Photons don't produce those statistics, my friend. It's COS^2(theta).

100% of the photon at one hand are strictly correlated (inverted) to those at the other end (at 90°). There is absolute correlation.
It is at 45° that the correlation is 0.5 (totally random)

If you're not sure, try taking a couple of polarized lenses and crossing them. 0% light transmitted.

Exactly, and I am happy you agree with that. But I am under the impression that you believe that Alice or Bob would get 0% light transmitted in a EPR experiment involving TWO entengled photon. That's not the case. They always get 50% light transmitted, whatever they try to do.

True, I use match % and not proper correlation %. However, on that basis, the proper correlation at 90 degree is -1. At 45 degrees it is 0. At theta=0 degrees, it is 1. You don't produce those either with your simulation.

I just dived the number of identical pair(fail) with the numbers of inverted pair(entangled) (making 0.5 the absolute lower value possible (random)). I will add another output of the correlation computation, if that help anybody.

 

[PeterDonis]

There is no function that take A and B as input. There are two function that take A(thus V) and the other take B(thus V). This is the same V.

You are missing the point: if you code the algorithm this way, it cannot reproduce the predictions of quantum mechanics. Your "V" is the equivalent of Bell's local hidden variables (he calls them $\lambda$). So your algorithm will produce results that must obey the Bell inequalities; but the predictions of QM (and actual experimental results) violate the Bell inequalities. The only way to have an algorithm that reproduces the predictions of QM (and violates the Bell inequalities) is for the function in the algorithm to take the settings at both A and B as inputs.

100% of the photon at one hand are strictly correlated (inverted) to those at the other end (at 90°). There is absolute correlation.
It is at 45° that the correlation is 0.5 (totally random)

But your algorithm has to give the correct correlations at all angles, not just at one. Of course if you just have the 90 degree angle, it's easy to write an algorithm your way that gives perfect anti-correlation. But that won't work for all angles.

 

[Boing3000]

No.

I was also under this impression thus you contradicted you previous statement.

And that is false: the wave function "computes" what happens at event A using input from other events as well as event A.

Because:

The wave function for a system of multiple particles is a function of the position of all the particles (if we are working in the position basis), not just the position of the particle you are interested in. So the wave function for the probability of a particular measurement result for the particle you are interested in does not depend only on quantities associated with the particle you are interested in.

And that exactly similar to V. After the preparation the wave function is non-local and consul-table by every event in the laboratory to get they (probability) value. Note that any individual event only need to specify its local-position to the wave function to get get the probability, nothing else.
That's identical to how the hidden variable work in the simulation.

 

[PeterDonis]

And that exactly similar to V.

Okay, at this point you're either trolling, shifting your ground, or waving your hands about the properties of an algorithm that you haven't actually constructed. Please give an explicit definition of V and your algorithm.

 

[PeterDonis]

Note that any individual event only need to specify its local-position to the wave function to get get the probability, nothing else.

If you think this is possible while still reproducing the predictions of QM, you need to think again. As I've already said before: the wave function of a two-particle system is a function of both particles' positions, not just one.

 

[Boing3000]

Your "V" is the equivalent of Bell's local hidden variables (he calls them $\lambda$).

No, it is not. My V can be local (two copy of the same Value), or non-local (a sharing of a variable)

So your algorithm will produce results that must obey the Bell inequalities; but the predictions of QM (and actual experimental results) violate the Bell inequalities.

It does both.

The only way to have an algorithm that reproduces the predictions of QM (and violates the Bell inequalities) is for the function in the algorithm to take the settings at both A and B as inputs.

This is false. And the way you propose is not possible, for reason explained previously.
It is not even sensible because one function cannot decide BOTH apparatus angle then pilot A and B at once (it is superdeterminism which is kind of lame)
Only a simulation, using a wavefunction (which does NOT use A and B as input for a result, but only A local position as input for the result (and only A & B as preparation/initial WF value/state).
Or another type of non-local Value/state

But your algorithm has to give the correct correlations at all angles, not just at one. Of course if you just have the 90 degree angle, it's easy to write an algorithm your way that gives perfect anti-correlation. But that won't work for all angles.

Of course

 

[PeterDonis]

Of course

In other words: you agree that your algorithm only reproduces the correct QM correlations for one angle--90 degrees--not all angles?

 

[PeterDonis]

My V can be local (two copy of the same Value), or non-local (a sharing of a variable)

What's the difference? You're saying there's this variable V that can be taken as an input by the "function" that gives you the measurement result for either A or B. What does it matter how the variable is stored?

If you would explicitly write out your algorithm, showing how V is calculated and how the measurement results at A and B are calculated, it would be a lot easier to understand what you are saying. Right now, as I said, it looks like you are either trolling, shifting your ground, or waving your hands about the properties of an algorithm that you haven't actually constructed.

one function cannot decide BOTH apparatus angle then pilot A and B at once

The function doesn't have to "decide" both measurement angles. It has to take them as inputs. It also doesn't have to output both A's and B's results; it just has to output either A's or B's--you can have one function for each.

 

[Boing3000]

Okay, at this point you're either trolling, shifting your ground, or waving your hands about the properties of an algorithm that you haven't actually constructed. Please give an explicit definition of V and your algorithm.

Have you ever consider that there is another possibility ? That you simply never tried to understand what I say instead of trolling me and threatening me ?
The explicit definition is at line 13 and both locality (or not) done at lie 14 here

If you think this is possible while still reproducing the predictions of QM, you need to think again. As I've already said before: the wave function of a two-particle system is a function of both particles' positions, not just one.

A OK. I though it was only the case in BM. So Alice or Bod cannot compute anything from the WF.

 

[Boing3000]

In other words: you agree that your algorithm only reproduces the correct QM correlations for one angle--90 degrees--not all angles?

Why don't you test it, to check...

 

[Boing3000]

What does it matter how the variable is stored?

Non-locality makes all the difference. Once you understand what Bell is actually proving, you know how to conserve correlation, without allowing any possibility for A to "influence" B (or the contrary)

Right now, as I said, it looks like you are either trolling, shifting your ground, or waving your hands about the properties of an algorithm that you haven't actually constructed.

I have posted it maybe a year ago in another thread.

The function doesn't have to "decide" both measurement angles. It has to take them as inputs. It also doesn't have to output both A's and B's results; it just has to output either A's or B's--you can have one function for each.

If I have one function for each, then each of those functions will only use one input. But that is right, you may create two local functions from one(non-local) which is globally aware of the two. That is the closest thing you have written that resemble the even more simply algorithm that I use.

 

[PeterDonis]

The explicit definition is at line 13 and both locality (or not) done at lie 14

I can read Javascript, but you can hardly expect everyone here to do so. When I asked for an explicit description I meant a description in words or math, something that everyone here could be expected to understand.

 

[PeterDonis]

When I asked for an explicit description I meant a description in words or math, something that everyone here could be expected to understand.

Actually, rather than wait for @Boing3000 to do this, I'm going to give such a description of a straightforward algorithm using the QM math. Here, schematically, is such a description:

The algorithm assumes a number of "trials", each of which consists of making a polarization measurement on each of a pair of entangled photons. Each pair of photons is assumed to be prepared identically in the "PP" state (i.e., their polarizations are 100% correlated if measured at the same angle, and 100% anti-correlated if measured at angles 90 degrees apart). The photons in each pair are labeled A and B, corresponding to the locations of the polarizers that measure them. (Strictly speaking, each polarizer either passes its photon or not, and a photon detector after the polarizer either detects the photon or doesn't.)

The algorithm provides two functions, $f_A$ and $f_B$, each of which takes defined inputs (given below) and outputs a measurement result for its corresponding photon for that trial. All of the information about the preparation procedure (identical for each trial) is encoded in these functions. So the only variables for each trial are the measurement settings (polarizer angles) at A and B; everything else is known in advance. Each measurement result is a boolean value: "1" means the photon was detected (i.e., passed the polarizer), "0" means the photon was not detected (did not pass the polarizer).

The inputs provided to the algorithm are the measurement settings (A, B) for each trial. These can be determined by any means desired, but they are external to the algorithm; the algorithm does not compute them, it just takes them as inputs. The input also, implicitly, determines the number of trials (by the number of pairs of settings that are provided).

According to Bell's Theorem, in order to properly reproduce the QM predictions (and the actual experimental results), each function, $f_A$ and $f_B$, must take as inputs the measurement settings for that trial at both A and B. It is impossible to have $f_A$ only take as input the settings for A, and $f_B$ only take as input the settings for B, and still reproduce the QM predictions.

I'll hold off on saying what the functions $f_A$ and $f_B$ actually are, for the case under discussion, since the above might already be enough to clarify what, exactly, the algorithm in question needs to compute and what inputs it takes.

 

[PeterDonis]

Alice or Bod cannot compute anything from the WF.

If they only know their own measurement setting, that's true. They need to know both measurement settings in order to compute probabilities from the WF.

 

[PeterDonis]

They need to know both measurement settings in order to compute probabilities from the WF.

Actually, I need to clarify this. If, for example, all Alice wants to know is the probability of a single photon passing her polarizer, if she knows the photons are all prepared in the PP state, then she already knows the answer to that question: 50%.

However, as I described in post #55, the algorithm we have been talking about has to produce an actual sequence of measurement results, not just the probability of a single photon passing the polarizer. The sequence of measurement results has to satisfy all of the predictions of QM, not just its prediction for what fraction of Alice's photons pass her polarizer. Those predictions include the correlation between Alice's and Bob's results; and producing results that satisfy the QM predicted correlations is what the algorithm cannot do unless the functions it uses to output Alice's and Bob's results at each trial take as input both Alice's and Bob's measurement settings for that trial.

 

[PeterDonis]

Why don't you test it, to check...

I have. Its results seem obviously wrong; for example, it's giving 100% correlation with Alice's angle at 0 and Bob's angle at 90.

 

[Mentz114]

Non-locality makes all the difference. Once you understand what Bell is actually proving, you know how to conserve correlation, without allowing any possibility for A to "influence" B (or the contrary) I have posted it maybe a year ago in another thread.If I have one function for each, then each of those functions will only use one input. But that is right, you may create two local functions from one(non-local) which is globally aware of the two. That is the closest thing you have written that resemble the even more simply algorithm that I use.

It is necessary and sufficient to model entanglement to assume that the entangled pair always have the same value for the entangled property.
Writing $P(xy|\alpha\beta) = \tfrac{1}{2}(P(x|\alpha)P(y|\alpha\beta) + P(y|\beta)P(x|\alpha\beta)$ to reflect the fact that whichever projection happens first, the setting is known by the other wing. $\alpha$ and $\beta$ are the binary variables representing the polarizer settings on the two wings, and 'xy' is the four possible outcomes (00,01,10,11).

From Malus law $P(x|\alpha) = \cos(\theta_0-\alpha)^2$ so we can write
$P(11|\alpha\beta)=\tfrac{1}{2}\cos(\alpha-\beta)^2\left[ \cos(\theta_0-\alpha)^2+\cos(\theta_0-\beta)^2 \right]$
$P(00|\alpha\beta)=\tfrac{1}{2}\sin(\alpha-\beta+\pi/2)^2\left[ \sin(\theta_0-\alpha)^2+\sin(\theta_0-\beta)^2 \right]$
and so
$P(11\ or\ 00) = \cos(\alpha-\beta)^2$
The only assumptions are that whichever photon is projected first is irrelevant and that the photons always have the same polarization. It also shows that that only the probability of a coincidence ( or anticoincidence) is estimable.

 

[PeterDonis]

only the probability of a coincidence ( or anticoincidence) is estimable

I don't think this is correct; QM also predicts, for the entangled state in question, that, for each wing taken by itself, the probability of a photon passing its polarizer is 50%.