Quantum Mechanics with a Hidden Variable!

[FrediFizzx]

It is an absolutely trivial derivation. There is nothing "mysterious" or "unexplained" about it at all. So why someone should try a different and much more convoluted route is beyond me.

But we also see, without referring to Bell's theorem at all, that in general all four combinations will eventually be produced for almost any given pair of settings (they all have non-zero probability). But a binary hidden variable could only produce two of these combinations. So something is clearly wrong with the paper.

There are a few different ways to derive the QM correlation for EPR-Bohm. There is nothing really "standard" about any of them. Ours just happens to follow the "standard" derivation in the appendix of the paper.

Now, you make a statement without proof that a binary HV could only produce two of the four combinations. Please prove your assertion. Thanks.

[Heinera]

Now, you make a statement without proof that a binary HV could only produce two of the four combinations. Please prove your assertion. Thanks.

How about this: "binary" means two? And given that a hidden variable in the sense of Bell should contain all relevant information in order to predict the outcomes, there can only be two possible combinations of outcomes if the variable is binary, right?

To borrow a phrase from Joy Christian, I think any schoolboy should be able to follow this argument, since it simply involves the capacity to distinguish two from four.

[FrediFizzx]

Now, you make a statement without proof that a binary HV could only produce two of the four combinations. Please prove your assertion. Thanks.

How about this: "binary" means two?

So what? How about this: 2 x 2 = 4. Please prove your assertion. But I don't think you can because for the right handed result we will still have the four outcome possibilities and for the left handed result we will still have the four outcome possibilities.
.

[Heinera]

Now, you make a statement without proof that a binary HV could only produce two of the four combinations. Please prove your assertion. Thanks.

How about this: "binary" means two?

So what? How about this: 2 x 2 = 4. Please prove your assertion. But I don't think you can because for the right handed result we will still have the four outcome possibilities and for the left handed result we will still have the four outcome possibilities.
.

takes binary values, let's call them "" and "" respectively.

Then the predicted combinations of outcomes can only be

or


In other words, two.

[FrediFizzx]

Now, you make a statement without proof that a binary HV could only produce two of the four combinations. Please prove your assertion. Thanks.

How about this: "binary" means two?

So what? How about this: 2 x 2 = 4. Please prove your assertion. But I don't think you can because for the right handed result we will still have the four outcome possibilities and for the left handed result we will still have the four outcome possibilities.
.

takes binary values, let's call them "" and "" respectively.

Then the predicted combinations of outcomes can only be

or


In other words, two.

Sorry, but that is complete nonsense and not a proof. You are making a false assumption. Do you even know what your false assumption is?
.

[Heinera]

Sorry, but that is complete nonsense and not a proof. You are making a false assumption. Do you even know what your false assumption is?
.

And the false assumption is exactly what?

[FrediFizzx]

Sorry, but that is complete nonsense and not a proof. You are making a false assumption. Do you even know what your false assumption is?
.

And the false assumption is exactly what?

You are making the assumption that the hidden variable has something to do with the 4 outcome possibilities. It doesn't. The HV simply determines if you have a right handed singlet or a left handed singlet.

[Heinera]

You are making the assumption that the hidden variable has something to do with the 4 outcome possibilities. It doesn't. The HV simply determines if you have a right handed singlet or a left handed singlet.

I am simply making the assumption that the hidden variable is a hidden variable in the sense of Bell, where it has everything to do with the four outcome possibilities. After all, that's why he introduced the functions and , which determines the outcomes.

[FrediFizzx]

You are making the assumption that the hidden variable has something to do with the 4 outcome possibilities. It doesn't. The HV simply determines if you have a right handed singlet or a left handed singlet.

I am simply making the assumption that the hidden variable is a hidden variable in the sense of Bell, where it has everything to do with the four outcome possibilities. After all, that's why he introduced the functions and , which determines the outcomes.

Determining the outcomes is not the same as the determining the 4 outcome possibilities and Bell only specified that the outcomes will be +/-1. He did not specify that if the HV is +1 the outcome will be +1, etc.

Well, not entirely your fault as we haven't shown the measurement functions yet. Jay is still working on possible further developments for that. But you can't get around the fact that at each detection station, you will have spin up or spin down 50-50 chance. So it doesn't matter if the HV is plus or minus. You will still get all 4 outcome possibilities.
.

[Yablon]

You are making the assumption that the hidden variable has something to do with the 4 outcome possibilities. It doesn't. The HV simply determines if you have a right handed singlet or a left handed singlet.

I am simply making the assumption that the hidden variable is a hidden variable in the sense of Bell, where it has everything to do with the four outcome possibilities. After all, that's why he introduced the functions and , which determines the outcomes.

Determining the outcomes is not the same as the determining the 4 outcome possibilities and Bell only specified that the outcomes will be +/-1. He did not specify that if the HV is +1 the outcome will be +1, etc.

Well, not entirely your fault as we haven't shown the measurement functions yet. Jay is still working on possible further developments for that. But you can't get around the fact that at each detection station, you will have spin up or spin down 50-50 chance. So it doesn't matter if the HV is plus or minus. You will still get all 4 outcome possibilities.
.

I really don't want to jump into the middle of this discussion, because I prefer to lay everything out systematically step by step in my "do you agree?" thread. But let me at least say this:

No matter what else we may or may not know about "hidden variables," including whether local hidden variables do or even can exist in nature, one thing we know is this: if they exist, they are HIDDEN. They can never be detected by human observational equipment, by definition. So, to suggest that some experiment which uses the 2x2=4 combinations Alice's measurement together in some way with Bob's measurement to ferret out the actual value of a hidden variable, is entirely contradicted by the hidden variable being HIDDEN. If you were able to deduce some physical variable of any sort out of Alice's in combination with Bob's , whatever else that physical variable might be, it sure as heck would not be hidden any more. All you would prove is that the variable you deduced was not hidden. You would not have deduced a thing about any variables -- if they exist -- which truly are "hidden."

Jay

[Heinera]

No matter what else we may or may not know about "hidden variables," including whether local hidden variables do or even can exist in nature, one thing we know is this: if they exist, they are HIDDEN. They can never be detected by human observational equipment, by definition. So, to suggest that some experiment which uses the 2x2=4 combinations Alice's measurement together in some way with Bob's measurement to ferret out the actual value of a hidden variable, is entirely contradicted by the hidden variable being HIDDEN. If you were able to deduce some physical variable of any sort out of Alice's in combination with Bob's , whatever else that physical variable might be, it sure as heck would not be hidden any more. All you would prove is that the variable you deduced was not hidden. You would not have deduced a thing about any variables -- if they exist -- which truly are "hidden."

Jay

Who in the heck ever talked about ferreting out the value of the hidden variable?

[Yablon]

Who in the heck ever talked about ferreting out the value of the hidden variable?

But we also see, without referring to Bell's theorem at all, that in general all four combinations will eventually be produced for almost any given pair of settings (they all have non-zero probability). But a binary hidden variable could only produce two of these combinations. So something is clearly wrong with the paper.

Heine:

I see no basis for you to assert that "a binary hidden variable could only produce two of these combinations." To prove this point, you would need to postulate a binary hidden variable, then show how it only produces two combinations, and show how the combinations have not somehow revealed i.e. unknowingly ferreted out that hidden variable. And if you succeeded, you would only have proved that the particular hidden variable you postulated only produces two outcomes and was actually hidden. That would prove nothing about a different, binary hidden variable that someone else might attempt to use.

Jay

[Yablon]

Who in the heck ever talked about ferreting out the value of the hidden variable?

But we also see, without referring to Bell's theorem at all, that in general all four combinations will eventually be produced for almost any given pair of settings (they all have non-zero probability). But a binary hidden variable could only produce two of these combinations. So something is clearly wrong with the paper.

Heine:

I see no basis for you to assert that "a binary hidden variable could only produce two of these combinations." To prove this point, you would need to postulate a binary hidden variable, then show how it only produces two combinations, and show how the combinations have not somehow revealed i.e. unknowingly ferreted out that hidden variable. And if you succeeded, you would only have proved that the particular hidden variable you postulated only produces two outcomes and was actually hidden. That would prove nothing about a different, binary hidden variable that someone else might attempt to use.

Jay

And, by the way, all it would take for someone else to contradict and thereby disprove your assertion (not saying that the rigorous proof is simple), is to find a binary hidden variable compatible with producing all four outcomes. So what you are really saying is that that cannot be done period, ever, so don't even bother trying.

[Heinera]

I see no basis for you to assert that "a binary hidden variable could only produce two of these combinations." To prove this point, you would need to postulate a binary hidden variable, then show how it only produces two combinations, and show how the combinations have not somehow revealed i.e. unknowingly ferreted out that hidden variable. And if you succeeded, you would only have proved that the particular hidden variable you postulated only produces two outcomes and was actually hidden. That would prove nothing about a different, binary hidden variable that someone else might attempt to use.
Jay

What? This is of course exactly what I proved. Let me repeat:

takes binary values, let's call them "" and "" respectively.

Then the predicted combinations of outcomes can only be

or


In other words, two.

[Yablon]

What? This is of course exactly what I proved. Let me repeat:

takes binary values, let's call them "" and "" respectively.

Then the predicted combinations of outcomes can only be

or


In other words, two.

Please do us all a favor, and put an equal sign into both of your expressions, then give us the numeric values and which these "outcomes" are equal to. That is, please write:


or


Then tell us the numbers which each of and actually is or can be.

[FrediFizzx]

I see no basis for you to assert that "a binary hidden variable could only produce two of these combinations." To prove this point, you would need to postulate a binary hidden variable, then show how it only produces two combinations, and show how the combinations have not somehow revealed i.e. unknowingly ferreted out that hidden variable. And if you succeeded, you would only have proved that the particular hidden variable you postulated only produces two outcomes and was actually hidden. That would prove nothing about a different, binary hidden variable that someone else might attempt to use.
Jay

What? This is of course exactly what I proved. Let me repeat:

takes binary values, let's call them "" and "" respectively.

Then the predicted combinations of outcomes can only be

or


In other words, two.

Repeating a false proof doesn't all of a sudden mean it is true.


or


You forgot to include what those are equal to.
.

[Heinera]

Please do us all a favor, and put an equal sign into both of your expressions, then give us the numeric values and which these "outcomes" are equal to. That is, please write:


or


Then tell us the numbers which each of and actually is or can be.

Oh, that is simple. They can be and

[FrediFizzx]

Please do us all a favor, and put an equal sign into both of your expressions, then give us the numeric values and which these "outcomes" are equal to. That is, please write:


or


Then tell us the numbers which each of and actually is or can be.

Oh, that is simple. They can be and

Well of course they can be that, but they also can be n_1 = (+1, -1), etc.

[Yablon]

Who in the heck ever talked about ferreting out the value of the hidden variable?

Please do us all a favor, and put an equal sign into both of your expressions, then give us the numeric values and which these "outcomes" are equal to. That is, please write:


or


Then tell us the numbers which each of and actually is or can be.

Oh, that is simple. They can be and

Well then, if you observe +1 then and if you observe -1 then . So you have ferreted out the hidden variable, just as I thought! Congratulations!!!

[Heinera]

So you have ferreted out the hidden variable, just as I thought! Congratulations!!!

Yes, but only because a simple binary hidden variable is so easy to ferret out. So what?