Thermal interpretation and Bell's inequality

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  • #51
DarMM
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OK, thanks, let me try again. Let's call the 2 top equations on p.198: (14a) and (14b).

So, to me, (14a) is true; it is simply a definition of Bell's terms.

But (14b) looks false to me; and this view is backed by the fact that it (by plain mathematics) leads to Bell's famous eqn (15): which IS false under QM.
So the first equation has:
$$-A(a, \lambda)A(b, \lambda) + A(a, \lambda)A(c, \lambda)$$
Extract ##A(a, \lambda)##:
$$A(a, \lambda)\left[A(c, \lambda) - A(b, \lambda)\right]$$
Since ##A(b,\lambda)^{2} = 1## by the equation (1) in the paper:
$$A(a, \lambda)\left[A(b,\lambda)^{2}A(c, \lambda) - A(b, \lambda)\right]$$
Then just extract ##A(b,\lambda)##:
$$A(a, \lambda)A(b,\lambda)\left[A(b,\lambda)A(c, \lambda) - 1\right]$$

So it's just basic algebra. No physical assumptions. The assumption QM breaks comes earlier in the paper. It's the assumption that ##A, B, C## are in fact random variables over some common space of polarization settings and variables ##\lambda##.
 
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  • #52
N88
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So the first equation has:
$$-A(a, \lambda)A(b, \lambda) + A(a, \lambda)A(c, \lambda)$$
Extract ##A(a, \lambda)##:
$$A(a, \lambda)\left[A(c, \lambda) - A(b, \lambda)\right]$$
Since ##A(b,\lambda)^{2} = 1## by the equation (1) in the paper:
$$A(a, \lambda)\left[A(b,\lambda)^{2}A(c, \lambda) - A(b, \lambda)\right]$$
Then just extract ##A(b,\lambda)##:
$$A(a, \lambda)A(b,\lambda)\left[A(b,\lambda)A(c, \lambda) - 1\right]$$

So it's just basic algebra. No physical assumptions. The assumption QM breaks comes earlier in the paper. It's the assumption that ##A, B, C## are in fact random variables over some common space of polarization settings and variables ##\lambda##.
But can't we test your basic algebra by stopping at the second equation and completing the integral there?

Since the result is nothing like Bell's inequality, it looks like Bell's difficulties are already in the second equation?
 
  • #53
PeterDonis
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can't we test your basic algebra by stopping at the second equation and completing the integral there?

Since the result is nothing like Bell's inequality

Please show your work. It might not look like Bell's inequality to you, but if you want to claim it's mathematically inconsistent with Bell's inequality, in the face of two people now showing you the basic algebra that converts on integrand into the other, you are going to have to show us the math in detail. Either that or we can just close this thread since obviously you are simply refusing to accept what anyone tells you.
 
  • #54
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Please show your work. It might not look like Bell's inequality to you, but if you want to claim it's mathematically inconsistent with Bell's inequality, in the face of two people now showing you the basic algebra that converts on integrand into the other, you are going to have to show us the math in detail. Either that or we can just close this thread since obviously you are simply refusing to accept what anyone tells you.
Please, I am not seeking to refuse anything. I am seeking to understand what I am being told. I understood you to say that Bell's inequality is based on mathematics unrelated to QM.

In asking me to show my work in detail: by observation it appears that completion of the integral over DarMM's second equation equals zero? Reason: The first term takes the values ±1 randomly, so (under integration over λ) its average value will be zero.

That was all.
 
  • #55
PeterDonis
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I understood you to say that Bell's inequality is based on mathematics unrelated to QM.

That's correct.

by observation it appears that completion of the integral over DarMM's second equation equals zero? Reason: The first term takes the values ±1 randomly, so (under integration over λ) its average value will be zero.

If this is true of any version of the integral, it's true of all of them; all of the integrands are simple algebraic transformations of each other, so they must all result in the same value on integration. So your reasoning here must be wrong.
 
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  • #56
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N88 said: I understood you to say that Bell's inequality is based on mathematics unrelated to QM.
That's correct.
N88 said: by observation it appears that completion of the integral over DarMM's second equation equals zero? Reason: The first term takes the values ±1 randomly, so (under integration over λ) its average value will be zero.
If this is true of any version of the integral, it's true of all of them; all of the integrands are simple algebraic transformations of each other, so they must all result in the same value on integration. So your reasoning here must be wrong.

Are you in a position to help me find my error?

It looks like a straight-forward integral whose first term (a function of λ) takes the values ±1 randomly?
 
  • #57
PeterDonis
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Are you in a position to help me find my error?

No, since you're unwilling to do even the simple work of writing down the integral in question and explaining, term by term, what you think it should give and why. Either show your work or this thread will be closed.
 
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  • #58
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No, since you're unwilling to do even the simple work of writing down the integral in question and explaining, term by term, what you think it should give and why. Either show your work or this thread will be closed.

From the second eqn in Post #51 above, completing the integral in line with Bell (1964):​
$$\smallint d\lambda \rho(\lambda)A(a, \lambda)\left[A(c, \lambda) - A(b, \lambda)\right]=0$$
since ##A(a, \lambda)## is a function of ##\lambda## and takes the values ±1 randomly.​
 
  • #59
PeterDonis
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since ##A(a, \lambda)## is a function of ##\lambda## and takes the values ±1 randomly

But the integrand is not just ##A(a, \lambda)##. It contains two other factors, both of which also vary with ##\lambda##. You have to look at how the entire integrand varies with ##\lambda##, not just one factor.
 
  • #60
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Bell assumes no retrocausality, no nonlocal effects, all variables come from a common sample space and that there is a single world.
In his theorem.

Beyond the theorem, he was (at that time essentially the only one) supporter of Bohmian mechanics, which is nonlocal, and the aim of the theorem was to prove that the nonlocality of BM is not a valid argument against it, because it is unavoidable for any realistic interpretation of quantum theory.
 
  • #61
N88
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For PeterDonis: Of course! Many thanks and my apologies! The problem occurs lower down! Let me redefine it using {.} to keep related terms together:

From the first eqn in Post #51 above, including the integrals in line with Bell (1964):​
$$E(a,b)-E(a,c)=-\smallint d\lambda \rho(\lambda)\left[\{A(a, \lambda)A(b, \lambda)\}-\{A(a, \lambda)A(c, \lambda)\}\right]\;\;(1)$$
$$=-\smallint d\lambda \rho(\lambda)\{A(a, \lambda)A(b, \lambda)\}\left[1-\{A(a, \lambda)A(b, \lambda)\}\{A(a, \lambda)A(c, \lambda)\}\right].\;\;(2)$$
$$So, since \;A(a, \lambda)A(b, \lambda)\leq1: \;\;(3)$$
$$ |E(a,b)-E(a,c)|\leq\smallint d\lambda \rho(\lambda)\left[1-\{A(a, \lambda)A(b, \lambda)\}\{A(a, \lambda)A(c, \lambda)\}\right]\;\;(4)$$
$$\leq1-E(a,b)E(a,c).\;\;(5)$$

Now, in eqn (4), if I used ##A(a, \lambda)A(a, \lambda)=1##, I would get Bell's inequality; just like you and DarMM and many others do using ##A(b, \lambda)A(b, \lambda)=1## in the alternative formulation in Post #51.

But then I would have converted (5) -- by an improper separation of terms* -- from an inequality that is never false into Bell's inequality which is frequently false.

* By an improper separation I mean this: I would have converted ##E(a,b)E(a,c)## into ##E(b,c)##; which is unlikely to be true. That a function ##F(b,c)## should equal a function ##F(a,b)F(a,c)## over all ##a,b,c##.

The above should show why I am still seeking to understand the mathematics behind Bell's theorem.

My question: Given the above, does Bell's inequality ##
|E(a,b)-E(a,c)|\leq1+E(b,c)
## arise from an improper mathematical separation of terms?
 
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  • #62
DarMM
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From the first eqn in Post #51 above, including the integrals in line with Bell (1964):
$$E(a,b)-E(a,c)=-\smallint d\lambda \rho(\lambda)\left[\{A(a, \lambda)A(b, \lambda)\}-\{A(a, \lambda)A(c, \lambda)\}\right]\;\;(1)$$
$$=-\smallint d\lambda \rho(\lambda)\{A(a, \lambda)A(b, \lambda)\}\left[1-\{A(a, \lambda)A(b, \lambda)\}\{A(a, \lambda)A(c, \lambda)\}\right].\;\;(2)$$
$$So, since \;A(a, \lambda)A(b, \lambda)\leq1: \;\;(3)$$
$$ |E(a,b)-E(a,c)|\leq\smallint d\lambda \rho(\lambda)\left[1-\{A(a, \lambda)A(b, \lambda)\}\{A(a, \lambda)A(c, \lambda)\}\right]\;\;(4)$$
$$\leq1-E(a,b)E(a,c).\;\;(5)$$

Now, in eqn (4), if I used ##A(a, \lambda)A(a, \lambda)=1##, I would get Bell's inequality; just like you and DarMM and many others do using ##A(b, \lambda)A(b, \lambda)=1## in the alternative formulation in Post #51.

But then I would have converted (5) -- by an improper separation of terms* -- from an inequality that is never false into Bell's inequality which is frequently false.

* By an improper separation I mean this: I would have converted ##E(a,b)E(a,c)## into ##E(b,c)##; which is unlikely to be true. That a function ##F(b,c)## should equal a function ##F(a,b)F(a,c)## over all ##a,b,c##.
I'm genuinely having a hard time understanding this. Where have you shown that ##E(a,b)E(a,c)## can be converted into ##E(b,c)##? I'm not following.

Also how do you get from your equation (1) to your equation (2)? Your equations are not mine or those given by Bell. Where are you getting them from?
 
  • #63
N88
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I'm genuinely having a hard time understanding this. Where have you shown that ##E(a,b)E(a,c)## can be converted into ##E(b,c)##? I'm not following.

In (4), if we let ##\{A(a, \lambda)A(b, \lambda)\}\{A(a, \lambda)A(c, \lambda)\}## reduce to ##A(b, \lambda)A(c, \lambda)## by allowing ##A(a, \lambda)A(a, \lambda)=1,## (as stated) then the integral gives ##-E(b,c) ##. It is similar to the way you get the same final result.

You and I and Bell start and finish at the same point IF we allow that reduction. Noting that such a reduction moves us from a totally valid inequality to Bell's partially valid inequality.

Also how do you get from your equation (1) to your equation (2)? Your equations are not mine or those given by Bell. Where are you getting them from?

Eqn (1) = eqn (2) by algebra, by way of ##\{A(a, \lambda)A(b, \lambda)\}\{A(a, \lambda)A(b, \lambda)\} = 1.##

Peter stressed that Bell's inequality is mathematical. So I am trying to understand how the mathematics goes from a physically valid start to a result that is false under QM.
 
  • #64
PeterDonis
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I am trying to understand how the mathematics goes from a physically valid start

You continue to miss the point: since the whole derivation is just a mathematical proof, if the conclusion doesn't match QM, the starting point doesn't either. In other words, what Bell is showing is that his starting point is not physically valid. It looks like it ought to be physically valid, but it isn't.

Since you appear unable to grasp this simple point despite numerous attempts to explain it, there is no point in continuing this thread. Thread closed.
 
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