Thermal interpretation and Bell's inequality

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SUMMARY

The forum discussion centers on the interpretation of Bell's inequality as presented in his 1964 paper, specifically the transition from the first to the second equation on page 198. Participants debate the assumptions underlying this transition, particularly in relation to Quantum Mechanics (QM) and the Thermal Interpretation (TI). It is established that Bell's equation (3) represents a QM expectation value that violates the assumptions used in the earlier equations, highlighting the distinction between classical interpretations and QM. The discussion concludes that while TI may offer alternative insights, it does not fundamentally alter the mathematical framework established by Bell.

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  • #61
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
N88 said:
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
DarMM said:
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.

DarMM said:
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
N88 said:
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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