Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I How does Bell make this step in his proof?

  1. Oct 4, 2016 #1

    N88

    User Avatar

    From drchinese website http://www.drchinese.com/David/Bell_Compact.pdf on page 406 there are 2 equations at the top. How do you get from the top one to the second one? There is a hint about using (1) but I think it cannot be done. You might be able to do it with other assumptions but I think they will be wrong too. Any suggestions?
     
  2. jcsd
  3. Oct 4, 2016 #2

    Nugatory

    User Avatar

    Staff: Mentor

    I have taken the liberty of changing the thread title to reflect the question that's actually being asked.
     
  4. Oct 5, 2016 #3

    zonde

    User Avatar
    Gold Member

    ##A(\vec{a},\lambda)## is either 1 or -1 so ##[1-A(\vec{b},\lambda)A(\vec{c},\lambda)]## is either 0 or 2. If for particular ##\lambda## it's 0 then expression in the first integral for the same ##\lambda## is 0 as well. If it's 2 then in first integral it's 2 or -2. So there is no way how module of the first integral can get bigger than the second integral.
     
  5. Oct 5, 2016 #4

    N88

    User Avatar

    Are you discussing how to get from the second equation to the third equation? My question is how to get from the first to the second.
     
  6. Oct 5, 2016 #5

    zonde

    User Avatar
    Gold Member

    You mean how to see that ##A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)=-A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)-1]## ?
     
  7. Oct 5, 2016 #6

    N88

    User Avatar

    Thanks. Yes.
     
  8. Oct 5, 2016 #7

    zonde

    User Avatar
    Gold Member

    It's easy to see in opposite direction. Just notice that from (1) you get that ##A(\vec{b},\lambda)^2=1##
     
  9. Oct 5, 2016 #8

    N88

    User Avatar

    But in the second equation, the first ##A(\vec{b},\lambda)## was tested by Bob when Alice's detector was set to ##\vec{a}##. The second ##A(\vec{b},\lambda)## in the second equation can be tested by Bob when Alice's detector is at setting ##\vec{c}##. How does Bob come to have the same particles to test to get ##A(\vec{b},\lambda)^2=1##? Because doesn't that relation ##A(\vec{b},\lambda)^2=1## only hold for tests on the same λ?

    Looks like there's another assumption somewhere?
     
  10. Oct 5, 2016 #9

    zonde

    User Avatar
    Gold Member

    ##\lambda## is independent of measurement settings. If follows from assumption that is stated at the top of page 404 (Alice's result does not depend on Bob's measurement setting and vice versa).
     
  11. Oct 5, 2016 #10

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    If (and this sequence works in either direction):
    ##-[A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)]=A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)-1]##

    Then:
    ##-[A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)]=A(\vec{a},\lambda)A(\vec{b},\lambda)[-1 + A(\vec{b},\lambda)A(\vec{c},\lambda)]##

    Then:
    ##A(\vec{a},\lambda)A(\vec{c},\lambda)=A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)]##

    Then we are finally saying:
    ##1[A(\vec{a},\lambda)A(\vec{c},\lambda)]=A(\vec{b},\lambda)A(\vec{b},\lambda)[A(\vec{a},\lambda)A(\vec{c},\lambda)]=A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)]##

    Which I think Bell took for granted we would follow, I guess with some reference to (1) - not sure. As zonde already mentioned, ##A(\vec{b},\lambda) A(\vec{b},\lambda)## is 1 classically. The above is a relationship that should be true if ##\vec{a}, \vec{b}, \vec{c}## are all simultaneously well defined. That assumption is essentially the assumption of realism - that ##\vec{a}, \vec{b}, \vec{c}## are well-defined at all times. If they are not, it is like dividing by zero and you get results that themselves are suspect. Ergo a Bell Inequality, which does not always hold.
     
  12. Oct 5, 2016 #11

    N88

    User Avatar

    But here you have assumed on the RHS the equality that I want to derive from the LHS. My earlier post remains valid:

    On RHS, the first ##A(\vec{b},\lambda)## was tested by Bob when Alice's detector was set to ##\vec{a}##. The second ##A(\vec{b},\lambda)## in the second equation can be tested by Bob when Alice's detector is at setting ##\vec{c}##. How does Bob come to have the same particles to test to get ##A(\vec{b},\lambda)^2=1##? Because doesn't that relation ##A(\vec{b},\lambda)^2=1## only hold for tests on the same λ?

    Looks like there's another assumption somewhere?[/QUOTE]
     
  13. Oct 5, 2016 #12

    N88

    User Avatar

    I agree. But to make sense of their test results, they need to be comparing paired results from tests on the same particle pair. And Bell and QM do not allow that you can test the same pair twice.
     
  14. Oct 6, 2016 #13

    zonde

    User Avatar
    Gold Member

    We can do that in Bell's model. In his model outcome of measurement is fully determined by measurement angle and ##\lambda## and of course we can copy value (or set of values) of lambda.
    Of course we can't rewind back reality, change something and then look what is different. But our models are not copies of reality, they are just models. And we can do such things with models.
     
    Last edited: Oct 6, 2016
  15. Oct 6, 2016 #14

    rubi

    User Avatar
    Science Advisor

    We can do it only in classical models. In quantum models, non-commuting observables don't simultaneously have well-defined outcomes, so that it's not possible to consistently assign values to unobserved quantities. This phenomenon is called contextuality. It's very apparent in the GHZ state, where any assignment of values to unbserved quantities is inconsistent.
     
  16. Oct 6, 2016 #15

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm not sure I understand your objection. Bell is assuming (for the purpose of showing that the assumption leads to a contradiction with the predictions of quantum mechanics) that when a twin-pair of particles is created, a value of [itex]\lambda[/itex] is associated with each particle, and that Alice's measurement result is a deterministic function of [itex]\lambda[/itex]. She deterministically gets the result [itex]A(\vec{a}, \lambda)[/itex], where [itex]\lambda[/itex] is the hidden variable associated with the twin pair, and [itex]\vec{a}[/itex] is the axis she uses for the measurement. Similarly, Bob's measurement result is a different function of [itex]\lambda[/itex]: He gets the result [itex]B(\vec{b}, \lambda)[/itex], where [itex]\vec{b}[/itex] is the axis he uses.

    So [itex]A(\vec{a}, \lambda)[/itex] and [itex]B(\vec{b}, \lambda)[/itex] are assumed to be ordinary functions of two parameters. His proof then just amounts to showing that there are no such functions reproducing the predictions of quantum mechanics.

    In reasoning about [itex]A[/itex] and [itex]B[/itex], you can forget that they correspond to measurements, and just think of them as mathematical functions.
     
  17. Oct 6, 2016 #16

    zonde

    User Avatar
    Gold Member

    Quantum contextuality basically means that certain measurement results of individual particles can't be predetermined by hidden variables. This is the same conclusion that Bell reaches with his theorem.
    However it does not mean that we can't calculate statistical predictions for alternative scenarios.
     
  18. Oct 6, 2016 #17

    rubi

    User Avatar
    Science Advisor

    It means that we cannot calculate statistical predictions for scenarios in which non-commuting observables have simultaneously well-defined outcomes. Hence the answer to post #12 is that this is where Bell's assumptions contradict QM, which eventually allows for the inequality violation.
     
  19. Oct 6, 2016 #18

    zonde

    User Avatar
    Gold Member

    OP has doubts whether Bell reasoning does not make additional assumption. So what is your answer? Is there additional assumption that does not follow from assumptions he has stated?
     
  20. Oct 6, 2016 #19

    rubi

    User Avatar
    Science Advisor

    Yes, Bell's paper is not very explicit about the assumptions he makes. This is the reason for the big confusion about his assumptions. Bell never states explicitely that the existence of values for unobserved variables is an assumption. On the contrary, he believes that this follows from the EPR argument. That's why I was pointing at the GHZ state, which shows that the EPR argument itself is inconsistent.
     
  21. Oct 6, 2016 #20

    zonde

    User Avatar
    Gold Member

    Right at the start Bell says: "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."
    Then later on he provides his reasoning how he arrives at "more complete specification of the state" described as lambda.
    Then again he gives his assumption at the top of page 404 (that B's result does not depend on A's measurement setting and vice versa).

    But as I understand you maintain position that Bell's reasoning is faulty and there can be independent measurements on two distant particles that with certainty will give (UP,DOWN) or (DOWN,UP) but not (UP,UP) nor (DOWN,DOWN) and yet it does not follow that we can meaningfully talk about UP or DOWN measurement for each particle separately. So it is reasonable option that combination of two independent parts taken together is meaningful while each independent part taken separately is not.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted