Notices
Results 1 to 20 of 20

Thread: Bell Inequalities question

  1. #1 Bell Inequalities question 
    Senior Member
    Join Date
    Oct 2013
    Location
    UK
    Posts
    997
    Do the Bell Inequalities say that there can be no hidden variables or that there can be no inherent variables at all when describing a quantum particle?
    Reply With Quote  
     

  2. #2  
    KJW
    KJW is offline
    Super Moderator
    Join Date
    Jun 2013
    Posts
    861
    Bell inequalities are logical relationships that objects possessing definite properties must satisfy. Quantum mechanics violates this, indicating that quantum objects need not possess definite values for particular variables. Bell's theorem says that quantum mechanics cannot be both local and counterfactually definite.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
    Reply With Quote  
     

  3. #3  
    Senior Member
    Join Date
    Oct 2013
    Location
    UK
    Posts
    997
    Quote Originally Posted by KJW View Post
    Bell inequalities are logical relationships that objects possessing definite properties must satisfy. Quantum mechanics violates this, indicating that quantum objects need not possess definite values for particular variables. Bell's theorem says that quantum mechanics cannot be both local and counterfactually definite.
    So no inherent variables at all then, not just hidden ones?
    Reply With Quote  
     

  4. #4  
    KJW
    KJW is offline
    Super Moderator
    Join Date
    Jun 2013
    Posts
    861
    The violation of Bell's inequalities shows that the quantum mechanics cannot be described in terms of classical statistics.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
    Reply With Quote  
     

  5. #5  
    KJW
    KJW is offline
    Super Moderator
    Join Date
    Jun 2013
    Posts
    861
    Quote Originally Posted by Jilan View Post
    So no inherent variables at all then, not just hidden ones?
    It doesn't say no inherent variables. It just says that variables do not necessarily have definite values.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
    Reply With Quote  
     

  6. #6  
    KJW
    KJW is offline
    Super Moderator
    Join Date
    Jun 2013
    Posts
    861
    It isn't that quantum mechanics must violate Bell's inequalities. It is that quantum mechanics can violate Bell's inequalities.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
    Reply With Quote  
     

  7. #7  
    Senior Member
    Join Date
    Oct 2013
    Location
    UK
    Posts
    997
    Quote Originally Posted by KJW View Post
    It doesn't say no inherent variables. It just says that variables do not necessarily have definite values.
    Does that come about mathematically from the requirement to square the amplitudes to get the probabilities?
    Reply With Quote  
     

  8. #8  
    KJW
    KJW is offline
    Super Moderator
    Join Date
    Jun 2013
    Posts
    861
    Quote Originally Posted by Jilan View Post
    Does that come about mathematically from the requirement to square the amplitudes to get the probabilities?
    I can't answer that question definitively, but it seems to come from (1): the complex-valued (or perhaps just non-positive) nature of the quantum wavefunction, and (2): the nature of the quantum superposition of multi-particle states.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
    Reply With Quote  
     

  9. #9  
    Senior Member
    Join Date
    Oct 2013
    Location
    UK
    Posts
    997
    Quote Originally Posted by KJW View Post
    I can't answer that question definitively, but it seems to come from (1): the complex-valued (or perhaps just non-positive) nature of the quantum wavefunction, and (2): the nature of the quantum superposition of multi-particle states.
    Thanks for this. I've just been wondering about it as it appears that if you were just looking at probability amplitudes (or moduli of them if complex) rather than the squares of them, the inequality holds. But once you square the amplitudes and try to equate them with some inherent property you get the inequality. I don't know though if this is universally true or just is the case with the 60 degree spin type situation that was given as an example of the inequality breaking down. I would be very interested to know as I can see why set theory could be applied to the amplitude (which is a complete description of the system) but don't see how it could be legitimately applied it to the square of them ( which I interpret as the joint probability of something meeting something else coming the other way).
    In this post please read square to mean the multiplication by the complex conjugate. Thanks for reading this... I hope it makes some sense.
    Reply With Quote  
     

  10. #10  
    KJW
    KJW is offline
    Super Moderator
    Join Date
    Jun 2013
    Posts
    861
    Quote Originally Posted by Jilan View Post
    Thanks for this. I've just been wondering about it as it appears that if you were just looking at probability amplitudes (or moduli of them if complex) rather than the squares of them, the inequality holds. But once you square the amplitudes and try to equate them with some inherent property you get the inequality. I don't know though if this is universally true or just is the case with the 60 degree spin type situation that was given as an example of the inequality breaking down. I would be very interested to know as I can see why set theory could be applied to the amplitude (which is a complete description of the system) but don't see how it could be legitimately applied it to the square of them ( which I interpret as the joint probability of something meeting something else coming the other way).
    In this post please read square to mean the multiplication by the complex conjugate. Thanks for reading this... I hope it makes some sense.
    You are quite possibly right with regards to the "square" of the wavefunction, so maybe I should make this (3) in post #8.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
    Reply With Quote  
     

  11. #11  
    Senior Member
    Join Date
    Oct 2013
    Location
    UK
    Posts
    997
    Quote Originally Posted by KJW View Post
    You are quite possibly right with regards to the "square" of the wavefunction, so maybe I should make this (3) in post #8.
    It does appear that set theory is being applied to something it shouldn't be applied to at all so it's not surprising it doesn't work. Perhaps I'm oversimplifying it, I don't know. I'm coming round to thing that all of QM involves joint probabilities of two things (particle and detector) being in the same place at the at the same time. So set theory is pretty much irrelevant.
    Reply With Quote  
     

  12. #12  
    KJW
    KJW is offline
    Super Moderator
    Join Date
    Jun 2013
    Posts
    861
    Quote Originally Posted by Jilan View Post
    It does appear that set theory is being applied to something it shouldn't be applied to at all so it's not surprising it doesn't work. Perhaps I'm oversimplifying it, I don't know.
    It isn't just set theory. It is also about the Boolean logic associated with objects possessing definite properties. That is, Bell's inequality can be recast as a Boolean logic identity involving three Boolean variables: "particle has property 1", "particle has property 2", and "particle has property 3". The violation of this identity by quantum mechanics indicates that one cannot associate these variables with either a TRUE or FALSE value. That is why local hidden variables cannot work: because a hidden variable implies that one can say in principle, even if it cannot be determined as fact, that the hidden variable has definite value X. That is, the statements "hidden variable has value X", "hidden variable has value Y", etc are either definitely true or definitely false, in violation of quantum mechanics. The non-locality "loophole" basically is saying that the quantum entangled particles are somehow remaining in physical communication and informing each other of their quantum mechanical decisions, a notion that in my opinion is totally absurd.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
    Reply With Quote  
     

  13. #13  
    KJW
    KJW is offline
    Super Moderator
    Join Date
    Jun 2013
    Posts
    861
    Quote Originally Posted by Jilan View Post
    I'm coming round to thing that all of QM involves joint probabilities of two things (particle and detector) being in the same place at the at the same time. So set theory is pretty much irrelevant.
    But quantum wavefunction values are not probabilities. Probability values are positive real numbers and quantum wavefunction values are neither. Note that probability values are ultimately about counting objects, so a quantum wavefunction goes against the notion of objects that can be counted. A better way to view quantum mechanics is to consider a real-valued vector space (the vector space of quantum mechanics is complex-valued, but I wish to downplay the significance of complex numbers). Let the orthonormal basis vectors of this vector space correspond to independent distinct "realities" (outcomes of a measurement). Now consider a unit vector in this space. This unit vector can be expressed as a weighted vector sum of the basis vectors:



    Because is a unit vector, Pythagoras' theorem says:



    The unit vector is a weighted superposition of the "realities" corresponding to the basis vectors. Suppose it represents the probabilities of finding the actual "reality" to be in each of the "realities" corresponding to the basis vectors. Since the actual "reality" must be in one of the basis "realities", the sum of the probabilities must be equal to 1. But the above Pythagorean expression says that the sum of the squares of the vector components must be equal to 1. Therefore, it is the square of the vector component that is the measure of probability of the actual "reality" being the corresponding basis "reality".
    Last edited by KJW; 11-09-2013 at 12:07 PM.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
    Reply With Quote  
     

  14. #14  
    Senior Member
    Join Date
    Oct 2013
    Location
    UK
    Posts
    997
    Quote Originally Posted by KJW View Post
    Since the actual "reality" must be in one of the basis "realities", the sum of the probabilities must be equal to 1. But the above Pythagorean expression says that the sum of the squares of the vector components must be equal to 1. Therefore, it is the square of the vector component that is the measure of probability of the actual "reality" being the corresponding basis "reality".
    Thanks for this, I can appreciate why you need to square the amplitudes to arrive at a probability in real time, but which amplitudes? You say that the actual reality must be in one of the basis realities, (based on squaring the amplitude of the entity), but I thought that this was only the case once it was measured. My issue is that this must have as much to do with the detector as with entity being measured so applying set theory to the entity alone would be illogical. I have been wondering whether the Born rule is more a probability of a particle meeting another particle of the same type (electrons say being detected by another electron -is there another way?). To first approximation in the centre of mass frame one would be the conjugate of the other.
    Reply With Quote  
     

  15. #15  
    KJW
    KJW is offline
    Super Moderator
    Join Date
    Jun 2013
    Posts
    861
    Quote Originally Posted by Jilan View Post
    You say that the actual reality must be in one of the basis realities, (based on squaring the amplitude of the entity), but I thought that this was only the case once it was measured.
    Oh! I didn't explicitly mention the measurement, did I? In fact, the choice of basis vectors is determined by the observable being measured. Without measurement, the unit vector representing the quantum state is itself a "reality" in its own right. That is, it can be one of the basis vectors of an orthonormal basis. While we as observers tend to focus on the position basis, quantum mechanics doesn't have a preferred basis. In this way, the principle of relativity carries over into quantum mechanics.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
    Reply With Quote  
     

  16. #16  
    KJW
    KJW is offline
    Super Moderator
    Join Date
    Jun 2013
    Posts
    861
    Quote Originally Posted by Jilan View Post
    I have been wondering whether the Born rule is more a probability of a particle meeting another particle of the same type (electrons say being detected by another electron -is there another way?)
    I tend to think that the Born rule is a statement that the interaction of a quantum state with the detector leads to orthogonal detector states, the orthogonality of the detector states being a natural consequence of the high-dimensionality of the Hilbert space of this macroscopically large multi-particle state. Measuring devices are macroscopic for a reason. If one interacts a microscopic particle with another microscopic particle, then all one gets is a microscopic entangled two-particle state, with almost all the uncertainties of the original particles, and therefore not constituting what we would regard as a measurement.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
    Reply With Quote  
     

  17. #17  
    Senior Member
    Join Date
    Oct 2013
    Location
    UK
    Posts
    997
    Quote Originally Posted by KJW View Post
    Oh! I didn't explicitly mention the measurement, did I? In fact, the choice of basis vectors is determined by the observable being measured. Without measurement, the unit vector representing the quantum state is itself a "reality" in its own right. That is, it can be one of the basis vectors of an orthonormal basis. While we as observers tend to focus on the position basis, quantum mechanics doesn't have a preferred basis. In this way, the principle of relativity carries over into quantum mechanics.
    Yes, that's my point. If the quantum state is the reality you can apply set theory to it, however to apply it to the product of the state multiplied by its compex conjugate and then to say the result rules out hidden variables seems a bit random to me.
    Reply With Quote  
     

  18. #18  
    Senior Member
    Join Date
    Oct 2013
    Location
    UK
    Posts
    997
    Quote Originally Posted by KJW View Post
    I tend to think that the Born rule is a statement that the interaction of a quantum state with the detector leads to orthogonal detector states, the orthogonality of the detector states being a natural consequence of the high-dimensionality of the Hilbert space of this macroscopically large multi-particle state. Measuring devices are macroscopic for a reason. If one interacts a microscopic particle with another microscopic particle, then all one gets is a microscopic entangled two-particle state, with almost all the uncertainties of the original particles, and therefore not constituting what we would regard as a measurement.
    Macroscopic or not doesn't the process of detection always start with a microscopic interaction which then gets amplified in some way to produce a non-reversible process?
    Reply With Quote  
     

  19. #19  
    KJW
    KJW is offline
    Super Moderator
    Join Date
    Jun 2013
    Posts
    861
    Quote Originally Posted by Jilan View Post
    Yes, that's my point. If the quantum state is the reality you can apply set theory to it, however to apply it to the product of the state multiplied by its compex conjugate and then to say the result rules out hidden variables seems a bit random to me.
    When I say that the quantum state is the reality, that is within the context of the multiverse of the many-worlds interpretation, where every eigenstate of every observable is a reality. The notion of hidden variables implicitly rejects the existence of the multiverse by seeking to explain quantum mechanics in terms of the observed reality that we are familiar with. It is here that we encounter the obstruction of Bell's inequality, for we cannot attribute a definite value to every observable. This is not a problem for the many-worlds interpretation because a quantum state is a superposition of different values and that is taken to be the reality. Technically, the problem arises because different observables generally do not share eigenstates, and therefore a definite value for one observable is a superposition of different values for another observable. This is unavoidable for it is an inherent property of the observables themselves, not the quantum state.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
    Reply With Quote  
     

  20. #20  
    Senior Member
    Join Date
    Oct 2013
    Location
    UK
    Posts
    997
    Quote Originally Posted by KJW View Post
    When I say that the quantum state is the reality, that is within the context of the multiverse of the many-worlds interpretation, where every eigenstate of every observable is a reality. The notion of hidden variables implicitly rejects the existence of the multiverse by seeking to explain quantum mechanics in terms of the observed reality that we are familiar with. It is here that we encounter the obstruction of Bell's inequality, for we cannot attribute a definite value to every observable. This is not a problem for the many-worlds interpretation because a quantum state is a superposition of different values and that is taken to be the reality. Technically, the problem arises because different observables generally do not share eigenstates, and therefore a definite value for one observable is a superposition of different values for another observable. This is unavoidable for it is an inherent property of the observables themselves, not the quantum state.
    Ah, I get it now. The difference is how you interpret the state.
    Reply With Quote  
     

Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •