3 ms·

Yes! (and good question) In this case we use Einstein's definition of local hidden variable (which he called "element of reality"): "If, without in any way dis

Yes! (and good question) In this case we use Einstein's definition of local hidden variable (which he called "element of reality"): "If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of reality corresponding to that quantity."

pmoriarty 5y ago
If I may infer that this statement has been interpreted as not a conditional as it is stated verbatim (in the form A -> C where A is the antecedent and C is the consequent), but instead is a biconditional (in the form A <-> C), then it seems to be stating that to exist is to be certainly predictable. That's a curious definition of existence. And now that I think about it a bit more, it seems like it's not really talking about what existence is, but rather the consequences of existence and the consequences of certain prediction. That's not quite the same as a definition of existence, is it?
- ziofill 5y ago
  What the statement says is that if a property is certainly predictable, then it exists. It doesn't say that all existing things are certainly predictable.
  - pmoriarty 5y ago
    But that doesn't say what it means by existence. It leaves that part unsaid. To go a bit deeper in to this, it's not just saying certainly predictable properties are certainly predictable. It wants to say something else about them: that they exist. What does it mean by that? The statement as it stands is more about certainly predictable properties than it is about existence itself.
    - ziofill 5y ago
      I think I understand your point: we don't define existence per se. But here we only need to define the existence of the value of a property. And we define it by saying that if the we can predict the value before measuring it, then the value "exists". Take it as an axiom. When I said "the very value of the entangled properties does not exist locally" in my first post I meant that there is no way to predict the value (or of consistently assigning a value to the property) before performing a measurement.

pmoriarty 5y ago

If I may infer that this statement has been interpreted as not a conditional as it is stated verbatim (in the form A -> C where A is the antecedent and C is the consequent), but instead is a biconditional (in the form A <-> C), then it seems to be stating that to exist is to be certainly predictable. That's a curious definition of existence. And now that I think about it a bit more, it seems like it's not really talking about what existence is, but rather the consequences of existence and the consequences of certain prediction. That's not quite the same as a definition of existence, is it?

ziofill 5y ago
What the statement says is that if a property is certainly predictable, then it exists. It doesn't say that all existing things are certainly predictable.
- pmoriarty 5y ago
  But that doesn't say what it means by existence. It leaves that part unsaid. To go a bit deeper in to this, it's not just saying certainly predictable properties are certainly predictable. It wants to say something else about them: that they exist. What does it mean by that? The statement as it stands is more about certainly predictable properties than it is about existence itself.
  - ziofill 5y ago
    I think I understand your point: we don't define existence per se. But here we only need to define the existence of the value of a property. And we define it by saying that if the we can predict the value before measuring it, then the value "exists". Take it as an axiom. When I said "the very value of the entangled properties does not exist locally" in my first post I meant that there is no way to predict the value (or of consistently assigning a value to the property) before performing a measurement.