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B Is QE knowing how to correlate particles' measurements?

  1. Apr 18, 2017 #1
    source = http://scienceblogs.com/principles/2012/03/14/entanglement-is-not-that-magic/
    It's not clear to me from the above whether this experiment is repeated on the same entangled pair of photons, or a different pair each time. Also, thanks to SR, I have trouble with the "when."


    Would the following correctly paraphrase the above?

    Particles are considered entangled if we know how to correlate their measurements, and not if we don't. In other words, there is still/always a correlation, we just don't know what it is.
     
  2. jcsd
  3. Apr 18, 2017 #2

    Nugatory

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    A different pair every time. This will be completely clear if you look at the mathematical description of the system state before and after the first measurement, instead of relying on the (inevitably somewhat imprecise) natural-language description.

    It may or may not be a correct paraphrase, but either way it is somewhere between seriously misleading and flat-out wrong. Particles are considered entangled if the wave function describing the quantum system has a specific mathematical property (called "non-factorizable"). This definition will also address your other concern: "Also, thanks to SR, I have trouble with the 'when.'".

    Without going into more detail than will work in a B-level thread, here is an example:
    Consider a quantum system with two commuting observables, which I will call ##\hat{A}## and ##\hat{B}##. A measurement of ##\hat{A}## can yield the values ##\alpha_1## or ##\alpha_2## and a measurement of ##\hat{B}## can yield the values ##beta_1## or ##\beta_2##; in the lingo we would say that the ##\alpha_i## are the two eigenvalues of ##\hat{A}## and the ##\beta_i## are the two eigenvalue of ##\hat{B}##.

    One of the possible quantum states of this system is the state "A measurement of ##\hat{A}## will yield ##\alpha_1## and a measurement of ##\hat{B}## will yield ##\beta_1##"; we would write that state as ##|\alpha_1\beta_1\rangle##. It is not an entangled state. Nor is the state "A measurement of ##\hat{A}## will yield ##\alpha_2## and a measurement of ##\hat{B}## will yield ##\beta_2##", which we would write as ##|\alpha_2\beta_2\rangle##. In both these states, the fact that knowing one value tells us the other is no more mysterious than the fact that looking at one face of a coin tells us what the other face is.

    However, there are also superpositions of these states such as ##|\alpha_1\beta_1\rangle+|\alpha_2\beta_2\rangle##. This is an example of a state that is not factorizable, and where we would consider ##\hat{A}## and ##\hat{B}## to be entangled. If we measure either ##\hat{A}## or ##\hat{B}##, the superposition will go away (if you like you could say the wave function collapses, but that is just about guaranteed to get you in trouble with special relativity so I suggest that you don't) and the system will be in either the state ##|\alpha_1\beta_1\rangle## or ##|\alpha_2\beta_2\rangle##.

    Now, suppose that ##\hat{A}## is "the spin measured at Alice's detector" and ##\hat{B}## is "the spin measured at Bob's detector".... and that superposition state describes the quantum system that is informally described as "two entangled particles". Note some key properties of this system:
    - The correlation happens because the superposition ends.
    - It doesn't matter whether both ##\hat{A}## and ##\hat{B}## are measured; any measurement ends the superposition.
    - There's no entanglement once the superposition is ended. That's why you only get one correlated measurement on the pair.
    - It doesn't matter which measurement happens "first"; either way the system ends up in one of the two unsuperimposed states. This is why it doesn't matter that relativity tells us that there is no meaningful definition of "first".
     
  4. Apr 18, 2017 #3
    Thanks for all the clarification here Nugatory. It's going to take me a bit to digest, and to rationalize with (int that it goes beyond) what I thought I read in that article.
     
  5. Apr 18, 2017 #4

    Nugatory

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    Do compare the wave function in the piece you quoted with the example that I gave:
    ##|\alpha_1\beta_1\rangle+|\alpha_2\beta_2\rangle##
    ##|H>1|H>2+|V>1|V>2##
    They're slightly different notations for the same concept: A state that is superposition (that's the ##+## sign, and it's what makes the state non-factorizable and entangled) of two states, in both of which two observables have definite values if measured.
     
  6. Apr 18, 2017 #5
    I don't think I'm familiar with any of the operators here. For me,
    > is greater than
    + is addition
    | is a bitwise or
     
  7. Apr 18, 2017 #6

    Nugatory

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    ##+## is addition here, no surprises although the things we're adding are vectors not numbers.

    You have Dirac to blame thank for the rest of this notation (and, kidding aside, it is very convenient). Loosely speaking, ##|Q\rangle## is a vector in some Hilbert space; the vertical bar and the ##\rangle## are part of the symbol for the vector the same way that the arrow in ##\vec{V}## is part of the way we were writing vectors in high school. The ##Q## or whatever is inside the ##|## and ##\rangle## is just a convenient label for the vector in question; there's nothing wrong with writing ##|frodo\rangle## and ##|bilbo\rangle## as long as the context makes it clear which vector I've decided to call Frodo and which vector I've decided to call Bilbo.

    The things with right-angle-brackets like ##|Q\rangle## are called "kets". ##|AB\rangle## and ##|A\rangle|B\rangle## are different notations for the same ket.

    You don't need to worry about it now, but the name will make more sense if you know that associated with every ket is another mathematical object called a "bra" written as ##\langle{Q}|##. The inner product of two vectors (kets) ##|A\rangle## and ##|B\rangle## is ##\langle{A}|B\rangle##, a "bra-ket" or "bracket", and that's where the notation comes from.
     
    Last edited: Apr 18, 2017
  8. Apr 19, 2017 #7
    I really do appreciate your taking the time and trouble to explain some of QM's terms and math to me. I'm afraid, as of now, my mathematical toolbox doesn't contain tools for performing operations on Hilbert space (infinite dimensional) vectors, or even a clear understanding of them (though I do remember reading about Nash's proofs involving partial differentials in his biography). To be honest, I respect and am impressed by, but not that interested in, the math. I think I know my limitations enough to say I'm extremely unlikely to ever make any contribution to it. And it doesn't really address my overarching curiosity. Personally I'm much more interested in the philosophical implications of QE. Like, what does it say about matter, the universe and existence that distant particles are able to "know" anything about each other? What underlying physical mechanism/law/construct allows Alice and Bob, a billion lightyears from and travelling at 0.99 c relative to each other, to derive from entangled particles the same long encryption key in their respective nows?
     
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