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Time frame of entangled particles

  1. Oct 12, 2015 #1
    I was thinking about entanglement today and found myself struggling to envision entanglement and relativity working together. If I send entangled electrons out in opposite directions, their states are not determined until one of them is observed. I know that. What if one of them passes through an extreme gravity well so that to an observer on earth, one electron is a year old and the other is a billion years old.

    When you observe the electron that's been traveling through normal space, is the state of the other known? For when and from what reference frame? If you observe the electron that travelled through the gravity well, would the result be identical?
  2. jcsd
  3. Oct 12, 2015 #2
    There's much simpler time paradoxes than this, concerning entanglement. See https://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser

    "If the experimental apparatus is changed while the photon is in mid‑flight, then the photon should reverse its original "decision" as to whether to be a wave or a particle. Wheeler pointed out that when these assumptions are applied to a device of interstellar dimensions, a last-minute decision made on earth on how to observe a photon could alter a decision made millions or even billions of years ago."
  4. Oct 12, 2015 #3

    Simon Phoenix

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    Let's suppose we have our usual Alice and Bob and give them each a single spin-1/2 particle from a (maximally) entangled pair.

    Formally we'd describe the state of Alice's particle as ρ = ½(|0><0| + |1><1|)

    This is exactly the same state Alice would have if she was given just a single particle prepared in either |0> or |1> entirely at random. In other words there's no experiment Alice can do on her particle alone to tell the difference between whether she has one partner of an entangled pair, or whether she has just a single un-entangled particle prepared in up or down states entirely at random.

    Let's now suppose that the particle on its way to Bob undergoes some unitary process - this does not change the (local) state for Alice. Again there's no experiment she can perform on her particle alone that will tell her whether such a unitary process has occurred (this is effectively the basis of the no-signalling theorem which tells us that Alice and Bob cannot use their entangled pairs to communicate faster than light).

    The correlation only becomes apparent when we compare measured observables, which requires Alice and Bob to compare results - so Alice is going to have to wait a long time in your example.

    But to answer your question then according to standard QM yes if the initial state was a singlet state of the form ~ |01> + |10> if Alice measured her particle and obtained the result 1 then in a billion years in Alice's frame, Bob, measuring spin in the same direction, would obtain the result 0 (provided that the gravitational well did not effect a spin rotation - in which case Alice's result would be perfectly correlated with some other spin direction of Bob's).
  5. Oct 12, 2015 #4


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    Actions on separated entangled particles commute. It doesn't matter what order the actions happen in; the same outcome expectations happen either way.

    Because who-went-first doesn't matter (in terms of observable results), relativity of simultaneity doesn't come into the picture at all. Some interpretations of quantum mechanics have to care, e.g. collapse "occurring instantaneously everywhere", but the way they care always has no observable consequences.
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