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

Measurement of entangled particles causes dechorence at a distance?

  1. Oct 3, 2013 #1
    If we entangle particles and seperate them by a large distance, can the action of measuring one cause decoherence at the other's location?

    If yes then does this violate relativisitic causality? Could we not use this proccess to transmit information instantaneously?

    If no then why not? Is there a fundamental difference between the process of direct measurement of the local particle and the indirect measurement of the distant particle? Is decoherence any observer relative phenomena?
     
    Last edited: Oct 3, 2013
  2. jcsd
  3. Oct 3, 2013 #2

    Nugatory

    User Avatar

    Staff: Mentor

    You don't even need to appeal to decoherence to find an apparent faster-than-light influence. We could just as easily say that when either particle is measured the wave function of both collapses, and indeed that's how it was often described before decoherence was understood.

    However, there is no violation of relativistic causality - because there's no causality involved.

    I measure my particle and get spin-up, you measure yours and get spin-down. Is that because you went first, lucked into a spin-down measurement and collapsed my particle's wave function into the spin-up state, or because I went first, lucked into a spin-up measurement and collapsed your particle's wave function into the down state? That question is completely meaningless if the two measurement events are separated by a space-like interval because there's no ordering between them - neither one can be said to have happened first. All we know is that when we get together afterwards and compare notes, I measured spin-up and you measured spin-down, and that result is independent of which measurement "happened first".
     
  4. Oct 3, 2013 #3
    The difference is that in the standard EPR paradox that you describe, you don't know when or indeed if, I have measured my particle. We can confer later using sub-luminal information transfer to work it out, but you don't know instantenously.
     
    Last edited: Oct 3, 2013
  5. Oct 3, 2013 #4

    Nugatory

    User Avatar

    Staff: Mentor

    Ah - I'm sorry, I did misunderstand the question you were asking.
     
  6. Oct 3, 2013 #5
    No, there is no way to know instantaneously whether the other particle was measured. That would convey information FTL (faster than light).
     
  7. Oct 3, 2013 #6
    Are you saying that we can indirectly measure distant particles without causing decoherence? If so, why is the local measurement different to the distant, indirect measurement?
     
  8. Oct 3, 2013 #7

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    Measuring one places the other in a state in which it is now separate from the combined pair. So the decoherence is immediate and non-local. But there is no effective method to determine that change has occurred.
     
  9. Oct 4, 2013 #8
    If this is true then why is non-local decoherence impossible to observe when local decoherence is observable?
     
  10. Oct 4, 2013 #9
    because it is non-local. Rule # 1: no information can be transmitted FTL period.
     
  11. Oct 4, 2013 #10

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    Regardless of distance, you cannot detect decoherence by looking at a single particle (or system).
     
  12. Oct 4, 2013 #11
    I don't doubt it.

    The question is designed to establish why that is true in the case of decoherence with entanglement.
     
  13. Oct 4, 2013 #12
    I don't doubt it.

    The question is designed to establish why that is true in the case of decoherence with entanglement.

    Are you sure about this?

    A single particle or system can interfere with itself and can be detected by checking for this interference, right?
     
  14. Oct 4, 2013 #13

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    Generally, entangled particles do not self-interfere. After they decohere, they still don't particularly look any different unless you do something to make them such. For example, you could focus and therefore collect an entangled particle into a very small point and then send it through a double slit. You would see interference. But that would be true of any photon, entangled or unentangled.
     
  15. Oct 5, 2013 #14
    yes the single particle can (and always will) interfere. however the interference pattern is not clear. it's a mixture of various patterns.

    a co-incidence counter is required to filter/extract the interference pattern.


    to make the interference pattern clearer we need to make the photon more coherent.
    as we start making the photon more coherent, the entanglement starts weakening.

    Uncertainty principle (complementarity to be more precise) says - we cannot have your cake (neat interference pattern) and eat it too (strong entanglement)

    for more details you can see post 52 in the below link

    https://www.physicsforums.com/showthread.php?t=320334&page=3
     
    Last edited: Oct 5, 2013
  16. Oct 5, 2013 #15
    Suppose the distant particle is in a superposition of spin states and entangled with the local particle. A distant observer can check at any time if this superposition is still present using an interference based observation.

    The local observer can, at any time measure the spin of the local particle ending the distant superposition. The distant observer now knows that the local particle has already been observed.

    This situation can't be consistent with relativistic causality.

    How do we resolve this?
     
    Last edited: Oct 5, 2013
  17. Oct 5, 2013 #16

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    There are a couple of specific elements you need to be aware of. Suppose we have distantly separated entangled particles called Alice and Bob:

    a. Coherence (and lack thereof) is a function of position and momentum distribution. Which means that Alice and Bob start out with sufficiently wide a spread that there is no interference to be had. This is just a restatement of what we have already said. There is no measurement uou can perform on Alice or Bob at this point that will detect if either is still in a superposition. They are still entangled, and there is no interference. This too is just a restatement of what we have already said.

    b. So what you are thinking is that there is some manipulation which can be performed on Alice which changes the above and you will get interference for Bob. That is not correct. The only way to get interference from Bob is to first change Bob to be coherent. And there is no way for Alice to do that! It is true that measuring Alice may cast Bob into a known state, but none of those are coherent for Bob.

    c. To be more specific: suppose you measure Alice's spin. You now know Bob's spin. But Bob is still not coherent, nothing has changed in the position or momentum bases.

    d. A rule you should be aware of: after measuring Alice's spin per c. above, Alice and Bob are still entangled! That is because collapse has only occurred on the polarization basis. Commuting bases are not affected. You can entangle particles on all or just some bases. For example, a single Type I PDC crystal produces partially entangled photon pairs. Because they are NOT polarization entangled. Vice versa, measuring the polarization of of fully entangled photons does not necessarily eliminate all entanglement on them. So they can retain their superposition.

    e. So what happens if we measure Alice as to position? We cast Bob into a known position too. Which is tantamount to knowing which slit Bob will traverse. So there is no interference for Bob. If we don't measure Alice, Bob is not coherent and there is no interference. Either way, no interference.

    Nature is quite amazing!
     
  18. Oct 6, 2013 #17
    No you won't, a single particle can not create an interference pattern, ever!
     
  19. Oct 6, 2013 #18

    Demystifier

    User Avatar
    Science Advisor

    No.

    One should distinguish wave-function collapse from decoherence. Unlike the former, the latter can be derived from the Schrodinger equation, which cannot involve non-local change of the state.
     
  20. Oct 6, 2013 #19
    Coherence is not a property of a single particle. You need at least two particles to talk of coherence. It is a global property of at least 2 particles. Therefore anything which disturbs the relevant aspect of the one of the particles immediately destroys the coherence of the system of particles globally. But you won't know that until you have informasyion from both particles.

    This does not violate causality because you are dealing with non-local global information to start with. Its like you had a long string attached to the two particles. Then you can say they are LINKED, which is a global property of both particles. Cutting the string locally at one end does not cut the string at the other instantaneously but it immediately destroys their LINK no matter how long the string is. At the other end, you won't know that the link was destroyed until that information travels to you.
     
  21. Oct 6, 2013 #20
    in non linear quantum mechanics yes.
    unlike of standard quantum mechanics (a linear one).


    .
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook