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

Quantum entanglement

  1. Sep 29, 2006 #1
    If you have two particles 'a' and 'b' which are entangled with each other and you observe the state of 'a' so consequently you know the state of 'b', if you destroy 'a' will the state of 'b' become unknown again?

    Cheers
    Jake
     
  2. jcsd
  3. Sep 29, 2006 #2
    I doubt it. Basically, imagine equation:
    a + b + 1 = 0;
    here we know what a equals to: a = -b - 1, same applies to b.
    If we would measure b, for example, and find out that b=3, then we would know what a equals to: a + 3 + 1 = 0; a = -4.
    Now the relationship is broken, so we can set b to anything -- a doesn't care as long as it exists independently.

    The fun part is that when relationship (entanglement) persists, values "swing" back and forth and so it's impossible to predict what will happen.
     
  4. Sep 29, 2006 #3
    Hi Chipset,

    I like your way of explaining entanglement. It seems like you have reached the heart of the matter without using big words.

    Do you have an equally intuitive description for Bell's theorem?
     
  5. Sep 29, 2006 #4

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    In your example: NO, because the superposition of states has collapsed. (At least with respect to the non-commuting operators that were measured.) They are essentially no longer entangled.
     
  6. Sep 29, 2006 #5
    Have you carefully thought through how an optical experiment of entanglement actually works?
    With photon ‘a’ & ‘b’ how did you detect the state of ‘a’ ???
    Do you think you still have a photon to do anything with, even after you gain whatever information you could from it from the photo-detector. Hasn’t it been destroyed as its energy was used to stimulate the detector.

    Even when testing electrons after doing any one test on a electron you cannot do second test on that same electron and consider it to “be the same” as in expecting the second test to guarantee giving you the same results as if the first test had not been preformed.
     
  7. Sep 29, 2006 #6
    The simplest experiment would involve two entangled photons, A and B, and a polarized filter, set to a specific value, let's say 90 degrees. Since they are entangled, the two photons will show the same polarization when they are measured, though the polarization is undetermined until one of them is measured.

    If photon A passes through the filter and is detected, then we know that photon B will also pass through a 90 degree filter. Photon A has been destroyed at this point, since it either was stopped by the filter or by the detector; Photon B will retain the polarization value forever, unless it is altered or destroyed by another event.
     
  8. Sep 30, 2006 #7

    wm

    User Avatar

    Question: Was photon A polarized at 90 degrees before it was measured?
     
  9. Oct 1, 2006 #8
    Not sure why your directing your comments to me - aren’t just repeating what I just said

    Did you mean to address this tojnan014 on how the OP answers its own question.

    Not sure why you are here WM; shouldn’t you be fixing the links on your crackpot site?; you promised to be done by last month.
     
  10. Oct 1, 2006 #9
    Sorry about that, yes, it intended for the originator.
     
  11. Oct 1, 2006 #10
    No, the fact that it went through a 90 degree filter doesn't mean it was 90 before it hit the filter. On average, half of the A photons will go through the filter, and they can do so with any polarization.

    And, no matter what the angle is, if A goes through a filter at that angle, then so will B.
     
  12. Oct 2, 2006 #11
    No offense to chipset, but he hasn't exactly 'explained' quantum entanglement. As far as I'm aware, nobody has been able to do that yet. :smile:

    Events (or the quantum probabilities related to those events) at, say, two detectors, A and B, are called quantum entangled when their combined detection rate, or probability thereof, P(AB) can't be expressed as the product of the individual detection rates, or probabilities), P(A)P(B).

    But the 'nature' of quantum entanglement is unknown.

    An intuitive description of Bell's Theorem might be along these lines:
    In the archetypal optical Bell test setup, if the polarizer at A is offset from the horizontal (0 degrees in a unit circle) setting at a = 30 degrees, and the polarizer at B is offset in the other direction from the horizontal at b = 30 degrees, then the angular difference of this joint setting, (a,b), is 60 degrees (or two times the offset of either a or b). The angular difference is usually called Theta.

    One representation of Bell's Theorem says, in effect, that the correlation at a 60 degree angular difference between a and b (at Theta = 60 degrees) should be exactly twice what it is at Theta = 30 degrees. But (according to qm, and apparently experimentally) it isn't.

    A simple Bell-type inequality might be,

    N(a, not b) + N(b, not c) >= N(a, not c),

    where a, b and c are three values or properties (or not) of objects in some grouping of objects. For example, the group of objects might be the words in this post and the values or properties might be letters (say, a, b and c). So that, for example, N(a, not b) refers to the number of words in this post that contain the letter a but not the letter b, and so on.

    But the objects, and values or properties exhibited (or not) by those objects, can be virtually anything -- and the above inequality is always true.
     
  13. Oct 2, 2006 #12
    Thanks, mgelfan - I will think on this and get back to you
     
  14. Oct 2, 2006 #13

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    In principle: I think it would be possible to take 2 entangled particles and perform a spin measurement on them. That would collapse their spin states. But I don't think it would collapse the states of other non-commuting observables... would it?
     
  15. Oct 5, 2006 #14
    This is pretty neat:

     
  16. Oct 5, 2006 #15
    Well, I will wait for the paper with all the details which will probably reveal that local realism is as alive as ever. I am not impressed with a fidelity of 60 percent (of the measured coincidences I presume, no mentioning of efficiencies and so on here), the distance between alice and bob is 0.5 meter, so the locality loophole is wide open too...
    Anyway, given the historical victorious cries about the Aspect experiment as well as the local realist follow-up papers, I presume it is best to remain calm ...

    I do not mean to imply that the experiment is not an accomplishment, just that one has to be cautious about the conclusions one draws from this.

    Careful
     
    Last edited: Oct 5, 2006
  17. Oct 6, 2006 #16
    Well, I didn't really set my goal to explain what's within quantum entanglement or otherwise I'd probably be preparing my Nobel prize speech right now :)

    Why should it?

    I believe that mgelfan has explained that but you can imagine Bell's theorem as simple lack of determinism when in quantum world. That is, you cannot expect coin to be thrown and 50% of the times land on tail as would be expected.
     
  18. Oct 6, 2006 #17
    Not true
    QM does expect 50% and experiment shows this. Just like the "entangled" coin being tested at another time and place also has 50%. What is unexpected is that the matched "entangled" coins give these random results in coordination with each other. That is they are correlated with no apparent means of using a pre-determined unknown hidden variable of a HVT to account for it. Addressing this fixed and determined HV is all that Bell deals with; it has nothing to do with the philosophy of determinism.

    In fact Bell does not care if classical physics is deterministic or non-deterministic. It only cares about finding a HV to show that reality is “LOCAL” or if not implying that reality must therefore be “NON-LOCAL” this is all Bell can do. Even as the experiments show that Non-Local must be the case for reality; Bell offers no means to help decide between Non-Local Theories.
    Non-Local theories (QM, BM, MWI, Strings, M, etc,) each must look for a way to prove there case Bell will not help. (Note: some of the Non-Local theories have their own version of “local” within their theory – don’t confuse that with Bell Local)
     
  19. Oct 6, 2006 #18

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    Careful, I still get confused about your position.

    The only reason the items you mention would matter is IF the actual values, with "perfect" measurements (to your satisfaction), were different from the QM predicted values. Is this what you are imagining to going to result? I thought your opinion was that even if existing Bell tests are confirmed to your satisfaction in the future, you argue that local reality is not excluded because Bell's assumptions are a bit too simplistic to lead to an absolute rejection of LR.
     
  20. Oct 6, 2006 #19
    Right, but there are subtleties involved in these scenario's too (actually I have to check out all the details of the paper before I comment any further). :smile:

    Careful
     
  21. Oct 6, 2006 #20

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

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

Have something to add?



Similar Discussions: Quantum entanglement
  1. Quantum Entanglement (Replies: 5)

  2. Quantum Entanglement (Replies: 6)

  3. Quantum entanglement (Replies: 15)

Loading...