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Some questions on quantum teleportation

  1. Nov 24, 2012 #1
    Hello I just have some questions about quantum teleportation.

    So generally you see the Alice and Bob explaination for it with the entangled particles a, b and the c which is being teleported.

    Firstly in regards to entanglement lets say polarization of a photon, some places say both entangled particles are the same so if you measured vertical polarization on a then you know b has vertical polarization, but I have read other articles that say they are opposite so if you measure vertical polarization on a then you know b has horizontal polarization. Which one is correct? I'm guessing they are the same as I have read many more that say they are the same.

    Alot of articles using it say in it that after the Bell measurement of ac both particles are destroyed and b then contains that information just that it is not useful information or randomized so it hasn't broken the FTL rule.

    Even if it is randomized though isn't that still breaking the FTL rule since randomized information is still... well... information?

    In other articles I see they say that information has not been instantly transfered only a quantum state has been instantly transfered (so no FTL breaking) but has it?

    E.g. if I measured b before the Bell measurement of ac I would get result X. If after the Bell measurement on ac I measure b (but I don't modify it from the information they sent about ac) would I get the same X result that I would have gotten had I measured it before the ac measurement, meaning nothing (not even a quantum state) has been instantly transfered or would I get a different result (but still not nkowing if it is the correct one or not) meaning that something has?

    Some articles lead me to believe that just measuring b would always yield the same result no matter if you have done anything to ac or not but then others imply that it would infact be different just that you have no way of knowing if it's the result that they were sending or not.

    Now if you do get the same result no matter when you measure b (without modifying it) then could you measure b before the ac measurement and record it then say fly off on holiday, they perform the Bell measurement on ac and send you the result of that while your on holiday, and you use that reults and the result of your measurement of b to come up with the same result as if you were in the lab and used the ac information to modify b and get a result?

    Hence my name, it's just so confusing when you read what should be the same story from different people but some are saying a different story but saying it's the same one.
     
    Last edited: Nov 24, 2012
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  3. Nov 24, 2012 #2

    mfb

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    It depends on the production of the entangled photon pair, but does not really matter for teleportation. You can even shift the polarization if you do it carefully.

    It is information without a 1:1-relation to the state of "a" before the measurement. You need the measurement result of (ac) to get it.

    If you measure b, you break entanglement, and b will just stay whatever you measured.

    That is not possible.
     
  4. Nov 25, 2012 #3
    It would stay whatever was measured but would what was measured be the same as if you waited till after you do the ac measurement, but you do not modify b with the ac information you got you just measure it like you would have before the ac measurement?

    I.e b will only ever give a different result IF you modify it using the infromation you got from ac, IF you don't modify it from that measurement then whenever you measure b regardless of if you have done ac or not would have the same result?

    Or does doing the ac measurement mean b is actually changed so if you did measure (but don't take into account the information from ac) then it would be a different result then measuring it before ac just that the result wouldn't match c which is what it would be if you took the ac information into account when measuring it?

    Even though entanglement is broken, are you not able to still physically modify b so that it becomes the same as c? Isn't that what your doing after the ac measurement anyway, just modifying b so that it is the same as c, doesn't the measurement of ac break the entanglement of ab?
     
  5. Nov 25, 2012 #4

    mfb

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    It does not help to wait. The measurement of b does not give any interesting result on its own ever. You need to know the result of the ac measurement to "interpret" the measurement of b. And even then, you ruined the quantum state (you wanted to teleport) by your direct measurement. You could have measured c and transmitted the result directly instead. That is not quantum teleportation, it is classical data transfer.

    You don't know what c was ;).

    That would work with classical states only.
     
  6. Nov 25, 2012 #5

    DrChinese

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    Welcome to PhysicsForums!

    One of the usual ways to generate entangled photon pairs is with Parametric Down conversion. Type I yields pairs with the same polarization. Type II yields pairs with opposite / orthogonal polarization.
     
  7. Nov 26, 2012 #6
    I'm not after an interesting result or trying to interpret b, I just want to know if it is any different.

    Pretend I have a device and when i press the button it measures a particle and returns a numerical value.

    Say I have b and if before location A does anything if I press the button it would return a value of say 7.

    Now say on the other hand if when I pressed the button location A had taken the measurement of ac (and say i'm 100 light years away so I don't have their results for ac to use to interpret b) would the result be the same "7" that I would of gotten had they not taken the ac measurement or would because they took the ac measurement the result be different to 7 (even though I don't have the ac information to use to interpret the reading of b)?

    In this case i'm not trying to teleport c from location A to B, to get any information or anything, I havn't taken a measurement of b before the ac measurement and then 2nd one after, i'm only trying to see if the result for the measurement of b would be different or not after an ac measurement without having that ac information to interpret it?

    I hope it doesn't sound like i'm just repeating myself since I understand your answers, just don't think you got exactly what I am trying to ask with this, thank you for entertaining my laymeness... if thats a word....

    DrChinese, thanks it would of been nice if "any" of the stuff i've read on entanglement/teleportation over the last decade even made slight mention that engtangled particles can be either rather then making it sound like they are stating they are always the same (or opposite depending on the source of the article).
     
  8. Nov 26, 2012 #7

    mfb

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    Different from what? You cannot repeat the same experiment in a perfect way and ask "will it change?", you will run into non-determinism or similar concepts (depending on your interpretation) of quantum mechanics.
    You will measure both possible results for b with 50% probability in all setups, independent of any measurement of ac. As I said, the measurement of b on its own is not interesting.
     
  9. Nov 26, 2012 #8

    DrChinese

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    The above is accurate. Teleportation is also called entanglement swapping.

    I don't follow which photons are which in your example, but I will assume that you mean:

    First a and b are originally entangled. Then b's entanglement is swapped to c. Now a and c are entangled. Is this what you mean?

    If so: the swapping process will not work if you learn b's polarization. Further, it matters not the order of observing a, b or c. The results are completely independent of ordering. In fact, you can entangle particles AFTER they are detected. Sounds weird, but is quite true. (However, you cannot generally entangle particles in this case on demand.)
     
  10. Nov 27, 2012 #9
    Okay that's the answer I was looking for, I wasn't sure if measuring b before measuring ac would yield a specific result or if there was that probability of it being any result.

    Some of the things I read made it sound like even though b was in a superposition of both states there is still a specfic "correct" one of those states that it would always be in when you measured it rather then there being a 50/50 chance of it being either one.

    From which of my posts the 1st, 2nd or 3rd? as I meant something a bit different in one of them.
     
  11. Nov 29, 2012 #10
    If b is in a superposition of two states, then yes, there is a specific basis under which the probability of finding b in that state is one. Let me use light as an example. Say b is in the superposition [itex]\frac{1}{\sqrt{2}}|H> + \frac{1}{\sqrt{2}} |V>[/itex], where H means horizontally polarised and V means vertically polarised. Notice that if I orient my polariser at 45 degrees to the vertical, then 100% of the light will pass through.

    However, in this case, b is not in a superposition of two states. Rather, the superposition is of the total system ab, which is different from the subsystems a and b. This is the essence of entanglement. If I only measure b, I end up with a mixture instead. Let me illustrate this with the light example again. Say the light is 50% horizontally polarised, and 50% vertically polarised. Now, I orient a polariser at [itex]\theta[/itex] to the horizontal. By Malus' Law, the amount of light passing through is
    [tex]\frac{1}{2}cos^{2}\theta + \frac{1}{2}sin^{2}\theta = \frac{1}{2}[/tex]
     
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