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Thread: FTL communication by using entangle keys.

  1. #1 FTL communication by using entangle keys. 
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    This is just my idea. Correct me, if I am wrong
    In 2010, I read that one photon is spited in to 3 photon in the experiment. definitely all three will be entanglement with each other.
    http://www.nature.co...s.2010.381.html
    Now in 2017, I find many papers where more than 3 photons are entangle with each other used for communication keys.

    From above this is clear that three entangle keys of photons can be created & if one key is read, other two will break up from entangle. (Please correct me if I am wrong)
    Now, this can be used for FTL & quantum scanner.

    Idea For FTL communication :-Now, there are 3 points A & B on ground & C is Go- synchronize satellite.
    Let, distance AC & BC are exactly same.
    Now, from Satellite C, three entangles keys are fired. one at point A & two at point B.
    Now, if I read key at point A, entangle between three particle will break up instantaneously & will find that two keys at point B is not entangle. ( Digitally I call this as 1)
    Now, if I not read key at point A, entangle between three particle will not break up instantaneously & will find that two keys at point B is perfectly entangle. ( Digitally I call this as 0)
    Means, affecting key at point A, entangle of two keys at point B is affected instanteniously.
    This is FTL communication.
    & 0,1 0,1 can be transmitted from point A to point B instantaneously.


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  3. #2  
    Forum Sophomore Karsus's Avatar
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    The problem with entanglement is that you can only know if entanglement was successful ~after~ all parties communicate classically (speed of light or slower) to each other and compare results. When you transmit entangled qubits they are in a superposition, and from the perspective of each party individually the result they measure is probabilistic and not encoded with a distinct message.


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  4. #3  
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    I may be wrong but if we use more than 2 entangle photons keys then it may be possible.
    Let, consider A, B & C
    If we have three observable commuting A, B, and C each taking on either of two possible values, then an eigenstate will be written as something like | + − − ⟩ where the three signs represent the values of A, B, C in that order.
    A state like √ 2 /2 ( | + − − ⟩ + | − + + ⟩ or √ 2/ 2 ( | + − − ⟩ + | − + − ⟩ is entangled;
    Measurement of A will cause the wave function to collapse to either | + − − ⟩ or | − + + ) states in which we know more about what will happen when and if we subsequently measure B or C.
    So, if B, C pair is at long distance from A then also we can know that possible B & C states by measuring A.
    As entangle between A, B & C break down.
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  5. #4  
    Forum Sophomore Karsus's Avatar
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    I am not really sure what you mean. Which particle does the message sender get? A, B or C? And which does the receiver get?
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  6. #5  
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    Moved to Personal Theories.
    "[Dywyddyr] makes a grumpy bastard like me seem like a happy go lucky scamp" - PhDemon
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  7. #6  
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    OK, Let, A, B, C are three entangle photons fired from geo-synchronize satellite to two equidistant points 1 & 2 on ground.
    Let, Point 1 gets A photon & point 2 gets pair A, B entangled photons.
    From above it is clear that if A is measure then
    Measurement of A will cause that the wave function to collapse to either | + − − ⟩ or | − + + ) states.
    OR
    if I was at point 2 & just measure states at point 2 of B, C pair photons are + - or - + then we will be 100 % sure that measurement at point 1 is not done already.

    This is the information, we received at point 2 from point 1 that photon A is not measured already.
    -----------------------------------------------------------------------------
    Now, by using group of entangle triplicate of photons, we can make sure that probability of information transfer from points 1 to 2 goes to 100% because at least one pair will give + - result if measurement at point 1 is not done.
    -------------------------------------------------------------------------------
    Main problem is that if there are N entangle photons & one kept at some other location then by just measuring remaining (N-1) photons states, can we decided that remaining single photon is already measure or not?
    If answer is yes then information can be transfer by FTL.
    Otherwise,
    it is not possible.
    Thanks
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  8. #7  
    KJW
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    Suppose one has the entangled state √2/2 ( | + − − ⟩ + | − + + ⟩ ). If A is not measured, then B and C will be the entangled state √2/2 ( | − − ⟩ + | + + ⟩ ), which will produce either | − − ⟩ or | + + ⟩ when measured. If A is measured, then A, B and C will be either | + − − ⟩ or | − + + ⟩, B and C will be a non-entangled state, either | − − ⟩ or | + + ⟩, producing either of these when measured. Thus, the result of measuring B and C will be the same regardless of whether or not A is measured.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  9. #8  
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    Ok, thanks.
    but suppose one has the entangled state √ 2/ 2 ( | + − − ⟩ + | − + − ⟩ then what happen
    Measurement of A will cause the wave function to collapse to either | + − − ⟩ or | − + + ) states
    or not.


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  10. #9  
    KJW
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    Quote Originally Posted by Mahesh khati View Post
    Ok, thanks.
    but suppose one has the entangled state √ 2/ 2 ( | + − − ⟩ + | − + − ⟩ then what happen
    Measurement of A will cause the wave function to collapse to either | + − − ⟩ or | − + + ) states
    or not.
    No. Measurement of A will cause the wavefunction to "collapse" to either | + − − ⟩ or | − + − ⟩ (C is the | − ⟩ state in both cases).
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  11. #10  
    KJW
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    Quote Originally Posted by KJW View Post
    Quote Originally Posted by Mahesh khati View Post
    √ 2/ 2 ( | + − − ⟩ + | − + − ⟩
    (C is the | − ⟩ state in both cases).
    I should remark that for the quantum state √2/2 ( | + − − ⟩ + | − + − ⟩ ), C is not entangled with A and B:

    √2/2 ( | + − − ⟩ + | − + − ⟩ ) = √2/2 ( | + − ⟩ + | − + ⟩ ) | − ⟩
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  12. #11  
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    It is typing mistake, entangled state before collaping is √ 2/ 2 ( | + − + ⟩ + | − + − ⟩
    then, according to you after A is read. It will collapse to
    √2/2 ( | - + ⟩ + | + - ⟩ )
    I just today read that by using three entangle photons research is going on to
    developed a system that will let physicists to send beam quantum information to and from space. It is nice.
    https://www.aps.org/publications/apsnews/201305/entangledphoton.cfm


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  13. #12  
    KJW
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    Quote Originally Posted by Mahesh khati View Post
    √2/2 ( | + − + ⟩ + | − + − ⟩ )

    according to you after A is read. It will collapse to √2/2 ( | - + ⟩ + | + - ⟩ )
    I don't know why you say this, but no. After A is measured, the three-particle state will either be | + − + ⟩ or | − + − ⟩, with the B and C two-particle state as either | − + ⟩ or | + − ⟩. The measurement of A not only breaks the entanglement between it and the other two particles, but also breaks the entanglement between B and C (this is not true in general, but depends on the initial three-particle state).


    Quote Originally Posted by Mahesh khati View Post
    I just today read that by using three entangle photons research is going on to developed a system that will let physicists to send beam quantum information to and from space. It is nice. https://www.aps.org/publications/aps...gledphoton.cfm
    It is unlikely that the researchers are intending to use entanglement to communicate information. It should be noted that the no-communication theorem says that information can't be transmitted by entanglement.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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  14. #13  
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    Thanks
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