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Special relativity violation using entanglement

  1. Feb 18, 2015 #1
    As I understand it, faster than light communication is not possible, but I have a specific example which concludes that it is and I'm trying to find the mistake.

    The scheme uses two things
    1) An entangled Bell pair ## | \phi \rangle = | 0 0 \rangle + | 1 1 \rangle## ( neglecting normalization )
    2) The fact that orthogonal states can be distinguished.

    To begin with ## |\phi\rangle ## is given to Alice and Bob each having a qubit.

    Bob has in his positions a copying unitary that can copy states ## |0\rangle ## and ## |1\rangle ## to an ancillary qubit and he can then measure this qubit to distinguish them, explicitly the unitary is [tex] U = |0\rangle\langle 0 | \otimes |0\rangle\langle 0 | + | 1\rangle\langle 1 | \otimes | 1\rangle\langle 0 | [/tex]

    The unitary acts as, ## U|00\rangle = | 00\rangle## and ##U|10\rangle = | 11\rangle ## thus a measurement of the ancillary gives the state of the qubit in Bob's possession as long as it is either ##|0\rangle## or ##|1\rangle##.

    Now suppose that Bob keeps applying ##U## and measuring the ancilla. The reduced density matrix on Bob's side is, ## \rho_b = \frac{1}{2} \left( |0\rangle\langle 0 | + | 1 \rangle\langle 1| \right) ##, the maximally mixed state.
    So Bob will measure 0 half the time and 1 half the time. Hence he can determine that
    ##P(0) = 1/2## and ## P(1) = 1/2.##

    Suppose now that Alice makes a measurement on her qubit giving either 0 or 1. The collapsed state
    ##| \phi^{\prime} \rangle ## is now ## | 00 \rangle ## or ## | 11 \rangle ## making Bob's state ## | 0 \rangle ## or ## | 1 \rangle ##.

    Now as Bob keeps measuring, he will determine either,
    ##P(0) = 1 ## and ## P(1) = 0 ## or ##P(0) = 0## and ##P(1) = 1 ##

    Hence asymptotically he can determine that Alice has made the measurement. So the information that a measurement has been made can be operationally determined by Bob even if he is light years away.
     
    Last edited: Feb 18, 2015
  2. jcsd
  3. Feb 18, 2015 #2

    mfb

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  4. Feb 18, 2015 #3
    The no cloning theorem states that you cannot build a unitary to copy an arbitrary state. However you can still create one that copies known orthogonal states.
     
  5. Feb 18, 2015 #4

    atyy

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    Try checking if the copying unitary will entangle the ancilla and Bob's qubit,m so that after Bob's first measurement on the ancilla he will collapse his qubit.
     
  6. Feb 19, 2015 #5
    Yes! This seems to be the answer. If you see the effect of the unitarty on the entire system,
    [tex] (\mathbb{1}_A\otimes U_{BC}) ( \rho_{AB} \otimes |0\rangle\langle 0 |_C ) (\mathbb{1}_A\otimes U_{BC}^{\dagger}) = \frac{1}{2} (|000\rangle\langle 000| + |111\rangle\langle 111| )[/tex] and a measurement of Bob's ancilla collapses the state and destroys the entanglement. Thank you :)
     
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