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B Decoherence thought experiment

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  1. Oct 21, 2016 #1
    Let's say a memory qubit inside a quantum computer is in state

    ## α \left|1\right>+β\left|0\right> ##

    This computer is equipped with a device that emits photons that carry the same qubit as the aforementioned memory location.

    Alice and Bob, that are very far from each other, receive and measure those photons.

    Let's say Bob measures this sequence of bits:
    01010111011011111011111111111111111111

    And Alice measures this sequence of bits:
    010100100010000001000000000000000000000

    I tried to make those sequences realistic looking in such case when probability of measuring either 1 or 0 is the same at the beginning.

    Now Bob contains a classical bit 1, and Alice contains a classical bit 0.

    Who has decohered here? The computer can still send out qubits to Joe and Jill, who can go to opposite classical states, so I guess the computer is not decohered. So it's more like Bob and Alice became decohered by their own measurements. Right?
     
    Last edited: Oct 21, 2016
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  3. Oct 21, 2016 #2

    DrClaude

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    Staff: Mentor

    If I understand you right, what you are proposing is not possible.
    This violates the no-cloning theorem. You could have a device that would spew out a pair of qubits in a predefined state, but not copy an arbitrary state.

    Also, I see no entanglement in what you propose, so why do you need two qubits and two observers?
     
  4. Oct 21, 2016 #3

    I want Bob and Alice to measure the state of the memory location. I don't care if the state is predefined. I would also like there to be no collapse to 1 or 0 when Bob and Alice make the first measurement.

    So how about if we entangle lot of photons with the memory location, send the photons to Bob and Alice, who are equipped with measuring devices that produce results with lot of uncertainty.

    Now I hope that as Bob makes more and more measurements, Bob's state becomes gradually the state "I know the memory location is in state 1". And as Bob is using the not so good measuring device, a good measuring device located nearby changes so that it only measures photons to be in the state that correspond state 1 of the memory location.
     
  5. Oct 21, 2016 #4

    DrClaude

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    I really think this violates the no-cloning theorem.
     
  6. Oct 21, 2016 #5

    PeterDonis

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    2016 Award

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    So do I. Thread closed.
     
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