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I Is entanglement commutative?

  1. Apr 5, 2017 #1
    I've seen diagrams of quantum computer components at a high level that discusses multiplexing laser reflections over many qubits, and I have to believe that entanglement as a hardware operation has to be scaled to the many qubits by means of some operation that is applied to each of them simultaneously. That being said, if I remember correctly, there were examples of the slit experiment at a microscopic level to give readers at a introductory level an impression of what the hardware was doing. But I don't think that that was strictly what was actually at that level. Perhaps I working with a very vague understanding, but what I want to know is, if you have light that is entangled, and you strike a super cooled qubit of any kind, does that mean that that qubit is also suspended in entanglement? In other words, is entanglement commutative?
  2. jcsd
  3. Apr 5, 2017 #2


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    If you have a pair of entangled photons A and B, and one of those photons B interacts with a qubit C, the amount of entanglement between A and the joint system BC remains the same (assuming no additional environmental interaction).

    The amount of entanglement between A and B may change due to B interacting with C, but if A has no further interaction with B or C, the total entanglement between A and BC must remain constant.
  4. Apr 5, 2017 #3
    Ok, so it's as though A now shares a total entanglement with all three, but BC sort of share a subspace determined by the metrics of their interaction, is that correct?
  5. Apr 6, 2017 #4


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    If I understand you correctly, yes.

    A shares entanglement with BC, and the amount of entanglement between A and BC is the same before and after B and C interact.
    What's different is how much entanglement A shares with just B, or with just C.

    There's a useful concept called the monogamy of entanglement that says the amount of entanglement B shares with AC cannot be less than the sum of the entanglement between A and B, and between B and C.
    [itex]E(A:BC)\geq E(A:B) + E(A:C)[/itex]
    So, as A becomes less entangled with B, A must be more entangled with C, (or at least, the maximum possible entanglement between A and C increases.
  6. Apr 6, 2017 #5
    Fascinating. Thank you.
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