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I Triplet State Entanglement

  1. Jul 31, 2016 #1
    Hi all! I know that the singlet state is entangled. Is the center triplet state also entangled? At first blush it seems like it should be as the state expression has the same structure with the minus sign replaced by a plus.

    Thanks in advance!
     
  2. jcsd
  3. Jul 31, 2016 #2

    Strilanc

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    What's the center triplet state?

    I know that the singlet state is ##|01\rangle - |10\rangle## and that the triplet states are the other three Bell basis states ##|00\rangle + |11\rangle## and ##|01\rangle + |10\rangle## and ##|00\rangle - |11\rangle## but I'm not sure what you mean by the center one.
     
  4. Jul 31, 2016 #3

    Haha ya sorry that was a little confusing. I meant ##|01\rangle + |10\rangle##. I didn't know how to use LaTeX on here.
     
  5. Jul 31, 2016 #4

    Strilanc

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    Yes, it's also entangled.

    In fact, you can turn one state into the other with local operations: just rotate either one of the involved qubits 180 degrees around its Z axis.

    (Actually, that's not quite the right operation if you're doing these things conditionally and the global phase factor matters. In that case one of the sides would be applying ##XZX## instead of ##Z##.)
     
  6. Jul 31, 2016 #5
    Ok, thanks! So the ##|00\rangle+|11\rangle## and ##|00\rangle-|11\rangle## are entangled as well? Now that I'm looking at them there doesn't seem to be a way to factor them either.
     
  7. Jul 31, 2016 #6

    Strilanc

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    You might find this blog post useful.

     
  8. Jul 31, 2016 #7
    Wow thanks, that was a crazy article. So because the singlet and the triplet states are all examples of these "shared diagonal states" i.e. unitary matrices multiplied by root 2, they are all entangled states?

    I am wondering about the statement at the end "This all breaks down if you have more than two parties." Why though?
    I was wondering if you could maybe treat a 3-particle system as a rank 3 tensor and then instead of pre/post multiplying by ##U/U^T## you would have three different operations to hook into the appropriate index of the tensor in 3 dimensions?
     
  9. Jul 31, 2016 #8

    Strilanc

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    Right. Their singular value decomposition is the same, and the specific grid layout is chosen so that that means you can get from one to the other with local operations.

    You might be right, but there's probably a bunch of complications that make it not nearly as nice. For example, clearly it's not going to be equivalent to a unitary matrix anymore since a matrix is a 2d thing. The equivalent of a singular value decomposition is also probably a lot more gross (and would have to involve more than a single diagonal's worth of values).
     
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