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How to calculate density matrix for the GHZ state

  1. May 19, 2015 #1
    The GHZ state is:

    [itex] |\psi> = \frac{|000> + |111>}{\sqrt2} [/itex]

    To calculate density matrix we go from:

    [itex] GHZ = \frac{1}{2}(|000> + |111>)(<000| + <111|) [/itex]
    [itex] GHZ = \frac{1}{2}( |000><000| + |111><111| + |111><000| + |000><111|) [/itex]

    To:

    [itex] GHZ
    = 1/2[
    \left( \begin{array}{cc}
    1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    \end{array} \right)
    +
    \left( \begin{array}{cc}
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
    \end{array} \right)
    +
    \left( \begin{array}{cc}
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    \end{array} \right)
    +
    \left( \begin{array}{cc}
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    \end{array} \right)

    ] [/itex]

    And finally to:

    [itex]
    GHZ = 1/2\left( \begin{array}{cc}
    1 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    1 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
    \end{array} \right)
    [/itex]

    But I see another author (p2) separates the Hilbert space into two subsystems GHZA⊗GHZBC and gets a "reduced" density matrix:

    [itex]
    GHZ_A⊗GHZ_{BC} = 1/4\left( \begin{array}{cc}
    1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
    \end{array} \right)
    [/itex]

    Can anyone explain what this final matrix represents, and how one calculates it?
     
    Last edited: May 19, 2015
  2. jcsd
  3. May 23, 2015 #2

    jfizzix

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    Gold Member

    That density matrix represents the mixed state:
    [itex]\frac{1}{4}\big(|0,0,0\rangle\langle 0,0,0|+|0,1,1\rangle\langle 0,1,1|+|1,0,0\rangle\langle 1,0,0|+|1,1,1\rangle\langle 1,1,1|\big)[/itex]
    or more simply:
    [itex]\frac{1}{4}\big(|0\rangle\langle 0|+|1\rangle\langle 1|\big)\otimes\big(|0,0\rangle\langle 0,0|+|1,1\rangle\langle 1,1|\big)[/itex]
    To figure that out, system [itex]A[/itex] separates the density matrix into four equal square blocks (of size 4x4)
    Of each block, system [itex]B[/itex] separates the density matrix into four equal square sub-blocks (of size 2x2)
    Of each sublock, the state of system [itex]C[/itex] associated to those particular amplitudes of A and B are given.
     
  4. May 25, 2015 #3
    Thanks. I can (roughly, without yet drawing it up) see how you've constructed this expression from the final matrix I drew up.
    Hmmm, I'm trying to work through this equation but am getting the wrong result:
    [itex]
    1/4\left( \begin{array}{cc}
    1 & 0\\
    0 & 1\\
    \end{array} \right)
    \otimes
    1/4\left( \begin{array}{cc}
    1 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 1\\
    \end{array} \right)
    = 1/4\left( \begin{array}{cc}
    1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
    \end{array} \right)
    [/itex]
    But perhaps I'm just making a simple mistake somewhere. At least, I can kind of see what you've done: you've tried to find a density matrix that corresponds to system A, and a density matrix corresponding to system BC, such that putting them together (##\otimes##) gives you the final matrix from my first post.
    Unfortunately I don't quite follow you here. In particular, the talk of systems A and B separating the matrix into blocks, and the talk of the state of C being given (but not A or B (or A versus BC)?). Does it relate in some way to the calculation I just made above?
    My main confusion is this: take the original density matrix for the pure state. I understand that the diagonal terms represent probabilities for measurement outcomes: 0.5 and 0.5. I also understand (to a lesser degree) that the off-diagonal terms represent the system's ability to exhibit interference. So the density matrix for a pure state like GHZ has a clear interpretation. But how does one interpret the "reduced" density matrix for GHZA⊗GHZBC?
     
  5. May 25, 2015 #4

    jfizzix

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    It is a simple mistake, easily rectified.
    What's here the kronecker product in reverse order. If [itex]X[/itex] is the first matrix you have here, and [itex]Y[/itex] is the second matrix, then to get [itex]X\otimes Y[/itex], you should make a big matrix out of (in this case 4) copies of [itex]Y[/itex], and then multiply each copy of [itex]Y[/itex] by the corresponding entry in [itex]X[/itex].
    It looks like that you've done instead is make a big matrix out of... 16 copies of [itex]X[/itex], and then multiplied them by the corresponding entries of [itex]Y[/itex].
    In short, you've found [itex]Y\otimes X[/itex] instead of [itex]X\otimes Y[/itex].
     
  6. May 27, 2015 #5
    Yes I see the mistake now, thanks. What I still don't understand is the relationship between original density matrix and the reduced density matrix. In particular, we know that the density matrix of the GHZ state is:
    [itex]\frac{1}{2}( |000><000| + |111><111| + |111><000| + |000><111|) [/itex]
    We now want to break the system down into two systems (A and BC). But how do we know that this:
    [itex]\frac{1}{4}\big(|0\rangle\langle 0|+|1\rangle\langle 1|\big)\otimes\big(|0,0\rangle\langle 0,0|+|1,1\rangle\langle 1,1|\big)[/itex]
    is a valid way to decompose the system? Why, for example, do we associate ##\frac{1}{4}\big(|0\rangle\langle 0|+|1\rangle\langle 1|\big)## with subsystem A? How do we know that this expression in any way represents subsystem A?
     
  7. May 29, 2015 #6

    jfizzix

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    The product
    [itex]\frac{1}{4}\big(|0\rangle\langle 0|+|1\rangle\langle 1|\big)\otimes\big(|0,0\rangle\langle 0,0|+|1,1\rangle\langle 1,1|\big)[/itex]

    is not a valid decomposition of the GHZ state:
    [itex]\frac{1}{2}( |000><000| + |111><111| + |111><000| + |000><111|) [/itex]

    The GHZ state is not separable into such a product or any other product. It's about as entangled as three-particle spin states get.

    If one were to measure particle A of the GHZ state (in the 0/1 basis), the state one would get as a result would be:
    [itex]\frac{1}{2}\big(|0\rangle\langle 0|\otimes|0,0\rangle\langle 0,0|+|1\rangle\langle 1|\otimes |1,1\rangle\langle 1,1|\big)[/itex]
     
  8. May 31, 2015 #7
    But in decoherence theory, isn't it standard to separate an entangled state into system state and environment state? I thought that something like that must be going on here. For example, I thought that GHZA⊗GHZBC amounted to something like treating systems B&C as the environment. Is that not right? I'm trying to figure out how to calculate the (reduced) density matrix called 'GHZA⊗GHZBC' from just the density matrix of the GHZ state.

    (In case it helps, I'm ultimately trying to understand a particular definition of (integrated) information of a quantum state (see e.g. page 2), which is calculated via a function of the density matrix of the state and the reduced density matrices of its decompositions.)

    That's useful, thanks. I've calculated it as follows:
    [itex]
    1/2\left( \begin{array}{cc}
    1 & 0\\
    0 & 0\\
    \end{array} \right)
    \otimes
    1/2\left( \begin{array}{cc}
    1 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    \end{array} \right)
    +
    1/2\left( \begin{array}{cc}
    0 & 0\\
    0 & 1\\
    \end{array} \right)
    \otimes
    1/2\left( \begin{array}{cc}
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 1\\
    \end{array} \right)
    = 1/2\left( \begin{array}{cc}
    1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
    \end{array} \right)
    [/itex]
    I interpret the diagonal values as representing probabilities (0.5 chance for |000> and 0.5 chance for |111>) and the off-diagonal values as capacity for interference, which is zero given that we now have an eigenstate (either |000> or |111>).
    But then I wonder what the two additional 1's on the diagonal of GHZA⊗GHZBC represent; and why the diagonal contains all ones in the case of GHZA⊗GHZB⊗GHZC = I/8.
     
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