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Why is the integrated information of a Bell state = 0?

  1. Apr 13, 2015 #1
    In section IIIA (p11) Max Tegmark tries to prove that the integrated information Φ of a bell state is zero.

    The definition of Φ that Tegmark uses is given by the mutual information I minimized over all possible factorizations.
    The bell state has I=2 when written in the usual basis.
    Tegmark then appears to argue that we can move to a basis in which the entire bell state is given by a single basis vector (and not a superposition of basis vectors), which is a completely factorizable state (which he apparently proves in equation 10) yielding Φ=0.

    What I don't understand is how that counts as a factorization? Surely the valid bases are the infinity of spin-space bases, none of which allow for Φ to be zero. Or am I confusing factorizations with bases somehow?

    Would love to hear from someone with a better grasp of the mathematical details!
     
  2. jcsd
  3. Apr 13, 2015 #2

    wle

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  4. Apr 14, 2015 #3
    I am not sure what the confusion is. I have not read the entire article, but the Bell states are two vectors in a perfectly good basis (I'm at work so I don't have time to type it, but you can google it). Now, this may not be a basis in which it is feasible to actually make a measurement, but mathematically it works. Maybe I am missing the question. Now, the state is NOT able to be factored in the computational basis which is why it is an entangled state in the computational basis.
     
  5. Apr 15, 2015 #4
    A qubit contains one bit of information since measuring yields either |0> or |1>, both of which contain only one bit of information. The bell state contains two bits of mutual information, since measuring yields either |0>|0> or |1>|1>. To derive the integrated information of the bell state we derive the mutual information minimized over all possible factorizations. Here is where I get a bit lost.

    I take it that a possible factorization is this: find some basis to represent the state, then break the Hilbert space down into separable subspaces. Is that not right?
    Assuming that's right, the next worry is the factorization. Tegmark does all this with density matrices (whereas I'm more comfortable with Dirac notation). He thus calculates the density matrix for the bell state:

    [itex] \frac{|00> + |11>}{\sqrt{2}} \rightarrow \rho = \left( \begin{array}{cc}
    1/2 & 0 & 0 & 1/2 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    1/2 & 0 & 0 & 1/2\\
    \end{array} \right) [/itex]

    Now he goes for a basis change, one where the entire bell state is a basis vector:

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

    I think I understand this bit: there are four basis vectors corresponding to the four singlet states. That's why we have a four-by-four matrix. These basis vectors all correspond to matrices with ones in a corner zero elsewhere? Okay but now we are told that:

    [itex] \rho' = \left( \begin{array}{cc}
    1 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0\\
    \end{array} \right) =
    \left( \begin{array}{cc}
    1 & 0 \\
    0 & 0 \\
    \end{array} \right)

    \left( \begin{array}{cc}
    1 & 0 \\
    0 & 0 \\
    \end{array} \right) [/itex]
    Which is meant to show that we have factorised a bell state into two subsystems that are completely independent (separable?).
    I don't understand what these two separable matrices correspond to physically.
    What would they correspond to in Dirac notation for state vectors?
     
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