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Can we look upon entanglement as measurement?

  1. Sep 3, 2013 #1
    Hi,

    I have a question about how to interpret the state of an entangled photon

    H = horisontal, V = vertical polarization



    The global state is |Phi> = (|H>|V> + |V>|H>)/sqrt2.

    By density operator formalism:

    rho = |Phi><Phi|
    = (1/2) ( |H>|V><H|<V| + |H>|V><V|<H| + |V>|H><H|<V| + |V>|H><V|<H| ).

    in order to find the state on the right side, trace out the left side :

    rho_right = Tr_left(rho) = (1/2) (|H><H|+|V><V|),

    How should this state be interpreted? 1) as a photon EITHER in H OR V, or 2) a mixed state (superposition) containing components of both the left and the right side.

    This would help me to understand this paper:

    http://arxiv.org/abs/1301.1673
     
  2. jcsd
  3. Sep 3, 2013 #2

    vanhees71

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    Science Advisor
    2016 Award

    First of all, you start with a two-photon state with entangled photons. It's a pure state given by a normalized state ket [itex]|\Phi \rangle[/itex]. The corresponding statistical operator of the pure state is the projection operator [itex]\hat{\rho}_{2 \gamma}=|\Phi \rangle \langle \Phi|[/itex]. This gives you the statistical properties on the polarization of both photons according to Borns rule.

    Now you want to forget about the fact that you deal with a two-photon state and only know about the polarization state state of one photon. Thus you trace over the other photon's polarization states to end up with your reduced density operator [itex]\hat{\rho}_{\text{right}}=1/2 \mathbb{1}[/itex]. This shows that an observer only looking at this photon's polarization measures an totally unpolarized beam of photons, i.e., he finds with probability 1/2 either a horizontally or a vertically polarized photon, and that's it.
     
  4. Sep 3, 2013 #3
    Thanks for your answer, but when you say a "totally unpolarized beam of photons" does that imply a statistical mixture of states? Under the discussion: "What is the difference between entangled and "normal" photons?" another Sci Advisor said that:

    "When you discard information about the other particle, you are left with a statistical mixture. (mathematically, this amounts to taking a partial trace)"

    A statistical mixture of states is not equivalent to a linear superposition of states.
     
  5. Sep 3, 2013 #4
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