Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can we look upon entanglement as measurement?

  1. Sep 3, 2013 #1

    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:

  2. jcsd
  3. Sep 3, 2013 #2


    User Avatar
    Science Advisor
    Gold Member
    2017 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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook