Mordred

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Mordred last won the day on November 18

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About Mordred

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    http://www.Cosmology101.wikidot.com

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    University of the Caribou
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  1. Mordred, I have a technical question here.... my work is a bit more complicated and requires a bit more depth than I am used to - in the case of a projective space, the identity P^2 = P is said to hold -  what does it mean?

    In my case I am looking at, I have an antisymmetric matrix that requires to be squared in the projected space to yield the identity/unity. The projective space looks like

     


    [math]P = \frac{\mathbf{I} + n \cdot \sigma}{2} = |\psi><\psi|[/math]

     

    The square of the Pauli matrix should yield an identity [math](n \cdot \sigma)^2 = \mathbf{I}[/math] (unit vectors naturally square into unity). What is the square of the dyad in such a case? 

    1. Show previous comments  2 more
    2. Mordred

      Mordred

      [math]\rho^2=\rho[/math] is the projector condition [math]\rho[/rho is only a projector if and only if [math]\rho^2=\rho[/math]

      wiki explains it a bit better

      In operator language, a density operator is a positive semidefinite, Hermitian operator of trace 1 acting on the state space.[7] A density operator describes a pure state if it is a rank one projection. Equivalently, a density operator ρ describes a pure state if and only if

      ρ=ρ2{\displaystyle \rho =\rho ^{2}}{\displaystyle \rho =\rho ^{2}},

      i.e. the state is idempotent. This is true regardless of whether H is finite-dimensional or not.

      https://en.wikipedia.org/wiki/Density_matrix

       

    3. Dubbelosix

      Dubbelosix

      Oh I see. Thanks. 

      Ahh yes of course, and it satisfies pure states, as expected. 

    4. Mordred

      Mordred

      correct 

      lol messed up the closing tag above but can't edit ah well ya got the answer P