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About Mordred

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    University of the Caribou
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    cosmology and particle physics

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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


      [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.


    3. Dubbelosix


      Oh I see. Thanks. 

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

    4. Mordred



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