# Mordred

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1. ## Dubbelosix    Mordred

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

$P = \frac{\mathbf{I} + n \cdot \sigma}{2} = |\psi><\psi|$

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

1. Show previous comments  2 more
2. $\rho^2=\rho$ is the projector condition $\rho[/rho is only a projector if and only if [math]\rho^2=\rho$

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}},

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

3. Oh I see. Thanks.

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

4. correct

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