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Status Replies posted by Mordred
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Successfully defended my viva. Guess i have to get a real job now.
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Welcome back
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Welcome back
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Welcome back
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Hello , mordred. Why SUSY are still not found on the collider
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Hello , mordred. Why SUSY are still not found on the collider
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I'm gonna be a Grandpa!
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Welcome back and Happy New Year Mordred.
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Welcome back and Happy New Year Mordred.
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Long time no see.
Welcome back mate.
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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?
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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?
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[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}},
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
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Isn't it nice to have an honest, educational discussion? I see so much trash on the science forums and had very negative responses at some other sites with people calling my work rubbish and work that I do not understand.
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Thanks all its nice to see my methodology recieve positive results. The thread under discussion should however be creditted to Dubbelsox.
He is abiding by the proper methodology of a toy universe model to the letter. Makes it far easier to engage in a proper scientific discussion.
Yes Stringy I do occassionally vist physicsforum however I find I make greater contributions here. Mainly due simply because this site has a controllable Speculations forum.
Strange as that may sound lol
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I got a national rank of 3506 in an exam.
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I got a national rank of 3506 in an exam.
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After a few more pages, I'll be done writing my proof of the Collatz conjecture. I know it has been almost two months since I proved it, but I had to do a lot of research to write this paper. Plus, I still have to work and take care of day to day life. However, I will be posting new software that I'm writing that uses my equation to list all possible trajectories of a given height. This stuff is really awesome, and I even have a reporter from KOCO that wants to do an exclusive on me a...