
Posts
8300 
Joined

Last visited

Days Won
24
Content Type
Profiles
Forums
Calendar
Status Replies posted by Mordred

Welcome back

Welcome back

Welcome back

Hello , mordred. Why SUSY are still not found on the collider

Hello , mordred. Why SUSY are still not found on the collider

I'm gonna be a Grandpa!

Welcome back and Happy New Year Mordred.

Welcome back and Happy New Year Mordred.

Long time no see.
Welcome back mate.

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?

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?

[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 finitedimensional or not.
https://en.wikipedia.org/wiki/Density_matrix


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.

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


I got a national rank of 3506 in an exam.

I got a national rank of 3506 in an exam.

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...