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Mordred

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Status Replies posted by Mordred

  1. Welcome back

    1. Mordred

      Mordred

      well trying to insert an image screwed it this time around not time out.  this was the 4th time I had to correct the latex. LOL I was going to write up the relevent info on the CKM mixing matrix but I'm tired of it getting screwed up

    2. (See 8 other replies to this status update)

  2. Welcome back

    1. Mordred

      Mordred

      Well two nights in a row so I may also be hitting server maintenance while in edit mode. Not a biggie just now have to redo the fix I had completed and added to lmao.

    2. (See 8 other replies to this status update)

  3. Welcome back

    1. Mordred

      Mordred

      Pretty good what's the new method for posting latex for some reason all the [math] instructions I had in my last post disappeared 

    2. (See 8 other replies to this status update)

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

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

    1. Mordred

      Mordred

      Third possibility SUSY particles require a higher energy level than we can currently produce. The Higgs boson typically has less mass than the SUSY particles. (Total mass )

    2. (See 5 other replies to this status update)

  6. I'm gonna be a Grandpa! 

    1. Mordred

      Mordred

      Welcome to the authorized Spoiler club lmao grandkids are fun get them all hyped up and send em back home to their parents.

    2. (See 10 other replies to this status update)

  7. Welcome back and Happy New Year Mordred.

     

  8. Welcome back and Happy New Year Mordred.

     

    1. Mordred

      Mordred

      Thanks, glad to be back had to take a break due to RL problems that are now resolved. Happy New year to you as well

    2. (See 4 other replies to this status update)

  9. Long time no see.

     

    Welcome back mate.

     

    :)

    1. Mordred

      Mordred

      lol glad to be back as well, there are time when I need to take a break much like an unofficial summer break from school lol

    2. (See 2 other replies to this status update)

  10. 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. Mordred

      Mordred

      correct 

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

    2. (See 4 other replies to this status update)

  11. 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. 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

       

    2. (See 4 other replies to this status update)

  12. 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. 

     

    1. Mordred

      Mordred

      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

    2. (See 6 other replies to this status update)

  13. 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...

    1. Mordred

      Mordred

      Keep at it, it is always nice to see some of our advise being applied in a positive direction of advancement.

    2. (See 3 other replies to this status update)

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