QED Prerequisites : Geometric algebra 31 - More about spinors (corrected!)

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  • เผยแพร่เมื่อ 6 พ.ย. 2024

ความคิดเห็น • 7

  • @davidhand9721
    @davidhand9721 7 หลายเดือนก่อน +1

    Hey buddy. You doing alright? Working on the next one? I'd like to see more on Dirac spinors at some point, get a better handle on how they relate to geometry of the field derivatives, if that's on the agenda. Totally unrelated to this series, I also really want to get a sense of what the SU(3) generators are doing in color space; the lie algebra series ends just before you get to that, I think. You're a really gifted science communicator, and I've got you to thank for a lot of things I know. I hope you understand that you're doing something really valuable here.

    • @XylyXylyX
      @XylyXylyX  7 หลายเดือนก่อน +1

      Thank you for your kind comments. I am really trying to put out more content. Perhaps this weekend?
      Tell me more about your specific interests regarding Dirac spinors?

    • @davidhand9721
      @davidhand9721 6 หลายเดือนก่อน +1

      @@XylyXylyX I assume it's sort of like a tensor product space between a pauli spinor and something else. In the GA representation, pauli spinors are these rotors - bivectors + scalar. What's the second basis then? Given that it's about antimatter, and Feynman diagrams use matter going back in time to represent antimatter, _and_ the spinors context in the Dirac equation, I want to say that the second basis is something like a time derivative, differential form, something like that. I have no clue how that would be represented in GA even if I knew I was right, which I do not assume one bit.
      I think you once said that the Dirac gamma matrices had a geometric meaning, as well, and every time I try to work out what that is, I don't really trust my result very much. By analogy to the GA representation of the SU(2) group, i.e. the Pauli sigma matrices, I presume there's a direct correlation between the gamma STA basis and the gamma matrices. At least I expect the Lie bracket to be identical, but it's hard for me to picture how basis vectors are supposed to equate to what the gamma matrices are doing. These two things are related, so I'm pretty sure having the full set of Dirac spinor basis vectors will make it all click for me.

    • @XylyXylyX
      @XylyXylyX  6 หลายเดือนก่อน +1

      @@davidhand9721 ok. All those interests are easy to address. I’ll try to get to it asap.

    • @davidhand9721
      @davidhand9721 6 หลายเดือนก่อน +1

      @@XylyXylyX thanks man! I appreciate it.

  • @atlekulild3219
    @atlekulild3219 11 หลายเดือนก่อน +1

    Amazing work! Sorry you had to redo the video :(
    (I'll use x, y and z to represent sigma 1, 2 and 3 respectively) You didn't mention it, so I'll give a simple proof here for why operating on a state with i is represented by multiplying the spinor on the right with Iz. If we have a spinor S represented by the state |S>, then simple matrix multiplication will show us that i|S> is equivalent to xyz|S>, because the matrix resulting from xyz is just the identity matrix multiplied by i. But we know how to represent xyz|S> in Geometric Algebra, it's simply xyzSzzz = ISz = SIz.

  • @Pluralist
    @Pluralist 11 หลายเดือนก่อน