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Graph sol. of Dirac eq., type "graph paper" needed?

  1. Mar 16, 2015 #1
    What type of "graph paper" do I need to graph an arbitrary solution, Ψ, of the Dirac equation in 3+1 dimensional spacetime? Assume the "graph paper" has the minimum dimensions required to do the job.

    Would this work? At each point of spacetime we need a complex plane which takes care of the magnitude and phase of Ψ. The spinor part of Ψ is graphed on a pair of properly ruled three-spheres, S^3 (do we need to allow for the spin part of Ψ vary with spacetime?).

    So we need 4 dimensions for spacetime, 2 dimensions for the complex plane, 6 dimensions for the two three-spheres, 8 dimensions at each point in spacetime, so we need 12 dimensional graph paper?

    Or could we cheat and approximately graph an arbitrary solution of of the Dirac equation by having a set of 4 (or 3?) independent complex vectors at each point of spacetime?

    Thanks for any help!

    Edit, I think I may have made things more complicated then needed. I think I was told in a thread that a time varying phase times a spinor just moves a point in S^3 on some "orbit" in S^3 so that the above can be simplified a bit?
     
    Last edited: Mar 16, 2015
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  3. Mar 16, 2015 #2

    bhobba

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    Each component of the Dirac 4 vector has 5 dimensions - you do the math.

    Thanks
    Bill
     
  4. Mar 17, 2015 #3

    samalkhaiat

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    I could not make any sense out of the OP or your answer to it. Can you tell me, what is this "Dirac 4 vector"? And, how is it that "each component" of this "4 vector" has "5 dimensions"?
     
  5. Mar 17, 2015 #4

    bhobba

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    The solution to the Dirac equation is a four component vector of 4 wave-functions:
    http://www.nyu.edu/classes/tuckerman/quant.mech/lectures/lecture_7/node1.html

    Each wave-function requires 4 parameters - 3 of space and one of time to specify a complex number. But a complex number requires two parameters so you have 5 dimensions not 4. But you have 4 of them so it requires 8 numbers at each point in space time. Hence 12 dimensions.

    That's in so far as I interpreted the OP's question correctly - as you point out its rather obscure.

    Thanks
    Bill
     
  6. Mar 18, 2015 #5
    Thanks for the replys, nothing like asking a question to realize on further reflection that what little I knew was even smaller then that.

    I was trying to include three-spheres into my graphing problem because we know there is a relationship between them and spinors, namely a point on a three-sphere can be represented by a normalized two-component spinor, see post 3, in the following thread,

    https://www.physicsforums.com/threa...-by-a-2-component-spinor.603404/#post-3899072

    If the spinors are not normalized I think then I have problems? How if at all can I still use some three-spheres in my graphing problem?

    Also in the same thread I was told that multiplying a spinor times a simple time dependent phase produces an orbit of a point in the three-sphere. Not sure how that fact can help in my problem.

    Thanks for any and all help!
     
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