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Why this expression is time-reversal odd?

  1. Dec 13, 2013 #1
    P and k are four-momentum of two particles.
    I read in a paper which said that
    [itex][\slashed{P},\slashed{k}]=\slashed{P}\slashed{k} - \slashed{k}\slashed{P}[/itex]
    is time-reversal odd.
    Why?
     
  2. jcsd
  3. Dec 13, 2013 #2

    ChrisVer

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    could you please fix the Latex?
    What is P and k? momenta?
     
  4. Dec 13, 2013 #3

    ZapperZ

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    ... and please make a full citation of the paper.

    Zz.
     
  5. Dec 13, 2013 #4

    George Jones

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    Unfortunately, MathJax does not support the "slashed" command from the amsmath package, which is used to implement Feynman's slash notation. Try

    $$\left[\gamma \cdot P , \gamma \cdot k \right] = \gamma \cdot P \ \gamma \cdot k − \gamma \cdot k \ \gamma \cdot P, $$

    where ## \gamma \cdot P = \gamma^\mu P_\mu##.

    [edit]Or just use (the possible more useful in this case) ##\gamma^\mu P_\mu## and ##\gamma^\nu k_\nu##.[/edit]
     
    Last edited: Dec 13, 2013
  6. Dec 13, 2013 #5
    Try [t e x]/ \!\!\!\! P[/ t e x]: [itex]/ \!\!\!\! P[/itex]
    and [t e x]/ \!\!\! k[/ t e x]: [itex]/ \!\!\! k[/itex]
     
  7. Dec 13, 2013 #6
  8. Dec 13, 2013 #7
    Thanks a lot for the texing.
     
  9. Dec 13, 2013 #8

    ChrisVer

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    *wrong answer*
    Well, the reason why there is T-oddness I guess is because there is a [itex]i[/itex] in front, making that term complex..
     
    Last edited: Dec 13, 2013
  10. Dec 13, 2013 #9
    But ##\Phi(P,k,S|n)## on the left of the Eq. must be T-even according to it's physical meaning, so, if ##A_4## is T-odd, ##[\gamma \cdot P, \gamma \cdot k]## must be T-odd.
    Actually, I think it is ##[P, k]## be T-odd that infers ##A_4## be T-odd, not the opposite, because ##A_4## is an unknown coefficient function.
    So I wonder why.
    Thanks a lot!
     
  11. Dec 14, 2013 #10

    ChrisVer

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    lol, I changed my initial post because it was wrong, I guess that:
    "Well, the reason why there is T-oddness I guess is because there is a [itex]i[/itex] in front, making that term complex.."
    is your answer...
     
  12. Dec 14, 2013 #11
    Well, I think the ##i## in front is just for convenience. And the gamma matrices are not all real, for example, in Dirac representation, ##\gamma^2## is complex.
    So, I think the problem is how does ##\left[\gamma \cdot P , \gamma \cdot k \right]##change under time reversal operation.

    I know that, under time reversal, ##P=(P_0,\vec{P})##becomes ##(P_0,-\vec{P})##, and in QFT course I've learned that ##\gamma^{\mu}##becomes ##(-1)^{\mu}\gamma^{\mu}##(Peskin's book, Page71). But what about ##\left[\gamma \cdot P , \gamma \cdot k \right]##?
     
  13. Dec 14, 2013 #12

    Bill_K

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    No, it's not there just for convenience. It's there because i[γ·P, γ·K] = 2 σμν Pμ Kν. And if you look at the table on p.71 of Peskin, you'll see how σμν transforms.
     
  14. Dec 14, 2013 #13
    Year, but for the terms in ##\sigma_{\mu\nu}P^{\mu}k^{\nu}## like ##\sigma_{02}P_0k_2##, is T-even, because ##\sigma_{02}## odd, ##P_0## even, and ##k_2## odd.
    Regards!
     
  15. Dec 14, 2013 #14

    Bill_K

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    No, σ02 is even. In the sum, σ0i is even, σjk is odd, P0 and K0 are even, while Pj and Kj are odd. All the terms in the sum σμνPμKν wind up transforming the same way, namely they are all odd.
     
  16. Dec 14, 2013 #15
    Could you please explain why ##\sigma_{0i}## is T-even? isn't it ##-(-1)^{\mu}(-1)^{\nu}##?
     
  17. Dec 14, 2013 #16

    Bill_K

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    Peskin uses a very strange notation here! :eek: He says "(-1)μ" is supposed to mean 1 for μ=0 and -1 for μ=1, 2, 3. In the table, for T we're given - (-1)0(-1)i, which is interpreted to mean (-1)(1)(-1) which is +1.

    His derivation of the results in this table could use a little cleanup work too. In addition to complex conjugation ψ → ψ*, the time reversal operation includes a matrix acting on ψ. He says this matrix is γ1γ3, but the value actually depends on your choice of representation for the gamma matrices. He uses the Weyl representation Eq.(3.25), in which γ2 is imaginary and all the others real. In general, time reversal can be written ψ → Dψ* where D is a matrix having the property D-1γμD = γ'μ where γ'0 = - γ0 and γ'j = γj.
     
  18. Dec 14, 2013 #17
    WOW!:thumbs: Many thanks, I will check it.
     
  19. Dec 31, 2013 #18
    Your looking for a transformation of the equation... yes? or, an explanation?
     
  20. Dec 31, 2013 #19

    Bill_K

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    Which do you think has not already been given?
     
  21. Dec 31, 2013 #20
    The transformation, why it's T-odd. And I think it is solved already.
     
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