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Questions in Srednicki's book

  1. Feb 25, 2014 #1
    I have questions about chapters 5 & 9 in Srednicki's book.

    1.
    Below eq. (5.18), he says "$<p|\phi (0)|0>$ is a Lorentz-invariant number.", but I do not know why.

    2.
    Below eq. (5.26), he says "After shifting and rescailing (and renaming some parameters)".
    Are $m$ and $g$ in eq. (5.26) different from ones in eq. (5.27)?

    3.
    Below eq. (9.9), he says "as we will see, "$Y=O(g)$ and $Z_i =1+O(g^2)$", which means these g-dependence is just the expectation at this stage.
    In eq. (9.18), however, he used $Z_g =1+O(g^2)$ without showing this $Z_g$-behavior.
    Is my understanding incorrect?
     
  2. jcsd
  3. Feb 26, 2014 #2

    Avodyne

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    1. This is not really important. It's a number. Don't worry about whether it's Lorentz invariant or not.

    2. Yes.

    3. He's still assuming it. He shows it in a later chapter.
     
  4. Feb 26, 2014 #3
    Thank you very much!
    I understand 2 & 3, but sorry I do not fully understand 1.

    I think that the Lorentz-invariance of [itex]<p|\phi (0)|0>[/itex] is important.
    [itex]<p|\phi (0)|0>[/itex] should be just some number that is independent of [itex]p^\mu[/itex], and it is the Lorentz-invariance that makes [itex]<p|\phi (0)|0>[/itex] be, i.e., the Lorentz-invariance imposes the condition that [itex]<p|\phi (0)|0>[/itex] is a function of [itex]p^2=-m^2[/itex] (number).
    In other words, when we do not consider the Lorentz-invariance, is there no possibility that [itex]<p|\phi (0)|0>=p_x[/itex], for example?
     
  5. Feb 26, 2014 #4

    naima

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    Look at eq 3.14 and let x=0, you get [itex]\phi (0)[/itex] replace it in [itex]<p|\phi (0)|0>[/itex] by the integral
    put <p| and |0> in the integral. Look at p 26: f(k) is chosen so that the result is lorentz invariant.
     
    Last edited: Feb 26, 2014
  6. Feb 26, 2014 #5
    Thank you, naima.

    For "free" field, you are right.
    But now in chap. 9 we consider "interacting" field, and ##\phi (0)## is interacting one.
    Thus, I do not think we can use eqs. in chap. 3 to prove ##<p| \phi (0) |0>## is Lorentz-invariant.

    How about following one?
    For ##t=\pm \infty##, ##\phi (x)## becomes free field.
    Then, ##<p| \phi (x) |0>## is Lorentz-invariant as you explained.
    Therefore, the r.h.s ##e^{-ipx}<p| \phi (0) |0>## is Lorentz-invariant, too.
    Since ##e^{-ipx}## is of course Lorentz-invariant, we conclude that ##<p| \phi (0) |0>## also is.
     
  7. Feb 27, 2014 #6

    naima

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    Sredinski writes that he assumes this normalization condition (to have a coherent theory). This has not to be proven.
     
  8. Feb 27, 2014 #7

    Avodyne

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    The Lorentz transformations of the fields and the one-particle states are the same in the free and interacting theories, so the free-field calculation applies.
     
  9. Feb 28, 2014 #8
    Thank you, naima & Avodyne.
    I understand.
     
  10. Feb 28, 2014 #9

    naima

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    Take the Higgs in the ## \phi^4 ## theory. he is never free. he is always self interacting to keep its mass.
    When Srednicki studies this family of self interacting theories he says that he has to admit four normalization conditions. they cannot be derived.
     
  11. Mar 1, 2014 #10
    Thank you for your instructive comment!
     
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