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Srednicki 7.14 ND 7.16

  1. Nov 19, 2008 #1
    [tex]\int[/tex]dt' G(t-t') f(t') = 1/i [tex]\delta[/tex]/[tex]\delta[/tex]f(t)

    where G(t-t') = i/ 2w exp (iw (t-t'))

    I thought the RHS of the first equation is f(t). Can someone explain?

    thank you
     
  2. jcsd
  3. Nov 19, 2008 #2

    Avodyne

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    This doesn't make sense, and I'm not sure where you got it. Something like it that is correct and that is used to get the 2nd line of (7.16) is

    [tex]{1\over i}{\delta\over\delta f(t_1)}\left[{i\over2}\int dt\,dt'\,f(t)G(t-t')f(t')\right] = \int dt'\,G(t_1 -t')f(t').[/tex]
     
  4. Nov 20, 2008 #3
    Well, I thought what I write was implied in the equations in the Srednicki book.

    But then I don't see how we get from the first line two the second in 7.16. with your equation.

    Is the term in the brackets equal to one? And you mean t_2 instead of t_1, right?
     
  5. Nov 20, 2008 #4

    Avodyne

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    What you wrote does not make sense. The functional derivative on your right-hand is not acting on anything.

    Let

    [tex]Z(f)=\langle 0|0\rangle_f[/tex]

    From 7.11,

    [tex]Z(f)=\exp K(f)[/tex]

    where

    [tex]K(f)={i\over 2}\int dt\,dt'\,f(t)G(t-t')f(t')[/tex]

    By the chain rule,

    [tex]{\delta\over\delta f(t_2)}Z(f)={dZ\over dK}\;{\delta\over\delta f(t_2)}K(f)[/tex]

    and since [tex]Z=\exp K[/tex], [tex]dZ/dK = \exp K = Z[/tex]. Now we use

    [tex]{1\over i}\,{\delta K(f)\over\delta f(t_2)}=
    {1\over i}\,{i\over 2}\int dt\,dt'\left[\left({\delta f(t)\over\delta f(t_2)}\right)G(t-t')f(t')+f(t)G(t-t')\left({\delta f(t')\over\delta f(t_2)}\right)\right][/tex]

    [tex]{}\qquad\qquad={1\over 2}\int dt\,dt'\Bigl[\delta(t-t_2)G(t-t')f(t')+f(t)G(t-t')\delta(t'-t_2)\Bigr][/tex]

    [tex]{}={1\over 2}\int dt'\,G(t_2-t')f(t')+{1\over 2}\int dt\,f(t)G(t-t_2)[/tex]

    [tex]{}=\int dt'\,G(t_2-t')f(t')[/tex]

    where, to get the last line, we use [tex]G(t-t_2)=G(t_2-t)[/tex], and change the dummy integration variable in the 2nd term from [tex]t[/tex] to [tex]t'[/tex], so that it is then the same as the first term.
     
  6. Nov 21, 2008 #5
    thank you!!!
     
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