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Srednicki equation 7.7

  1. Mar 2, 2010 #1
    Hello everyone,

    I will be glad if someone can explain how equation 7.7

    [tex]\tilde{x}(E)[/tex] = [tex]\tilde{q}(E)[/tex] + [tex]\frac{\tilde{f}(E)}{E^2-\omega^2+i\epsilon}[/tex]

    is a shift by constant, heres the link for the book

    http://www.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

    thanks
     
  2. jcsd
  3. Mar 2, 2010 #2
    Hi, I'm working through this book at the moment too, and have received lots of excellent help here, so would be glad to try and help (although being a newbie at this stuff too I may not be completley correct).

    I believe in chapter 7, the path integral has a measure Dq, i.e. it is with respect to variations of the configuration space (integrating over all possible configurations the particle could traverse, including the the classically realised one, but also every single other possible. Later these will be field configurations)

    If we make the change in 7.7 [tex] \tilde{x}(E)= \tilde{q}(E)+ \frac{\tilde{f}(E)}{E^2-\omega^2+i\epsilon} [/tex], then the last term is a constant as far as [tex] \int Dq [/tex] is concerned as it doesn't depend on the dynamical variable q. It's like shifting everyone of the infinite possible trajectories you're integrating over by a constant.

    This is more or less the same as changing variables in a normal integral [tex] \int dx [/tex], by having [tex] y \rightarrow x-\chi [/tex]. Clearly dy=dx, the same thing is going on in the path integral case, it's just complicated by integrating over all the paths.
     
    Last edited: Mar 2, 2010
  4. Mar 2, 2010 #3
    Thank you for replying, I am glad I am not the only one struggling trough qft,

    I thought of similar thing, but your affirmation is helpful..
     
  5. Mar 2, 2010 #4
    I wil be glad if you can also explain why eq 7.17 is not just

    [tex] \frac{1}{i^2}[G(t_2-t_1)G(t_4-t_3)][/tex]
     
  6. Mar 2, 2010 #5
    Well it ultimately comes down to the Leibniz product rule (and making sure you don't set f=0 until after the differentiation, so you have to differentiate the exponential as part of the product when you go onto the next derivative etc). Acting with the three [tex] \delta [/tex] derivatives on [tex] \langle 0 \mid 0 \rangle_f [/tex] (as given explicitly in 7.11), you will get a range of terms, but then you set f to 0 only the ones in 7.17 survive.

    I hope that helps, it's hard to explain without writing it out explicitly which would probably take forever in latex. But it really is just the product rule, and making sure you only set f=0 at the end, so you don't lose any terms.
     
  7. Mar 2, 2010 #6
    I will do part of the process for the 3 point function (which will be zero but will show the details of how this works anyway)

    [tex] \langle 0 \mid TQ(t_1)Q(t_2)Q(t_3) \mid 0 \rangle= \frac{1}{i}\frac{\delta}{\delta f(t_1)}\frac{1}{i}\frac{\delta}{\delta f(t_2)}\frac{1}{i}\frac{\delta}{\delta f(t_3)} \langle0\mid0\rangle_f [/tex]

    [tex] \langle 0 \mid TQ(t_1)Q(t_2)Q(t_3) \mid 0 \rangle= \frac{1}{i}\frac{\delta}{\delta f(t_1)}\frac{1}{i}\frac{\delta}{\delta f(t_2)}\frac{1}{i}\frac{\delta}{\delta f(t_3)} \langle0\mid0\rangle_f [/tex]

    [tex]\langle0\mid0\rangle_f=exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'})] [/tex]

    So acting with first derivative:

    [tex] \frac{1}{i}\frac{\delta}{\delta f(t_3)}exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'})]= \{\frac{1}{i} \frac{i}{2} \int dt dt^{'} \delta (t-t_3) G(t-t^{'})f(t^{'})+\frac{1}{i}\frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})\delta (t^{'}-t_3) ) \}exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ] [/tex]

    This is equal to

    [tex] \{\frac{1}{2} \int dt^{'} G(t_3-t^{'})f(t^{'})+\frac{1}{2} \int dt f(t)G(t-t_3) \}exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ] [/tex]

    Now since [tex] t, t^{'} [/tex] are dummy variables and since G is symmetric in it's arguments (because of the modulo) we can write this simply as


    [tex] \int dt^{'} G(t_3-t^{'})f(t^{'})exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ] [/tex]

    This is how Srednicki gets line 2 of 7.16.

    Now you want to act on this in the same way with the [tex] \frac{1}{i}\frac{\delta}{\delta f(t_2)} [/tex]

    So now you just use the product rule again not forgetting exp term yet which hasn't gone to one as we haven't yet set f=0.

    This leads you to the next line:

    [tex] \frac{1}{i}\frac{\delta}{\delta f(t_2)} \int dt^{'} G(t_3-t^{'})f(t^{'})exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ] = \{ G(t_3-t_2)+\int dt^{'} G(t_3-t^{'})f(t^{'})\int dt^{''} G(t_2-t^{''})f(t^{''})\}exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ] [/tex]

    Note this is equation 7.16 (only with t2, t3 instead of t1,t2) explcitly, my second term being what he calls "term with fs", which dies when we set f=0 at the end.

    Finally we can hit this with the last derivative as follows:

    [tex] \frac{1}{i}\frac{\delta}{\delta f(t_1)}\{ G(t_3-t_2)+\int dt^{'} G(t_3-t^{'})f(t^{'})\int dt^{'} G(t_2-t^{'})f(t^{'})\}exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ] [/tex]
     
    Last edited: Mar 2, 2010
  8. Mar 2, 2010 #7
    Which gives:
    [tex] \{ G(t_3-t_2)\int dt^{'} G(t_1-t^{'})f(t^{'})+G(t_3-t_1)\int dt^{'} G(t_2-t^{'})f(t^{'}) [/tex]

    [tex]+G(t_2-t_1)\int dt^{'} G(t_3-t^{'})f(t^{'})+\int dt^{'} G(t_3-t^{'})f(t^{'})\int dt^{''} G(t_2-t^{''})\int dt^{'} G(t_1-t^{'}) \} exp[ \frac{i}{2} \int dt dt^{'} f(t)G(t-t^{'})f(t^{'}) ] [/tex]

    Notice that setting f to zero now, will kill everything, this is why all odd number correlation functions are vanishing. Hopefully I haven't made too many mistakes and you can carry out this process to get the 4 point correlation function. You should be left with Srednicki's three terms as the only ones without f's then you can set f=0 to kill all others.
     
    Last edited: Mar 2, 2010
  9. Mar 2, 2010 #8
    wonderful, thankyou very much...

    I am grateful you helped me in both the problems
     
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