Derivation in Srednicki's which got me puzzled.

  1. Hi. on page 95 , I am not sure how did he derive the second term on the RHS of equation (13.16).

    http://books.google.co.il/books?id=...V4QSnhYFA&ved=0CBsQ6AEwAA#v=onepage&q&f=false

    I mean if I plug back I should get:

    [tex] i\theta(x^0-y^0)\int_{4m^2}^\infty \rho(s) \int \tilde{dk} e^{ik(x-y)} + i\theta(y^0-x^0)\int_{4m^2}^\infty \int \tilde{dk} e^{-ik(y-x)} [/tex]

    And since [tex] \int \tilde{dk} e^{\pm ik(x-y)} = \int \frac{d^3 k}{(2\pi)^3 2k^0} e^{\pm ik(x-y)} = \frac{\delta(x-y)}{2k^0}[/tex]

    But I don't see how do we arrive from all of the above to the epression: [tex] \frac{1}{k^2+s-i\epsilon}[/tex] in the integral, anyone cares to elaborate?

    Thanks.
    P.S
    I forgot to mention that: [tex] k^0 =\sqrt{\vec{k}^2 +s} [/tex]
    where s is one of Mandelstam parameters, I believe it's standard in the literature.
     
    Last edited: Jul 15, 2014
  2. jcsd
  3. Wait a minute I have a mistake.

    It should be: [tex] \int \frac{d^3 k }{(2\pi)^3 2k^0} e^{\pm ik(x-y)} = \frac{\delta^3(\vec{x}-\vec{y})}{2k^0} e^{\pm (-ik^0)(x^0-y^0)}[/tex]

    I still don't see how did he get to this term in equation 13.16. :-(
     
  4. ChrisVer

    ChrisVer 2,298
    Gold Member

    Let's have a look at the first term of [itex]<0|T \phi(x) \phi(y)|0>[/itex]

    It is:

    [itex] \theta(x^0 - y^0) <0|\phi(x) \phi(y)|0>[/itex]

    Making use of 13.12

    [itex] \theta(x^0 - y^0) (\int \bar{dk} e^{ik(x-y)} + \int_{4m^{2}}^{∞} ds \rho(s) \int \bar{dk} e^{ik(x-y)}) [/itex]

    Similiarly for the 2nd term:

    [itex] \theta(y^0 - x^0) (\int \bar{dk} e^{-ik(x-y)} + \int_{4m^{2}}^{∞} ds \rho(s) \int \bar{dk} e^{-ik(x-y)}) [/itex]

    Now add them... the 1st terms will give you what you have from equation 13.15
    Also you can put the thetas in the integrals and use 13.15 also for the second terms, getting the integral of rho(s) out...

    So you will have something like (it's a sketching not the result):
    [itex] \int F(k^{-2}) + \int F(k'^{-2}) \int \rho [/itex]
    from where you can take out a common factor... of course the 2 k's you have, also as the author points out, are not the same- the one has "mass" m, while the other has "mass" s... So the two "propagators" appearing will be different.
     
    Last edited: Jul 15, 2014
  5. ChrisVer

    ChrisVer 2,298
    Gold Member

    Also about the integral you ask....
    It's easier to try to evaluate:
    [itex] \int \frac{d^{d}k}{(2 \pi)^{d}} \frac{1}{k^{2}+m^{2}-i \epsilon} e^{ik(x-y)} [/itex]

    using residues to end up in the RHS of 13.15...
    In fact the residues are going to give you some positive energies moving forward and some negative energies moving backward.
     
  6. Ok, I think this expression is derived in the previous sections, I just didn't notice that the second term was included in the square brackets. My bad, sorry for that.

    Thanks.
     
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