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Computing vacuum expectation values

  1. Aug 3, 2010 #1
    I have small question computing vacuum expectation values here http://www.cns.gatech.edu/FieldTheory/extras/SrednickiQFT03.pdf" [Broken] from Mark Srednicki.

    My problem is with equation 210 on the pdf page 69. In the second line of 210, where does the second term come from?

    Z(J) and W(J) are defined one page 62-63 with equations 196 and 197, and the computation for the vacuum expectation value of a single field is given in 198, which makes sense to me.

    But not the second term in the second line of 210!

    thank you
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 3, 2010 #2


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    Re: propagator

    it's just the chain rule and the product rule. treat [itex]\delta_i[/itex] as an ordinary derivative and work it out.

    There is one small typo: in the first line there should be a 1/Z[J=0]. This then cancels when you write it in terms of W.
  4. Aug 4, 2010 #3
    Re: propagator

    Thanks Blechman!
  5. Aug 6, 2010 #4
    Re: propagator

    Ok, bit ashamed to come back, but ..

    So I want to compute (d/dJ_1)(d/dJ_2)exp(iW(J(1,2))

    (d/dJ_1)(d/dJ_2)exp(iW(J(1,2)) = (d/dJ_1)((d/dJ_2)iW(J(1,2)exp(iW(J(1,2)) (chain rule)

    since at the end we set J=0 and W(0)=0 defined, exp(iW(J(0))=1, exp(iW(J(1,2)) drops out

    so I got (d/dJ_1)((d/dJ_2)iW(J(1,2)

    now I should apply the product rule to get to the second line of 210, but sadly I can't see how
  6. Aug 6, 2010 #5


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    Re: propagator

    You may only do that at the very end of the calculation (after having applied all derivatives). Apply the derivative wrt J_1 on everything and *then* set J=0. You will get his expression.
  7. Aug 6, 2010 #6
    Re: propagator

    Now got it!

    Thanks nrqed, thanks blechman!
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