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Confused about Wick & Zee

  1. Feb 5, 2012 #1


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    reading Zee's QFT Theory in a Nutshell and on p 13 on the explanation of the Gaussian integral he loses me going from (9) to (10) by jumping from a relatively straightforward derivation of [itex]\int e^{-1/2ax^2}dx [/itex]
    by then saying "acting on this with -2(d/da)

    to get [itex]<x^{2n}> ={\int e^{-1/2ax^2}x^{2n}dx}/{\int e^{-1/2ax^2}dx}= \frac{1}{a^n}(2n-1)!![/itex]

    not sure how he gets to the double factorial expression - how do you know this integral would not diverge?

    also not clear on the meaning of -2(d/da) as an operator

    assuming the notation <x> means the expectation of some arbitrary variable

    the wiki article and other internet sources on Wick's theorem refers to ordering of creation and annihilation operators, which Zee has not mentioned yet

    sorry for the rambling post, but any help would be greatly appreciated
  2. jcsd
  3. Feb 5, 2012 #2


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    The idea is to calculate

    [tex]I_0(a) = \int_{-\infty}^{+\infty}dx\,e^{-ax^2}[/tex]

    Then you observe that

    [tex]x^2\,e^{-ax^2} = -\frac{\partial}{\partial a}e^{-ax^2}[/tex]

    Now you rewrite the integral as follows

    [tex]I_2(a) = \int_{-\infty}^{+\infty}dx\,x^2\,e^{-ax^2} = \int_{-\infty}^{+\infty}dx\,\left(-\frac{\partial}{\partial a}\right)e^{-ax^2} = -\frac{\partial}{\partial a} \int_{-\infty}^{+\infty}dx\,e^{-ax^2} = -\frac{\partial}{\partial a} I_0(a)[/tex]

    Of course this is a formal trick and requires a proof in principle.

    From now one you can continue with

    [tex]I_{2n}(a) = -\frac{\partial}{\partial a} I_{2n-2}(a)[/tex]


    [tex]I_{2n}(a) = \left(-\frac{\partial}{\partial a}\right)^{2} I_{0}(a)[/tex]
    Last edited: Feb 5, 2012
  4. Feb 5, 2012 #3
    But what does [tex] \int_{-\infty}^{+\infty}dx\,x^2\,e^{-1/2\,ax^2}[/tex] give? Does someone have a Gaussian table handy to look it up?

    From then on it should be easy. Just hit on the result repeatedly with -2(d/da) and after you have done so n times, divide by (2pi/a)^1/2.
    Last edited: Feb 5, 2012
  5. Feb 5, 2012 #4

    Physics Monkey

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    The point is that you can figure out the x^2 integral only knowing the value of the [itex] \int dx \exp{(- a x^2)} = \sqrt{\pi/a}[/itex]. Tom's post explains how.
  6. Feb 5, 2012 #5
    What x^2 integral? Also, it's not [itex] \int dx \exp{(- a x^2)}[/itex], but [tex] \int dx\,x^2\,e^{-1/2\,ax^2}[/tex] and Tom did not explain (as far as I understand his post) that we have to take the differential on both sides to get this factioral expression on the RHS, which was the OP question about.

    Anyway, here is a crystal clear explanation of what goes on:

  7. Feb 5, 2012 #6


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    in order to keep things exciting I hesitated to provide all the details and the final result ;-)

    start with the well-known result for

    [tex]I_0(a) = \int_{-\infty}^{+\infty}dx\,e^{-ax^2}[/tex]

    and calculate

    [tex]I_{2n}(a) = \left(-\frac{\partial}{\partial a}\right)^{2} I_{0}(a);\;\;n=1,2,\ldots[/tex]

    that's all you need
    Last edited: Feb 5, 2012
  8. Feb 5, 2012 #7


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    thanks everyone for the responses and the link to the weylmann.com site, it was all very helpful
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