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Integrating by parts in path integral (Zee)

  1. Jun 7, 2010 #1
    Hi all,

    I have an exceptionally basic question, taken from P21 of Zee. Eq. 14 is

    [tex]Z=\int D\psi e^{i\int d^4x(\frac{1}{2}[(\partial\psi )^2-m^2\psi^2] + J\psi)}[/tex]

    The statement is then made that 'Integrating by parts under the [tex]\int d^4x[/tex]' leads to Eq. 15:

    [tex]Z=\int D\psi e^{i\int d^4x[-\frac{1}{2}\psi(\partial^2+m^2)\psi + J\psi]}[/tex].

    Now, I am being supremely thick, but I don't see how this follows. Could somebody please spell it out in small words?
  2. jcsd
  3. Jun 7, 2010 #2
    Consider the Green's identity equation (4) in the link which is basically a 3D version of integrating by parts:


    and taking the surface integral to be zero implies that

    [tex] \int dV \;\nabla \phi \nabla \psi = -\int dV \;\phi \nabla^2 \psi [/tex]

    and so in this example, one can extend that to 4D and take the volume boundary term to zero
  4. Jun 8, 2010 #3


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    "Partially integrating" means here that you use partial differentation to rewrite the integrand.

    Here the relevant term becomes

    \int_{\Omega} \partial \phi \partial \phi = \int_{\Omega} \partial(\phi\partial\phi) - \int_{\Omega} \partial^2 \phi

    The first term on the RHS becomes

    \int_{\Omega} \partial(\phi\partial\phi) = \int_{\partial\Omega} \phi\partial\phi

    by Stokes theorem. Imposing boundary conditions, this term vanishes.
  5. Jun 8, 2010 #4
    Aha yes. Thank you both.
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