Integrating by parts in path integral (Zee)

[GreyBadger]

Hi all,

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

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

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

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

Now, I am being supremely thick, but I don't see how this follows. Could somebody please spell it out in small words?

 

[waht]

Consider the Green's identity equation (4) in the link which is basically a 3D version of integrating by parts:

http://mathworld.wolfram.com/GreensIdentities.html

and taking the surface integral to be zero implies that

$$\int dV \;\nabla \phi \nabla \psi = -\int dV \;\phi \nabla^2 \psi$$

and so in this example, one can extend that to 4D and take the volume boundary term to zero

 

[haushofer]

"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.

 

[GreyBadger]

Aha yes. Thank you both.