1. Jan 13, 2015 #1

   Lapidus

   

   In Schwartz QFT book, while explaining the Faddeev-Popov procedure, he presents this following observation at (25.99):
   
   
   Can someone help understanding me why this expression equals one?
   
   THANK YOU!

    

   Lapidus, Jan 13, 2015
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3. Jan 13, 2015 #2

   Einj

   

   This is the functional integral extention of the usual relation for a delta functions. Suppose you have an n-dimentional vector field which is a function of another vector, say ##\vec v\equiv \vec v(\vec x)##. Then you have:
   
   $$
   1=\int d^nv \delta^n(\vec v)=\int d^nx\det\left(\frac{\partial v^i}{\partial x^j}\right) \delta^n(\vec v(\vec x))
   $$
   
   It's quite intuitive that the functional integral extension is the one you wrote.

    

   Einj, Jan 13, 2015

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