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Deriving Faddeev-Popov

  1. Jan 13, 2015 #1
    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?

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