Deriving Faddeev-Popov

  • Thread starter Lapidus
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  • #1
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Main Question or Discussion Point

In Schwartz QFT book, while explaining the Faddeev-Popov procedure, he presents this following observation at (25.99):
upload_2015-1-13_13-16-6.png


Can someone help understanding me why this expression equals one?

THANK YOU!
 

Answers and Replies

  • #2
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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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