- #1
ChrisPhys
- 6
- 0
I can't convince myself whether the following functional derivative is trivial or not:
##\frac \delta {\delta \psi(x)} \big[ \partial_x \psi(x)\big],##
where ##\partial_x## is a standard derivative with respect to ##x##.
One could argue that
## \partial_x \psi(x) = \int dx' [\partial_{x'} \psi(x')] \delta (x - x') = - \int dx' \psi(x') \partial_{x'} \delta (x - x'),##
assuming there is no boundary term in integration by parts.
In this case, the functional derivative would give
##\frac \delta {\delta \psi(x)}\Big[ - \int dx' \psi(x') \partial_{x'} \delta (x - x') \Big] = - \partial_{x'} \delta (x - x')\Big|_{x'=x} = 0.##
Any thoughts? Is this rigorous?
##\frac \delta {\delta \psi(x)} \big[ \partial_x \psi(x)\big],##
where ##\partial_x## is a standard derivative with respect to ##x##.
One could argue that
## \partial_x \psi(x) = \int dx' [\partial_{x'} \psi(x')] \delta (x - x') = - \int dx' \psi(x') \partial_{x'} \delta (x - x'),##
assuming there is no boundary term in integration by parts.
In this case, the functional derivative would give
##\frac \delta {\delta \psi(x)}\Big[ - \int dx' \psi(x') \partial_{x'} \delta (x - x') \Big] = - \partial_{x'} \delta (x - x')\Big|_{x'=x} = 0.##
Any thoughts? Is this rigorous?