- #1
BlackHole213
- 34
- 0
If I have the operator, ##e^{a\partial_p}## acting on ##f(p)##, I know that $$e^{a\partial_p}f(p)=f(p+a)\,.$$
If I have ##e^{a\partial_p^2}f(p)##, this is just the Weierstrass transform of ##f(p)##. However, what happens if I have a general operator, ##e^{g(p)\partial_p}## or ##e^{g(p)\partial_p^2}##. How would I know what ##e^{g(p)\partial_p}## or ##e^{g(p)\partial_p^2}## does to ##f(p)##?
If I have ##e^{a\partial_p^2}f(p)##, this is just the Weierstrass transform of ##f(p)##. However, what happens if I have a general operator, ##e^{g(p)\partial_p}## or ##e^{g(p)\partial_p^2}##. How would I know what ##e^{g(p)\partial_p}## or ##e^{g(p)\partial_p^2}## does to ##f(p)##?