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I Generalizing the translation operator

  1. Jan 25, 2017 #1
    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)##?
     
  2. jcsd
  3. Jan 29, 2017 #2

    DCN

    User Avatar

    Write it out in a Taylor series.
     
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