3

I have following exercise: If $[C,D]$ is a c-number and $f(x)$ is a well-behaved function (i.e. all derivatives exist and are finite), show that: $$[C, f(D)]=[C,D]f'(D)$$ where $f'(D) = \frac{d}{dx}f(x)$

I used expansion: $$f(D) = \sum_{k} a_k D^k $$ that way: $$\sum_k a_k [C,D^k]=\sum_k a_k k [C,D]D^{k-1}$$ I don't know how to show that $$[C,D^k]=k[C,D]D^{k-1} $$

Do I need to expand function from operator in another way?

Brandon Enright
  • 11,937
misimik
  • 29
  • 1
    Related: http://physics.stackexchange.com/q/78222/2451 and links therein. – Qmechanic Dec 13 '13 at 21:25
  • 1
    Yeah. That's enough. Thank you. Shame, that I couldn't find before I posted the question, but I've reach my goal, so that's nice. – misimik Dec 13 '13 at 22:28
  • @misimik +1 even though simple and you did work it out, this is a good example of a derivation at work so it's a pretty example to keep in mind when talking about this stuff (shows how a kind of chain rule comes from Leibnitz rule). The linked example also (which you need to put ladder operators in their normal ordering). – Selene Routley Dec 14 '13 at 02:11

0 Answers0