Commutator proof

by cahill8
Tags: commutator, proof
 P: 31 1. The problem statement, all variables and given/known data Show $$\left[x,f(p)$$$$\right)]$$ = $$i\hbar\frac{d}{dp}(f(p))\right.$$ 2. Relevant equations I can use $$\left[x,p^{n}$$$$\right)]$$ = $$i\hbar\\n\right.$$$$p^{n}\right.$$ f(p) = $$\Sigma$$ $$f_{n}$$$$p^{n}$$ (power series expansion) 3. The attempt at a solution I started by expanding f(p) to the power series which makes $$\left[x,\Sigma\\f_{n}\\p^{n}$$$$\right)]$$ and I know I must use the commutator identity [A, BC] = [A,B]C + B[A,C] but the power series cannot be split up into two products(BC) ? So I'm not sure how to go on
 P: 31 In a text book it says it can be shown using that equation Trying a different method: [x, f(p)] = [x,$$\sum_{n}\\f_{n}p^{n}$$] = [x,fnpn + $$\sum_{n-1}\\f_{n}p^{n}$$] using [A, B+C] = [A,B] + [A,C] = [x, fnpn] + [x, $$\sum_{n-1}\\f_{n}p^{n}$$] using [A, BC] = C[A,B] + B[A,C] = fn[x, pn] + pn[x, fn] + [x, $$\sum_{n-1}\\f_{n}p^{n}$$] using [x, pn] = i$$\hbar$$npn-1 = fni$$\hbar$$npn-1 + pn[x, fn] + [x, $$\sum_{n-1}\\f_{n}p^{n}$$] [x, fn] = 0 as fn is a const. = fni$$\hbar$$npn-1 + [x, $$\sum_{n-1}\\f_{n}p^{n}$$] am I on the right track?