How to prove $[x,p]=i$ $\implies$ $[x,p^n]=inp^{n-1}$?

Question

How to prove $[x,p]=i$ $\implies$ $[x,p^n]=inp^{n-1}$?

I can do this using $p=i\frac{d}{dx}$, but my book hasn't introduced this yet so is there another proof without using this ? These are just linear operators acting on a hilbert space.

Answer 1

Here's a hint: based upon the dependence upon $n$ in your answer, a proof by induction is probably the way to go. Your induction hypothesis should be: $[x,p^n] = inp^{n-1}$ and you want to use this property to prove that $[x,p^{n+1}] = i(n+1)p^n$. To see how these are related, note that

$$\begin{eqnarray}[x,p^{n+1}] &=& xp^{n+1}-p^{n+1}x\\ &=& (xp^n)p-p(p^nx) \\ &=& (xp^n)p-(p^nx)p+(p^nx)p-p(p^nx) \\ &=& (xp^n-p^nx)p+p^n(xp-px).\end{eqnarray}$$