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.
$\endgroup$ 12 Answers
$\begingroup$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
$$\begin{eqnarray}[x,p^{n+1}] &=& xp^{n+1}-p^{n+1}x\\ &=& (xp^n)p-p(p^nx) \\ &=& (xp^n)p-(p^nx)p+(p^nx)p-p(p^nx) \\ &=& (xp^n-p^nx)p+p^n(xp-px).\end{eqnarray}$$
$\endgroup$ 4 $\begingroup$$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ Use $\ds{\bracks{A,BC} = B\bracks{A,C} + \bracks{A,B}C}$.
\begin{align} \bracks{x,p^{n}}&=\bracks{x,pp^{n - 1}} = p\bracks{x,p^{n - 1}} + \bracks{x,p}p^{n - 1}=p\bracks{x,p^{n - 1}} + ip^{n - 1} \end{align}
\begin{align} \bracks{x,p^{n}}&=p\braces{p\bracks{x,p^{n - 2}} + \ic p^{n - 2}} + ip^{n - 1} =p^{2}\bracks{x,p^{n - 2}} + 2\ic p^{n - 1} \\[3mm]&=\cdots=p^{k}\bracks{x,p^{n - k}} + k\ic p^{n - 1} \end{align}
$\endgroup$With $\ds{k = n - 1\,,\quad \bracks{x,p^{n}} = p^{n - 1}\bracks{x,p} + \pars{n - 1}\ic p^{n - 1}}$: $$ \color{#00f}{\large\bracks{x,p^{n}} = \ic np^{n - 1}} $$
