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This seems like a really straightforward problem that nevertheless stumps me.

We have $\frac{i}{\hbar}[p,px^n] = \frac{i}{\hbar}(p[p,x^n] + [p,p]x^n)$

Now I have previously found that $[p,x^n] = - i \hbar n x^{n-1}$ and we know that $[p,p] = 0$ by axiom.

So we should have $\frac{i}{\hbar}(p[p,x^n] + [p,p]x^n) = \frac{i}{\hbar}p(-i \hbar n x^{n-1}) = pnx^{n-1}.$ But I know this cannot be the right answer because it should not be the partial derivative of $px^n$ with respect to x. I'm obviously missing something, but what?

Qmechanic
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LC Nielsen
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2 Answers2

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It looks right to me. A function $f$ satisfies $\frac{i}{\hbar}[p,\,f]=[\partial_x,\,f]=\partial_x f$ since $[\partial_x,\,f]g=\partial_x (fg)-f\partial_x g = (\partial_x f)g$.

J.G.
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In general, for any function of $f(x)$, the action of $\hat x$ and $\hat p$ is given as $$ \hat xf(x)=xf(x)\, ,\qquad \hat pf(x)=-i\hbar \partial_xf(x)=-i\hbar f'(x) $$ so, for $\hat p\hat x^n$, you simply apply in proper sequence.

Thus \begin{align} [p,px^n]f(x)&=-i\hbar \partial_x (-i\hbar \partial_x) x^n f(x) - (-i\hbar \partial_x)x^n(-i\hbar \partial_x)f(x)\, ,\\ &=-\hbar^2 \partial_x^2(x^n f(x))+\hbar^2(\partial_x x^n f'(x))\, . \end{align} Since you can use the replacements $$ f'(x)\to \frac{i}{\hbar}\hat pf(x)\qquad f''(x)\to -\frac{1}{\hbar^2}\hat p^2f(x) $$ and $f(x)$ is arbitary, you should be able to eliminate $f(x)$ at the very end to get a "clean" answer in terms of operators only (which will not be a simple derivative.)

ZeroTheHero
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