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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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    Related (if not duplicates of): https://physics.stackexchange.com/q/78222/25301, https://physics.stackexchange.com/q/98372/25301, https://physics.stackexchange.com/q/87038/25301, https://physics.stackexchange.com/q/143503/25301 – Kyle Kanos Sep 13 '17 at 12:51
  • Thanks, but none of those seem to apply. I think this is because $px^n$ is a function of two variables (operators)... – LC Nielsen Sep 13 '17 at 13:04

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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  • The answer to the question should supposedly show that in general, $\frac{i}{\hbar}[p,G(p,x)] \neq \frac{\partial}{\partial x} G(p,x)$. As opposed to if G is a function of x only. Also, the practice system I feed the answers into doesn't accept it... – LC Nielsen Sep 13 '17 at 13:02
  • Got it - all I needed to do was to expand to $- i \hbar n (n-1) x ^{n-2} + n x ^{n-1} p$...

    I feels silly now.

     – LC Nielsen Sep 13 '17 at 14:13
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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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  • Aha, so let's see...

    $\begin{align} -\hbar^2 \partial_x^2(x^n f(x))+\hbar^2(\partial_x x^n f'(x)), . \end{align}$

    Well, we have $f'(x) = \partial_x f(x)$ so the answer should be...

    $\hbar^2(-\partial_x^2 x^n + \partial_x x^n \partial_x) = \hbar^2(-n (n-1) x^{n-2} + nx^{n-1} \partial_x)$

    Is that right?

     – LC Nielsen Sep 13 '17 at 13:17
  • You can further simplify but you get the idea. – ZeroTheHero Sep 13 '17 at 13:20
  • It actually will be the simple derivative. $n p x^{n-1}$ I mean – OON Sep 13 '17 at 13:20
  • All I needed to do was expand it to $- i \hbar n (n-1) x ^{n-2} + n x ^{n-1} p$... I feel really silly now. – LC Nielsen Sep 13 '17 at 14:18
  • when you don't know it's more formidable than it looks. Best of luck. – ZeroTheHero Sep 13 '17 at 14:22