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I was trying to show that: $$[\hat{x}^n,\hat{p}]= i \hbar n \hat{x}^{n-1}.$$

I know from the definition of a pair of a commutator in QM they act on a wave function like this:

$$[\hat A, \hat B] = \hat A \hat B - \hat B \hat A$$

So.. in order to prove this, I apply it some arbitrary wave function $\Psi$ and pull the lever.. (albeit apparently improperly)

$$(\hat p \hat x^n - \hat x^n \hat p )\Psi$$

$$-i \hbar \frac{d}{dx}(x^n\Psi) + ix^n\hbar \frac{d}{dx}(\Psi)$$

Now, the reason why I can't put the $x^n$ in what the derivative operator is applying on the term on the right.. is that because we're assuming that the commutator is not commutative? I thought this applied to the operators themselves, not the resultant function due to those operators. Basically, I thought that while

$$[\hat A, \hat B] = \hat A \hat B - \hat B \hat A$$

is not commutative, i.e. that $\hat A \hat B \ne \hat B \hat A$

it isn't apparent to me, then, that

$$-i \hbar \frac{d}{dx}(x^n\Psi) + ix^n\hbar \frac{d}{dx}(\Psi)$$

is also not commutative.

The first are operators, the second are elements of a field where associativity is a thing. Let me know if anything about my question is confusing.

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  • $\begingroup$ You simply need to apply the Leibniz rule for the first term. See, after substituting $\hat{x}=x$ and $\hat{p}=-i\hbar \frac{d}{dx}$ (which means you are using position representation), the only operator remains is the derivative operator. It is now ordinary differentiation on functions. $\endgroup$ – Oktay Doğangün Apr 24 '18 at 20:15
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/87038/2451 and links therein. $\endgroup$ – Qmechanic May 21 '18 at 18:04
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You need to apply the derivative to both terms inside the braces. The differential operator does not commute with $x$.

$$x \frac{d}{dx} \neq \frac{d}{dx}x = \left(\frac{d}{dx}x\right)+x \frac{d}{dx} $$

This means that

$$-i \hbar \frac{d}{dx}(x^n\Psi) = -i\hbar nx^{n-1}\Psi -i\hbar x^n \left(\frac{d}{dx}\Psi\right)$$

If this doesn't solve your problem, please clarify your question.

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Here is another hint to help you out. Write the commutator as \begin{eqnarray*} [p, x^n] &=& p x^{n-1} x - x^{n - 1} x p \\ &=& p x^{n-1} x - x^{n-1}([x, p] + p x) \\ &=& p x^{n-1} x - x^{n-1} p x - x^{n-1}[x, p] \\ &=& [p, x^{n - 1}] x - x^{n - 1}[x, p] \\ &=& [p, x^{n - 1}] x - i \hbar x^{n - 1}. \end{eqnarray*} You have just represented the commutator $[p, x^n]$ with the commutator one power lower. Try to see if you can construct a recursive proof using a similar technique.

FYI: I think you might have written the commutator in your professor's problem backwards. $[p, x^n] = - [x^n, p] = - i \hbar n x^{n - 1}$.

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You can prove the formula by induction, using the propriety of commutators (it is easy to be proven) that $$\left[A B,C\right]=A\left[ B,C\right]+\left[A ,C\right]B $$. Indeed

$$\left[x,p\right] = i\hbar$$ $$\left[x^2,p\right] =x\left[x,p\right]+\left[x,p\right]x = 2 i\hbar x$$ $$\left[x^3,p\right] =x^2\left[x,p\right]+\left[x^2,p\right]x = 3 i \hbar x^2$$

and finally $$\left[x^n,p\right] =x^n\left[x,p\right]+\left[x^{n-1},p\right]x = i \hbar n x^{n-1}$$

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