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Computate the commutator of $[p^n,x]=-i\hbar np^{n-1}$ with $p=-i\hbar \frac{\delta}{\delta x}$ the impulse operator. $\hbar$ stands for $\frac{h}{2\pi}$.

Answer: I do it with induction over $n$. For $n=1$ it is clear (from Lecture). Now $n\to n+1$: For a commutator is $[AB,C]=A[B,C]-[A,C]B$.By applying this on $[p^n,x]$ it follows: $$[p^n,x]=p[p^{n-1},x]-[p,x]p^{n-1}=-p(n-1)i\hbar p^{n-2}+i\hbar p^{n-1}$$ and that is $-i\hbar (n-2)p^{n-1}$.

So what is the mistake?

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  • $\begingroup$ Are you sure about the formula $[AB,C]=A[B,C]-[A,C]B$? $\endgroup$ – Braindead May 8 '16 at 17:22
  • $\begingroup$ @TheLedge Sorry for clicking the wrong button. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 11 '18 at 20:23
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I suspect your main problem is the commutator formula.

Let's take a look:

$$[AB,C] = (AB)C - C(AB) = ABC - CAB$$

Now, let's look at the right-hand side. The RHS has two terms: $A[B,C]$ and $[A,C]B$.

$$A[B,C] = A(BC - CB) = ABC - ACB$$

$$[A,C]B = (AC-CA)B = ACB - CAB$$

So in fact, the correct formula looks like:

\begin{align} A[B,C]+[A,C]B&= ABC - ACB + ACB - CAB\\ & = ABC - CAB \\ & = [AB,C] \end{align}

So, it looks like the formula you should be using is:

$$[AB,C] = A[B,C]+[A,C]B$$

I believe you can complete the problem from here.

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  • $\begingroup$ omg, yes. I see. Thank you. $\endgroup$ – serge May 8 '16 at 18:10

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