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?

  • $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.