0

I'm struggling trying to prove $[\overrightarrow{x}, f(\overrightarrow{p})]= i\hbar\frac{\partial f(\overrightarrow{p})}{\partial \overrightarrow{p}}$.

I already proved that $[x_i,p_i^n]=i\hbar \frac{\partial p_i^n}{\partial p_i}$ and I need to express $f$ in a power series but I don't know how to do that and derive the result.

Can someone help me?

Qmechanic
  • 201,751
Abel Gutiérrez
  • 65

1 Answers1

1

I will not provide the full derivation, but will help you start your journey.

You can use the power series:

\begin{equation} f(p) = c^{(0)} + c^{(1)}p \;+\; ...c^{(n)}\frac{p^n}{n!} \end{equation}

Start looking at the commutator:

\begin{equation} \left[x, c^{(0)} + c^{(0)}p \;+\; ...\frac{c^{(n)}}{n!}p^n\right] \end{equation}

maybe look at it like:

\begin{equation} \sum_n \left[x, \frac{c^{(n)}}{n!}p^n\right] \end{equation}

user18764
  • 554
  • I already read that in other pages, but I get confused because my teacher claimed: $f(p)=\sum_n \left(\overrightarrow{p}\overrightarrow{\nabla}\right)^nf(\overrightarrow{p})/n!$ so when I expressed the nabla operator in terms of momentun but I didn't arrive to the result. – Abel Gutiérrez Apr 03 '21 at 13:39
  • Right so applying $\nabla^{n}$ to the Maclaurian Series should give you the first expression in my comments.

    In other words, the power series of expression 1 is the Maclaurian Series for some generic function of the momentum operator.

    If your teacher does not care whether or not you use the power-series approach, the link in the comments is more straight-forward.

     – user18764 Apr 05 '21 at 19:12