# Commutation relation between function and operator

Show that $[p, V(x)] = V'(x)[p, x]$ , where $p$ is an operator, and $V(x)$ is an analytic function. I don't quite understand what the brackets [] do to an operator, but for matrices we use them to mean [A,B] = AB - BA.

The actual question I'm trying to answer is 'show that $[p, V(x)] = i\hbar V'(x)T$ ', which is satisfied if this: $[p, V(x)] = V'(x)[p, x]$ is true, because I'm also given that $[p, x] = i \hbar T$ and $[S,T]=0$. There's probably an alternative way of proving this relation, but I unfortunately haven't manage to prove either. Thanks for any help or hints!

Attempt to answer the actual question I've been asked: $$[p_x, V(a)] \psi = (p_xV(a) - V(a)p_x)\psi$$ $$=-i\hbar \frac{\partial({V(a) \psi})}{\partial{x}}+i\hbar V(a) \frac{\partial{\psi}}{\partial{x}}$$ $$= -i\hbar \frac{\partial{V}}{\partial{x}}$$ Not 100% sure what to do from there. Very close, but I have to get T from somewhere.

• Are we supposed to guess what all of these objects are? – Tobias Kildetoft Jan 26 '17 at 19:43
• Ah sorry, I'll add what I know! – user13948 Jan 26 '17 at 19:44
• An operator on what? An analytical function from where to where? Are these Lie brackets? – Tobias Kildetoft Jan 26 '17 at 19:46
• I don't know what the operator operates on, and it's analytical in the sense that the Taylor expansion exists and converges I think, although that also isn't specified. The brackets are commutators but I only know how they apply to matrices, so [A,B] = AB - BA. – user13948 Jan 26 '17 at 19:48
• Think of $V(x)$ not as of a function but rather as of an operator of multiplication by $V(x)$, i.e. $V(x)[f(x)] \equiv V(x) \cdot f(x)$. – mavzolej Jan 26 '17 at 20:11

Now, there's something that bothers me in this solution. Namely, if I do \begin{gather} [\hat{p},\hat{x}^k] = [\hat{p},\hat{x} \cdot \hat{x}^{k-1}] = [\hat{p},\hat{x}] \hat{x}^{k-1} + \hat{x}[\hat{p}, \hat{x}^{k-1}] = \ldots = k \hat{x}^{k-1} (i \hbar \hat{T}) \quad, \end{gather} it leads to the wrong result, $(i \hbar \hat{T}) V'(\hat{x})$.