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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.

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    Are we supposed to guess what all of these objects are? – 2017-01-26
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    Ah sorry, I'll add what I know! – 2017-01-26
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    An operator on what? An analytical function from where to where? Are these Lie brackets? – 2017-01-26
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    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. – 2017-01-26
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    Think of $V(x)$ not as of a function but rather as of an operator of multiplication by $V(x)$, i.e. $V(x)[f(x)] \equiv V(x) \cdot f(x)$. – 2017-01-26
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    @mavzolej I can expand and apply to some test function $f(x)$ to get $pVf(x) - Vpf(x)$ and I think that's treating it like an operator of multiplication, but where does differentiation come into it? I can't get any further. – 2017-01-26
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    Since your potential is a function of $x$, you are using the usual representation of the position and momentum operators, in which $\hat{x}$ is just the multiplication by $x$, while the momentum is $\hat{p} =\dfrac{\hbar}{i} \dfrac{\partial} {\partial x}$. – 2017-01-26
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    @mavzolej I have actually tried substituting in the momentum operator. I get a missing $T$ if I follow it through. Shall I add my working to the original post? – 2017-01-26
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    Missing $T$ if I apply my result to the actual question I'm trying to answer, I mean. – 2017-01-26
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    @mavzolej Added some workings. Treating V like an operator of multiplication. – 2017-01-26

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\begin{gather} [\hat{p},\hat{x}^k] = [\hat{p},\hat{x}^{k-1}\cdot \hat{x}] = [\hat{p},\hat{x}^{k-1}] \hat{x} + \hat{x}^{k-1}[\hat{p}, \hat{x}] \\ = [\hat{p},\hat{x}^{k-2}\cdot\hat{x}] \hat{x} + \hat{x}^{k-1} (i \hbar \hat{T}) = \ldots = k \hat{x}^{k-1} (i \hbar \hat{T}) \quad; \end{gather} \begin{gather} [\hat{p},V(\hat{x})] =[\hat{p},\sum_{k=0}^\infty \dfrac{1}{k!} V^{(k)}(0)\hat{x}^k] = \sum_{k=0}^\infty \dfrac{1}{k!} V^{(k)}(0)[\hat{p},\hat{x}^k]\\ = \sum_{k=0}^\infty \dfrac{1}{k!} V^{(k)}(0)k \hat{x}^{k-1} (i \hbar \hat{T}) = V'(\hat{x}) (i \hbar \hat{T}) \quad. \end{gather}

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, $(i \hbar \hat{T}) V'(\hat{x})$.

Anyways, the truth is out there.

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    No that's brilliant! T isn't an operator it's a constant, it can go wherever! Amazing, thank you so much! – 2017-01-27
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    For the constant - true. But still, I'm a little puzzled about the general case. – 2017-01-27
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    Actually my mistake, it isn't a constant. However x and T commute so the result holds, although presumably only for the case when xT-Tx = 0. – 2017-01-28