Is it possible to calculate in a general way the commutator of an operator $O$ which depends on some variable $x$ and the derivative of this $O$ with respect to $x$?
$${O}={O}(x)\\
\left[\partial_x{O}(x), O(x^{\prime})\right]=?
$$
To be honest I don't quite understand, what the derivative of an operator is. When one plugs in the standard definition of the derivative of a function, it looks as though the operator commutes with its derivative, but I don't quite know how to feel about that "proof".
In practice I need this to calculate the commutator of the field operator of a free scalar field and any of its four derivatives:
$$\left[\partial_\mu\phi(\mathbf{x},t),\phi(\mathbf{x}^{\prime},t)\right]$$
(I'm looking at scalar field theory described by a Lagrangian density $\mathcal{L}=\partial_\mu \phi^*\partial^\mu\phi-m^2\phi\phi^*$)
One can of course simply compute this by plugging in the field operator, but I was wondering about the general situation.
And what about the more general case of two operators, whose commutator is known?
$$\left[O(x),U(x^{\prime})\right]=A(x,x^{\prime})\\
\left[\partial_x{O}(x),U(x^{\prime})\right]=?
$$
Best Answer
The derivative of an operator: Let $X(t),\;\mathbb{R}\rightarrow \mathcal{X}$, where $\mathcal{X}$ is some normed linear space, say a Banach or Hilbert space. Then we can define the derivative in the usual way: $$\partial _{t}X(t)=\lim_{\delta \rightarrow 0}\frac{X(t+\delta )-X(t)}{\delta}.$$ However, on $\mathcal{X}$ different topologies exist, strong, weak, uniform, etc. Hille and Phillips, "Functional Analysis and Semi-Groups" contains a quite readable discussion of these matters.
In general the commutator does not vanish. Consider
$$X\left(\xi\right)=U\cos\xi+V\sin\xi\Rightarrow\dfrac{dX}{d\xi}=-U\sin\xi+V\cos\xi$$ then one gets, after some algebras
$$\left[X\left(\xi\right),\dfrac{dX}{d\xi}\right]=\left[U,V\right]$$
Then it is clear that one recovers the usual result for commuting $U$ and $V$ operators.
Edit: the result is correct ! Thanks to the comments.