Consider the usual commutation relations of two scalar fields
\left[\phi_{m}\left(t,\boldsymbol{x}\right),\pi_{n}\left(t,\boldsymbol{y}\right)\right]=\boldsymbol{i}\delta_{mn}\delta\left(\boldsymbol{x}-\boldsymbol{y}\right),
\left[\phi_{m}\left(t,\boldsymbol{x}\right),\phi_{n}\left(t,\boldsymbol{y}\right)\right]=\left[\pi_{m}\left(t,\boldsymbol{x}\right),\pi_{n}\left(t,\boldsymbol{y}\right)\right]=0.
What's the commutator of \left[\partial_{i}\phi_{m}\left(t,\boldsymbol{x}\right),\phi_{n}\left(t,\boldsymbol{y}\right)\right], where \partial_{i}\equiv\partial/\partial x^{i} is one of the three spatial derivatives?
What about \left[\partial_{i}\phi_{m}\left(t,\boldsymbol{x}\right),\pi_{n}\left(t,\boldsymbol{y}\right)\right] ?
Attempt 1:
\begin{array}{cl} \left[\partial_{i}\phi\left(t,\boldsymbol{x}\right),\phi\left(t,\boldsymbol{y}\right)\right] & =\partial_{i}\left[\phi\left(t,\boldsymbol{x}\right),\phi\left(t,\boldsymbol{y}\right)\right]+\left[\partial_{i},\phi\left(t,\boldsymbol{y}\right)\right]\phi\left(t,\boldsymbol{x}\right)\\ & =\left[\partial_{i},\phi\left(t,\boldsymbol{y}\right)\right]\phi\left(t,\boldsymbol{x}\right)\\ & =\left(\partial_{i}\phi\left(t,\boldsymbol{y}\right)\right)\phi\left(t,\boldsymbol{x}\right)-\phi\left(t,\boldsymbol{y}\right)\partial_{i}\phi\left(t,\boldsymbol{x}\right)\\ & =? \end{array}
Answer
Since we're not taking time derivatives, this is actually a pretty simple thing, but something that, for some reason, doesn't really pop out on a first viewing of a problem like this.
The confusion perhaps arises from the fact that you have two types of operators acting on different spaces. You have the derivative operator \partial_i acting on the space of functions from \mathbb{R}^n to some general algebra of fields. You also have the field operators themselves, acting on your Hilbert space \mathcal{H}. Since these two operators act on different spaces, then we have
\left[\frac{\partial}{\partial x^i}\phi(\textbf{x},t),\phi(\textbf{y},t)\right]=\frac{\partial}{\partial x^i}\left[\phi(\textbf{x},t),\phi(\textbf{y},t)\right]=0.
That is to say, you can pull out the derivative since only the first term in the commutator depends on \textbf{x}. Similarly, we have
\left[\frac{\partial}{\partial x^i}\phi(\textbf{x},t),\pi(\textbf{y},t)\right]=\frac{\partial}{\partial x^i}\left[\phi(\textbf{x},t),\pi(\textbf{y},t)\right]=i\frac{\partial}{\partial x^i}\delta(\textbf{x}-\textbf{y}).
I don't know what it is about this question that trips people up (including myself the first time I was faced with something like this), but it's a lot simpler than it's made out to be.
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