The question is basically in the title.

My naive thought was that when a commutation relation holds for all field operators $\Psi(\vec{x})$ (by "all" I mean "at all positions $\vec{x}$") on a fixed time-slice, then it should automatically also apply to the spatial derivatives of those field operators, $\partial_{\vec{x}} \Psi$.

My reasoning to that is that a derivative is the limit of the difference of two operators at two proximate positions. Since for the spatial derivatives, all operators in question lie in the same time-slice, they all commute / anticommute, and hence their difference should do so as well.

Is that true, or are there some issues with infinities that I didn't take into account here?

EDIT: I'm talking about commutators/ anticommutators like $$ \{ \Psi_a(\vec{x}), \partial_i \Psi_b(\vec{x}')\} $$ as well as $$ \{\partial_i \Psi_a(\vec{x}), \partial_j \Psi_b(\vec{x}')\} $$ I would assume both of them to still be $0$.


Yes, in quantum field theory a equal-time (super)commutator relation, say, $$[\hat{\phi}({\bf x},t),\hat{\pi}({\bf y},t)]=i\hbar \hat{\bf 1}~\delta^3({\bf x}\!-\!{\bf y}),$$ can be differentiated on both sides wrt. spatial derivatives.


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