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I am self studying Quantum Field Theory and I am using the book An Introduction to Quantum Field Theory by Peskin and Schroeder. Currently I am working on problem 2.2 (a). In the textbook problem, the authors provide us with a Hamiltonian, whose Hamiltonian density contains a $\nabla \phi^*(y)\cdot \nabla \phi(y)$. One part of the question asks us to compute the Heisenberg equation of motion, which involves computing $[\phi(x), H]$, and this involves computing $[\phi(x), \nabla \phi^*(y)\cdot \nabla \phi(y)]$. I know that for operators $A$, $B$, $C$ that $$[A,BC] = B[A,C] + [A,B]C.$$ This, however, involves the product of operators, but does a similar formulae hold for dot products of vector operators? I.e. is this true: $$[\phi(x), \nabla \phi^*(y)\cdot \nabla \phi(y)] = \nabla \phi^*(y)\cdot[\phi(x),\nabla \phi(y)] + [\phi(x),\nabla \phi^*(y)]\cdot \nabla \phi(y).$$

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    $\begingroup$ Why don't you expand it in modes? $\endgroup$ Jun 30 at 12:08
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You have the usual scalar product $\nabla\Phi^\ast\nabla\Phi=\sum_i\partial_i\Phi^\ast\partial_i\Phi$ and since the commutator is bilinear, your equation at the end of your post is true.

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