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Let $\Phi_1,\ \Phi_2$ be two operators on a Hilbert space $\mathcal{H}$ such that they commute $$[\Phi_1(x),\ \Phi_2(x)]=0\qquad \forall x\in M$$ at all points in the Minkowski space $M$ and let them be differentiable to any order so that $\partial_\mu\Phi_i(x)=\frac{\partial \Phi_i(x)}{\partial x^\mu}$ and $\partial_\mu\partial_\nu\Phi_i(x)$ and so on exist for $i=1,\ 2$. Then, does the following commutator hold? $$[\partial_\mu\Phi_1(x),\ \partial^\mu\Phi_2(x)]\stackrel{?}{=}0.$$ I've tried to expand derivatives such as $\partial_\mu[\Phi_1,\ \Phi_2]=0$ and rearrange terms but don't seem to obtain the result above.

MrPillow
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1 Answers1

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Let $\Phi_1,\ \Phi_2$ be two operators on a Hilbert space $\mathcal{H}$ such that they commute $$[\Phi_1(x),\ \Phi_2(x)]=0\qquad \forall x\in M$$ at all points in the Minkowski space $M$ and let them be differentiable to any order so that $\partial_\mu\Phi_i(x)=\frac{\partial \Phi_i(x)}{\partial x^\mu}$ and $\partial_\mu\partial_\nu\Phi_i(x)$ and so on exist for $i=1,\ 2$. Then, does the following commutator hold? $$[\partial_\mu\Phi_1(x),\ \partial^\mu\Phi_2(x)]\stackrel{?}{=}0.$$

In this case, it looks like the best you can do is: $$ 0 = [\partial^\nu\Phi_1,\partial_\mu\Phi_2] +[\partial_\mu\Phi_1,\partial^\nu\Phi_2] +[\partial_\mu\partial^\nu\Phi_1,\Phi_2] +[\Phi_1,\partial^\nu\partial_\mu\Phi_2]\;. $$

But, in the special case of a free field where, say, $\partial^2\Phi_i = m_i^2\Phi_i$, you can set $\mu=\nu$ and sum to find $$ 0=2[\partial_\mu\Phi_1\partial^\mu\Phi_2]+m_1^2[\Phi_1,\Phi_2]+m_2^2[\Phi_1,\Phi_2]\tag{free case} $$ or, in other words, in this special case $$ [\partial_\mu\Phi_1,\partial^\mu\Phi_2]=0\tag{free case} $$

hft
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