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Given a boson field described by $\psi(\vec{x})$, conserved momentum from the Lagrangian (which isn't relevant here) is $\vec{P} = \frac{\hbar}{2i} \int d^3 x \left( \psi^\dagger \nabla \psi - \nabla \psi^\dagger \psi \right)$. Question claims the commutator of $\vec{P}$ and $\psi(\vec{x})$ is $\frac{\hbar}{i} \nabla \psi (\vec{x})$ but my math doesn't lead me there. Expanding the commutator results in:

$$\left[ \vec{P}, \psi (\vec{x}) \right] = \frac{\hbar}{2i} \int d^3 x \left[ \psi^\dagger (\vec{x}') \nabla \psi (\vec{x}') - \nabla \psi^\dagger (\vec{x}') \psi (\vec{x}'), \psi(\vec{x} ) \right]$$ $$= \frac{\hbar}{2i} \int d^3 x' \left[ \psi^\dagger (\vec{x}'), \psi (\vec{x} ) \right] \nabla \psi (\vec{x}') + \frac{\hbar}{2i} \int d^3 x' \psi^\dagger (\vec{x}') \left[ \nabla \psi (\vec{x}') , \psi (\vec{x} ) \right] - \frac{\hbar}{2i} \int d^3 x' \left[\nabla \psi^\dagger (\vec{x} '), \psi (\vec{x}) \right] \psi (\vec{x}') - \frac{\hbar}{2i} \int d^3 x' \nabla \psi^\dagger (\vec{x} ') \left[\psi (\vec{x}'), \psi (\vec{x})\right] $$

The first term turns into a delta function that becomes $-\frac{\hbar}{2i} \nabla \psi (\vec{x})$ and the last term is zero, but I'm not sure how to evaluate the middle two terms.

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For the middle terms you use that the nabla differential acts on fields indexed by $\vec x^\prime$. It sees the field operator $\psi(\vec x)$ as a constant due to its independence of $\vec x^\prime$. Thus, we can pull the differential operator out of the commutator which yields for example $$\int d^3\vec x^\prime \psi(\vec x^\prime) \nabla_{\vec x^\prime} [\psi(\vec x^\prime), \psi(\vec x)]$$ for the second term. The subscribt $\vec x^\prime$ of $\nabla_{\vec x^\prime}$ is added to indicade on which variable it acts. Now, what seperates us from the solution you are searching for, is just a partial integration to shift the differential to the first field in the integral. This should give you the wished result.

I hope this could help you. Cheers!

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