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In first quantization, the operator $J$ assumes the form $J=\sum_{i}j(x_i)$. In Fock space, it is instead written as $J=\int dx \psi^\dagger(x)j(x)\psi(x)$, where $\psi^\dagger, \psi$ are the field operators.

Is $j(x)$ an operator in Fock space or simply a function of x?

If $j(x)=\nabla^2_x$, then does it commute with $\psi^\dagger$ and $\psi$?

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    $\begingroup$ Please see our FAQ on writing good titles. $\endgroup$ – DanielSank Sep 10 '16 at 18:08
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    $\begingroup$ $j(x)$ is the first quantized position representation form, which makes it x-dependent. If it is a differential form, then it acts on $\psi(x)$, $\psi^\dagger(x)$, depending on definition. So what you want to consider are commutators $[\psi(x), \partial_x\psi(x')]$, not $\psi(x), \partial_x]$. Regarding the latter, keep in mind that the state vectors $|\phi\rangle$ on which the operators act are elements of the Fock space, not wave functions, which means something like $\partial_x|\phi\rangle$ does not exist. $\endgroup$ – udrv Sep 11 '16 at 3:29
  • $\begingroup$ For how the differential of a field operator is defined see for instance physics.stackexchange.com/questions/103464/…. $\endgroup$ – udrv Sep 11 '16 at 3:29

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