Commutators of differential and field operators

In $p+ip$ superconductivity, a term in the Hamiltonian (in polar coordinates) is $$H=\int \mathrm{d}^2\vec{r} \left[ \left( \partial_r -\frac{i}{r} \partial_{\theta} \right) \psi^\dagger(\vec{r}) \right] \psi^\dagger(\vec{r})$$ where $\psi^\dagger$ is the field operator that creates a spinless fermion. To follow convention and arrange the term as a matrix element, it should be possible to put it on the form: $$H=\int \mathrm{d}^2\vec{r} \; \psi^\dagger(\vec{r}) \left( -\partial_r +\frac{i}{r} \partial_{\theta} \right) \psi^\dagger(\vec{r})$$ Which is to say the two factors anti-commute, or at least the commutator is something that becomes a constant when integrated? My question is how to show this is possible? I have tried expanding $\psi^\dagger = \sum \psi_k a_k^\dagger$, but I am not sure if the basis is supposed to depend on the boundary conditions or not. I am also not quite sure how to calculate the commutator between the differential operators and the creation operator, $a_k^\dagger$, either.

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i) The transition from the first to the second line could be seen as an integration by part. ii) The anti-commutation of the fermionic operator is a property you require from the outset, not something you can prove with commuting operators : in short, the transition between the first to the second line is a definition. iii) There is no rule for the commutation of the creation/annihilation operators and the derivative : in short, see i). In any case, the derivative is no longer a derivative in the momentum-$k$-space onto which $a_k$ projects. –  FraSchelle Mar 25 '14 at 10:49