Discrete Fourier transforms of Hubbard hoppings: a contradiction?

Question

It's well known that condensed matter Hamiltonians of the form $$\mathcal{H} = t\sum_{\langle i j \rangle} a_i^\dagger a_j + a_j^\dagger a_i$$ where $a_i$ are bosonic creation/annihilation operators, are readily solved by the discrete Fourier transformed operators $$a_k = \frac{1}{\sqrt{N}}\sum_i e^{ikj}a_j$$

It can be shown that these operators obey canonical bosonic commutation relations, i.e.

$$[a_k, a_q^\dagger] = \delta_{k,q}, [a_k, a_q] = [a^\dagger_k,a^\dagger_q]=0$$

I have a frustrated ferromagnetic system that I am solving by a first-order Holstein-Primakoff transformation on the spin operators, which naturally gives rise to terms in the Hamiltonian of the form $a_i a^\dagger_{i+1}$. As these correspond to different real-space sites, I would assume that they commute. This would seem to imply

$$ \sum_j a_j a^\dagger_{j+1} = \frac{1}{N}\sum_{j,k,q}e^{ikj}a_ke^{-ik'j-ik'}a_{k'}^\dagger = \sum_k a_k a^\dagger_ke^{-ik}$$ $$ = \sum_j a^\dagger_{j+1}a_j = \frac{1}{N}\sum_{j,k,q}e^{-ik'j-ik'}a_{k'}^\dagger e^{ikj}a_k = \sum_k a^\dagger_k a_ke^{-ik}$$

The right hand sides of these two lines differ by a factor of $N$, which is a contradiction. Where is the mistake in this argument?

EDIT: part b) of the question - how does this influence the interpretation of the summand as a dispersion relation for the ground state? If arbitrary factors of e^ik can be added, does this mean that cos(k) can be added to and subtracted from the ground state energy arbitrarily?

Answer 1

I hope that I do not misunderstand something about your question, but $e^{-ik}$ goes once around the unit circle, so $$0=\sum_k e^{-ik} = \sum_k (a_k a_k^\dagger-a_k^\dagger a_k) e^{-ik}=\sum_k a_k a_k^\dagger e^{-ik}-\sum_ka_k^\dagger a_k e^{-ik}$$ and thus, $$\sum_k a_k a_k^\dagger e^{-ik}=\sum_ka_k^\dagger a_k e^{-ik}$$