Second quantization: periodicity of annihilation and creation operators in momentum space, originally on a lattice

Question

I have a Hamiltonian $H$ on a periodic lattice, which is expressed as, say:

$$H = \sum_{n} (A_n a^\dagger_n a_n + B_n a^\dagger_{n+1} a_n + h.c.)$$

where $A_n$ and $B_n$ are periodic in space (over the lattice) with a period of $\beta$. Now, in order to get the momentum space representation of the Hamiltonian, I express the annihilation (and creation) operators as:

$$a_n = \sum_{k=-\infty}^{\infty}e^{-ikn}a_k$$

Instead of having the sum over $k$ running from $-\infty$ to $+\infty$, I would like to have the sum running only over the Brillouin zone. I am not sure how to go about doing that. Any help would be appreciated!

Answer 1

As the size of your system is infinite, then the range of $k$ becomes continuous. The correct form is $$ a_n = \int_{-\pi}^{\pi}\! \frac{dk}{\sqrt{2\pi}}e^{-i k n} a(k) $$ where we implicitly took the lattice spacing to be $1$, which makes $k$ dimensionless. Note that the commutation relations that we want to preserve are $[a^{\dagger}_n, a_m]=\delta_{m,n}$, and this integral works. Note, however, that the commutation relations of $[a^{\dagger}(k), a(q)] \propto \delta(k-q)$.

In general, I would recommend to introduce dimensions and starting with a finite lattice site, taking the limit $N\to\infty$ while keeping lattice spacing $\alpha$ fixed, to get the feeling of how the sums become integrals and how, in fact, $a(k)$ must get dimensions of square root of inverse length.