Discrete to continuous quantum operator

Question

Let's say that we have a discrete lattice with $N$ sites. Let's label the site by the index $i$.

Let's say that we have the operators $a_i$ and $a_i^\dagger$ which correspond to the creation and annihilation of magnons. We have

$$\hat{N}=\sum_i a_i^\dagger a_i,$$ where $\hat{N}$ is the number operator.

Let's say we want to use a continuum limit for the Hamiltonian (rather than a discrete sum. The Hamiltonian consists of terms that are quadratic in these operators) by using continuous creation/annihilation operators $\phi^\dagger(x)$ and $\phi(x)$ respectively, which also satisfy the commutation relations for bosons. Then, we need to make the following substitutions: $$a_i^\dagger \to \sqrt{a} \phi^\dagger(x),$$ $$a_i \to \sqrt{a} \phi(x),$$ where the $\sqrt{a}$ is a result of the normalization by conserving the total particle number:

$$\hat{N}=\sum_i a_i^\dagger a_i = \int dx \phi^\dagger(x)\phi(x).$$

I still don't understand this normalization factor of $\sqrt{a}$. How does the conservation of total particle number when moving from sum to integral exactly require this normalization factor of $\sqrt{a}$? It still doesn't make quite sense to me, even with the above explanation.

$a$ is the lattice constant/spacing here. Furthermore, we have:

$\phi(x+a) \approx \phi(x) + a\frac{d\phi}{dx}$

Answer 1

If $x= na$ then write $\sqrt a\phi(na)= a_n$ etc, $$ \sum_n a^\dagger_n a_n \sim \int \phi^\dagger (na) \phi(na) adn= \int \phi^\dagger( x) \phi(x) dx, $$ the "$\sim$" being the usual approximation of a sum of slowly varying things by an integral.