Bose-Einstein condensation: Bogoliubov Approximation

Question

I'm trying to understand the Bogoliubov approximation from "Statistical Mechanics" by Pathria and Beale. First of all they say

Since $a_0^{\dagger}a_0=n_0=O(N)$ and $(a_0a_0^{\dagger}-a_0^{\dagger}a_0)=1<<N$, it follows that $a_0a_0^{\dagger}=(n_0+1)\simeq a_0^{\dagger}a_0$

and this part is clear. It's not clear the following logic step

The operators $a_0$ and $a_0^{\dagger}$ may, therefore, be treated as c-numbers, each equal to $n_0^{1/2}\simeq N^{1/2}$

Can someone explain me why we can treat these operators as c-number?

Answer 1

The first statement reformulated gives you $[a_0, a^\dagger_0] \simeq 0$ (there is an assumption that operators involving $a^\dagger_0a_0$ are replaced by their expectation values). Then, the approximation is that since the operators commute, they can be replaced by classical objects.

So $a_0$ is a number equal to $n^{1/2}_0$ and its conjugate is $a^*_0 = \left(n^{1/2}_0\right)^*$.

EDIT : Here is a short justification of the replacement of operators by numbers. You can replace an operator by a number when states are always eigenvectors of the operator in question. In this case, we suppose that the number of particles is well defined, so we can replace $N$ by a number. Then, within the approximation that $a$ and $a^\dagger$ commute, they also commute with $N$. Since states are eigenvectors of $N$, they’re also eigenvectors of $a$ and $a^\dagger$, so we replace them by numbers.

Answer 2

Perhaps, it will serve you a fully analized example to get an understanding of this "approximation":

L. Banyai, About the c-number approximation of the Macroscopical Boson Degrees of Freedom within a Solvable Model. phys. stat.sol. (b), 234, 14 (2002)