# Bose-Einstein condensation: Bogoliubov Approximation

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 $$(a_0a_0^{\dagger}-a_0^{\dagger}a_0)=1<, it follows that $$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?

• Well, you're saying that (within the approximation) they commute, right? Aug 28 '19 at 10:27

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.