# Expanding the interaction term in Hamiltonian for weakly interacting Bose gas

Let the second quantization Hamiltonian

and the Bogoliubov prescription

Could you explain me how to obtain the following expression

I know that the second term in this expression is equivalent to putting $$k=k'=q=0$$, but I don't know the following step.

These expressions can be found in ch 3.1 here

To obtain this form of the Hamiltonian you just need to rewrite the sum $$\sum_{k,k',q}V(q) a^\dagger_{k+q}a^\dagger_{k'-q}a_{k'}a_k$$ in the form $$V(0)a^\dagger_0a^\dagger_0a_0a_0 + 2\sum_{k\neq 0}(V(0)+V(k))a^\dagger_k a_k a^\dagger_0 a_0 + \sum_{k\neq 0} V(k)(a^\dagger_k a^\dagger_{-k} a_0 a_0 + a^\dagger_0 a^\dagger_0 a_{-k} a_k)+$$ $$+ 2\sum_{k,q\neq0}V(q)(a^\dagger_0 a^\dagger_{k+q}a_ka_q + a^\dagger_{k+q}a^\dagger_{-q}a_ka_0) + \sum_{k,k',q\neq0}V(q)a^\dagger_{k+q}a^\dagger_{k'-q}a_{k'}a_k$$ and then to change operators $$a^\dagger_0$$ and $$a_0$$ by c-number $$\sqrt{N_0}$$.

By the way, I think there is a misprint in the article in a term containing only one $$a_0$$ or $$a^\dagger_0$$ operator.

Upd. I just made use of simple formulas like: $$\sum_{k} f_k = f_0 + \sum_{k\neq0} f_k, \qquad \sum_{k,k'} F_{k,k'} = F_{0,0} + \sum_{k\neq0} F_{k,0} + \sum_{k'\neq0}F_{0,k'} +\sum_{k,k'\neq0}F_{k,k'}$$ and so on. You need to select terms containing different combinations of $$a^\dagger_0$$ and $$a_0$$ operators.

• That is just my doubt. I don't know how to rewrite that sum. For instance, I know that, if $k=k'=q=0$, we obtain $$V(0)a^\dagger_0a^\dagger_0a_0a_0.$$ But, what about this term $$\sum_{k\neq 0}V(k)a^\dagger_k a_k a^\dagger_0 a_0 ?$$ $q=k$? – Dinesh Shankar Mar 28 at 20:14
• @Dinesh Shankar, I will update my answer. – Gec Mar 28 at 20:19