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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.
