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Let the second quantization Hamiltonian

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and the Bogoliubov prescription

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Could you explain me how to obtain the following expression

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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

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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.

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  • $\begingroup$ 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 $$\sum_{k\neq 0}V(k)a^\dagger_k a_k a^\dagger_0 a_0 ?$$ $q=k$? $\endgroup$ – Dinesh Shankar Mar 28 at 20:14
  • $\begingroup$ @Dinesh Shankar, I will update my answer. $\endgroup$ – Gec Mar 28 at 20:19

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