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I'm reading The nuclear many-body problem. In the Appendix (sec. C), they cover how to write a given single-particle operator $$ F=\sum_{i=1}^A f_i $$ where $i$ runs over all the ($A$) particles in the system, in second quantization form: $$ F =\sum_{kk'} f_{kk'}c^\dagger_k c_{k'}. $$ Where $f_{kk'} = \langle k'|f|k\rangle$, $c^\dagger_k, c_k$ are the one-particle creation and destruction operators and $k$, $k'$, run over all possible single-particle states (these states form a complete basis). Up to here I find no problem. However, later in the Appendix (sec. E), when introducing operators in quasiparticle space, they write a general Hermitian one-particle operator as:

$$ F =\sum_{kk'} f_{kk'}c^\dagger_k c_{k'} + \frac{1}{2}(g_{kk'}c^\dagger_k c^\dagger_{k'} + \text{h.c.}) $$

To be clear, $c^\dagger_k, c_k$ are still in the particle basis, quasiparticles have not entered the game yet. I have found nowhere in the book why this second term is added and no definition of the coefficients $g_{kk'}$ is given (or at least I can't find it). To add to my confusion, the general expression they use for the Hamiltonian, a two particle operator, remains unchanged:

$$ H=\sum_{l_1 l_2} \epsilon_{l_1 l_2} c_{l_1}^{\dagger} c_{l_2}+\frac{1}{4} \sum_{l_1 l_2 l_3 l_4} \bar{v}_{l_1 l_2 l_3 r_4} c_{l_1}^{\dagger} c_{l_2}^{\dagger} c_{l_4} c_{l_3} $$

The questions are: Why is this additional term needed in the single-particle operator, where does it come from and how can I find the $g_{kk'}$ coefficients?

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  • $\begingroup$ Nothing prevents us from writing a general hermitian one particle operator in this form. Thus, the form has to do not with second quantization, but with the consideration of specific presentation that follows in the book. $\endgroup$
    – Roger Vadim
    Dec 19, 2022 at 16:49
  • $\begingroup$ I don't know the book, but the difference between both cases is that the first form preserves the particle number while the second does not. The Hamiltonian in your last equation, again, preserves the particle number. $\endgroup$
    – Tobias Fünke
    Dec 19, 2022 at 16:55
  • $\begingroup$ @TobiasFünke Why do you say that the second form does not preserve the number? I see a term creating two particles of momentum $k$ and $k'$ but the h.c. term contains the destruction of the same particles. Am I wrong? $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Dec 19, 2022 at 17:25
  • $\begingroup$ @GiorgioP Yes, but there is a sum in between. Take for example a vector $\psi \in H_N \subset \mathcal F$, where $H_N$ denotes the space of $N$ indistinguishable particles and $\mathcal F$ the corresponding Fock space. Acting with the first operator we have that $F_1\psi \in H_N \subset \mathcal F$, but with the second form this no longer holds, i.e. $F_2\psi \not\in H_N$; rather, I'd say that $F_2\psi \sim \psi_N + \psi_{N+2} + \psi_{N-2}$, where $\psi_k \in H_K$. Put differently, I think that $F_1$ commutes with the number operator $N$, while $F_2$ does not, no? $\endgroup$
    – Tobias Fünke
    Dec 19, 2022 at 17:36
  • $\begingroup$ @GiorgioP See for example the case of superconductors $\endgroup$
    – Tobias Fünke
    Dec 19, 2022 at 18:08

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