In the book Bose-Einstein Condensation by Pitaevski, Lev; Petrovitch, and Sandro Stringari (Oxford University Press), the Hamiltonian for weakly interacting Bose gas reads as, $$H=\sum\dfrac{p^2}{2m}\hat{a}_p^\dagger \hat{a}_p+\dfrac{1}{2V}\sum_{p_1,p_2,q} V_q \hat{a}_{p_1+q}^\dagger \hat{a}_{p_2-q}^\dagger \hat{a}_{p_1} \hat{a}_{p_2}\tag{1}\label{1}$$ In the book they have assumed $V_q=V_0$ to be independent of $q$. Expanding the second term one can write the following, $$\dfrac{V_0}{2V} \hat{a}^\dagger_0 \hat{a}^\dagger_0 \hat{a}_0 \hat{a}_0+\dfrac{V_0}{2V}\sum_{q\neq0}(\hat{a}^\dagger_q\hat{a}_q+\hat{a}^\dagger_{-q}\hat{a}_{-q}+\hat{a}^\dagger_{q}\hat{a}^\dagger_{-q}+\hat{a}_{q}\hat{a}_{-q})\\ =\dfrac{V_0}{2V} \hat{a}^\dagger_0 \hat{a}^\dagger_0 \hat{a}_0 \hat{a}_0+\dfrac{V_0}{2V}\sum_{q\neq0}(2\hat{a}^\dagger_q\hat{a}_q+\hat{a}^\dagger_{q}\hat{a}^\dagger_{-q}+\hat{a}_{q}\hat{a}_{-q}) $$

However, in a thesis I saw $V_q$ was a function of $q$, although it was not specified what kind of function it was. In that case, the second term was expanded as follows, $$\begin{eqnarray} &&\dfrac{V_0}{2V} \hat{a}^\dagger_0 \hat{a}^\dagger_0 \hat{a}_0 \hat{a}_0+\dfrac{1}{2V}\sum_{q\neq0}[V_q(\hat{a}^\dagger_q\hat{a}_q+\hat{a}^\dagger_{-q}\hat{a}_{-q}+\hat{a}^\dagger_{q}\hat{a}^\dagger_{-q}+\hat{a}_{q}\hat{a}_{-q})]\tag{2}\label{2}\\ &=&\dfrac{V_0}{2V} \hat{a}^\dagger_0 \hat{a}^\dagger_0 \hat{a}_0 \hat{a}_0+\dfrac{1}{2V}\sum_{q\neq0}[V_q(2\hat{a}^\dagger_q\hat{a}_q+\hat{a}^\dagger_{q}\hat{a}^\dagger_{-q}+\hat{a}_{q}\hat{a}_{-q})]\tag{3}\label{3} \end{eqnarray} $$ My question: Is it valid to go from \eqref{2} to \eqref{3} even when $V_q$ may be of indefinite parity(neither purely even nor purely odd)? In short, is the following true if the operators are bosonic? $$\sum_{q\neq0}V_q(\hat{a}^\dagger_q\hat{a}_q+\hat{a}^\dagger_{-q}\hat{a}_{-q})=2\sum_{q\neq0}V_q\hat{a}^\dagger_q\hat{a}_q$$


In the book "Bose-Einstein Condensation" (page 27) the author wrote that $V(\vec{r})$ is a two-body potential.

We can therefore assume that this potential only depends on the distance between particles $|\vec{r}|$ (I don't have in mind another possible kind of two-body potential). So we have $V(\vec{r})=V(-\vec{r})$. And for the Fourier transform (I do it in 1D for simplicity of notations) :

$\begin{align}V_{-q}&=\int\limits_{-\infty}^{+\infty}dxe^{+iqx}V(x)\\ &=\int\limits_{+\infty}^{-\infty}(-dx)e^{-iqx}V(-x) \\&= \int\limits_{-\infty}^{+\infty}dxe^{-iqx}V(x) \\ &=V_{q} \end{align}$

where I did the change of variable $x\rightarrow -x$ in the second line


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