# commutation relations when calculating Hamiltonian

I am reading Topics on Superfulidity of Walter Greiner Book Titled "Quantum:Mechanics Special Chapters" In Exercise page no 200, Hamiltonian has been discussed and derived throughly using commutation relation . I understood very nicely but equation 4 and 5 is where I am stucked on extra v[squared] Term. Here is the equation

and Using the relation

Equation 5 is derived from above equation as below

I was wondering where does extra

term appeared in 2nd term of equaiton 5.

If you take the new bosonic operators \begin{align*} \begin{aligned} \hat{a}_{k} &=u(k) \hat{A}_{k}+v(k) \hat{A}_{-k}^{\dagger} \\ \hat{a}_{k}^{\dagger} &=u(k) \hat{A}_{k}^{\dagger}+v(k) \hat{A}_{-k} \end{aligned} \end{align*} and plug them into \begin{align*} \begin{aligned} \hat{H}_{1}=& \frac{N^{2}}{2 L^{3}} \mathcal{V}(0)+\sum_{k} \frac{(\hbar \boldsymbol{k})^{2}}{2 m} \hat{a}_{\boldsymbol{k}}^{\dagger} \hat{a}_{\boldsymbol{k}} \\ &+\frac{n_{0}}{2 L^{3}} \sum_{\boldsymbol{k}} \mathcal{V}(k)\left(\hat{a}_{\boldsymbol{k}}^{\dagger} \hat{a}_{-\boldsymbol{k}}^{\dagger}+\hat{a}_{\boldsymbol{k}} \hat{a}_{-\boldsymbol{k}}+2 \hat{a}_{\boldsymbol{k}}^{\dagger} \hat{a}_{\boldsymbol{k}}\right) \end{aligned} \end{align*} you will get from \begin{align*} a_{k}a_{-k} = u^2 A_k A_{-k} + v^2 A_{-k}^{\dagger} A_{k}^{\dagger} + uv (A_kA_{k}^{\dagger} + A_{-k}^{\dagger}A_{-k} ) \end{align*} the first $$v^2$$ and from $$a_{k}^{\dagger} a_{-k}^{\dagger}$$ the second $$v^2$$ term in these equations.