Why divide diagonal elements in Bogoliubov transform? In order to simplify the following Hamiltonian:
$$H=\sum_k(A b_k^\dagger b_{-k}^\dagger+A^* b_k b_{-k}+Bb_k^\dagger b_k)\\=\sum_k \Psi_k^\dagger H(k)\Psi_k$$
where $b$ is the bosonic operator, and $$\Psi_k=\left(\begin{array}{l}{b_{k}} \\ {b_{-k}^{\dagger}}\end{array}\right)$$we need to apply Bogoliubov transform, or, equivalently, equation of motion, which means diagonalize the dynamical matrix $D(k)$ directly:
$$i \frac{d}{d t}\left(\begin{array}{l}{b_{k}} \\ {b_{-k}^{\dagger}}\end{array}\right)=D(k)\left(\begin{array}{c}{b_{k}} \\ {b_{-k}^{\dagger}}\end{array}\right)$$
we know the form of $D(k)$ is related to $H(k)$. However, if we use the initial form of $H$, the $H(k)$ is:
$$H(k)=\left(\begin{array}{cc}{B} & {A^{*}} \\ {A} & {0}\end{array}\right)$$
On the other hands, we can rewrite the initial Hamiltonian as(just like most textbook):
$$H=\sum_k(A b_k^\dagger b_{-k}^\dagger+A^* b_k b_{-k}+\frac{B}{2}b_k^\dagger b_k+\frac{B}{2}b_{-k}^\dagger b_{-k})$$
now the $H(k)$ is:
$$H(k)=\left(\begin{array}{cc}{B/2} & {A^{*}} \\ {A} & {B/2}\end{array}\right)$$
It seems that the  diagonalization results are different if use these two kinds of $H(k)$. 
I think I may make some mistakes when using the first way? 
 A: If you do the diagonalisation, the end result doesn't change. 
Let's assume that the commutation relations among the $b$s are
$$[b_k, b_{k'}]=[b^\dagger_k, b^\dagger_{k'}]=0$$
$$[b_k, b^\dagger_{k'}] = \delta_{k, k'}$$
Let 
$\Psi_k=\left(\begin{array}{l}{b_{k}} \\ {b_{-k}^{\dagger}}\end{array}\right)$
as in the OP's question. Then we can write
$$H=\sum_k(A^* b_k^\dagger b_{-k}^\dagger+A b_k b_{-k}+Bb_k^\dagger b_k)=\sum_k \Psi_k^\dagger H_1(k)\Psi_k$$
where
$$H_1(k)=\left(\begin{array}{cc}{B} & {A^{*}} \\ {A} & {0}\end{array}\right)$$
Note that we swapped $A$ and $A^*$ in order to get the same form of $H_1$ as in the OP's question (this appears to be a mistake in the OP's question).
Assuming that the coefficient $B$ depends on $k$ only through its modulus, so that $B(k)=B(-k)$, then we can also write
$$H=\sum_k(A^* b_k^\dagger b_{-k}^\dagger+A b_k b_{-k}+\frac{B}{2}b_k^\dagger b_k+\frac{B}{2}b_{-k}^\dagger b_{-k})$$
In order to write this as a $2\times 2$ matrix as in the previous case, we need the last term to be of the form $b_{-k} b_{-k}^\dagger$. Using the commutation relations for the $b$s we get
$$H=\sum_k(A^* b_k^\dagger b_{-k}^\dagger+A b_k b_{-k}+\frac{B}{2}b_k^\dagger b_k+\frac{B}{2}b_{-k}b_{-k}^\dagger)-\sum_k \frac{B}{2}=\\
=-\sum_k \frac{B}{2}+\sum_k \Psi_k^\dagger H_2(k)\Psi_k$$
dove
$$H_2(k)=\left(\begin{array}{cc}{B/2} & {A^{*}} \\ {A} & {B/2}\end{array}\right)$$
Now we can diagonalise both hamiltonians,
$$H_1 = U_1^\dagger D_1 U_1\\
H_2 = U_2^\dagger D_2 U_2$$
where $D_1, D_2$ are diagonal and the unitary matrices $U_1, U_2$ define new bosonic operators, 
$$\Phi_{k1}=\left(\begin{array}{l}{a_{k,1}} \\ {a_{-k,1}^{\dagger}}\end{array}\right)=U_1 \Psi_k$$
and similarly $\Phi_{k2}$. The new operators $a$ should satisfy the same commutation relations as the $b$s. A simple calculation shows that assuming that $A$ and $B$ depend on $k$ only through its modulus the first two commutation relations are automatically satisfied. In order for the last commutation relation to be satisfied, we require that $U_1, U_2 \in SU(2)$, i.e. they have unit determinant.
The eigenvalues of $H_1, H_2$ are
$$E_{1,\pm}=\frac{B}{2}\pm \sqrt{\left(\frac{B}{2}\right)+|A|^2}\\
E_{2,\pm}=\frac{B}{2}\pm |A|$$ 
So after the diagonalisation, we get
$$H=\sum_k \Psi_k^\dagger H_1(k)\Psi_k= \sum_k \Phi_{k,1}^\dagger D_1(k)\Phi_{k,1}=\sum_k \left(E_{1,-}a_{k,1}^\dagger a_{k,1} + E_{1,+}a_{-k,1} a_{-k,1}^\dagger \right)$$
Now we can use the commutation relations for the $a$s and then send $k \to -k$ in the second term. This brings the second term to the same form as the first one, 
$$H=\sum_k \left(E_{1,-}+E_{1,+}\right)a_{k,1}^\dagger a_{k,1}+\sum_k E_{1,+}=\sum_k B a_{k,1}^\dagger a_{k,1}+\sum_k E_{1,+}$$
The same exact procedure can be done on the second hamiltonian to get
$$H=\sum_k B a_{k,2}^\dagger a_{k,2}+\sum_k E_{2,+}-\sum_k \frac{B}{2}$$
Therefore the two methods lead to Hamiltonians which differ by a constant; the operators $a_{k,1}$ and $a_{k,2}$ are different, but they obey the same commutation relations, and therefore the resulting spectrum is the same. If I had to speculate as to why authors (sometimes) prefer the second method is that it allows one to simply read off the eigenvalues directly from the hamiltonian.
