Some questions about Bogoliubov transformation

consider a Hamiltonian, after mean-field approximation, can be written as
$$hat{H}=sum_{p}(epsilon_{p}+gn_{0})a_{p}^{dagger}a_{p}+frac{1}{2}gn_{0}(a_{k+p}a_{k-p}+hat{a}^{dagger}_{k+p}hat{a}^{dagger}_{k-p})$$
where $g$ is the coupling strength between particles and $n_{0}$ is the particle density.

Denote $hat{H}=UMU^{dagger}$ and $U={a_{k-p},a^{dagger}_{k+p}}$. Suppose the new basis is written as

$$alpha_{p}=u_{p}a_{k-p}+v_{p}a^{dagger}_{k+p}$$ and $$beta_{p}=v_{k}a_{k-p}+u_{k}a^{dagger}_{k+p}$$

Then the Bogoliubov spectrum would be the spectrum of $sigma_{z}M$, where $sigma_{z}={1,0;0,-1}$

We can write $hat{H}$ as:

$hat{H}=frac{1}{2}sum_{p} [(epsilon_{k+p}+gn_{0})a_{k+p}^{dagger}a_{k+p}+(epsilon_{k-p}+gn_{0})a_{k-p}^{dagger}a_{k-p}]+frac{1}{2}gn_{0}(a_{k+p}a_{k-p}+hat{a}^{dagger}_{k+p}hat{a}^{dagger}_{k-p})$,

in this case,

$M={frac{1}{2}(epsilon_{k+p}+gn_{0}),frac{1}{2}gn_{0};frac{1}{2}gn_{0},frac{1}{2}(epsilon_{k-p}+gn_{0})}tag{1}$

However if we write $hat{H}$ as

$hat{H}=sum_{p} (epsilon_{k+p}+gn_{0})a_{k+p}^{dagger}a_{k+p}+frac{1}{2}gn_{0}(a_{k+p}a_{k-p}+hat{a}^{dagger}_{k+p}hat{a}^{dagger}_{k-p})$

$M$ would be:

$M={epsilon_{k+p}+gn_{0},frac{1}{2}gn_{0};frac{1}{2}gn_{0},0}tag{2}$

After calculation, one can see (1) and (2) give different spectrum.

And in (1), the energy corresponding to $beta_{p}$ is equal to that of $alpha_{-p}$. But this relation does not hold in (2).

Generally, I see most textbooks tackle this problem like (1). However, I did not see where is wrong in (2). Is there symmetry forbidding writing $hat{H}$ as (2)?