Fermionic vacuum under a Bogoliubov transformation Context:
Consider a Bogoliubov-de Gennes Hamiltonian,
\begin{align}
\hat{H}_{BdG} = \sum_{j,k} \hat{\Psi}_j^{\dagger}H_{jk}\hat{\Psi}_k,
\end{align}
where $\hat{\Psi}$ is a $2n$-dimensional vector of fermionic creation operators and its annihilation counterparts, and $H$ is a $2n\times2n$ Hermitian matrix that obeys particle-hole symmetry: $\hat{P}\hat{H}\hat{P} = -\hat{H}^*$. The vacuum of the operators contained in $\hat{\Psi}$ can be defined as the state $\vert 0 \rangle$ such that $\hat{\Psi}_j\vert 0 \rangle = 0$ for all $j=1,...,n$.
In general this Hamiltonian can be put in its diagonal form by diagonalizing $\hat{H}$. If $U$ is the unitary matrix that does so, and we define $\hat{d} = U \hat{\Psi}$, then
\begin{align}
\hat{H}_{BdG} &= \sum_{j} \left( \epsilon_j \hat{d}_j^{\dagger}\hat{d}_j - \epsilon_j \hat{d}_j \hat{d}_j^{\dagger} \right) \\
&= \sum_{j} \epsilon_j \left( 2 \hat{d}_j^{\dagger}\hat{d}_j - 1 \right),
\end{align}
where $\epsilon_j$ are the positive eigenvalues of $H$.
Here comes the question:

*

*When does the vacuum of the new operators $\hat{d}_j$ exist?

By vacuum I mean the state such that $\hat{d}_j \vert 0 \rangle_{BdG} = 0 \quad \forall ~ j=1,...,n$.
By existing I mean being an element of the Fock space generated by the orignal vacuum $\vert 0 \rangle$ and the original fermionic operators $\hat{\Psi}^{\dagger}_j$.


*When it does exist (if it ever does), how can I write it in the original Fock basis, that is, using the original vacuum and the original operators?

*Is this vacuum the groundstate of the system?

I believe that these questions might be equivalent to answering
How does the Bogoliubov transformation affect the vacuum of a system?
 A: I'm not sure whether you want just the linear algebra, or a more detailed functional analysis that worries about operator convergence. If it's just the former, the following extract from my notes may help:
Let
$$
 \hat H_{\rm Bogoliubov}= a^\dagger_i H_{ij}a_j +\frac 12 \Delta_{ij} a^\dagger_i a^\dagger_j +\frac 12 \Delta^{\dagger}_{ij} a_i a_j\nonumber\\
= \frac12 \left(\matrix{ a^\dagger_i &a_i}\right)\left(\matrix{ H_{ij}& \phantom {-}\Delta_{ij}\cr
                                                                           \Delta^{\dagger}_{ij}& -H^T_{ij}}\right)
                  \left(\matrix{ a_j\cr a^\dagger_j}\right) +\frac 12 {\rm tr}(H).   \nonumber                                                           
$$
If we arrange for the positive eigenvalues of the BdG operator to be those for  $(u,v)^T$ and set
$$
 a_i= u_{i\alpha}b_\alpha +v^*_{i\alpha}b^\dagger_\alpha\nonumber\\
 a^\dagger_i= v_{i\alpha} b_\alpha +u^*_{i\alpha}b^\dagger_\alpha.\nonumber
$$
the  mutual orthonormality  and completeness of the eigenvectors
ensure  that the $b_\alpha$, $b^{\dagger}_\alpha$ have the same anti-commutation relations as the $a_i$ $a^\dagger_i$. In terms of the $b_\alpha$ $b^\dagger_\alpha$, the second-quantized Hamiltonian becomes
$$
 \hat H_{\rm Bogoliubov} =\sum_{\alpha=1}^N E_\alpha b^\dagger_\alpha b_\alpha -\frac 12 \sum_{\alpha=1}^N E_\alpha +\frac 12\sum_{i=1}^N E^{(0)}_i.
$$
Here the $E^{(0)}_i$ are the eigenvalues of $H$. Unlike the $E_\alpha$, these can be of either sign.
If all the $E_\alpha$ are strictly positive, the  new ground state is non degenerate and is the unique state $|{0}\rangle_b$   annihilated by all the $b_\alpha$.
If we could find a unitary operator ${\mathcal U}$ that acts on the $2^N$-dimensional Fock space
such that
$$
b_\alpha = a_iu^*_{i\alpha}+ a^\dagger_i v^*_{i\alpha}=  {\mathcal U}a_i{\mathcal  U}^{-1},\nonumber\\
b^\dagger_\alpha = a^\dagger_iu_{i\alpha}+ a_i v_{i\alpha}=  {\mathcal U}a^\dagger_i {\mathcal U}^{-1},\nonumber
$$
then we would have
$
|{0}\rangle_b={\mathcal U}|{0}\rangle_a
$, where $|{0}\rangle_a$ is the no-particle vacuum state annihilated by all the $a_i$.
Except in the simplest cases, it is not easy to find a closed-form expression for ${\mathcal U}$.
An alternative strategy for obtaining $|{0}\rangle_b$  begins by  noting   that if   that the matrix $u_{i\alpha}$ is invertible then
the condition  $b_i |{0}\rangle_b=0$  is equivalent to
$$
(a_i+a^\dagger_k  v^*_{k\alpha}(u^*)^{-1}_{\alpha i})|{0}\rangle_b=0, \quad  i=1,\ldots N.
$$
We therefore introduce  the skew-symmetric  matrix
$$
S_{ij}=   v^*_{i\alpha}(u^*)^{-1}_{\alpha j}
$$
which satisfies
$$
\exp\left\{\frac 12 a^\dagger_ia^\dagger_jS_{ij}\right\} a_k \exp\left\{-\frac 12 a^\dagger_ia^\dagger_jS_{ij}\right\} 
 =a_k+a^\dagger_iS_{ik}.
 $$
From this we conclude that we can take $ |{0}\rangle_b$ to be
$$
 |{0}\rangle_b ={\mathcal N}  \exp\left\{\frac 12 a^\dagger_ia^\dagger_jS_{ij}\right\}|{0}\rangle_a
 $$
where $|{0}\rangle_a$ is the original no-particle state.  This expression explicitly  displays the superconducting ground state  as a coherent  superposition of Cooper-pair states, and allows us to identify $S_{ij}$ with the (unnormalized)  pair wavefunction.
By assuming that the $E_\alpha$ are positive  we have swept a potential problem under the rug.  When we  arrange for the positive energy BdG eigenvectors to be the $(u,v)^T$ and the negative eigenvectors to be $(v^*,u^*)^T$ we may have  to interchange columns in the $2N$-by-$2N$ matrix
$$
 U= \left[\matrix{u &v^*\cr v&u^*}\right].
 $$
Each interchange  has the effect of changing the sign of ${\rm det} [U]$ and one can show that  a negative sign for ${\rm det} [U]$ precludes the invertibility of the $N$-by-$N$ matrix $u$, and hence denies us the skew matrix  $S_{ij}$.  To  avoid this issue we can  keep  ${\rm det} [U]$ positive, but  at the price  that one of the $E_\alpha$ ---  let us call it  $E_{\alpha_0} $ --- may have to remain    negative. If so,  the lowest energy state has the quasiparticle level $E_{\alpha_0}$ occupied
$$
 |{0}\rangle_{\rm ground} \propto b^\dagger_{\alpha_0}    \exp\left\{\frac 12 a^\dagger_ia^\dagger_jS_{ij}\right\} |{0}\rangle_a.
$$
The state $|{0}\rangle_{\rm ground}$ is therefore  a superposition of states with an odd number of particles, one of which is always unpaired.
To see that a negative determinant for $U$ prevents $u$ from the being invertible we  consider some  properties of $2N$-by-$2N$ unitary matrices  of the form
$$
 U=\left[\matrix{u &v^*\cr v&u^*}\right], \quad U^\dagger= \left[\matrix{u^\dagger &v^\dagger\cr v^T&u^T}\right].
 $$
The equations $U U^\dagger=1=U^\dagger U$ give us
$$ 
uu^\dagger+v^*v^T=1= u^\dagger u+v^\dagger v,\nonumber\\
uv^\dagger+v^*u^T=0= u^\dagger v^*+v^\dagger u^*,\nonumber\\
vu^\dagger+u^*v^T=0= v^Tu+u^Tv,\nonumber\\
vv^\dagger +u^* u^T=1= v^Tv^*+u^Tu^*.\nonumber
 $$
These equations are symmetric under the interchange $u\leftrightarrow v$.
To  get  $U^*$ from $U$  we need to exchange   even number of rows and columns; consequently    ${\rm det}[U]= {\rm det}[U^*]$ is a real number. Further   $1={\rm det}[U]{\rm det}[U^*]$ tells us that
${\rm det}[U]=\pm 1$.  Under the interchange of $u$ and $v$, however, we have
$$
 \left|\matrix{u &v^*\cr v&u^*}\right| = (-1)^N \left|\matrix{v &u^*\cr u&v^*}\right|.
 $$
If $u$ is invertible, Schur's determinant identity
$$
 \left|\matrix{A &B\cr C&D}\right| ={\rm det}[A] {\rm det}[D- CA^{-1}B]
 $$
tells us that
$$
 {\rm det}[U]= {\rm det}[u] {\rm det}[u^*-v u^{-1} v^*]\nonumber\\
 = {\rm det}[u] {\rm det}[u^*+v v^\dagger   (u^T)^{-1}]\nonumber\\
 = {\rm det}[u] {\rm det}[u^*+(1-u^*u^T)(u^T)^{-1}]\nonumber\\
 ={\rm det}[u] {\rm det}[(u^T)^{-1}]\nonumber\\
 =1.
 $$
Similarly, if $v$ is invertible the $u\leftrightarrow v$ symmetry converts the above algebra to give
$$
(-1)^N {\rm det}[U]=\left|\matrix{v &u^*\cr u&v^*}\right| = {\rm det}[v] {\rm det}[(v^T)^{-1}]=1.
 $$
We see that when  $N$ is even  and  ${ \rm det}[U]=-1$ neither $u$ nor $v$ can be inverted. When $N$ is odd  ${ \rm det}[U]=-1$ precludes $u$ from being inverted, while   ${ \rm det}[U]=+1$ precludes    $v$ from    being inverted.
When $N$ is odd and  $v$ is invertible we can define a "full"  state that obeys $a^\dagger_i|{\rm full}\rangle=0$ for all $i$ and construct  the odd-particle-number ground state $|{0}\rangle_{\rm ground}$ as a paired state of holes.
