Consider a set of boson/fermion creation and annihilation operators satisfying the canonical (anti-)commutation rules (CCR/CAR):
$$
[a_i, a_j]_\eta
= [a^\dagger_i, a^\dagger_j]_\eta = 0,
\quad
[a_i, a^\dagger_j]_\eta
= \delta_{ij}
\quad (i,j = 1,…,N)
$$
where $\eta$ is the statistical sign ($+1$ for boson, and $-1$ for fermion), $[A,B]_\eta = AB – \eta BA$, and $N$ is the dimension of one-particle Hilbert space. A general canonical transformation mixes the creation and annihilation operators to produce a new set of bosons/fermions $\{b_j\}_{j=1}^N$, which I call the Bogoliubov quasi-particles:
$$
b_i = \sum_{j=1}^N
(u_{ij} a_j + v_{ij} a^\dagger_j)
, \quad
b^\dagger_i = \sum_{j=1}^N
(v^*_{ij} a_j + u^*_{ij} a^\dagger_j)
$$
and the $\{b_i\}$ operators should also satisfy the CCR/CAR:
$$
[b_i, b_j]_\eta
= [b^\dagger_i, b^\dagger_j]_\eta = 0,
\quad
[b_i, b^\dagger_j]_\eta
= \delta_{ij}
\quad (i,j = 1,…,N)
$$
One can show that to preserve the CCR/CAR, the matrices $U = \{u_{ij}\}$, $V = \{v_{ij}\}$ should satisfy the requirement
$$
U V^{\mathsf{T}} = \eta V U^{\mathsf{T}} \quad \text{and} \quad
U U^\dagger – \eta V V^\dagger = 1
$$
Question: Let $|0_a\rangle, |0_b\rangle$ be the vacuum states of the $a, b$ particles respectively, i.e.
$$
\forall i = 1,…,N: \quad
a_i |0_a\rangle = 0, \quad
b_i |0_b\rangle = 0
$$
How to express $|0_b\rangle$ in terms of states and operators of the $a$ particles?
Best Answer
The following is summarized from two notes (both can be found here) by Prof. Michael Stone (@mikestone). Confusing typos in these notes are corrected.
From a general perspective, the two set of operators $\{a\}, \{b\}$ are two realizations of the CCR/CAR. By the Stone-von Neumann theorem, there is a unitary operator $\mathcal{U}$ that acts on the many-body Hilbert space such that
$$ b_i = \mathcal{U} a_i \mathcal{U}^\dagger \ \Rightarrow \ b^\dagger_i = \mathcal{U} a^\dagger_i \mathcal{U}^\dagger $$
Then we find $|0_b\rangle \propto \mathcal{U}|0_a\rangle$, since
$$ a_i |0_a\rangle = \mathcal{U}^\dagger b_j \mathcal{U} |0_a\rangle = 0 $$
However, as noted by Prof. Stone, the general form of $\mathcal{U}$ is very complicated. Thus we take another approach.
Assume that $U$ is invertible (this is not an "innocent assumption"). Then $b_i |0_b \rangle = 0$ is equivalent to
$$ 0 = U^{-1}_{ij} b_j |0_b\rangle = (a_i - S_{ij} a^\dagger_j) |0_b \rangle, \quad S \equiv - U^{-1} V $$
Using the constraint $U V^\mathsf{T} = \eta V U^\mathsf{T}$, one can show that
$$ S^\mathsf{T} = \eta S $$
i.e. the matrix $S$ is symmetric for bosons, and anti-symmetric for fermions. Then we construct the bilinear
$$ Q \equiv \frac{1}{2} \sum_{i,j} S_{ij} a^\dagger_i a^\dagger_j $$
and use the BCH formula to calculate
$$ \begin{equation*} e^Q a_i e^{-Q} = a_i + [Q,a_i] + \frac{1}{2}[Q,[Q,a_i]] + \cdots \end{equation*} $$
Fortunately, the commutator
$$ [Q, a_i] = - S_{ij} a^\dagger_j $$
commutes with $Q$ (for both bosons and fermions). Therefore we simply get
$$ \begin{equation*} e^Q a_i e^{-Q} = a_i + [Q,a_i] = a_i - S_{ij} a^\dagger_j \end{equation*} $$
and $b_i |0_b \rangle = 0$ is equivalent to
$$ e^Q a_i e^{-Q} |0_b\rangle = 0 $$
This can be satisfied by all $a_i$ if and only if
$$ e^{-Q} |0_b \rangle \propto |0_a \rangle $$
The "if" is obvious. To show "only if", suppose that $|\psi\rangle \equiv e^{-Q} |0_b \rangle \ne |0_a \rangle$; then $e^Q$, which consists of only $\{a^\dagger_i\}$, cannot annihilate $|\psi\rangle$. Finally, up to a normalization constant $\mathcal{N}$, the vacuum $|0_b\rangle$ is
$$ |0_b\rangle = \mathcal{N} e^Q |0_a \rangle $$
(When I have time I will update on finding $\mathcal{N}$.)