  Let $X$ be an underlying space with a reference measure $\sigma$. Let $K$ be
an integral operator in $L^2(X,\sigma)$ with integral kernel $K(x,y)$. A point
process $\mu$ on $X$ is called determinantal with the correlation operator $K$
if the correlation functions of $\mu$ are given by
$k^{(n)}(x_1,\dots,x_n)=\operatorname{det}[K(x_i,x_j)]_{i,j=1,\dots,n}$. It is
known that each determinantal point process with a self-adjoint correlation
operator $K$ is the joint spectral measure of the particle density
$\rho(x)=\mathcal A^+(x)\mathcal A^-(x)$ ($x\in X$), where the operator-valued
distributions $\mathcal A^+(x)$, $\mathcal A^-(x)$ come from a gauge-invariant
quasi-free representation of the canonical anticommutation relations (CAR). If
the space $X$ is discrete and divided into two disjoint parts, $X_1$ and $X_2$,
by exchanging particles and holes on the $X_2$ part of the space, one obtains
from a determinantal point process with a self-adjoint correlation operator $K$
the determinantal point process with the $J$-self-adjoint correlation operator
$\widehat K=KP_1+(1-K)P_2$. Here $P_i$ is the orthogonal projection of
$L^2(X,\sigma)$ onto $L^2(X_i,\sigma)$. In the case where the space $X$ is
continuous, the exchange of particles and holes makes no sense. Instead, we
apply a Bogoliubov transformation to a gauge-invariant quasi-free
representation of the CAR. This transformation acts identically on the $X_1$
part of the space and exchanges the creation operators $\mathcal A^+(x)$ and
the annihilation operators $\mathcal A^-(x)$ for $x\in X_2$. This leads to a
quasi-free representation of the CAR, which is not anymore gauge-invariant. We
prove that the joint spectral measure of the corresponding particle density is
the determinantal point process with the correlation operator $\widehat K$.
