Consider a QFT in flat spacetime from the point of view of an inertial observer with coordinates $(t,{\vec x})$. Suppose this guy expands a quantum field $\hat{\phi}$ in two distinct complete sets of orthonormal modes $\{f_i(t,\vec{x})\}$ and $\{g_\alpha(t,\vec{x})\}$ as follows $$\hat{\phi}(t,{\vec x})=\sum_i\left(\hat{a}_if_i(t,\vec{x})+\hat{a}_i^\dagger f_i^*(t,\vec{x})\right)=\sum_\alpha\left(\hat{b}_\alpha g_\alpha(t,\vec{x})+\hat{b}_\alpha^\dagger g_\alpha^*(t,\vec{x})\right)$$ with the following commutation relations $$[\hat{a}_i,\hat{a}_j^\dagger]=\delta_{ij}, [\hat{a}_i,\hat{a}_j]=[\hat{a}_i^\dagger,\hat{a}_j^\dagger]=0, \\ [\hat{b}_\alpha,\hat{b}_\beta^\dagger]=\delta_{\alpha\beta}, [\hat{b}_\alpha,\hat{b}_\beta]=[\hat{b}_\alpha^\dagger,\hat{b}_\beta^\dagger]=0.$$
In the above, the '$i$'-type indices could be the wavevector indices ${\vec k}$ (as in the case of plane wave modes), and '$\alpha$'-type indices could be something different.
Thus the operators $\hat{b}_\alpha$ will be related to the operators $\hat{a}_i$ by bogoliubov transformation, and if $\hat{a}_i|0_f\rangle=0$ and $\hat{b}_\alpha|0_g\rangle=0$, it is easy to show that $$\langle 0_f|\hat{b}^\dagger_\alpha\hat{b}_\alpha|0_f\rangle\neq 0.$$
Therefore, even for an inertial observer in flat spacetime, the notion of vacuum does not seem to be unique. However, if we say that the vacuum state is one for which $\langle H\rangle$ is minimum, we have a unique vacuum state and a unique mode expansion. In flat space, this turns out to be an expansion in terms of plane wae modes. See the answer Interpretation of the nonuniqueness of vacuum of QFT in flat spacetime for a given inertial observer; No Lorentz transformation; No accelerated motion by @ChiralAnomaly.
Now, whether a state contains particles or not is an experimental question. Suppose we have agreed to define our vacuum as a state in which $\langle H\rangle$ is minimized i.e. we have agreed to use the plane wave modes. But how does the detector know what definition of vacuum to use?
