Variance of an interacting quantum field in its vacuum state

A non-interacting quantum field $\hat{\phi}(x)$ can be decomposed into $a_{\textbf{k}}$ and $a_{\textbf{k}}^\dagger$. This enables us to calculate the variance of a free field. For example, the variance of the free real scalar field, in the vacuum $|0\rangle$ of the theory, is computed to be (without a momentum cut-off) $${\rm Var}(\phi)_0=\langle0|\phi^2|0\rangle-\big(\langle0|\phi|0\rangle\big)^2\\=\int\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{\textbf{k}^2+m^2}}\rightarrow \infty.$$ Now, consider an interacting quantum field theory described by the Hamiltonian $H$ and $|\Omega\rangle$ is the vacuum state of the interaction theory i.e. $$H|\Omega\rangle=0|\Omega\rangle\hspace{0.5cm}\big(\text{also,}~P^\mu|\Omega\rangle=0|\Omega\rangle\big).$$ Decomposing the field into creation and annihilation operators is no longer possible. So how does one compute the variance $${\rm Var}(\phi)_\Omega=\langle\Omega|\phi^2|\Omega\rangle-\big(\langle\Omega|\phi|\Omega\rangle\big)^2$$ in $\lambda\phi^4$ theory?