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When we quantize a free field theory, we set $\phi(x)$ to be the operators and we take the Fourier transform to determine the creation and annihilation operators $a_\omega,a^\dagger_\omega$ such that $\sum_\omega (a^\dagger_\omega)^{n_\omega}|0\rangle$ be a base for the Fock space of the theory. This base contains the eigenvectors of hamiltonian $H_0$.

Then we introduce the interaction hamiltonian $H_{int}$ but we still work in the base of $H_0$.

Does $H_{int}$ change the eigenvectors and shift the eigenvalues of the system? (I'm asking in the sense of nonpertubative.

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Non-perturbatively, it most cases it does. However, we work quite a lot in perturbation theory. In perturbation theory the basic assumption is that $H_{\text{int}}$ should not change the spectrum of the Hilbert space. Further, in perturbation theory, all of QFT is done in the "interaction picture" where all fields evolve w.r.t. $H_0$. Thus it is consistent to use the original formulae from free field theory even when $H_{\text{int}}$ is included.

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