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When we have a system with hamiltonian $H = H_{0} + V$, we can expand the ground state wavefunction $\Psi_{0}$ using the wavefunction of the non-interacting system $\phi_{0}$, that is an eigenfunction of $H_{0}$. In the Rayleigh-Schrodinger perturbation theory, we choose the normalization of $\Psi_{0}$ such that $\langle \phi_{0} | \Psi_{0} \rangle = 1$. My question is: How do we know that these wave functions are not orthogonal? In this case, this product would be zero. I read in the book "A Guide to Feynman Diagrams in the Many-Body Problem" that, when the interacting system has a different symmetry from the non-interacting system, we will have necessarily $\langle \Psi_{0} | \phi_{0} \rangle = 0$, but I don't know what means "have a different symmetry", and I also would like a explanation about when we can do this normalization.

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It’s built in the physics of the perturbation approach. Perturbation theory makes sense when the dominant term to the exact ground state is the unperturbed ground state, else it is not a perturbation. More to the point, one expects that the largest overlap $\vert \langle \Psi_0\vert \phi_k\rangle\vert$ occurs for $k=0$. if this is NOT the case, it’s no longer a perturbation of the system described by $H_0$.

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This is a "how-to-make-my-QM-textbook-mathematically-rigorous" question. I usually refer these to Kato's Perturbation Theory for Linear Operators.
For "reasonable" $H$ and $V$ it can be shown that $H+\lambda V$ has an eigenfunction $\Psi(\lambda)$ that varies smoothly with $\lambda$ for "small" $\lambda$.
Whether $\langle \Psi(0) \Psi(\lambda) \rangle \ne 0$ for $\lambda \in [0,1]$ depends on properties of $H$ and $V$.

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