# Normalization in perturbation theory

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.

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 $$H_0$$.
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 $$V$$.