So the argument goes that for a slightly perturbed Hamiltonian $$ H = H_0 + V, $$ there will be some exactly known states, $\left|\phi\right>$, solving $$ H_0\left|\phi\right> = E\left|\phi\right>, $$ and some later-time states that we are interested in, $\left|\psi\right>$, solving $$ (H_0 + V)\left|\psi\right> = E\left|\psi\right>, $$ such that we can write the LSE, $$ \left|\psi\right> = \left|\phi\right> + \frac{1}{E - H_0}V\left|\psi\right>, $$ which can be checked by multiplying through with $E - H_0$.

This is all fine to me, except the assumption that the energy $E$ for both states are equal. When we add a perturbation to the system, shouldn't the energies shift?

It seems Schwartz argue that, since the energies shift continuously as we turn on $V$, there will always be a $\left|\psi\right>$ with the same $E$ as some $\left|\phi\right>$. But how can we be sure that the potential doesn't shift the energy of some of the $\left|\psi\right>$'s up above the energies of all the $\left|\psi\right>$'s? Additionally, if we don't know which $\left|\phi\right>$ the $\left|\psi\right>$ we are interested in has equal energy to, what can the equation even be used for?

And also; why is the derivation $$ H_0\left|\psi\right> + V\left|\psi\right> = E\left|\psi\right> \quad\Rightarrow\quad H_0\left|\psi\right> - E\left|\psi\right> = - V\left|\psi\right> \quad\Rightarrow\quad \left|\psi\right> = \frac{1}{E - H_0}V\left|\psi\right> $$ not equally valid? Both this and LSE can't be right, can they?

Bonus question: If i use $\left<x\right|$ on the LSE, I get $$ \psi(x) = \phi(x) + \left<x\right|\frac{1}{E - H_0}V\left|\psi\right>. $$ If both $\psi(x)$ and $\phi(x)$ are (presumably) normalised wave-functions, how can one normalised wave-function be equal to another normalised wave-function plus something, and still be normalised? It seems like the probabilistic interpretation breaks here.

So the argument goes that for a slightly perturbed Hamiltonian $$ H = H_0 + V, $$ there will be some exactly known states, $\left|\phi\right>$, solving $$ H_0\left|\phi\right> = E\left|\phi\right>, $$ and some later-time states that we are interested in, $\left|\psi\right>$, solving $$ (H_0 + V)\left|\psi\right> = E\left|\psi\right>, $$ such that we can write the LSE, $$ \left|\psi\right> = \left|\phi\right> + \frac{1}{E - H_0}V\left|\psi\right>, $$ which can be checked by multiplying through with $E - H_0$.

This is all fine to me, except the assumption that the energy $E$ for both states are equal. When we add a perturbation to the system, shouldn't the energies shift?

It seems Schwartz argue that, since the energies shift continuously as we turn on $V$, there will always be a $\left|\psi\right>$ with the same $E$ as some $\left|\phi\right>$. But how can we be sure that the potential doesn't shift the energy of some of the $\left|\psi\right>$'s up above the energies of all the $\left|\psi\right>$'s? Additionally, if we don't know which $\left|\phi\right>$ the $\left|\psi\right>$ we are interested in has equal energy to, what can the equation even be used for?

And also; why is the derivation $$ H_0\left|\psi\right> + V\left|\psi\right> = E\left|\psi\right> \quad\Rightarrow\quad H_0\left|\psi\right> - E\left|\psi\right> = - V\left|\psi\right> \quad\Rightarrow\quad \left|\psi\right> = \frac{1}{E - H_0}V\left|\psi\right> $$ not equally valid? Both this and LSE can't be right, can they?

Bonus question: If i use $\left<x\right|$ on the LSE, I get $$ \psi(x) = \phi(x) + \left<x\right|\frac{1}{E - H_0}V\left|\psi\right>. $$ If both $\psi(x)$ and $\phi(x)$ are (presumably) normalised wave-functions, how can one normalised wave-function be equal to another normalised wave-function plus something, and still be normalised? It seems like the probabilistic interpretation breaks here.