# Details in the derivation of the Lippmann-Schwinger equation

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 $$\left on the LSE, I get $$\psi(x) = \phi(x) + \left.$$ If both $$\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.

• It seems Schwartz argue that - please give an exact reference... Jul 29, 2023 at 13:07
• Isnt this just asuming the scattering to be elastic? Jul 30, 2023 at 8:49

The usual situation is that V is short-ranged, so that at large separations $$H\to H_0$$, and $$\psi(x) \to \phi(x)$$. But the Schrodinger equation should be true for all $$x$$, so the energy is the same at short and long distances. The issue with normalization can be tricky since scattering states will well-defined momentum aren’t normalizable. But assuming we’re working with normalizable wave packets, the second term in the Lippman-Schwinger equation can produce something proportional to $$\phi$$. Specifically, it will cause destructive interference, pulling flux out of $$\phi$$ and putting it into other channels.