Green function in scattering theory

Question

I'm having a bit of trouble with a step in scattering theory.

Context:

The Schrödinger equation for a two-body scattering problem can be written as: $$ (E - H_0) |\psi\rangle = V |\psi\rangle. $$ Here, $ V $ is the interaction potential, and $ H_0 $ is the free particle Hamiltonian.

By rearranging this equation, we obtain: $$ (E - H_0) |\psi\rangle = V |\psi\rangle. $$ Thus: $$ |\psi^\pm\rangle = \frac{1}{E - H_0 \pm i \varepsilon}V |\psi^\pm\rangle. $$ By adding the homogeneous solution, we get: $$ |\psi^\pm\rangle = |\psi_0\rangle + \frac{1}{E - H_0 \pm i \varepsilon} V|\psi^\pm\rangle. $$

Problem:

Now comes my problem. From what I've seen, in order to calculate $$ \langle \mathbf{r}'|\frac{1}{E - H_0 \pm i \varepsilon}|\mathbf{r}\rangle $$ we end up with: $$ \frac{1}{(2\pi)^3} \int \mathrm{d}\mathbf{k}' \langle \mathbf{r} | \frac{1}{E - H_0 \pm i\varepsilon} | \mathbf{k}' \rangle \exp\left(-i \mathbf{k}' \cdot \mathbf{r}'\right). $$

It is said here that we can replace $ E $ by $ \frac{\hbar^2 k^2}{2m} $, and of course $ H_0^{-1} | \mathbf{k}' \rangle = \frac{2m}{\hbar^2 k'^2} | \mathbf{k}' \rangle $.

What I don't understand is why we can replace $ E $ by $ \frac{\hbar^2 k^2}{2m} $. It doesn't make sense to me because $ E $ is the eigenvalue of the whole Schrödinger's equation, not just the free part.

Answer 1

What I don't understand is why we can replace $ E $ by $ \frac{\hbar^2 k^2}{2m} $. It doesn't make sense to me because $ E $ is the eigenvalue of the whole Schrödinger's equation, not just the free part.

If $E$ is positive, and if $m$ is positive (which I assume it is), and if $k$ is real, then you can always write: $$ E = \frac{\hbar^2 k^2}{2m}\;, $$ since $\hbar^2$ is positive and the square of a real number is positive.

You are basically just redefining the variable of interest from $E$ to $k$.