I'd like to know the derivation of Lippmann-Schwinger equation (LSE) in operator formalism and on what assumptions it is based. I consulted the Ballentine book as advised in this Phys.SE post, but I still don't understand it.

It states that if LSE holds: $$ |\psi\rangle = |\phi\rangle + G_0(E^+) V|\psi\rangle \tag{1}$$

then $H|\psi\rangle = E|\psi\rangle$ or alternatively $$ V|\psi\rangle = (E-H_0)|\psi\rangle \tag{2}$$

Here $H_0 |\phi\rangle = E |\phi\rangle$ and $G_0(E^+)$ signifies a limit $G_0(E+i\eta)$ as $\eta \to 0$.

Now, it is easy to verify that $(1) \Rightarrow (2)$. But, as the book states, $(1)$ contains more information and I think that $(2) \nRightarrow (1)$. So how is LSE truly derived? If we consider problem $(2)$ then obviously $ |\psi\rangle = G_0(E^+) V|\psi\rangle $ is also a solution. Why do we need $|\phi\rangle$?

And lastly how is $|\psi\rangle$ defined? If it is defined by $(2)$ then why does it have the same eigenvalues as $|\phi\rangle$ has for $H_0$? Why is it even true that the spectral problem $(2)$ has a solution?


Most of the time, an (elastic) scattering problem can be reduced in :

  • An incoming initial wave / quantum state $|\phi\rangle$, which most of the time is taken to be a plane wave / free state $|\textbf{k}\rangle$ eigenstate of the free hamiltonian : $$ \hat{\mathcal{H}}_0\,|\phi\rangle=E\,|\phi\rangle\quad\text{with}\quad\hat{\mathcal{H}}_0=-\frac{\Delta}{2} $$

  • A scattering potential $\hat{V}(\hat{\textbf{x}})$, which a priori can be anything (spherically symmetric, disordered, etc).

Obviously, in that case $|\phi\rangle$ is no longer an eigenstate of the full hamiltonian : $$ \hat{\mathcal{H}}=\hat{\mathcal{H}}_0+\hat{V} $$

Of course what you want is to find such eigenstate that you noted $|\psi\rangle$, so that : $$ \hat{\mathcal{H}}\,|\psi\rangle=E\,|\psi\rangle\quad\text{i.e.}\quad (E-\hat{\mathcal{H}}_0)|\psi\rangle=\hat{V}|\psi\rangle $$ Such equation is nothing but a differential equation. To solve it, you have first to find the homogene solution, i.e. without the second member $\hat{V}|\psi\rangle$. It's easy to see that the homogene solution $|\psi\rangle_{h}$ is $|\phi\rangle$.

The particular solution $|\psi\rangle_p$ can be foud by playing a little bit with the formulas. By definition of the free (retarded) Green function $$ \hat{G}_0(\epsilon)=\frac{1}{\epsilon-\hat{\mathcal{H}}_0+\mathrm{i}\eta} $$ you have : $$ \lim_{\eta\rightarrow 0}\hat{G}_0^{-1}(\epsilon)=E-\hat{\mathcal{H}}_0=\hat{G}_0^{-1}(E)\quad\text{with}\quad E=\lim_{\eta\rightarrow 0}\epsilon+\mathrm{i}\eta $$ Then follows $$ \hat{G}_0^{-1}(E)|\psi\rangle=\hat{V}|\psi\rangle\quad\text{i.e.}\quad|\psi\rangle=\hat{G}_0(E)\hat{V}|\psi\rangle=|\psi\rangle_p $$

As always, the general solution of a differential equation is the sum of the homogene and particular solutions : $$ |\psi\rangle=|\psi\rangle_h+|\psi\rangle_p=|\phi\rangle+\hat{G}_0(E)\hat{V}|\psi\rangle $$ which is the so called Lippmann Schwinger equation.

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    $\begingroup$ So why does $| \psi \rangle$ and $| \phi \rangle$ have the same eigenenergies $E$? $\endgroup$ – Minethlos Jul 17 '15 at 11:13
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    $\begingroup$ Because such derivation is only valid for elastic scattering, i.e. the energy $E$ is conserved through the all process. $\endgroup$ – dolun Jul 17 '15 at 11:19
  • $\begingroup$ @dolun: Excellent answer. But if you compare $(E-H_0)|\psi\rangle = V|\psi\rangle$ in position space with the form of a nonhomogeneous ODE, then they don't actually match. $\endgroup$ – rainman Jan 11 '19 at 11:16

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