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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?

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2 Answers 2

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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
    Commented Jul 17, 2015 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
    Commented Jul 17, 2015 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
    Commented Jan 11, 2019 at 11:16
  • $\begingroup$ I do not believe in this answer. $\endgroup$
    – user151522
    Commented Jan 19 at 15:43
  • $\begingroup$ This answer, has begun to look, almost like nonsense! $\endgroup$
    – user151522
    Commented Jan 20 at 14:14
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For a proof, of what might be thought of as, the naive form of the Lippmann-Schwinger equation, see at https://math.stackexchange.com/a/4850513/553318

Other Information

Sakurai$^1$ Section 7.1, 'The Lippmann-Schwinger Equation', discusses this naive form of the equation, he say's that on it's own it has no meaning, but that this may be addressed by making $E$ slightly complex.

He then gives the Lippmann-Schwinger equation in the form

\begin{equation*} |\psi^{(\pm )} \rangle =|\phi \rangle+\frac{ 1 }{ E-H_0\pm i\epsilon }V|\psi^{(\pm )}\rangle \tag{7.1.6} \end{equation*}


I would like to explain why I do not believe in dolun's answer, dated July 17, 2015.

The answer of dolum, dated July 17, 2015, uses the idea expressed in the word equation, given below, but I don’t think this idea should be used here.

\begin{equation*} \begin{pmatrix} { A ~solution ~to ~ an~ inhomogenous~ differential~ equation } \end{pmatrix}= \begin{pmatrix} { A~ solution~ to~the~ associated~ homogenous~differential ~equation }\end{pmatrix}+\begin{pmatrix} { A~ particular ~solution~ to~ the~ inhomogenous ~differential ~equation }\end{pmatrix}\end{equation*}

The idea makes sense when we have a linear operator, $L$ say, and the following situation,

\begin{align*} L\phi_h(x)&=0 \\ L\psi_p(x)&=f(x) \end{align*} but we have, with $(E-H_0)=L$, $V=f(x)$, \begin{align*} L\phi_h(x)&=0 \\ L\psi_p(x)&=f(x)\psi_p(x) \end{align*}

I am assuming one spatial dimension in the above.


To be explicit.

When we have the "inhomogenous" equation, (1), given below \begin{equation*} (E-H_0)\psi =V\psi \tag{1} \end{equation*} and we know for non-zero $\phi_h$ and $\psi_p$ \begin{align} (E-H_0)\phi_h &=0 \tag{2}\\ (E-H_0)\psi_p &=V\psi_p \tag{3} \end{align} $[ \phi_h+\psi_p]$ does not satisfy ($\mathbf{1}$)!

You can see this by adding ($\mathbf{2}$) and ($\mathbf{3}$), note that $(E-H_0)$ is a linear operator, giving

\begin{equation*} (E-H_0)[ \phi_h+\psi_p] =V\psi_p \end{equation*} so ($\mathbf{1}$) is not satisfied.

What would be required for ($\mathbf{1}$) to be satisfied, is \begin{equation*} (E-H_0)[ \phi_h+\psi_p ] =V[ \phi_h+\psi_p ] \end{equation*}

Reference:

1, J.J.Sakurai, Ed San Fu Tuan, Modern Quantum Mechanics Revised Edition, Addison-Wesley Publishing Company (1994).

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    $\begingroup$ Thanks. Your point about dolun's answer makes sense. I recall reading Sakurai's explanation and didn't get the "need to make E slightly complex" part. Sounds like hand waving. Do you understand it? $\endgroup$
    – Minethlos
    Commented Jan 19 at 21:36
  • $\begingroup$ @Minethlos I guess it will make, at least some expressions in the theory well defined, in terms of taking limits as $\epsilon~ \to 0$. $\endgroup$
    – user151522
    Commented Jan 20 at 11:56
  • $\begingroup$ For a proof, of what you might think of as the naive form of the Lippmann-Schwinger equation, see at math.stackexchange.com/a/4850513/553318 $\endgroup$
    – user151522
    Commented Jan 24 at 19:11

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