Deriving the Lippmann-Schwinger Equation

Hello! Welcome to the fourth module of our course. We've spent the last several lessons exploring powerful approximation methods for solving the Schrödinger equation, culminating in our study of the WKB method for bound states. Now, we're going to pivot from particles trapped in potentials to particles that are free to move through space and interact with a potential for a finite time. This is the realm of Quantum Scattering Theory.

This topic is absolutely central to modern physics. It's the theoretical framework behind particle accelerator experiments, but it's equally crucial in your field of condensed matter physics, where it describes how electrons scatter off impurities, phonons, or other electrons, determining the transport properties of materials.

Our first step in building this framework is to reformulate the Schrödinger equation in a way that is specially adapted for scattering problems. By the end of this lesson, you will be able to derive the Lippmann-Schwinger equation for scattering states. This integral equation is the cornerstone of formal scattering theory, and we will build upon it for the rest of this module.

1. From a Differential to an Integral Equation

The process of scattering involves an incident particle, described by a plane wave, which interacts with a potential $V(\mathbf{r})$ and then scatters off in some direction. Our starting point is the time-independent Schrödinger equation (TISE):

$$\left(-\frac{{\hslash }^2}{2m}{\nabla }^2+V(\mathbf{r})\right)\psi (\mathbf{r})=E\psi (\mathbf{r})$$

Let's define the free-particle Hamiltonian as $H_0=-\frac{{\hslash }^2}{2m}{\nabla }^2$ and the energy as $E=\frac{{\hslash }^2{k}^2}{2m}$. The TISE becomes:

$$(H_0+V)|\psi ⟩=E|\psi ⟩$$

A simple rearrangement casts the equation in a more suggestive form:

$$(E-H_0)|\psi ⟩=V|\psi ⟩$$

Here, we can think of the right-hand side, $V|\psi ⟩$, as a "source" term. The interaction potential $V$ takes the state $|\psi ⟩$ and creates a source that drives the equation. Without this source term (i.e., if $V=0$), the equation would just be $(E-H_0)|\psi ⟩=0$, whose solutions are free-particle plane waves.

This structure suggests a formal solution. We can "invert" the operator $(E-H_0)$ to get:

$$|\psi ⟩=|\varphi ⟩+\frac{1}{E-H_0}V|\psi ⟩$$

Here, $|\varphi ⟩$ is a solution to the homogeneous equation, $(E-H_0)|\varphi ⟩=0$. Physically, $|\varphi ⟩$ represents the incident wave that exists even without the potential—a plane wave $e^{i\mathbf{k}\cdot \mathbf{r}}$. The second term represents the scattered wave, which is generated by the potential.

This is the abstract form of the Lippmann-Schwinger equation. Notice that $|\psi ⟩$ appears on both sides. This is not an explicit solution but rather an integral equation, which contains the full dynamics of the problem.

The following video provides an excellent, intuitive walkthrough of this reformulation in coordinate space.

2. The Green's Function and Physical Boundary Conditions

The crucial object in our integral equation is the operator $(E-H_0{)}^{-1}$. Its coordinate-space representation, $G(\mathbf{r},{\mathbf{r}}^{\prime })$, is called the Green's function. It is the solution to the differential equation:

$$\left(E-H_0\right)G(\mathbf{r},{\mathbf{r}}^{\prime })=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$$

In our notation with $E={\hslash }^2{k}^2/2m$, this becomes:

$$\left({\nabla }^2+k^2\right)G(\mathbf{r},{\mathbf{r}}^{\prime })=-\frac{2m}{{\hslash }^2}\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$$

The Green's function represents the wave produced at $\mathbf{r}$ by a point source at ${\mathbf{r}}^{\prime }$.

However, the inverse operator $(E-H_0{)}^{-1}$ is not uniquely defined because $H_0$ has a continuous spectrum of eigenvalues, and for $E>0$, the operator $E-H_0$ has a zero eigenvalue. This means the operator isn't invertible on the entire space. This mathematical ambiguity is resolved by physics. For a scattering problem, we must impose the boundary condition that the scattered part of the wave propagates outwards from the scattering center. We are not interested in solutions that describe waves converging on the potential from infinity.

To enforce this condition, we derive the Green's function using a Fourier transform and a clever trick from complex analysis known as the $iϵ$ prescription.

The following reading from David Tong's lecture notes provides a clear and compact derivation.

Let's summarize the key steps of the derivation from the notes:

1. Fourier Transform: We find the momentum-space Green's function, ${\tilde{G}}_0(k;\mathbf{q})$, which is much simpler:
   ${\tilde{G}}_0(k;\mathbf{q})=-\frac{2m}{{\hslash }^2}\frac{1}{q^2-k^2}$
2. Inverse Transform and the Pole Problem: To get the real-space Green's function $G_0(k;\mathbf{x})$, we must compute the inverse Fourier transform, which involves an integral over $\mathbf{q}$. This integral has poles at $q=|\mathbf{q}|=k$, making it ill-defined.
3. The $iϵ$ Prescription: We resolve this by shifting the poles infinitesimally off the real axis. For scattering, we replace $k^2$ with $k^2+iϵ$ in the denominator (this is equivalent to the $k^2-iϵ$ in the notes, depending on the sign convention for the Fourier transform; the physics is the same). This corresponds to replacing the operator $(E-H_0{)}^{-1}$ with $(E-H_0+iϵ{)}^{-1}$ in the formal Lippmann-Schwinger equation.
4. Contour Integration: The integral can now be performed using contour integration. Closing the contour in the upper half of the complex $q$-plane (required for the integral to converge for $|\mathbf{x}|>0$), we pick up the residue from the pole that now lies inside the contour. This procedure specifically selects a solution that behaves like $e^{ik|\mathbf{x}|}$.
5. The Outgoing Green's Function: The final result is the Green's function that generates outgoing spherical waves, denoted $G_0^{+}$:
   $G_0^{+}(k;\mathbf{r}-{\mathbf{r}}^{\prime })=-\frac{2m}{{\hslash }^2}\frac{e^{ik|\mathbf{r}-{\mathbf{r}}^{\prime }|}}{4\pi |\mathbf{r}-{\mathbf{r}}^{\prime }|}$
   The time-dependence of a stationary state is $e^{-iEt/\hslash }$, so this spatial part corresponds to a full wavefunction dependence of $e^{i(k|\mathbf{r}|-\omega t)}$, which clearly describes a wave moving outwards. The alternative choice, $-iϵ$, would have led to an ingoing wave, $e^{-ik|\mathbf{r}-{\mathbf{r}}^{\prime }|}$, which is unphysical for this problem.

For a deeper dive into the mechanics of the contour integration, which leverages your mathematics background, the following video provides an exhaustive and excellent treatment.

3. The Lippmann-Schwinger Equation

Now we have all the pieces. We can write down the final, explicit form of the Lippmann-Schwinger equation by substituting the incident plane wave $|\varphi ⟩\to e^{i\mathbf{k}\cdot \mathbf{r}}$ and the Green's function $G_0^{+}$ back into our integral equation.

The state $|\psi ⟩$ that satisfies the TISE with outgoing-wave boundary conditions is denoted $|{\psi }^{(+)}⟩$. In coordinate space, the equation is:

$${\psi }_{\mathbf{k}}^{(+)}(\mathbf{r})=e^{i\mathbf{k}\cdot \mathbf{r}}+\int d^3{\mathbf{r}}^{\prime }{G}_0^{+}(k;\mathbf{r}-{\mathbf{r}}^{\prime })V({\mathbf{r}}^{\prime }){\psi }_{\mathbf{k}}^{(+)}({\mathbf{r}}^{\prime })$$

Substituting the expression for $G_0^{+}$:

$${\psi }_{\mathbf{k}}^{(+)}(\mathbf{r})=e^{i\mathbf{k}\cdot \mathbf{r}}-\frac{2m}{4\pi {\hslash }^2}\int d^3{\mathbf{r}}^{\prime }\frac{e^{ik|\mathbf{r}-{\mathbf{r}}^{\prime }|}}{|\mathbf{r}-{\mathbf{r}}^{\prime }|}V({\mathbf{r}}^{\prime }){\psi }_{\mathbf{k}}^{(+)}({\mathbf{r}}^{\prime })$$

This is the Lippmann-Schwinger equation. It is an exact reformulation of the time-independent Schrödinger equation, with the crucial physical boundary conditions for a scattering experiment built right in.

In the more abstract bra-ket notation, it is written as:

$$|{\psi }^{(+)}⟩=|\mathbf{k}⟩+\frac{1}{E-H_0+iϵ}V|{\psi }^{(+)}⟩$$

where $|\mathbf{k}⟩$ is the incident plane wave state.

Test your understanding!

Let's check the consistency of our derivation. Act on the Lippmann-Schwinger equation in its coordinate-space form with the operator $\left({\nabla }^2+k^2\right)$. Show that you recover the original Schrödinger equation, $\left({\nabla }^2+k^2\right)\psi (\mathbf{r})=\frac{2m}{{\hslash }^2}V(\mathbf{r})\psi (\mathbf{r})$.

(Hint: Remember the defining property of the Green's function.)

Show answer

Let's apply the operator $\left({\nabla }^2+k^2\right)$ to both sides of the Lippmann-Schwinger equation:

$$\left({\nabla }^2+k^2\right){\psi }_{\mathbf{k}}^{(+)}(\mathbf{r})=\left({\nabla }^2+k^2\right)e^{i\mathbf{k}\cdot \mathbf{r}}+\left({\nabla }^2+k^2\right)\int d^3{\mathbf{r}}^{\prime }{G}_0^{+}(k;\mathbf{r}-{\mathbf{r}}^{\prime })V({\mathbf{r}}^{\prime }){\psi }_{\mathbf{k}}^{(+)}({\mathbf{r}}^{\prime })$$

1. First term: The plane wave is a solution to the free-particle equation, so ${\nabla }^2{e}^{i\mathbf{k}\cdot \mathbf{r}}=-k^2{e}^{i\mathbf{k}\cdot \mathbf{r}}$. Thus, $\left({\nabla }^2+k^2\right)e^{i\mathbf{k}\cdot \mathbf{r}}=0$.

2. Second term: The operator ${\nabla }^2$ acts on the $\mathbf{r}$ coordinate, not ${\mathbf{r}}^{\prime }$, so we can bring it inside the integral:

   $$\int d^3{\mathbf{r}}^{\prime }\left[\left({\nabla }^2+k^2\right)G_0^{+}(k;\mathbf{r}-{\mathbf{r}}^{\prime })\right]V({\mathbf{r}}^{\prime }){\psi }_{\mathbf{k}}^{(+)}({\mathbf{r}}^{\prime })$$

   By definition, the Green's function satisfies $\left({\nabla }^2+k^2\right)G_0^{+}(k;\mathbf{r}-{\mathbf{r}}^{\prime })=-\frac{2m}{{\hslash }^2}\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$. Substituting this in:

   $$\int d^3{\mathbf{r}}^{\prime }\left[-\frac{2m}{{\hslash }^2}\delta (\mathbf{r}-{\mathbf{r}}^{\prime })\right]V({\mathbf{r}}^{\prime }){\psi }_{\mathbf{k}}^{(+)}({\mathbf{r}}^{\prime })$$

3. Evaluate the integral: The delta function makes the integral trivial. It collapses when ${\mathbf{r}}^{\prime }=\mathbf{r}$:

   $$-\frac{2m}{{\hslash }^2}V(\mathbf{r}){\psi }_{\mathbf{k}}^{(+)}(\mathbf{r})$$

Putting it all together, we get:

$$\left({\nabla }^2+k^2\right){\psi }_{\mathbf{k}}^{(+)}(\mathbf{r})=0-\left(-\frac{2m}{{\hslash }^2}V(\mathbf{r}){\psi }_{\mathbf{k}}^{(+)}(\mathbf{r})\right)$$

Multiplying by $-{\hslash }^2/2m$ would recover the original form $(H_0+V)\psi =E\psi$. The integral equation is indeed equivalent to the original differential equation.

Conclusion

In this lesson, we have taken a major step into the formal theory of scattering. We have transformed the time-independent Schrödinger equation, a differential equation, into the Lippmann-Schwinger equation, an integral equation that automatically incorporates the physics of a scattering experiment.

Key Takeaways:

• The Schrödinger equation for a scattering problem can be written as $(E-H_0)|\psi ⟩=V|\psi ⟩$, where the potential term acts as a source.
• This can be formally solved as an integral equation: $|\psi ⟩=|\varphi ⟩+(E-H_0{)}^{-1}V|\psi ⟩$, where $|\varphi ⟩$ is the incident wave.
• The operator $(E-H_0{)}^{-1}$ is the Green's function. Its ambiguity is resolved by imposing the physical boundary condition of purely outgoing scattered waves.
• This boundary condition is implemented via the $iϵ$ prescription in momentum space, which leads to the outgoing Green's function $G_0^{+}$.
• The final Lippmann-Schwinger equation provides an exact relationship between the incident wave, the potential, and the full scattered wavefunction.

Preview of the Next Lesson:
The Lippmann-Schwinger equation is exact, but it's still not an explicit solution. How do we solve it? Its true power lies in its suitability for iterative solutions. In our next lesson, we will develop this iterative solution, known as the Born series. The first and most important term of this series gives the famous Born approximation, which provides a direct way to calculate the scattering amplitude and cross-section for weak potentials. This will be our first practical tool for analyzing realistic scattering problems.