advanced35 min studyLesson 395 in Physics

Lippmann–Schwinger resolvent equation

Encode outgoing boundary conditions in the free resolvent and distinguish an exact integral equation from its Born iteration.

Before you begin

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The question for today

What you’ll understand

Derive the scattering-state integral equation from Schrödinger dynamics.
Interpret the outgoing +i0 prescription and its imaginary part.
Solve a finite-rank resummation without claiming a universal Born convergence test.

The stationary Schrödinger equation is a differential equation plus boundary conditions. The Lippmann–Schwinger equation packages both into an integral equation: a free incident state plus the waves generated by the potential acting on the complete unknown state. Its exactness does not make it an explicit solution, because the unknown appears on both sides. It does, however, expose repeated scattering and supply the starting point for Born expansions and operator methods.

Invert the free operator with a boundary prescription

Split the Hamiltonian into free kinetic energy and an interaction. Rearranging the stationary equation gives the free energy operator acting on the scattering state equal to the potential acting on that same state. Formally inverting the free operator is ambiguous on the continuous spectrum, because free solutions exist at the scattering energy. Add the specified incident homogeneous solution and select a limiting resolvent whose infinitesimal imaginary shift chooses outgoing waves.

Use time dependence $e^{-iEt/\hbar}$ . The outgoing free resolvent is the limit from positive imaginary energy. In three spatial dimensions its coordinate kernel has a negative mass prefactor times an outgoing spherical wave. Changing the sign of the infinitesimal prescription selects the opposite asymptotic radiation condition. The symbol is not a small fitted absorption in the potential; it specifies how a singular operator is approached and survives as a distribution when the artificial regulator is removed.

|\psi^{(+)}\rangle=|\mathbf k\rangle+G_0^{(+)}(E)V|\psi^{(+)}\rangle,\quad G_0^{(+)}=(E-H_0+i0)^{-1},\quad G_0^{(+)}(\mathbf r,\mathbf r^\prime)=-\frac{\mu}{2\pi\hbar^2}\frac{e^{ik|\mathbf r-\mathbf r^\prime|}}{|\mathbf r-\mathbf r^\prime|}

G0 as an operator has inverse-energy units; its coordinate kernel has inverse energy per volume. The kernel and +i0 prescription use the outgoing eⁱᵏʳ convention and a free nonrelativistic reduced mass μ.↔ Scroll wide equations horizontally.

Iterate without mistaking the iteration for the equation

Insert the integral equation into itself. The first correction contains one potential insertion, the next two, and so on, linked by free propagation. This generates the Born series. The exact integral equation can remain meaningful even when that particular series fails to converge. A strong resonance is a familiar example: a denominator associated with repeated scattering can become small, so no short truncation of its geometric-like expansion captures the result.

The transition operator collects the interaction insertions before they are projected onto chosen incoming and outgoing states. It satisfies its own integral equation and has an imaginary part even for a Hermitian potential because the outgoing free propagator carries open-channel information. The operator identity decomposing the resolvent into a principal value and an energy-shell delta function identifies that source of imaginary response. It is the same mechanism that allows an order-consistent optical theorem to emerge from real interactions.

T=V+VG_0^{(+)}T=V+VG_0^{(+)}V+\cdots,\qquad \frac1{x+i0}=\mathcal P\frac1x-i\pi\delta(x)

T has energy units as an abstract operator; matrix elements depend on state normalization. The distribution identity acts inside appropriate integrals. Its negative imaginary resolvent part is consistent with the stated outgoing convention.↔ Scroll wide equations horizontally.

Use a separable model to see the denominator

A rank-one interaction provides an exactly resumable illustration. Choose a normalized form-factor state and let the potential be a strength multiplying its projector. The transition operator has the same projector structure, leaving one scalar equation for its energy-dependent coefficient. The projected free resolvent contains the propagation through that form factor. Solving the scalar equation displays directly where a pole can occur and how a finite-order Born polynomial would miss it.

The form factor determines the energy dependence and ultraviolet behavior; it cannot be omitted when claiming a physical potential or cross section. A complex projected resolvent at positive energy encodes escape into the continuum, whereas a real projected resolvent below threshold can support a bound-state denominator zero. The worked calculation treats a specified projected value at one energy, without claiming it constructs a complete spatial potential. The geometric-series comparison in the lab is an analogy for iteration only, not an operator inverse or a scattering-state solver.

V=\lambda|g\rangle\langle g|,\quad T=|g\rangle\tau(E)\langle g|,\quad \tau=\frac{\lambda}{1-\lambda G_g(E)},\quad G_g=\langle g|G_0^{(+)}|g\rangle

The state |g⟩ is normalized here, λ and τ have energy units, and Gg has inverse-energy units. This rank-one exact result does not imply every potential has a scalar denominator or a convergent Born series.↔ Scroll wide equations horizontally.

Work it through

Resum a rank-one interaction at one energy

For a normalized rank-one form factor, suppose $\lambda=2.0\,\mathrm{eV}$ and $G_g=(0.25-0.25i)\,\mathrm{eV^{-1}}$ at the selected positive energy. Find the exact scalar transition coefficient and compare it with the first Born coefficient.

The dimensionless product is $\lambda G_g=0.50-0.50i$ . Therefore the repeated-scattering denominator is $1-\lambda G_g=0.50+0.50i$ , not its complex conjugate.
Rationalize $\tau=2/(0.50+0.50i)$ to get $2(0.50-0.50i)/0.50=(2-2i)\,\mathrm{eV}$ . The coefficient has the same energy units as the interaction strength.
The first Born truncation is just $\tau_B=\lambda=2.0\,\mathrm{eV}$ . It misses the imaginary component and the exact magnitude $2\sqrt2\,\mathrm{eV}$ , although it happens to match the real part at this one chosen energy.

The result

The resummed coefficient is τ = (2 − 2i) eV, while the first Born coefficient is 2 eV. No physical cross section follows without the form factors and flux normalization.

The living notebook

Explore the model

Interactive lab

Interactive lab · optimization

A limited iteration analogy

Reset and set BOTH Step size η = 0.50 and Initial position x₀ = 2; predict xₙ = 2(0.5)ⁿ. Reset and set η = 2.20 and x₀ = 2; predict alternating growth with ratio −1.20. Step each fresh run. This is gradient descent on x²/2, not Born iteration, a resolvent or a scattering solver. It illustrates only that an iteration can fail even when an underlying equation has a finite solution.

ObjectiveDiscrete iterate

Position x0: 2
Objective x²/2: 2
Iteration behavior: Stable step

Step size η0.3Initial position x₀2

n = 0 · t = 0 s

Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe convergence of a chosen iteration is a separate issue from existence of the equation’s solution.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The limiting inverse of E − H0 + iε as ε decreases to zero selects outgoing eⁱᵏʳ behavior with e⁻ⁱᴱᵗ/ℏ. It is a boundary prescription, not a finite material loss.

Your turn

Try it yourself

As a scalar repeated-scattering analogy, solve $X=1+cX$ for $c=0.20$ . Compare the exact answer with the sum $1+c+c^2$ . Then determine why the same partial sums fail for $c=1.20$ even though the algebraic equation has a solution.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Rearrange the equation before discussing its geometric expansion. The geometric series requires the magnitude of c to be below one.

Compare with the solution

For $c=0.20$ , the exact value is $X=1/(1-0.20)=1.25$ , while the three-term sum is $1+0.20+0.04=1.24$ , short by $0.01$ . For $c=1.20$ , the algebraic solution is $X=-5$ , but the positive terms of $\sum c^n$ grow and do not converge to it. This does not prove that every scattering resolvent is equivalent to this scalar equation. It isolates the logical distinction between solving an equation and obtaining its solution from one proposed perturbation series.

Keep the idea

Take it with you

The outgoing resolvent implements boundary data as well as an inverse free operator.
Lippmann–Schwinger is exact, while truncating its Born iteration is an approximation.
Finite-rank models expose poles but do not replace full spatial and flux information.

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