advanced30 min studyLesson 932 in Physics

Shooting and matching for bound eigenstates

Turn endpoint consistency into an energy root and distinguish genuine matching from numerical contamination.

On this page

What you’ll understand

  • Construct a shooting mismatch for a finite recurrence.
  • Match decaying bound-state solutions with a declared logarithmic derivative.
  • Separate root tolerance, integration error and boundary truncation.

Bound-state energies are selected by compatibility across an entire spatial domain, rather than by an arbitrary local initial amplitude. Shooting methods integrate a trial-energy solution and measure its failure to satisfy the other boundary. Matching methods compare solutions propagated from opposite ends. Both reduce a differential eigenproblem to a root search, but a small root residual is meaningful only when the underlying integration and boundary conditions accurately represent the intended quantum problem.

Use linearity without accepting the zero solution

For a one-dimensional real potential, the time-independent Schrödinger equation is second order. Fix one boundary condition and choose a nonzero starting slope or neighboring value to set an arbitrary amplitude scale. Linearity permits normalization afterward. If both the wavefunction and its derivative are set to zero at a regular initial point, the resulting identically zero solution supplies no eigenstate. Vary energy instead of mistaking a change of amplitude for a new bound level.

A finite-difference recurrence makes this logic explicit. With zero potential and homogeneous Dirichlet endpoints, the dimensionless trial energy determines each next sample from two previous samples. The final sample is a mismatch function. Its zeros select eigenvalues of that discrete boundary problem. For smooth continuum shooting, Runge-Kutta or Numerov integration has separate step error; the recurrence below is deliberately only the simplest centered finite stencil, with no claim of Numerov accuracy.

yj+1=(2−λ)yj−yj−1,λ=E/ϵ,y0=0, y1=1,F(λ)=yN+1y_{j+1}=(2-\lambda)y_j-y_{j-1},\qquad \lambda=E/\epsilon,\qquad y_0=0,\ y_1=1,\qquad F(\lambda)=y_{N+1}
Epsilon is hbar squared over twice mass times spacing squared. The arbitrary nonzero y-one sets scale; zeros of F enforce the far Dirichlet condition for this finite recurrence.↔ Scroll wide equations horizontally.

Match logarithmic derivatives with their domain restrictions

For a localized state on an unbounded line, the physical exterior solution decays. Outward integration can amplify a tiny growing exponential contamination and overwhelm that solution. Integrating inward from a sufficiently distant asymptotic boundary and matching to an interior solution can improve conditioning. One convenient mismatch compares logarithmic derivatives, eliminating the unknown amplitude factor when neither matched wavefunction vanishes.

At a node the logarithmic derivative has a pole, so a sign change across that pole is not automatically an energy root. A Wronskian or another nonsingular matching condition can avoid that particular division. Finite jumps of a finite potential require continuity of wavefunction and derivative, whereas an infinite hard wall or a delta potential has different matching rules. Using a remembered condition without checking the operator's domain can therefore manufacture or miss levels.

Flog⁡(E)=yL′(xm;E)yL(xm;E)−yR′(xm;E)yR(xm;E),yL(xm)yR(xm)≠0F_{\log}(E)=\frac{y_L'(x_m;E)}{y_L(x_m;E)}-\frac{y_R'(x_m;E)}{y_R(x_m;E)},\qquad y_L(x_m)y_R(x_m)\ne0
The logarithmic mismatch has inverse-length units. Its definition excludes nodes at the matching point and assumes both propagated solutions belong to the same differential equation.↔ Scroll wide equations horizontally.

Identify a level before claiming convergence

In regular one-dimensional bound-state problems, node counts and symmetry help distinguish neighboring eigenstates. A numerical search should preserve its intended branch as spacing, matching point and outer boundary are changed. A root finder can converge tightly to the wrong level, to a pole of the mismatch, or to the spectrum of an overly small box. Report which state was tracked and how its energy and normalized shape respond to independent controls.

The finite square well offers an exact matching benchmark. An even solution is a cosine inside and a decaying exponential outside. Equating their logarithmic derivatives at the positive edge gives k times tangent of k times the half-width equal to the exterior decay constant. This relation is not an empirical fitting rule: it follows directly from derivative continuity and exposes the sign of the exponentially decaying tail. Its singular tangent points still require the same bracketing care as other logarithmic mismatches.

ktan⁡(kb)=κ,k=2m(E+V0)ℏ,κ=−2mEℏ,−V0<E<0k\tan(kb)=\kappa,\qquad k=\frac{\sqrt{2m(E+V_0)}}{\hbar},\quad \kappa=\frac{\sqrt{-2mE}}{\hbar},\quad -V_0<E<0
The potential is minus V-zero for absolute x less than b and zero outside. This displayed condition selects even states; odd states obey a different matching relation.↔ Scroll wide equations horizontally.

Two eigenvalues from a three-step recurrence

For two interior grid values between Dirichlet endpoints, use y0=0,y1=1y_0=0,y_1=1 and the stated free recurrence. Compute the endpoint mismatch y-three, find its energy roots, and give the normalized interior vectors.

  1. The next value is y2=2−λy_2=2-\lambda. A further step gives y3=(2−λ)2−1y_3=(2-\lambda)^2-1. The far boundary requires this polynomial to vanish.

  2. The roots are 2−λ=12-\lambda=1 and minus one, hence λ=1,3\lambda=1,3. The interior samples are respectively (1,1)(1,1) and (1,−1)(1,-1).

  3. Each vector has squared norm two before normalization, so divide by square root two. The energies are epsilon and three epsilon for this finite two-interior-point problem; a tight polynomial root does not make them exact continuum box levels.

The result

The finite shooting roots give energies epsilon and three epsilon, with symmetric and antisymmetric normalized two-site states.

Explore the model

Interactive lab

Interactive lab · root-finding

A bracket solves its specified scalar equation

Reset and set target one, upper bound two; predict the positive root one after one Step. Compare a fresh Reset run at target four, upper bound two; predict endpoint root two and immediate zero bracket bound. This lab solves only x squared minus target on a positive interval; it has no energy-dependent wavefunction, node count or bound-state mismatch.

A bracket solves its specified scalar equationBisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. Current readings: Midpoint after 0 refinements: 1.5; Bracket error bound: 1.5; Computed residual · x² − c: 0.25. All parameters are adjustable with the labeled controls below.Bisection on x squared minus 2: bracket from 0 to 3 after 0 refinements.00.751.52.253−2.360.192.745.297.84x · bracket stays nonnegativef(x) = x² − cn = 0 · [ 0, 3 ]
Quadratic / midpointCurrent bracketPositive root
Midpoint after 0 refinements
1.5
Bracket error bound
1.5
Computed residual · x² − c
0.25
t = 0 s

Bisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. 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 forA converged scalar root must still be linked to the correct eigenproblem and branch.

Check your understanding

Why can a logarithmic-derivative mismatch change sign without identifying a bound eigenstate?

Try it yourself

Let ϵL=ℏ2/(2mL2)\epsilon_L=\hbar^2/(2mL^2) and take a finite well of depth 2ϵL2\epsilon_L and half-width b=πL/4b=\pi L/4. Test the even trial energy E=−ϵLE=-\epsilon_L by comparing interior and exterior logarithmic derivatives at b. Is the tail normalizable?

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

Give me a hint

Both the oscillatory interior wavenumber and the positive exterior decay constant follow from energy differences. The tangent is one at pi over four.

Compare with the solution

The interior kinetic energy is epsilon-L and the exterior energy is negative epsilon-L, giving k=κ=1/Lk=\kappa=1/L. Therefore kb=π/4kb=\pi/4 and the interior logarithmic derivative is −ktan⁡(kb)=−1/L-k\tan(kb)=-1/L, equal to the decaying exterior value −κ-\kappa. Continuity can choose the exterior amplitude, and its exponential decay makes the state normalizable. This verifies one even matching solution for the stipulated well; an overall normalization still requires integrating the complete squared wavefunction.

Take it with you

  • Shooting adjusts energy to meet a second boundary, with arbitrary nonzero amplitude scale.
  • Logarithmic matching is invalid at its division poles.
  • Root, integration and boundary errors must be controlled independently.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.