advanced30 min studyLesson 931 in Physics

Discretizing Schrödinger operators and boundaries

Construct a finite Hamiltonian while keeping normalization, boundary conditions and dispersion errors explicit.

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What you’ll understand

  • Build a Hermitian finite-difference Hamiltonian with declared endpoint conditions.
  • Distinguish grid amplitudes from sampled probability density.
  • Compare a discrete eigenvalue or group speed with its continuum counterpart.

A computer evolves finitely many numbers, whereas a particle wavefunction on a line belongs to an infinite-dimensional space with a specified operator domain. Discretization chooses how those descriptions are connected. Grid spacing, box size, endpoint conditions and normalization are therefore part of the modeled problem. A matrix can be perfectly Hermitian and its propagation exactly unitary while still giving an inaccurate approximation to the continuum system one intended to study.

Specify the operator and the finite state norm

Use the nonrelativistic one-dimensional Hamiltonian with kinetic term minus hbar squared over twice the mass times the second derivative, plus a real potential. A centered uniform-grid stencil produces a tridiagonal matrix. Let a be the spatial spacing and define epsilon as the corresponding kinetic energy scale. For a uniform quadrature, use coefficients c-j equal to the sampled wavefunction multiplied by the square root of a, so their squared magnitudes sum to the discrete probability norm.

This coefficient convention makes ordinary matrix inner products match the chosen quadrature. Other bases, nonuniform grids or finite elements can introduce weights or an overlap matrix, which must then enter normalization and Hermiticity checks. A real symmetric stencil is Hermitian in its specified inner product; writing the same entries on an unrelated weighted grid does not preserve that property automatically. State the finite representation before using its eigenvectors as physical states.

ϵ=ℏ22ma2,(Hc)j=ϵ(2cj−cj−1−cj+1)+Vjcj,cj=a ψ(xj),∥c∥2=∑j∣cj∣2\epsilon=\frac{\hbar^2}{2ma^2},\qquad (Hc)_j=\epsilon(2c_j-c_{j-1}-c_{j+1})+V_jc_j,\qquad c_j=\sqrt a\,\psi(x_j),\quad \|c\|^2=\sum_j|c_j|^2
Epsilon has energy units, a has length units and c is dimensionless. The norm is the adopted uniform-grid quadrature representation, not an assertion that point samples alone determine every continuum feature.↔ Scroll wide equations horizontally.

Declare what happens at the endpoints

Homogeneous Dirichlet conditions set the endpoint wavefunction to zero and produce a hard-wall problem. Periodic conditions instead connect the last grid point to the first and model a ring. These choices change the operator and its spectrum; they are not interchangeable implementation details. On a finite box meant to approximate a localized bound state on the full line, moving the walls farther away is a separate convergence control from decreasing the grid spacing.

The finite matrix also has its own kinetic dispersion. Applying the centered stencil to a grid plane wave gives an energy proportional to two minus twice the cosine of k times a. At small k times a it approximates the continuum quadratic energy, but near the grid scale the discrepancy is large. A stable wavepacket can consequently travel at the wrong group velocity. Norm conservation cannot reveal this phase and transport error by itself.

Eh(k)=2ϵ[1−cos⁡(ka)],E(k)=ℏ2k22m,vh(k)=1ℏdEhdk=ℏmasin⁡(ka)E_h(k)=2\epsilon[1-\cos(ka)],\qquad E(k)=\frac{\hbar^2k^2}{2m},\qquad v_h(k)=\frac1\hbar\frac{dE_h}{dk}=\frac{\hbar}{ma}\sin(ka)
The plane-wave identity applies to a uniform periodic or infinite lattice. For low k the centered stencil is second order; the continuum velocity is hbar k divided by mass.↔ Scroll wide equations horizontally.

Separate matrix verification from continuum approximation

Multiplying a tiny Hamiltonian by a proposed eigenvector checks signs, indexing and endpoint handling directly. A zero matrix residual confirms an eigenpair of that matrix, not an exact continuum eigenpair. The same state can have an appreciable energy bias relative to a continuum box. Refining a while keeping the physical domain fixed tests the spatial approximation; simply adding more points while also moving the walls changes two controls at once.

The Hilbert-space dimension, represented momentum band and physical potential resolution must all be adequate for the observables of interest. A sharp barrier or rapidly varying state may need more resolution than a smooth low-energy state. Closed finite Hermitian evolution is unitary, whereas absorbing boundaries and open-system equations will deliberately use different norm or density-matrix rules later. Preserving those distinctions prevents every loss of norm from being called numerical failure and every preserved norm from being called physical success.

rE=Hc−Ec,∥rE∥=0 verifies the stated finite-matrix eigenpair onlyr_E=Hc-Ec,\qquad \|r_E\|=0\ \text{verifies the stated finite-matrix eigenpair only}
The residual has energy units for normalized c. Continuum domain, boundary and grid errors remain even if this algebraic residual vanishes.↔ Scroll wide equations horizontally.

An exact eigenvector of a three-site box

Use three interior points between Dirichlet walls at zero and 4a4a, with V zero. The finite Hamiltonian is epsilon times the tridiagonal matrix with diagonal two and neighboring entries minus one. Test c=(1,0,−1)/2c=(1,0,-1)/\sqrt2 and compare its energy with the continuum second box level.

  1. The coefficient norm is one. Multiplication by the dimensionless matrix gives (2,0,−2)/2=2c(2,0,-2)/\sqrt2=2c, so the matrix eigenvalue is 2ϵ2\epsilon and its residual is zero.

  2. The corresponding continuum second-level wavenumber is 2π/(4a)=π/(2a)2\pi/(4a)=\pi/(2a). Its energy is ϵπ2/4≈2.467ϵ\epsilon\pi^2/4\approx2.467\epsilon, larger than the finite-matrix value.

  3. The discrepancy is approximately 0.467ϵ0.467\epsilon despite the exact finite eigenpair. The coarse grid resolves too little of the continuum derivative; normalization and a zero matrix residual cannot remove that spatial approximation error.

The result

The finite eigenvalue is twice epsilon, while the matched continuum level is about 2.467 epsilon; exact matrix arithmetic still leaves a grid bias.

Explore the model

Interactive lab

Interactive lab · function

A kinetic scale changes quadratically

Reset and set coefficient one, position one; predict output one. Compare a fresh Reset run at coefficient one, position two; predict output four. This quadratic scaling can represent doubling inverse spacing in the kinetic scale, but the lab does not build a Hamiltonian, apply boundaries, or calculate quantum eigenstates.

A kinetic scale changes quadraticallyThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 inverse-spacing-squared scale makes grid choices important, while the quadratic instrument is only an algebraic analogy.

Check your understanding

A normalized finite-grid eigenvector has exactly zero matrix residual. What can still be wrong with its continuum interpretation?

Try it yourself

For the stated lattice dispersion choose ka=π/2ka=\pi/2. Compare its group velocity with the continuum value at the same k, and compute their ratio. Does unitary finite-matrix propagation fix the difference?

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

Give me a hint

The sine is one, while continuum k is pi divided by twice a. Both velocities have the same hbar-over-mass-spacing scale.

Compare with the solution

The lattice velocity is vh=ℏ/(ma)v_h=\hbar/(ma), while continuum velocity is v=πℏ/(2ma)v=\pi\hbar/(2ma). Their ratio is 2/π≈0.6372/\pi\approx0.637. Thus this grid-scale component moves too slowly in the centered-stencil model. An exactly unitary propagator preserves the norm for that model but does not change its dispersion. Improving the continuum approximation requires resolving smaller k times a or using a different spatial representation with its own checked limitations.

Take it with you

  • The finite inner product and boundaries define the matrix problem.
  • A zero algebraic residual does not remove continuum grid error.
  • Kinetic dispersion can bias phase and packet velocity while norm remains conserved.

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