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Boundary Conditions Table

Boundary conditions complete a wave-mechanics model. A differential expression such as −ℏ2d2/(2m dx2)-\hbar^2d^2/(2m\,dx^2) becomes a physical Hamiltonian only after the allowed wavefunctions, endpoints, matching rules, and asymptotic conditions are specified.

This table is a quick diagnostic reference. The canonical explanation is Boundary Conditions; the rows below summarize the conditions most often used by the model pages in this volume.

SituationStandard conditionPhysical roleUsed inCommon mistake
Infinite wallψ=0\psi=0 at the wall; for a box, ψ(0)=ψ(L)=0\psi(0)=\psi(L)=0Excludes the particle from the forbidden region and gives zero flux through the wallInfinite Square Well, Three-Dimensional BoxImposing ψ=0\psi=0 at a finite wall
Finite step or finite barrierψ(x0−)=ψ(x0+)\psi(x_0^-)=\psi(x_0^+) and ψ′(x0−)=ψ′(x0+)\psi'(x_0^-)=\psi'(x_0^+) for finite V(x)V(x) and constant massMatches amplitude and current across an interface with no singular impulseFinite Square Well, Potential Step, Rectangular Barrier TunnelingTreating a finite jump like an infinite wall
Delta potentialψ(0−)=ψ(0+)\psi(0^-)=\psi(0^+) and ψ′(0+)−ψ′(0−)=2mλψ(0)/ℏ2\psi'(0^+)-\psi'(0^-)=2m\lambda\psi(0)/\hbar^2 for V(x)=λδ(x)V(x)=\lambda\delta(x)Encodes the singular short-range interaction as a derivative jumpDelta Function PotentialForcing ψ′\psi' to be continuous and losing the delta interaction
Periodic boundaryψ(x+L)=ψ(x)\psi(x+L)=\psi(x) and, for the kinetic operator, ψ′(x+L)=ψ′(x)\psi'(x+L)=\psi'(x)Identifies endpoints as the same physical point and quantizes momentumPeriodic Boundary Conditions, Landau Levels for finite-area degeneracy counting; finite-volume normalization throughout scattering and free-motion pagesForgetting derivative matching or confusing periodic boxes with hard boxes
Bound state at infinityψ∈L2\psi\in L^2 and the growing asymptotic solution is discardedSelects normalizable states and discrete energiesFinite Square Well, Delta Function Potential, Hydrogen AtomKeeping an exponential that grows at infinity
Scattering stateSpecify incoming, reflected, and transmitted asymptotic waves; normalize by flux or delta normalizationDefines the physical scattering experiment and reflection/transmission coefficientsPotential Step, Reflection and Transmission Coefficients, Rectangular Barrier TunnelingComputing probabilities from amplitude squares when currents differ
Radial originRequire regularity at r=0r=0; for the reduced radial function, u(r)=rR(r)u(r)=rR(r) usually satisfies u(0)=0u(0)=0Removes nonphysical singular solutions and makes the radial Hamiltonian well-definedRadial Schrödinger Equation, Hydrogen AtomAccepting a singular radial solution because it solves the local equation
Radial infinityBound radial states decay; continuum radial states have oscillatory or Coulomb-modified asymptoticsSeparates bound spectra from scattering continua in central potentialsRadial Schrödinger Equation, Degeneracy of the Hydrogen AtomNormalizing continuum states as if they were square-integrable bound states

The most useful first question is not “what is the differential equation?” but “what is the domain?” Two models can use the same expression p^2/(2m)\hat p^2/(2m) and have different spectra because their allowed wavefunctions differ.

Finite and singular interactions should not be conflated. A finite discontinuity in V(x)V(x) makes ψ\psi and ψ′\psi' continuous in the usual constant-mass Schrödinger problem. A delta-function term leaves ψ\psi continuous but creates a derivative jump.

Asymptotic conditions are boundary conditions. Bound states discard growing exponentials; scattering states choose a physical incoming channel and outgoing response. This is why the same potential can have both bound-state and scattering pages without contradiction.

A practical check is probability current. In one dimension,

j=ℏmIm⁡(ψ∗dψdx).j=\frac{\hbar}{m} \operatorname{Im}\left(\psi^*\frac{d\psi}{dx}\right).

Hard-wall bound problems should not leak probability through the wall. Scattering problems may carry flux to infinity, but the incoming and outgoing currents must be compared consistently. Periodic problems may support circulating current, but the wavefunction must match around the loop.

For a second-order kinetic operator on an interval, self-adjoint boundary conditions make the integration-by-parts boundary term vanish for all allowed states:

[ϕ∗(x)ψ′(x)−ϕ′∗(x)ψ(x)]ab=0.\left[ \phi^*(x)\psi'(x)-\phi'^*(x)\psi(x) \right]_{a}^{b}=0.

This condition is not usually the fastest way to solve an undergraduate model, but it explains why boundary conditions are not optional. See Hermitian vs Self-Adjoint Operators for the formal warning.

  • Treating ψ=0\psi=0 as the universal condition at every interface.
  • Forgetting that finite barriers allow evanescent tails.
  • Applying derivative continuity to a delta-function potential.
  • Forgetting that scattering coefficients are current ratios.
  • Ignoring the radial measure when deciding whether a singular solution is physical.
  • Memorizing a spectrum without the boundary conditions that make it true.
  • Changing geometry from an interval to a ring while keeping the old spectrum.
  • D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
  • R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
  • C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
  • M. Reed and B. Simon, Methods of Modern Mathematical Physics II: Fourier Analysis, Self-Adjointness, Academic Press, 1975.
  1. A particle moves in a finite square well. Explain why the wavefunction should not vanish at the two edges of the well.
Solution

A finite wall does not exclude the outside region completely. For a bound state, the wavefunction usually decays exponentially outside the well, so ψ\psi and ψ′\psi' are matched across each finite jump. Forcing ψ=0\psi=0 at the edges would instead describe an infinite square well and would give the wrong spectrum.

  1. For V(x)=λδ(x)V(x)=\lambda\delta(x), what matching rule distinguishes the model from a free particle on the line?
Solution

The wavefunction is continuous,

ψ(0−)=ψ(0+)=ψ(0),\psi(0^-)=\psi(0^+)=\psi(0),

but the derivative jumps:

ψ′(0+)−ψ′(0−)=2mλℏ2ψ(0).\psi'(0^+)-\psi'(0^-) =\frac{2m\lambda}{\hbar^2}\psi(0).

Without this derivative jump, the delta potential would not affect the wavefunction.