advanced30 min studyLesson 935 in Physics

Absorbing boundaries and reflection artifacts

Interpret controlled norm loss, artificial reflection and finite-domain sensitivity without inventing a physical detector.

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What you’ll understand

  • Derive the norm-loss identity for a positive absorbing potential.
  • Distinguish unnormalized propagation from a conditional normalized state.
  • Estimate a packet-center reflection return while acknowledging dispersive tails.

A finite computational box can reflect an outgoing wavepacket or let it wrap around through periodic boundaries. Absorbing layers attempt to remove that unwanted returning amplitude before it corrupts the region of interest. This changes the numerical evolution deliberately: norm can decrease because the represented domain is no longer closed. The practical question is whether the absorber removes outgoing content accurately over the relevant energy range, while leaving the desired observables insensitive to its details.

Declare the non-Hermitian absorbing model

A common model adds minus i times a nonnegative Hermitian operator W to an otherwise Hermitian Hamiltonian. For a spatial absorbing potential, W is a nonnegative real function supported near a boundary. Differentiate the ordinary squared state norm using the Schrödinger equation and its adjoint. The Hermitian terms cancel and the absorbing contribution is nonpositive. This identity explains intentional loss without calling the finite model unitary.

The lost norm is a numerical bookkeeping quantity unless a specific physical measurement or reservoir model justifies another interpretation. A complex absorbing potential is not, by itself, a trace-preserving Lindblad equation for a reduced density matrix. If the remaining state is renormalized, it becomes a conditional description of surviving represented amplitude; the removed weight must still be recorded when computing unconditional reflection or transmission probabilities.

Heff=H−iW,W=W†≥0,ddt∥ψ∥2=−2ℏ⟨ψ∣W∣ψ⟩H_{\rm eff}=H-iW,\quad W=W^\dagger\ge0,\qquad \frac{d}{dt}\|\psi\|^2=-\frac2\hbar\langle\psi|W|\psi\rangle
W has energy units. The state in this identity is unnormalized after loss, and the expectation is not divided by its surviving norm.↔ Scroll wide equations horizontally.

Test profile and energy dependence instead of assuming perfect removal

A very abrupt absorbing layer can itself reflect a wave, while a weak or short layer can leave substantial amplitude to reach the outer boundary. Smoothness, strength, thickness and the outgoing energy band therefore need joint testing. Increasing strength monotonically does not necessarily improve the reflection error. Transparent boundary conditions, complex scaling and masks use different constructions with their own assumptions; they should not be treated as identical algorithms merely because they suppress returning signals.

A multiplicative mask removes amplitude at discrete times. Its probability effect involves the square of its magnitude, and its relationship to a continuous W depends on timestep. Changing the timestep while keeping an arbitrary per-step mask fixed can therefore change the physical loss rate of the numerical model. A sensible convergence study either holds the intended generator fixed or explicitly studies the resulting change in absorber prescription.

ψ↦e−WΔt/ℏψfor a pure absorbing substep,M=e−WΔt/ℏ\psi\mapsto e^{-W\Delta t/\hbar}\psi\quad\text{for a pure absorbing substep},\qquad M=e^{-W\Delta t/\hbar}
This substep is exact for the absorbing term alone. If H does not commute with W, composing it with Hermitian evolution introduces a separate splitting approximation.↔ Scroll wide equations horizontally.

Separate arrival estimates from rigorous error bounds

The time at which a reflected packet center returns to an interaction region can be estimated from travel distance and a stipulated group speed. This gives a useful diagnostic when moving the outer boundary: a corresponding shift of a spurious feature can reveal its boundary origin. A nonrelativistic Schrödinger packet has dispersive tails, however, so a center-travel estimate is not a strict causal guarantee that every earlier sample is uncontaminated.

Compare calculations with changed box size and absorber profile while holding the desired incident state and interaction fixed. Monitor the region of interest, outgoing flux and accumulated loss separately. A tiny final norm does not prove negligible reflection; the lost amplitude could have included desired components or masked an error after it occurred. Benchmarks with known scattering solutions and parameter-independent observables are more informative than judging an absorber from an attractive fading animation.

tcenter,return≈2(xb−x0)vgfor a stipulated narrow right-moving packet and specular reflected centert_{\rm center,return}\approx\frac{2(x_b-x_0)}{v_g}\qquad\text{for a stipulated narrow right-moving packet and specular reflected center}
This is a group-center travel estimate on a line, not a finite-propagation theorem for the Schrödinger equation or a bound on all boundary error.↔ Scroll wide equations horizontally.

An exact two-site absorbing substep

For one substep set H zero and W=diag⁡(0,w)W=\operatorname{diag}(0,w), starting from plus. Choose wΔt/ℏ=ln⁡2w\Delta t/\hbar=\ln2. Find the unnormalized final amplitudes, surviving norm, removed probability and the conditional second-site probability.

  1. The absorbing exponential leaves the first amplitude unchanged and multiplies the second by one half. The state is (1,1/2)/2(1,1/2)/\sqrt2.

  2. Its squared norm is 1/2+1/8=5/81/2+1/8=5/8, so the removed probability is 3/83/8. The second site's unconditioned represented weight is one eighth.

  3. Dividing that weight by the surviving norm gives conditional probability (1/8)/(5/8)=1/5(1/8)/(5/8)=1/5. Reporting only the normalized state would hide the three-eighths loss, which remains necessary for unconditional probability accounting.

The result

The surviving norm is five eighths, removed weight three eighths, and conditional second-site probability one fifth.

Explore the model

Interactive lab

Interactive lab · heat

A decaying amplitude is a specified model

Reset and set mode one, diffusivity one, then Step once; predict amplitude exp of minus one half. Compare a fresh Reset run at mode one, diffusivity two, and one Step; predict exp of minus one. This is an exact periodic heat mode, not a quantum absorber, detector, reflection test or non-Hermitian wavepacket propagation.

A decaying amplitude is a specified modelThe exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. Current readings: Current amplitude: 1; Spatial L² norm: 1.772; Decay rate · κn²: 3.6 /s. All parameters are adjustable with the labeled controls below.Periodic heat equation mode 3 at time 0: the amplitude is 1.0π/2π3π/22π−1−0.500.51x · periodic intervalTemperature perturbation uExact mode · n = 3
Current temperature modeInitial sineCurrent amplitude
Current amplitude
1
Spatial L² norm
1.772
Decay rate · κn²
3.6 /s
t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. 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 forDecay can be checked algebraically, but its probability and boundary interpretation depends on the actual dynamical model.

Check your understanding

What must be retained when a state propagated with an absorber is renormalized?

Try it yourself

A stipulated narrow packet center leaves x0=2Lx_0=2L toward an outer boundary at xb=10Lx_b=10L, with group speed 2L/T2L/T. Estimate its reflected center return time to its starting point. Repeat for boundary 14L14L. Can these times certify a strictly uncontaminated earlier interval?

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 center travels out and back over twice the distance. The propagation equation still has dispersive tails.

Compare with the solution

The first round trip is 2(10−2)L/(2L/T)=8T2(10-2)L/(2L/T)=8T. Moving the boundary to fourteen L gives 2(14−2)L/(2L/T)=12T2(14-2)L/(2L/T)=12T. A feature shifting by roughly four T can suggest boundary reflection in this idealized center picture. These estimates do not certify a strict error-free earlier window: Schrödinger dispersion and tails, a spread of group speeds and imperfect absorption can contribute before the reflected center arrives. Independent box and profile comparisons are still required.

Take it with you

  • Absorbing models deliberately replace closed-system norm conservation with controlled loss.
  • Conditional normalization must not erase unconditional probability accounting.
  • Absorber reflection and box effects require independent parameter checks.

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