advanced30 min studyLesson 936 in Physics

Quantum wavepacket scattering numerics

Connect incident spectra, asymptotic flux ratios and discrete probability currents in finite-domain calculations.

Before you begin

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The question for today

What you’ll understand

Convert stationary transmission amplitudes into flux probabilities.
Explain why a packet averages energy-dependent scattering information.
Check a finite hopping current and account for probability at boundaries.

A wavepacket scattering calculation prepares an incident state, propagates it through an interaction, and separates outgoing channels. Each of those operations can introduce a numerical or interpretive error. Stationary reflection and transmission coefficients provide useful exact benchmarks, but they describe energy-resolved flux ratios. A finite-width packet contains a range of momenta, and probabilities measured before the channels separate need not equal the asymptotic coefficients of a single monochromatic state.

Prepare the incident state and resolve its spectrum

The packet center, spatial width, mean momentum and phase profile jointly determine its momentum distribution. Position localization broadens that distribution, so a narrow spatial packet can probe an appreciable range of scattering energies. Normalize the chosen finite representation and compare its Fourier spectrum with the intended incident band. Tails near the grid cutoff, nearby walls or an absorber can change the incoming state before it reaches the interaction.

For a static real potential with a single asymptotic incident channel, energy components scatter independently. When outgoing packets are sufficiently separated and the continuum assumptions hold, the total transmitted probability is an incident spectral average of energy-dependent transmission. It is generally not transmission evaluated at the mean energy. Near a threshold or narrow resonance, that distinction can dominate a numerically small propagation error.

P_T=\int_0^\infty dk\,|a(k)|^2T(k),\qquad \int_0^\infty dk\,|a(k)|^2=1

The formula assumes a right-moving incident spectrum on positive k, its asymptotic momentum normalization, and a static potential with the relevant open channels. The chosen packet and its energy spread are physical inputs, not numerical nuisances to discard.↔ Scroll wide equations horizontally.

Use flux rather than amplitude magnitude alone

At a potential step, matching the stationary wavefunction and its derivative gives reflected and transmitted amplitudes. If the asymptotic wavenumbers differ, so do their group velocities. Transmission probability is the transmitted-to-incident flux ratio, which multiplies the squared amplitude ratio by the outgoing-to-incoming wavenumber ratio for equal particle mass. A transmitted amplitude larger than one is therefore compatible with a transmission probability below one.

For a real step with both channels propagating, reflected and transmitted flux ratios sum to one. This is a valuable sign and normalization check. Below the outgoing threshold, an evanescent tail is present but does not carry an outgoing propagating flux in the same formula. Applying the above-threshold coefficient blindly across that change confuses spatial amplitude with an asymptotic current. A finite-time packet can also store transient probability in the interaction region.

r=\frac{k_L-k_R}{k_L+k_R},\quad t=\frac{2k_L}{k_L+k_R},\qquad R=|r|^2,\quad T=\frac{k_R}{k_L}|t|^2\qquad(k_L,k_R>0)

These amplitudes use unit incident right-moving amplitude at a static real step at x zero and equal mass on both sides. R and T are dimensionless flux ratios.↔ Scroll wide equations horizontally.

Monitor finite-grid current and channel separation

A nearest-neighbor Hermitian hopping matrix has an exact discrete continuity relation for the site probabilities. The transfer rate across one link depends on the imaginary part of the neighboring amplitude product, so two states with the same site probabilities can carry opposite currents. This discrete rate has inverse-time units when the coefficient vector has unit ordinary norm. It must be translated consistently before comparison with a continuum probability-current density.

In a packet calculation, record probability in left, interaction and right regions and any absorber loss. Signed flux measures net transfer and can be affected by interference and recrossing; it is not automatically a first-arrival count. Wait for the intended channel separation and test the detector locations, final time, grid, timestep and absorber. A reflection-plus-transmission check that silently discards trapped or removed weight can look reassuring while violating the actual probability accounting.

F_{j\to j+1}=\frac{2\epsilon}{\hbar}\operatorname{Im}(c_j^*c_{j+1}),\qquad \dot P_j=F_{j-1\to j}-F_{j\to j+1}

The formula uses neighboring Hamiltonian entries minus epsilon, with P-j equal to squared coefficient magnitude. It is an exact finite-matrix continuity identity, not a continuum flux approximation independent of grid normalization.↔ Scroll wide equations horizontally.

Work it through

A step with a transmission amplitude above one

For an above-threshold static step take $k_L=2/L$ and $k_R=1/L$ , with equal mass and unit incident amplitude. Find r, t, R and T, and check flux conservation.

Derivative and value matching give $r=(2-1)/(2+1)=1/3$ and $t=2(2)/(2+1)=4/3$ . The transmitted amplitude exceeds one.
The reflected flux ratio is $R=1/9$ . The outgoing velocity ratio is one half, so $T=(1/2)(16/9)=8/9$ , not sixteen ninths.
The sum is one, as required for this real two-channel step. A packet with a finite incident momentum band would average the energy-dependent result instead of inheriting this one-energy number without qualification.

The result

The amplitudes are one third and four thirds, while the physical flux ratios are one ninth reflected and eight ninths transmitted.

The living notebook

Explore the model

Interactive lab

Interactive lab · scattering

A schematic weight is not a probability

Reset and set coupling one, angle zero; predict illustrative weight two. Compare a fresh Reset run at coupling one, angle ninety degrees; predict weight one. The implemented g-to-the-fourth times one-plus-cosine-squared weight is schematic and is neither a step transmission probability nor a real process cross section or wavepacket scattering solver.

Incoming termsOutgoing termsAngle θ

Coupling factor g⁴: 0.24
Angular factor 1 + cos²θ: 1.329
Illustrative weight: 0.319

Coupling g0.7Scattering angle θ55°

t = 0 s

A schematic term diagram with dimensionless weight g⁴(1 + cos²θ). Lines denote terms, not particle paths; this is not a process cross-section or loop calculation. 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 forAn amplitude or illustrative weight needs its declared flux and normalization interpretation before it can be called a probability.

Pause & reason

Check your understanding

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

Read the checkpoint explanation

Probability compares currents, not bare amplitude magnitudes. The ratio of asymptotic wavenumbers supplies the velocity factor for equal mass.

Your turn

Try it yourself

For the finite two-site Hamiltonian $H=\epsilon\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$ take $c=(1,i)/\sqrt2$ . Find the instantaneous transfer rate from site zero to one and both probability derivatives. Explain why equal probabilities do not imply zero current.

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 neighboring product is imaginary. Use the stated minus-epsilon hopping convention and discrete continuity signs.

Compare with the solution

The product $c_0^*c_1=i/2$ has imaginary part one half. Thus $F_{0\to1}=\epsilon/\hbar$ , $\dot P_0=-\epsilon/\hbar$ and $\dot P_1=+\epsilon/\hbar$ . Their sum is zero. The initial probabilities are both one half, but relative phase drives a nonzero current. These are instantaneous derivatives in a finite Hermitian system; extrapolating them linearly for arbitrary time would not preserve the exact bounded oscillatory probabilities.

Keep the idea

Take it with you

Packet transmission depends on the incident spectrum and asymptotic channel separation.
Flux probabilities can differ from squared amplitude ratios.
Discrete current, trapped weight and absorber loss belong in one consistent probability budget.

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