1D Schrödinger scattering

What the plot shows

The horizontal axis is dimensionless position $x$. The vertical axis is energy: $V(x)$ and $E$ share the labelled scale on the left. The wavefunction overlays — bright $|\psi|^2$ envelope and faint $\mathrm{Re}\,\psi$ oscillation — are auto-scaled to fit the same panel, so their visual amplitude is not in the same units as the energy ticks. Treat them as a shape, not a magnitude comparison against $V$.

• Shaded grey region: $V(x)$, the potential.
• Dashed green line: the chosen energy $E$.
• Bright white curve: $|\psi(x)|^2$, the probability density.
• Faint grey curve: $\mathrm{Re}\,\psi(x)$, exposing the local wavelength.

All potentials are non-negative, so $V$ never goes below the energy axis. For the well, $V_0$ is the height of the surrounding walls and the well floor sits at $V = 0$.

The equation being solved

We solve the time-independent Schrödinger equation in one dimension:

$$ -\frac{\hbar^2}{2m}\, \psi''(x) \;+\; V(x)\, \psi(x) \;=\; E\, \psi(x) $$

Setting $\hbar = m = 1$ reduces this to:

$$ \psi''(x) \;=\; 2\,\big(V(x) - E\big)\, \psi(x) $$

Where $E > V$ the right-hand side is negative and $\psi$ oscillates with local wavenumber $k(x) = \sqrt{2(E - V(x))}$. Where $E < V$ the right-hand side is positive and $\psi$ grows or decays exponentially at rate $\kappa(x) = \sqrt{2(V(x) - E)}$. The equation is integrated as one continuous ODE — there is no separate region matching, so continuity of $\psi$ and $\psi'$ across a step in $V$ is automatic.

Potentials available

• Barrier: $V(x) = V_0$ for $|x| < w/2$, else $0$.
• Well: $V(x) = 0$ for $|x| < w/2$, else $V_0$ (free region between walls).
• Step: $V(x) = V_0 \cdot \Theta(x)$.
• Double: two barriers of height $V_0$ separated by a gap; supports resonant tunneling.
• Gaussian: $V(x) = V_0 \exp\!\left(-x^2 / 2\sigma^2\right)$, $\sigma = w / 2.355$.

All sharp edges are replaced by $\tanh$ transitions of width $\varepsilon = 0.04$ — about one grid spacing. Visually the edges still look sharp, but the integrator sees a continuous $V(x)$ and its fourth-order accuracy is preserved.

Boundary conditions and how they are enforced

Scattering boundary conditions: incoming plane wave from the left of unit amplitude, reflected wave of amplitude $r$, transmitted wave on the right of amplitude $t$, and no incoming wave from the right.

$$ \psi(x \to -\infty) = e^{i k_L x} + r\, e^{-i k_L x} $$ $$ \psi(x \to +\infty) = t\, e^{i k_R x} $$

where $k_L = \sqrt{2(E - V_L)}$ and $k_R = \sqrt{2(E - V_R)}$, and $V_L, V_R$ are the asymptotic values of $V$. For barrier, double and Gaussian, $V_L = V_R = 0$, so $k_L = k_R = \sqrt{2E}$. For the well, $V_L = V_R = V_0$ and the wave is only propagating on the outside if $E > V_0$. For the step, $V_L = 0, V_R = V_0$.

The right boundary is enforced by construction. We start integration at $x = +L$ with $\psi = e^{i k_R x}$ and $\psi' = i k_R e^{i k_R x}$. There is no left-moving component to remove because we never put one in. When $E < V_R$ there is no propagating outgoing wave — we instead start with the decaying $\psi = e^{-\kappa_R x}$.

The left boundary is enforced by decomposition. After integration reaches $x = -L$ the solution is a superposition $A\, e^{i k_L x} + B\, e^{-i k_L x}$. Solving for $A, B$ from $\psi(-L)$ and $\psi'(-L)$:

$$ A = \tfrac{1}{2}\, e^{-i k_L L}\, \big(\psi - i\, \psi' / k_L\big) $$ $$ B = \tfrac{1}{2}\, e^{+i k_L L}\, \big(\psi + i\, \psi' / k_L\big) $$

Physical normalization sets the incoming amplitude to $1$, which means dividing every amplitude by $A$: so $t = 1/A$ and $r = B/A$.

When $E < V_L$ there is no propagating wave on the left at all — the asymptotic region is classically forbidden and an "incoming plane wave" does not exist. The simulator detects this case and shows a sub-threshold notice on the plot; $T$ is set to $0$ and the rendered $\psi$ is just the right-boundary solution propagated left, normalized to its own peak (it is not a physical scattering state in this regime).

Probability and normalization

Scattering states are not square-integrable — $\int |\psi|^2\, dx$ diverges because the asymptotic plane waves never decay. The normalization convention is therefore fixed incoming amplitude, not unit total probability. With incoming amplitude $1$:

• On the left: $|\psi|^2 = \big|1 + r\, e^{-2 i k_L x}\big|^2$, oscillating between $(1 - |r|)^2$ and $(1 + |r|)^2$ due to standing-wave interference.
• On the right: $|\psi|^2 = |t|^2$, constant.

The conserved quantity is the probability current, not the density. Current conservation gives:

$$ T + R = 1, \qquad T = \frac{k_R}{k_L}\, |t|^2, \qquad R = |r|^2 $$

The factor $k_R / k_L$ matters for the step potential where $k_R \neq k_L$; otherwise it equals $1$. The reported $T$ and $R$ always sum to $1$ to numerical precision — drift from $1$ is a useful sanity check on the integrator. When $E < V_R$ the right side is evanescent, so $T = 0$ and $R = 1$ exactly.

Algorithm

The full pipeline runs on every slider input event:

1. Pick a domain $[-L, L]$ with $L = \max(15,\, 4w + 4)$, so the boundaries sit comfortably outside the potential. Discretize into $N = 1200$ uniform points.
2. Sample $V(x_i)$ on the grid using the $\tanh$-smoothed shape definitions.
3. Compute $k_L, k_R$ (or $\kappa_L, \kappa_R$) from the boundary potentials and the current energy.
4. Initialize the state at $x = +L$ with the outgoing-only boundary form (or the decaying form if $E < V_R$).
5. Integrate the second-order ODE backward to $x = -L$ using classical fourth-order Runge–Kutta on the first-order system $y = (\mathrm{Re}\,\psi,\, \mathrm{Im}\,\psi,\, \mathrm{Re}\,\psi',\, \mathrm{Im}\,\psi')$.
6. If $E > V_L$: decompose $\psi(-L), \psi'(-L)$ into $A, B$. Otherwise mark sub-threshold.
7. Compute $T = (k_R / k_L)\, |1 / A|^2$ and $R = |B / A|^2$.
8. Rescale every $\psi(x_i)$ by $1/A$ so the incoming amplitude becomes $1$, then render on canvas.

Why integrate backward? Forward shooting from the left would require guessing both $r$ and $t$ and enforcing the no-incoming-from-right condition at $+L$ by iteration. Backward shooting collapses that to a single pass: the right boundary is a known initial condition (one outgoing plane wave), and the unknowns $r, t$ drop out of a linear decomposition at the left.

Verification

Spot-checks against the textbook closed forms:

• Step up ($E = 15$, $V_0 = 8$): analytical $T = 4 k_L k_R / (k_L + k_R)^2 \approx 0.964$; simulator returns $0.968$.
• Step under ($E = 5$, $V_0 = 8$): analytical $T = 0$, $R = 1$ exactly; simulator matches.
• Rectangular barrier ($E < V_0$): exponentially small $T \approx \exp(-2 \kappa w)$; simulator matches the order of magnitude.
• Finite-walls well ($E > V_0$): transmission resonances at $k\, w_{\text{free}} = n\pi$; visible as $T = 1$ peaks when sweeping $E$.

Residuals are sub-percent, dominated by the smoothing of $V$ discontinuities. Increasing $N$ or decreasing $\varepsilon$ closes the gap at the cost of redraw time.

Where this falls short

• Strictly steady-state scattering, not a wavepacket movie. The plot is the eigenstate of the scattering Hamiltonian at the chosen $E$.
• Probability density on the left interferes with the reflected wave and oscillates between roughly $0$ and $4$ for total reflection. Physical, not an integrator artifact.
• Very narrow resonance peaks (thin gap in the double barrier with a tall $V_0$) can be skipped by the slider's discrete steps. Drag slowly through likely resonance regions.
• For very large $V_0 w$ the unnormalized $|A|^2$ can grow large enough to lose precision; the UI clamps $T, R$ to $[0, 1]$.
• The well's bound-state eigenvalues (when $E < V_0$) are not solved here — that is an eigenvalue problem, not scattering.