SCIENCE · INVESTIGATE AND EXPLAIN

Normalize both pieces of a continuous asymmetric amplitude

Let u=x/L on a finite support 0..L. A real triangular amplitude rises as N u/q for 0≤u≤q and falls as N(1−u)/(1−q) for q≤u≤1, with zero outside. Here q strictly between zero and one is a peak-position fraction, not a probability. The two pieces agree at the peak and vanish at the endpoints.

Learning outcome: Normalize the two pieces separately, verify continuity at the peak and compare two intervals that cross different branches. Move the peak without changing L and explain why N stays fixed while interval probabilities change. State which extra information would turn this into a stationary-state problem.

Starting recap

Review definite integrals, complex amplitudes, probability moments, elementary differential equations and energy-momentum units. State boundary and normalization conditions, and distinguish calibration from unused validation.

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Try

For L=1 nm and q=0.5, N=sqrt(3) nm^−1/2. The probability left of x=L/4 is (1/4)³/(1/2)²=1/16, not one quarter or the density height. For q=0.25, the probability to the left of the peak is 0.25 even though that point is the amplitude maximum.

Open the experiment

Predict

Which quantity must integrate to one over physical position?

Your prediction and reason

Experiment

Normalize the two pieces separately, verify continuity at the peak and compare two intervals that cross different branches. Move the peak without changing L and explain why N stays fixed while interval probabilities change. State which extra information would turn this into a stationary-state problem.

Normalize a continuous piecewise amplitude

Worked teaching model · synthetic values · Support0..L with L1 nm. Real amplitude N u/q before peak q and N(1−u)/(1−q) after it, u=x/L; zero outside. Normalize with physical dx. Continuous amplitude has derivative corners and is not asserted to be a stationary state of a smooth finite potential.

Peak position divided by support length: 0.5 ratio Upper integration position divided by length: 0.25 ratio

Reset keeps your prediction, captures and practice history.

Continuous amplitude is not probability density

Graph axes rescale independently; read units and values, not screen angles. 1: normalized real amplitude (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Continuous amplitude is not probability density

normalized real amplitude

u = x/L	amplitude times sqrt(L)
0	0
0.01666667	0.05773503
0.03333333	0.1154701
0.05	0.1732051
0.06666667	0.2309401
0.08333333	0.2886751
0.1	0.3464102
0.1166667	0.4041452
0.1333333	0.4618802
0.15	0.5196152
0.1666667	0.5773503
0.1833333	0.6350853
0.2	0.6928203
0.2166667	0.7505553
0.2333333	0.8082904
0.25	0.8660254
0.2666667	0.9237604
0.2833333	0.9814955
0.3	1.03923
0.3166667	1.096966
0.3333333	1.154701
0.35	1.212436
0.3666667	1.270171
0.3833333	1.327906
0.4	1.385641
0.4166667	1.443376
0.4333333	1.501111
0.45	1.558846
0.4666667	1.616581
0.4833333	1.674316
0.5	1.732051
0.5166667	1.674316
0.5333333	1.616581
0.55	1.558846
0.5666667	1.501111
0.5833333	1.443376
0.6	1.385641
0.6166667	1.327906
0.6333333	1.270171
0.65	1.212436
0.6666667	1.154701
0.6833333	1.096966
0.7	1.03923
0.7166667	0.9814955
0.7333333	0.9237604
0.75	0.8660254
0.7666667	0.8082904
0.7833333	0.7505553
0.8	0.6928203
0.8166667	0.6350853
0.8333333	0.5773503
0.85	0.5196152
0.8666667	0.4618802
0.8833333	0.4041452
0.9	0.3464102
0.9166667	0.2886751
0.9333333	0.2309401
0.95	0.1732051
0.9666667	0.1154701
0.9833333	0.05773503
1	0

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Probability integrates the squared amplitude

Graph axes rescale independently; read units and values, not screen angles. 1: absolute amplitude squared (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Probability integrates the squared amplitude

absolute amplitude squared

u = x/L	density per unit u
0	0
0.01666667	0.003333333
0.03333333	0.01333333
0.05	0.03
0.06666667	0.05333333
0.08333333	0.08333333
0.1	0.12
0.1166667	0.1633333
0.1333333	0.2133333
0.15	0.27
0.1666667	0.3333333
0.1833333	0.4033333
0.2	0.48
0.2166667	0.5633333
0.2333333	0.6533333
0.25	0.75
0.2666667	0.8533333
0.2833333	0.9633333
0.3	1.08
0.3166667	1.203333
0.3333333	1.333333
0.35	1.47
0.3666667	1.613333
0.3833333	1.763333
0.4	1.92
0.4166667	2.083333
0.4333333	2.253333
0.45	2.43
0.4666667	2.613333
0.4833333	2.803333
0.5	3
0.5166667	2.803333
0.5333333	2.613333
0.55	2.43
0.5666667	2.253333
0.5833333	2.083333
0.6	1.92
0.6166667	1.763333
0.6333333	1.613333
0.65	1.47
0.6666667	1.333333
0.6833333	1.203333
0.7	1.08
0.7166667	0.9633333
0.7333333	0.8533333
0.75	0.75
0.7666667	0.6533333
0.7833333	0.5633333
0.8	0.48
0.8166667	0.4033333
0.8333333	0.3333333
0.85	0.27
0.8666667	0.2133333
0.8833333	0.1633333
0.9	0.12
0.9166667	0.08333333
0.9333333	0.05333333
0.95	0.03
0.9666667	0.01333333
0.9833333	0.003333333
1	0

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Normalization constant for length one nanometre	1.732051	nm^−1/2
Probability left of chosen upper position	0.0625	ratio
Probability left of amplitude peak	0.5	ratio
Probability over full support	1	ratio

Normalization uses the physical integration measure; a density height is not an interval probability.

The two shape-square integrals are q/3 and (1−q)/3, so N² L/3=1. Integrate the square piecewise: P(0<u<w)=w³/q² before the peak and1−(1−w)³/(1−q)² after it. At the peak P=q, not necessarily one half. The continuous triangular state has derivative corners; it is not claimed to solve a stationary smooth-potential Schrödinger equation.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Normalization uses the physical integration measure; a density height is not an interval probability.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define the state, source and evidence

Derive the quantitative relation

Probability is integral |psi|² dx, and dx=L du. The shape-square integrals are q/3 and (1−q)/3, so N²L/3=1 and N=sqrt(3/L), independent of the peak position. Omitting L would give a normalization constant with the wrong physical dimensions. Wavefunction units are inverse square root of length, while density units are inverse length.

Check normalization, limits and inference

For an upper position w before the peak, integrate to obtain P(u from zero to w)=w³/q². For w after the peak, it is simpler to integrate the omitted right tail, giving P=1−(1−w)³/(1−q)². At w=q, both formulas give P=q. Thus the density maximum need not divide probability into equal halves.

Compare a changed premise

The amplitude is continuous but its derivative has corners. It is a supplied normalized state, not asserted to be an eigenfunction of a smooth finite potential. A stationary Schrödinger problem would require a Hamiltonian and compatible derivative/boundary conditions; normalization alone does not provide those.

Worked original case

Investigation

Changed-case practice

Try the changed case before feedback

With q=0.5 and L=1 nm, find probability to the left of x=0.75 nm.

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original derivations and virtual evidence checked 31 October 2026 against the primary explanation. All questions, numerical packets and diagrams are original. These models do not establish actual beam, radiation, vacuum, optical or electrical laboratory performance. Public feedback is practice, not authenticated independent mastery. Preserve first reasoning, criterion critique, revision and a genuinely new assigned packet in the course evidence workspace.

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Public practice checks a specific claim. It does not establish independent mastery.

Which state, calibration or boundary premise is independently justified, and what unused observation could distinguish an alternative?

Your explanation and limitations

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Previous: Connect a kinetic-energy inference to an allowed diffraction direction Next: Use probability moments rather than reading off a density peak

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