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SCIENCE · INVESTIGATE AND EXPLAIN

Use probability moments rather than reading off a density peak

Consider a supplied position density per unit u=x/L: f(u)=(p+1)(p+2)u^p(1−u) for 0≤u≤1, with positive integer p. Integrating u^p−u^(p+1) gives 1/[(p+1)(p+2)], which proves normalization. Physical position density is f(x/L)/L, not f alone.

Learning outcome: Derive normalization and both moments for two p values. Predict the direction of the mean-mode difference before revealing the graph. Compute one interval probability independently and explain what phase information remains absent.

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.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

For p=2, mean(u)=3/5, mean(u²)=2/5 and variance=1/25. The mode is 2/3, farther right than the mean. Probability to the left of u=1/2 is 4(1/2)³−3(1/2)⁴=5/16. None of these numbers is a classical trajectory.

Open the experiment

Predict

For p=2, which value is the mean position divided by L?

Experiment

Derive normalization and both moments for two p values. Predict the direction of the mean-mode difference before revealing the graph. Compute one interval probability independently and explain what phase information remains absent.

Integrate moments of an asymmetric density

Worked teaching model · synthetic values · Normalized position density per unit u is (p+1)(p+2)u^p(1−u),0≤u≤1, L1 nm. This is a supplied probability state, not a claim of one particular Hamiltonian or classical trajectory. Integrate moments rather than identifying mean with the density maximum.

Reset keeps your prediction, captures and practice history.

An asymmetric normalized density

density per unit uu = x/L011.7777780

Graph axes rescale independently; read units and values, not screen angles. 1: supplied beta-family state (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for An asymmetric normalized density

supplied beta-family state

u = x/Ldensity per unit u
00
0.016666670.003277778
0.033333330.01288889
0.050.0285
0.066666670.04977778
0.083333330.07638889
0.10.108
0.11666670.1442778
0.13333330.1848889
0.150.2295
0.16666670.2777778
0.18333330.3293889
0.20.384
0.21666670.4412778
0.23333330.5008889
0.250.5625
0.26666670.6257778
0.28333330.6903889
0.30.756
0.31666670.8222778
0.33333330.8888889
0.350.9555
0.36666671.021778
0.38333331.087389
0.41.152
0.41666671.215278
0.43333331.276889
0.451.3365
0.46666671.393778
0.48333331.448389
0.51.5
0.51666671.548278
0.53333331.592889
0.551.6335
0.56666671.669778
0.58333331.701389
0.61.728
0.61666671.749278
0.63333331.764889
0.651.7745
0.66666671.777778
0.68333331.774389
0.71.764
0.71666671.746278
0.73333331.720889
0.751.6875
0.76666671.645778
0.78333331.595389
0.81.536
0.81666671.467278
0.83333331.388889
0.851.3005
0.86666671.201778
0.88333331.092389
0.90.972
0.91666670.8402778
0.93333330.6968889
0.950.5415
0.96666670.3737778
0.98333330.1933889
10

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
QuantityValueUnit
Mean position divided by length0.6ratio
Second moment divided by length squared0.4ratio
Position variance divided by length squared0.04ratio
Most likely position divided by length0.6666667ratio
Probability left of chosen position0.3125ratio

Expectation, density mode and interval probability answer different questions.

Integrating u^p(1−u) gives1/[(p+1)(p+2)]. The first two normalized moments are(p+1)/(p+3) and(p+1)(p+2)/[(p+3)(p+4)]. Differentiating the density places its interior mode at p/(p+1). Variance is second moment minus square of mean, not square of mode. No individual trajectory follows from these ensemble moments.

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

Expectation, density mode and interval probability answer different questions.

Integrating u^p(1−u) gives1/[(p+1)(p+2)]. The first two normalized moments are(p+1)/(p+3) and(p+1)(p+2)/[(p+3)(p+4)]. Differentiating the density places its interior mode at p/(p+1). Variance is second moment minus square of mean, not square of mode. No individual trajectory follows from these ensemble moments.

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

Understand

Define the state, source and evidence

Consider a supplied position density per unit u=x/L: f(u)=(p+1)(p+2)u^p(1−u) for 0≤u≤1, with positive integer p. Integrating u^p−u^(p+1) gives 1/[(p+1)(p+2)], which proves normalization. Physical position density is f(x/L)/L, not f alone.

Derive the quantitative relation

The first moment is mean(u)=(p+1)/(p+3). The second is mean(u²)=(p+1)(p+2)/[(p+3)(p+4)]. Subtracting the square of the mean gives variance; multiplying by L or L² restores dimensions. These are expectations over repeated preparations, not the position of an individual particle.

Check normalization, limits and inference

Differentiate the density: its interior maximum is at u=p/(p+1), which generally differs from the mean. Integrating to w gives cumulative probability (p+2)w^(p+1)−(p+1)w^(p+2). A mode, a mean and a median are distinct summaries; one cannot substitute one for another merely because the curve has one peak.

Compare a changed premise

A normalized nonnegative density can be represented by an amplitude magnitude, but it does not by itself specify phase, momentum statistics or a Hamiltonian. The workbench deliberately reports position moments only. A separate momentum observation could distinguish states with the same position density and different phases.

Worked original case

For p=2, mean(u)=3/5, mean(u²)=2/5 and variance=1/25. The mode is 2/3, farther right than the mean. Probability to the left of u=1/2 is 4(1/2)³−3(1/2)⁴=5/16. None of these numbers is a classical trajectory.

Investigation

Derive normalization and both moments for two p values. Predict the direction of the mean-mode difference before revealing the graph. Compute one interval probability independently and explain what phase information remains absent.

Changed-case practice

Try the changed case before feedback

For p=1, find probability to the left of u=1/2.

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.

Apply

For p=2, which value is the mean position divided by L?

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?

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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