SCIENCE · INVESTIGATE AND EXPLAIN

Obtain a discrete spectrum from two spatial boundaries

Inside a one-dimensional infinite well from x=0 to L, the stationary Schrödinger equation is −hbar² psiDoublePrime/(2m)=E psi. For positive E, write k²=2mE/hbar² and solve psi=A sin(kx)+B cos(kx). The left boundary psi(0)=0 forces B=0.

Learning outcome: Solve the two boundary equations before selecting n. Compare n=1,2,3 at fixed L and then change L at fixed n. Integrate one interval that does not end at a node, and distinguish normalization, stationary phase and energy measurement.

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 n=2, energy is 4×0.3760301626≈1.5041206504 eV and there is one interior node. Probability in the left third is 1/3+sqrt(3)/(8π)≈0.402249445. Doubling L at fixed n lowers energy by a factor of four.

Open the experiment

Predict

Why is n=0 excluded from the normalized infinite-well eigenstates?

Your prediction and reason

Experiment

Solve the two boundary equations before selecting n. Compare n=1,2,3 at fixed L and then change L at fixed n. Integrate one interval that does not end at a node, and distinguish normalization, stationary phase and energy measurement.

Derive energies from two boundary conditions

Worked teaching model · synthetic values · Nonrelativistic electron in ideal infinite well0..L. En=.3760301626 n²/Lnm² eV is a rounded teaching coefficient. Real spatial eigenfunction sqrt(2/L)sin(npi x/L), normalized over the well; the overall time phase does not move its density.

Well length: 1 nm Stationary-state quantum number: 2 integer

Reset keeps your prediction, captures and practice history.

Boundary-selected spatial eigenfunction

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

Equivalent numerical coordinates for Boundary-selected spatial eigenfunction

normalized eigenfunction

u = x/L	real amplitude (nm^−1/2)
0	0
0.01666667	0.1478256
0.03333333	0.2940315
0.05	0.437016
0.06666667	0.5752125
0.08333333	0.7071068
0.1	0.8312539
0.1166667	0.9462936
0.1333333	1.050965
0.15	1.144123
0.1666667	1.224745
0.1833333	1.291948
0.2	1.344997
0.2166667	1.38331
0.2333333	1.406466
0.25	1.414214
0.2666667	1.406466
0.2833333	1.38331
0.3	1.344997
0.3166667	1.291948
0.3333333	1.224745
0.35	1.144123
0.3666667	1.050965
0.3833333	0.9462936
0.4	0.8312539
0.4166667	0.7071068
0.4333333	0.5752125
0.45	0.437016
0.4666667	0.2940315
0.4833333	0.1478256
0.5	1.731912e-16
0.5166667	-0.1478256
0.5333333	-0.2940315
0.55	-0.437016
0.5666667	-0.5752125
0.5833333	-0.7071068
0.6	-0.8312539
0.6166667	-0.9462936
0.6333333	-1.050965
0.65	-1.144123
0.6666667	-1.224745
0.6833333	-1.291948
0.7	-1.344997
0.7166667	-1.38331
0.7333333	-1.406466
0.75	-1.414214
0.7666667	-1.406466
0.7833333	-1.38331
0.8	-1.344997
0.8166667	-1.291948
0.8333333	-1.224745
0.85	-1.144123
0.8666667	-1.050965
0.8833333	-0.9462936
0.9	-0.8312539
0.9166667	-0.7071068
0.9333333	-0.5752125
0.95	-0.437016
0.9666667	-0.2940315
0.9833333	-0.1478256
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
Stationary energy	1.504121	eV
Interior node count	1	integer
Probability in left third	0.4022494	ratio
Full normalization	1	ratio
Energy divided by ground-state energy	4	ratio

Both walls quantize the wave number; the stationary overall phase leaves the density unchanged.

Inside, psiDoublePrime+k²psi=0. The left wall removes the cosine coefficient; the right wall requires sin(kL)=0, hence k=npi/L with n=1,2,...; n=0 gives only the zero function. Normalization gives sqrt(2/L), energy hbar²n²pi²/(2mL²), and interval integral F_n(u)=u−sin(2npi u)/(2npi). This ideal nonrelativistic confinement is not a literal infinitely high material wall.

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

Both walls quantize the wave number; the stationary overall phase leaves the density unchanged.

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

The right boundary requires sin(kL)=0, so k=nπ/L. The n=0 choice produces the identically zero function and cannot be normalized; allowed n are positive integers. Normalize integral0L |psi|²dx to obtain sqrt(2/L). Energies are En=hbar²π²n²/(2mL²), so length and quantum-number changes have different scaling.

Check normalization, limits and inference

For u=x/L the cumulative probability is Fn(u)=u−sin(2nπu)/(2nπ). There are n−1 interior nodes. A stationary time factor exp(−iEnt/hbar) multiplies the entire state, leaving its density unchanged. A plotted sinusoid is an amplitude, not the particle bouncing along a classical path.

Compare a changed premise

The rounded electron coefficient used here is 0.3760301626 eV nm². Finite material barriers, relativistic energies and interactions are outside this ideal model. Later comparison with finite confinement must keep the same spatial width and energy reference rather than mixing a well bottom with a barrier reference.

Worked original case

Investigation

Changed-case practice

Try the changed case before feedback

With L=2 nm and n=2, find energy in eV using coefficient 0.3760301626 eV nm².

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.

Apply

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: Track a Gaussian phase without changing its position density Next: Evolve interference while retaining fixed energy probabilities

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