University Quantum Mechanics I · unit UQMI-06
The Infinite Square Well
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
The well, the boundary conditions, and what they cost: infer an input with uncertainty
The infinite square well is the potential V = 0 for 0 < x < L and V = ∞ elsewhere; its states obey ψ(0) = ψ(L) = 0, with ψ continuous at each wall and ψ′ free to jump there. Use the supplied topic model Y_out = 0.0381*(pi*f/w)^2/cos(pi*f/2)^2, where Y_out is WALL HEIGHT V₀ NEEDED (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. f is Inside wavenumber k, in units of π/L, measured in π/L. Keep w (Well width L) = 1 nm fixed. For this exercise use only 0.552063 π/L ≤ f ≤ 0.937937 π/L on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 2.1711374 eV, with a bounded reading uncertainty ±0.2846 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000385875 π/L, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
WALL HEIGHT V₀ NEEDED against f
Predictions from the supplied The well, the boundary conditions, and what they cost model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): WALL HEIGHT V₀ NEEDED / eV
Horizontal axis (x): f: Inside wavenumber k, in units of π/L / π/L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Inside wavenumber k, in units of π/L / π/L | WALL HEIGHT V₀ NEEDED / eV |
|---|---|
| 0.5520625 | 0.2737902767957964 |
| 0.64853125 | 0.574982552224795 |
| 0.745 | 1.3726794046089223 |
| 0.84146875 | 4.383558838666285 |
| 0.9379375 | 34.91809082913382 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(f) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.552063) = 0.27379 eV and Y_out(0.937937) = 34.9181 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives f_line = 0.552063 + (2.17114 − 0.27379)(0.937937 − 0.552063)/(34.9181 − 0.27379) = 0.573196 π/L. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives f = 0.787446 π/L; substituting back gives Y_out = 2.1711374 eV.
- The interpolation error in the input is |f_line − f| = 0.214251 π/L. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.8865374 eV and 2.4557374 eV. Sort the roots even if the response decreases: 0.775059 π/L ≤ f ≤ 0.797861 π/L.
- The lower and upper input deviations are 0.0123875 π/L and 0.0104144 π/L. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer f ≈ 0.787446 π/L; compatible interval [0.775059, 0.797861] π/L. Endpoint interpolation gives 0.573196 π/L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Energy quantisation: the condition for a nontrivial solution: infer an input with uncertainty
An allowed energy of the infinite well is a value of E at which ψ″ = −(2mE/ħ²)ψ has a solution that vanishes at both walls without vanishing everywhere; these are Eₙ = n²π²ħ²/(2mL²) with n = 1, 2, 3, … Use the supplied topic model Y_out = 0.0381*pi^2/w^2, where Y_out is GROUND LEVEL E₁ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Well width L, measured in nm. Keep en (Trial energy E) = 1.2 eV fixed. For this exercise use only 0.70625 nm ≤ w ≤ 1.49375 nm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.26705317 eV, with a bounded reading uncertainty ±0.014779 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875 nm, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
GROUND LEVEL E₁ against w
Predictions from the supplied Energy quantisation: the condition for a nontrivial solution model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): GROUND LEVEL E₁ / eV
Horizontal axis (x): w: Well width L / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Well width L / nm | GROUND LEVEL E₁ / eV |
|---|---|
| 0.7062499999999999 | 0.7538896819364491 |
| 0.903125 | 0.4610297936397561 |
| 1.0999999999999999 | 0.3107701881665328 |
| 1.296875 | 0.22357770007017602 |
| 1.49375 | 0.16852676508896058 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(w) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.70625) = 0.75389 eV and Y_out(1.49375) = 0.168527 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.70625 + (0.267053 − 0.75389)(1.49375 − 0.70625)/(0.168527 − 0.75389) = 1.3612 nm. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives w = 1.18662 nm; substituting back gives Y_out = 0.26705317 eV.
- The interpolation error in the input is |w_line − w| = 0.174576 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.25227417 eV and 0.28183217 eV. Sort the roots even if the response decreases: 1.15509 nm ≤ w ≤ 1.22089 nm.
- The lower and upper input deviations are 0.0315317 nm and 0.0342634 nm. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer w ≈ 1.18662 nm; compatible interval [1.15509, 1.22089] nm. Endpoint interpolation gives 1.3612 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Normalisation, the ground state, and zero-point energy: infer an input with uncertainty
The zero-point energy of the infinite square well is its lowest allowed energy, E₁ = π²ħ²/(2mL²), held by the normalised ground state ψ₁ = √(2/L) sin(πx/L) and kept by the particle even at absolute zero. Use the supplied topic model Y_out = 2/(d*(1-4*d/3)), where Y_out is TRAPEZOID ⟨T⟩ (ħ²/(2mL²)). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Trapezoid ramp length, measured in L. Keep w (Well width L) = 1 nm fixed. For this exercise use only 0.0978125 L ≤ d ≤ 0.357687 L on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 11.853593 ħ²/(2mL²), with a bounded reading uncertainty ±0.17462 ħ²/(2mL²). Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000259875 L, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
TRAPEZOID ⟨T⟩ against d
Predictions from the supplied Normalisation, the ground state, and zero-point energy model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): TRAPEZOID ⟨T⟩ / ħ²/(2mL²)
Horizontal axis (x): d: Trapezoid ramp length / L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Trapezoid ramp length / L | TRAPEZOID ⟨T⟩ / ħ²/(2mL²) |
|---|---|
| 0.0978125 | 23.513887124156692 |
| 0.16278125 | 15.692313133692718 |
| 0.22774999999999998 | 12.611142259464794 |
| 0.29271874999999997 | 11.206173002109892 |
| 0.35768749999999994 | 10.689449744150528 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(d) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.0978125) = 23.5139 ħ²/(2mL²) and Y_out(0.357687) = 10.6894 ħ²/(2mL²). The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.0978125 + (11.8536 − 23.5139)(0.357687 − 0.0978125)/(10.6894 − 23.5139) = 0.334097 L. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives d = 0.256336 L; substituting back gives Y_out = 11.853593 ħ²/(2mL²).
- The interpolation error in the input is |d_line − d| = 0.077761 L. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 11.678973 ħ²/(2mL²) and 12.028213 ħ²/(2mL²). Sort the roots even if the response decreases: 0.248833 L ≤ d ≤ 0.264596 L.
- The lower and upper input deviations are 0.00750356 L and 0.00825974 L. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer d ≈ 0.256336 L; compatible interval [0.248833, 0.264596] L. Endpoint interpolation gives 0.334097 L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Excited states, half-wavelengths, and nodes: infer an input with uncertainty
An excited state of the infinite well, ψₙ = √(2/L) sin(nπx/L) with n ≥ 2, is a standing wave with n half-wavelengths between the walls and n − 1 interior nodes: points where ψₙ, and so the probability density, is exactly zero. Use the supplied topic model Y_out = a-sin(2*pi*n*a)/(2*pi*n), where Y_out is P(0 < x < aL) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. a is Strip edge x = aL, measured in L. Keep n (Quantum number n) = 3; ph (Phase Eₙt/ħ) = 60 ° fixed. For this exercise use only 0.10625 L ≤ a ≤ 0.89375 L on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.63957456, with a bounded reading uncertainty ±0.045353. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875 L, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
P(0 < x < aL) against a
Predictions from the supplied Excited states, half-wavelengths, and nodes model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): P(0 < x < aL)
Horizontal axis (x): a: Strip edge x = aL / L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Strip edge x = aL / L | P(0 < x < aL) |
|---|---|
| 0.10625 | 0.058071508283525315 |
| 0.303125 | 0.33172717636718674 |
| 0.5 | 0.5 |
| 0.6968749999999999 | 0.6682728236328133 |
| 0.8937499999999999 | 0.9419284917164745 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(a) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.10625) = 0.0580715 and Y_out(0.89375) = 0.941928. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 0.10625 + (0.639575 − 0.0580715)(0.89375 − 0.10625)/(0.941928 − 0.0580715) = 0.624358 L. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives a = 0.586625 L; substituting back gives Y_out = 0.63957456.
- The interpolation error in the input is |a_line − a| = 0.0377333 L. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.59422156 and 0.68492756. Sort the roots even if the response decreases: 0.550823 L ≤ a ≤ 0.736193 L.
- The lower and upper input deviations are 0.0358018 L and 0.149568 L. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer a ≈ 0.586625 L; compatible interval [0.550823, 0.736193] L. Endpoint interpolation gives 0.624358 L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.