University Physics IV · unit UPIV-07
Quantum Potentials
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.
Piecewise potentials and matching: infer an input with uncertainty
A piecewise-constant potential is flat inside each of a few regions and jumps between them, so the Schrödinger equation is solved separately in every region and the pieces are joined by demanding that ψ and ψ′ agree at each boundary. Use the supplied topic model Y_out = 5.1227*sqrt(en)*cos(2.56135*sqrt(en))+5.1227*sqrt(max(v0-en,0.02))*sin(2.56135*sqrt(en)), where Y_out is ψ′ MISMATCH AT STEP (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value; unmarked trigonometric functions use radians. en is Energy E, measured in eV. Keep v0 (Step height V₀) = 6 eV fixed. For this exercise use only 0.67625 eV ≤ en ≤ 2.3825 eV 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 = -8.8173564 nm⁻¹, with a bounded reading uncertainty ±0.554 nm⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown en by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000170625 eV, 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.
ψ′ MISMATCH AT STEP against en
Predictions from the supplied Piecewise potentials and matching 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): ψ′ MISMATCH AT STEP / nm⁻¹
Horizontal axis (x): en: Energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Energy E / eV | ψ′ MISMATCH AT STEP / nm⁻¹ |
|---|---|
| 0.67625 | 8.015408474004285 |
| 1.1028125 | 0.10937867828723302 |
| 1.529375 | -6.61431321527364 |
| 1.9559374999999999 | -10.873470809858201 |
| 2.3825 | -12.510709565756605 |
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(en) = 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(0.67625) = 8.01541 nm⁻¹ and Y_out(2.3825) = -12.5107 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives en_line = 0.67625 + (-8.81736 − 8.01541)(2.3825 − 0.67625)/(-12.5107 − 8.01541) = 2.07549 eV. 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 en = 1.71706 eV; substituting back gives Y_out = -8.8173564 nm⁻¹.
- The interpolation error in the input is |en_line − en| = 0.358425 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -9.3713564 nm⁻¹ and -8.2633564 nm⁻¹. Sort the roots even if the response decreases: 1.66553 eV ≤ en ≤ 1.77268 eV.
- The lower and upper input deviations are 0.0515336 eV and 0.0556148 eV. 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 en ≈ 1.71706 eV; compatible interval [1.66553, 1.77268] eV. Endpoint interpolation gives 2.07549 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The infinite square well: infer an input with uncertainty
The infinite square well is the model of a particle that moves freely inside a segment of length L and is forbidden outside it, so its stationary states are the standing sine waves that vanish at both walls. Use the supplied topic model Y_out = 0.376*(2*n+1)/max(w*w,0.0025), where Y_out is GAP TO n+1 (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. w is Well width L, measured in nm. Keep n (Quantum number n) = 2 fixed. For this exercise use only 0.46375 nm ≤ w ≤ 0.93625 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 = 3.3246891 eV, with a bounded reading uncertainty ±0.17699 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 0.00004725 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.
GAP TO n+1 against w
Predictions from the supplied The infinite square well 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): GAP TO n+1 / 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 | GAP TO n+1 / eV |
|---|---|
| 0.46375 | 8.74158136020517 |
| 0.5818749999999999 | 5.5526263872047785 |
| 0.7 | 3.8367346938775513 |
| 0.818125 | 2.8087851572325575 |
| 0.9362499999999999 | 2.144737709914956 |
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(0.46375) = 8.74158 eV and Y_out(0.93625) = 2.14474 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.46375 + (3.32469 − 8.74158)(0.93625 − 0.46375)/(2.14474 − 8.74158) = 0.851736 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 = 0.751975 nm; substituting back gives Y_out = 3.3246891 eV.
- The interpolation error in the input is |w_line − w| = 0.0997608 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.1476991 eV and 3.5016791 eV. Sort the roots even if the response decreases: 0.732725 nm ≤ w ≤ 0.772827 nm.
- The lower and upper input deviations are 0.0192504 nm and 0.0208521 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 ≈ 0.751975 nm; compatible interval [0.732725, 0.772827] nm. Endpoint interpolation gives 0.851736 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Superpositions and time evolution in a box: infer an input with uncertainty
A superposition in a box is a normalised sum Ψ = Σ cₙψₙ over the well's energy eigenfunctions, in which |cₙ|² is the probability that an energy measurement returns Eₙ and each term evolves by its own phase factor e^(−iEₙt/ħ). Use the supplied topic model Y_out = cosd(a)^2 + n^2*sind(a)^2, where Y_out is MEAN ENERGY ⟨H⟩ (E₁). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. a is Mixing angle θ, measured in °. Keep n (Upper level n) = 2; t (Elapsed time in beat periods) = 0 fixed. For this exercise use only 9.5625 ° ≤ a ≤ 80.4375 ° 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 = 2.9031906 E₁, with a bounded reading uncertainty ±0.1521 E₁. 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 0.0070875 °, 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.
MEAN ENERGY ⟨H⟩ against a
Predictions from the supplied Superpositions and time evolution in a box 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): MEAN ENERGY ⟨H⟩ / E₁
Horizontal axis (x): a: Mixing angle θ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Mixing angle θ / ° | MEAN ENERGY ⟨H⟩ / E₁ |
|---|---|
| 9.5625 | 1.082790930299683 |
| 27.28125 | 1.6302781778102293 |
| 45 | 2.4999999999999996 |
| 62.71875 | 3.3697218221897702 |
| 80.4375 | 3.9172090697003172 |
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(9.5625) = 1.08279 E₁ and Y_out(80.4375) = 3.91721 E₁. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 9.5625 + (2.90319 − 1.08279)(80.4375 − 9.5625)/(3.91721 − 1.08279) = 55.0818 °. 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 = 52.7962 °; substituting back gives Y_out = 2.9031906 E₁.
- The interpolation error in the input is |a_line − a| = 2.28558 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.7510906 E₁ and 3.0552906 E₁. Sort the roots even if the response decreases: 49.8182 ° ≤ a ≤ 55.8638 °.
- The lower and upper input deviations are 2.97809 ° and 3.06753 °. 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 ≈ 52.7962 °; compatible interval [49.8182, 55.8638] °. Endpoint interpolation gives 55.0818 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Expectation values in a box: infer an input with uncertainty
For a well eigenstate, the expectation value of an observable is the probability-weighted mean of the values a measurement could return, and its uncertainty is the standard deviation of that same distribution, σ_A = √(⟨A²⟩ − ⟨A⟩²). Use the supplied topic model Y_out = 0.3760*n*n/(w*w), where Y_out is ELECTRON ENERGY E_n (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 n (Quantum number n) = 1 fixed. For this exercise use only 0.30625 nm ≤ w ≤ 1.09375 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 = 0.6076484 eV, with a bounded reading uncertainty ±0.044001 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.
ELECTRON ENERGY E_n against w
Predictions from the supplied Expectation values in a box 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): ELECTRON ENERGY E_n / 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 | ELECTRON ENERGY E_n / eV |
|---|---|
| 0.30625 | 4.0089962515618485 |
| 0.503125 | 1.4853747926391727 |
| 0.7 | 0.7673469387755103 |
| 0.896875 | 0.4674379924486154 |
| 1.09375 | 0.314305306122449 |
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(0.30625) = 4.009 eV and Y_out(1.09375) = 0.314305 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives w_line = 0.30625 + (0.607648 − 4.009)(1.09375 − 0.30625)/(0.314305 − 4.009) = 1.03123 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 = 0.786625 nm; substituting back gives Y_out = 0.6076484 eV.
- The interpolation error in the input is |w_line − w| = 0.244601 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.5636474 eV and 0.6516494 eV. Sort the roots even if the response decreases: 0.759603 nm ≤ w ≤ 0.816752 nm.
- The lower and upper input deviations are 0.0270216 nm and 0.0301269 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 ≈ 0.786625 nm; compatible interval [0.759603, 0.816752] nm. Endpoint interpolation gives 1.03123 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.