University Quantum Mechanics I · unit UQMI-07
Finite Square Wells
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 finite square well: three regions and two energy sectors: infer an input with uncertainty
The finite square well is the potential V = 0 for |x| < a and V = V₀ for |x| > a; its energy eigenstates fall into a bound sector, 0 < E < V₀, with discrete energies and tails that decay outside, and an unbound sector, E > V₀, where every energy is allowed. Use the supplied topic model Y_out = sqrt(max(v0-en,0)/0.0381), where Y_out is OUTSIDE κ, IF E < V₀ (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. v0 is Wall height V₀, measured in eV. Keep en (Electron energy E) = 0.5 eV fixed. For this exercise use only 0.506375 eV ≤ v0 ≤ 1.46188 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 = 3.9326015 nm⁻¹, with a bounded reading uncertainty ±0.16379 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 v0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00009555 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.
OUTSIDE κ, IF E < V₀ against v0
Predictions from the supplied The finite square well: three regions and two energy sectors 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): OUTSIDE κ, IF E < V₀ / nm⁻¹
Horizontal axis (x): v0: Wall height V₀ / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v0: Wall height V₀ / eV | OUTSIDE κ, IF E < V₀ / nm⁻¹ |
|---|---|
| 0.506375 | 0.4090511394015054 |
| 0.74525 | 2.537125908191343 |
| 0.984125 | 3.5646448509474022 |
| 1.223 | 4.356188466165795 |
| 1.461875 | 5.024546048363572 |
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(v0) = 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(0.506375) = 0.409051 nm⁻¹ and Y_out(1.46188) = 5.02455 nm⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives v0_line = 0.506375 + (3.9326 − 0.409051)(1.46188 − 0.506375)/(5.02455 − 0.409051) = 1.23582 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 v0 = 1.08923 eV; substituting back gives Y_out = 3.9326015 nm⁻¹.
- The interpolation error in the input is |v0_line − v0| = 0.146591 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.7688115 nm⁻¹ and 4.0963915 nm⁻¹. Sort the roots even if the response decreases: 1.04117 eV ≤ v0 ≤ 1.13933 eV.
- The lower and upper input deviations are 0.0480599 eV and 0.0501041 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 v0 ≈ 1.08923 eV; compatible interval [1.04117, 1.13933] eV. Endpoint interpolation gives 1.23582 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Classically forbidden regions and exponential decay: infer an input with uncertainty
A classically forbidden region is any stretch where the potential energy V exceeds the particle's energy E; where V is flat there, ψ″ = κ²ψ, so ψ is built from real exponentials and its decaying part falls by a factor e over the penetration depth 1/κ = ħ/√(2m(V − E)). Use the supplied topic model Y_out = 1/sqrt(m*dv/0.0381), where Y_out is PENETRATION DEPTH δ = 1/κ (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. dv is Energy deficit V₀ − E, measured in eV. Keep en (Energy above the floor E) = 1 eV; m (Particle mass m) = 1 mₑ fixed. For this exercise use only 0.24875 eV ≤ dv ≤ 1.35125 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 = 0.20336108 nm, with a bounded reading uncertainty ±0.0053166 nm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dv by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00011025 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.
PENETRATION DEPTH δ = 1/κ against dv
Predictions from the supplied Classically forbidden regions and exponential decay 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): PENETRATION DEPTH δ = 1/κ / nm
Horizontal axis (x): dv: Energy deficit V₀ − E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dv: Energy deficit V₀ − E / eV | PENETRATION DEPTH δ = 1/κ / nm |
|---|---|
| 0.24875 | 0.3913640621540622 |
| 0.5243749999999999 | 0.2695513422383628 |
| 0.7999999999999999 | 0.21823152842795193 |
| 1.0756249999999998 | 0.1882053843687896 |
| 1.3512499999999998 | 0.16791698755219245 |
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(dv) = 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(0.24875) = 0.391364 nm and Y_out(1.35125) = 0.167917 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dv_line = 0.24875 + (0.203361 − 0.391364)(1.35125 − 0.24875)/(0.167917 − 0.391364) = 1.17637 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 dv = 0.921275 eV; substituting back gives Y_out = 0.20336108 nm.
- The interpolation error in the input is |dv_line − dv| = 0.255092 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.19804448 nm and 0.20867768 nm. Sort the roots even if the response decreases: 0.874929 eV ≤ dv ≤ 0.971403 eV.
- The lower and upper input deviations are 0.0463457 eV and 0.0501281 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 dv ≈ 0.921275 eV; compatible interval [0.874929, 0.971403] eV. Endpoint interpolation gives 1.17637 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Bound states and what normalisability demands of the tails: infer an input with uncertainty
A bound state is a solution of the time-independent Schrödinger equation whose ∫|ψ|² dx over all x is finite, so that it can be normalised; in a finite well this forces 0 < E < V₀ and tails that die away as e^(−κ|x|). Use the supplied topic model Y_out = cos(2*sqrt(v0*f/0.1524))+0.5*(sqrt((1-f)/max(f,0.001))-sqrt(f/max(1-f,0.001)))*sin(2*sqrt(v0*f/0.1524)), where Y_out is GROWING COEFFICIENT D (F = 1) (dimensionless). 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. f is Trial energy E, as a fraction of V₀, measured in × V₀. Keep v0 (Well depth V₀) = 2 eV fixed. For this exercise use only 0.122 × V₀ ≤ f ≤ 0.2228 × V₀ 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 = -0.96816101, with a bounded reading uncertainty ±0.026454. 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 0.00001008 × V₀, 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.
GROWING COEFFICIENT D (F = 1) against f
Predictions from the supplied Bound states and what normalisability demands of the tails 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): GROWING COEFFICIENT D (F = 1)
Horizontal axis (x): f: Trial energy E, as a fraction of V₀ / × V₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Trial energy E, as a fraction of V₀ / × V₀ | GROWING COEFFICIENT D (F = 1) |
|---|---|
| 0.122 | -0.1565831103293227 |
| 0.1472 | -0.5827573847272279 |
| 0.1724 | -0.8758681076922413 |
| 0.19759999999999997 | -1.0568656155720602 |
| 0.22279999999999997 | -1.1445215438618357 |
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(0.122) = -0.156583 and Y_out(0.2228) = -1.14452. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = 0.122 + (-0.968161 − -0.156583)(0.2228 − 0.122)/(-1.14452 − -0.156583) = 0.204806 × V₀. 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 = 0.183488 × V₀; substituting back gives Y_out = -0.96816101.
- The interpolation error in the input is |f_line − f| = 0.0213178 × V₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.99461501 and -0.94170701. Sort the roots even if the response decreases: 0.180043 × V₀ ≤ f ≤ 0.187216 × V₀.
- The lower and upper input deviations are 0.0034449 × V₀ and 0.0037283 × V₀. 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 ≈ 0.183488 × V₀; compatible interval [0.180043, 0.187216] × V₀. Endpoint interpolation gives 0.204806 × V₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Matching ψ and ψ′ at the walls of a finite well: infer an input with uncertainty
Matching at a finite wall means requiring ψ and ψ′ to be continuous there; their ratio ψ′/ψ, the logarithmic derivative, then takes the same value on both sides, and since the decaying tails fix it at −κ on the right and +κ on the left, each wall gives one amplitude-free equation tying the interior solution to E. Use the supplied topic model Y_out = clamp(sqrt(en/0.0381)*tan(0.4*sqrt(en/0.0381)+phi*pi/180)-sqrt((1.5-en)/0.0381),-99,99), where Y_out is MISMATCH AT x = −a (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; clamp(x,a,b) confines x to [a,b]; unmarked trigonometric functions use radians. en is Trial energy E, measured in eV. Keep phi (Interior phase φ) = 0 ° fixed. For this exercise use only 0.635025 eV ≤ en ≤ 1.32487 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 = -12.298931 nm⁻¹, with a bounded reading uncertainty ±0.64325 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 0.000068985 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.
MISMATCH AT x = −a against en
Predictions from the supplied Matching ψ and ψ′ at the walls of a finite 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): MISMATCH AT x = −a / nm⁻¹
Horizontal axis (x): en: Trial energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Trial energy E / eV | MISMATCH AT x = −a / nm⁻¹ |
|---|---|
| 0.6350250000000001 | -70.28553287809387 |
| 0.8074875 | -20.85397854361356 |
| 0.9799499999999999 | -13.98718247701727 |
| 1.1524124999999998 | -10.577799032816166 |
| 1.3248749999999998 | -8.0105947644767 |
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(0.635025) = -70.2855 nm⁻¹ and Y_out(1.32487) = -8.01059 nm⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives en_line = 0.635025 + (-12.2989 − -70.2855)(1.32487 − 0.635025)/(-8.01059 − -70.2855) = 1.27737 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 = 1.05583 eV; substituting back gives Y_out = -12.298931 nm⁻¹.
- The interpolation error in the input is |en_line − en| = 0.221538 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -12.942181 nm⁻¹ and -11.655681 nm⁻¹. Sort the roots even if the response decreases: 1.02478 eV ≤ en ≤ 1.08964 eV.
- The lower and upper input deviations are 0.0310563 eV and 0.0338108 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 ≈ 1.05583 eV; compatible interval [1.02478, 1.08964] eV. Endpoint interpolation gives 1.27737 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.