University Quantum Mechanics I · topic UQMI-07.07
The number of bound states, counted from z0
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.
Further practice
Work through a fresh case before tackling the extension.
The number of bound states, counted from z₀: predict a finite change
The number of bound states of a finite square well of depth V₀ and half-width a is fixed by z₀ = (a/ħ)√(2mV₀) alone: N = ⌊2z₀/π⌋ + 1, one more than the integer part of 2z₀/π. Use the supplied topic model Y_out = 0.0381*(pi*(floor(2*a*sqrt(v0/0.0381)/pi)+1)/(2*a))^2, where Y_out is DEPTH FOR THE NEXT STATE (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. a is Half-width a, measured in nm. Keep v0 (Well depth V₀) = 2 eV fixed. For this exercise use only 0.108437 nm ≤ a ≤ 0.195062 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. Calculate Y_out at a = 0.127495 nm and 0.139622 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DEPTH FOR THE NEXT STATE against a
Predictions from the supplied The number of bound states, counted from z₀ 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): DEPTH FOR THE NEXT STATE / eV
Horizontal axis (x): a: Half-width a / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Half-width a / nm | DEPTH FOR THE NEXT STATE / eV |
|---|---|
| 0.10843749999999999 | 7.994765631013062 |
| 0.13009374999999998 | 5.554588166520395 |
| 0.15175 | 4.082322684349851 |
| 0.17340624999999998 | 3.1263330137678764 |
| 0.19506249999999997 | 2.470683743669056 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of a separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a = 0.127495 nm, substitute into Y_out = 0.0381*(pi*(floor(2*a*sqrt(v0/0.0381)/pi)+1)/(2*a))^2: Y₀ = 5.78334 eV.
- At a = 0.139622 nm, the same fixed-parameter model gives Y₁ = 4.8223 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.961039 eV; the input change is Δa = 0.0121275 nm.
- Average sensitivity is ΔY_out/Δa = -79.2446 eV per nm. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 5.78334 eV, Y₁ = 4.8223 eV; ΔY_out = -0.961039 eV; average sensitivity = -79.2446 eV per nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
The number of bound states, counted from z₀: infer an input with uncertainty
The number of bound states of a finite square well of depth V₀ and half-width a is fixed by z₀ = (a/ħ)√(2mV₀) alone: N = ⌊2z₀/π⌋ + 1, one more than the integer part of 2z₀/π. Use the supplied topic model Y_out = 0.0381*(pi*(floor(2*a*sqrt(v0/0.0381)/pi)+1)/(2*a))^2, where Y_out is DEPTH FOR THE NEXT STATE (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. a is Half-width a, measured in nm. Keep v0 (Well depth V₀) = 2 eV fixed. For this exercise use only 0.108437 nm ≤ a ≤ 0.195062 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 = 3.6141854 eV, with a bounded reading uncertainty ±0.17153 eV. 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 0.0000086625 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.
DEPTH FOR THE NEXT STATE against a
Predictions from the supplied The number of bound states, counted from z₀ 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): DEPTH FOR THE NEXT STATE / eV
Horizontal axis (x): a: Half-width a / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Half-width a / nm | DEPTH FOR THE NEXT STATE / eV |
|---|---|
| 0.10843749999999999 | 7.994765631013062 |
| 0.13009374999999998 | 5.554588166520395 |
| 0.15175 | 4.082322684349851 |
| 0.17340624999999998 | 3.1263330137678764 |
| 0.19506249999999997 | 2.470683743669056 |
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(0.108437) = 7.99477 eV and Y_out(0.195062) = 2.47068 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a_line = 0.108437 + (3.61419 − 7.99477)(0.195062 − 0.108437)/(2.47068 − 7.99477) = 0.177131 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 a = 0.161279 nm; substituting back gives Y_out = 3.6141854 eV.
- The interpolation error in the input is |a_line − a| = 0.0158521 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.4426554 eV and 3.7857154 eV. Sort the roots even if the response decreases: 0.157583 nm ≤ a ≤ 0.165248 nm.
- The lower and upper input deviations are 0.00369611 nm and 0.00396901 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 a ≈ 0.161279 nm; compatible interval [0.157583, 0.165248] nm. Endpoint interpolation gives 0.177131 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.