University Quantum Mechanics I · topic UQMI-07.04
Matching conditions: psi and psi' continuous at each wall
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.
Matching ψ and ψ′ at the walls of a finite well: predict a finite change
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. Calculate Y_out at en = 0.786792 eV and 0.883371 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δen across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of en separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At en = 0.786792 eV, substitute into 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): Y₀ = -22.3546 nm⁻¹.
- At en = 0.883371 eV, the same fixed-parameter model gives Y₁ = -17.0018 nm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 5.35279 nm⁻¹; the input change is Δen = 0.096579 eV.
- Average sensitivity is ΔY_out/Δen = 55.4239 nm⁻¹ per eV. Its positive sign means that this response increases 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 = -22.3546 nm⁻¹, Y₁ = -17.0018 nm⁻¹; ΔY_out = 5.35279 nm⁻¹; average sensitivity = 55.4239 nm⁻¹ per eV.
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.
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.