University Quantum Mechanics I · topic UQMI-07.02
Classically forbidden regions and exponential decay
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.
Classically forbidden regions and exponential decay: predict a finite change
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. Calculate Y_out at dv = 0.4913 eV and 0.64565 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δdv across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of dv separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At dv = 0.4913 eV, substitute into Y_out = 1/sqrt(m*dv/0.0381): Y₀ = 0.278477 nm.
- At dv = 0.64565 eV, the same fixed-parameter model gives Y₁ = 0.24292 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0355565 nm; the input change is Δdv = 0.15435 eV.
- Average sensitivity is ΔY_out/Δdv = -0.230363 nm per eV. 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 = 0.278477 nm, Y₁ = 0.24292 nm; ΔY_out = -0.0355565 nm; average sensitivity = -0.230363 nm 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.
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.