University Quantum Mechanics I · topic UQMI-05.04
A continuous spectrum, degenerate above zero
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.
A continuous spectrum, degenerate above zero: predict a finite change
The spectrum of a free particle is the set of energies at which −(ħ²/2m)ψ″ = Eψ has a solution that stays bounded as x → ±∞: every E ≥ 0, with two independent solutions at each E > 0 and only the constant at E = 0. Use the supplied topic model Y_out = sqrt(max(-en,0)/0.0381), where Y_out is GROWTH CONSTANT κ (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. en is Energy E, measured in eV. Keep w (Window half-width W) = 6 nm fixed. For this exercise use only -0.225625 eV ≤ en ≤ -0.005125 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 = -0.177115 eV and -0.146245 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.
GROWTH CONSTANT κ against en
Predictions from the supplied A continuous spectrum, degenerate above zero 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): GROWTH CONSTANT κ / 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 | GROWTH CONSTANT κ / nm⁻¹ |
|---|---|
| -0.225625 | 2.4334987179981598 |
| -0.17049999999999998 | 2.115435089242376 |
| -0.11537499999999999 | 1.7401767792661504 |
| -0.06025 | 1.2575232917907824 |
| -0.005124999999999991 | 0.3667620968632634 |
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 = -0.177115 eV, substitute into Y_out = sqrt(max(-en,0)/0.0381): Y₀ = 2.15608 nm⁻¹.
- At en = -0.146245 eV, the same fixed-parameter model gives Y₁ = 1.9592 nm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.196885 nm⁻¹; the input change is Δen = 0.03087 eV.
- Average sensitivity is ΔY_out/Δen = -6.37787 nm⁻¹ 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 = 2.15608 nm⁻¹, Y₁ = 1.9592 nm⁻¹; ΔY_out = -0.196885 nm⁻¹; average sensitivity = -6.37787 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.
A continuous spectrum, degenerate above zero: infer an input with uncertainty
The spectrum of a free particle is the set of energies at which −(ħ²/2m)ψ″ = Eψ has a solution that stays bounded as x → ±∞: every E ≥ 0, with two independent solutions at each E > 0 and only the constant at E = 0. Use the supplied topic model Y_out = sqrt(max(-en,0)/0.0381), where Y_out is GROWTH CONSTANT κ (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. en is Energy E, measured in eV. Keep w (Window half-width W) = 6 nm fixed. For this exercise use only -0.225625 eV ≤ en ≤ -0.005125 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 = 1.5464802 nm⁻¹, with a bounded reading uncertainty ±0.13305 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 0.00002205 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.
GROWTH CONSTANT κ against en
Predictions from the supplied A continuous spectrum, degenerate above zero 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): GROWTH CONSTANT κ / 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 | GROWTH CONSTANT κ / nm⁻¹ |
|---|---|
| -0.225625 | 2.4334987179981598 |
| -0.17049999999999998 | 2.115435089242376 |
| -0.11537499999999999 | 1.7401767792661504 |
| -0.06025 | 1.2575232917907824 |
| -0.005124999999999991 | 0.3667620968632634 |
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(-0.225625) = 2.4335 nm⁻¹ and Y_out(-0.005125) = 0.366762 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives en_line = -0.225625 + (1.54648 − 2.4335)(-0.005125 − -0.225625)/(0.366762 − 2.4335) = -0.130989 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 = -0.09112 eV; substituting back gives Y_out = 1.5464802 nm⁻¹.
- The interpolation error in the input is |en_line − en| = 0.0398691 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.4134302 nm⁻¹ and 1.6795302 nm⁻¹. Sort the roots even if the response decreases: -0.107473 eV ≤ en ≤ -0.0761156 eV.
- The lower and upper input deviations are 0.0163533 eV and 0.0150044 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 ≈ -0.09112 eV; compatible interval [-0.107473, -0.0761156] eV. Endpoint interpolation gives -0.130989 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.