University Physics III · topic UPIII-09.07
The time-independent Schrodinger equation
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 time-independent Schrödinger equation: predict a finite change
The time-independent Schrödinger equation, −(ħ²/2m)ψ″ + U(x)ψ = Eψ, is the eigenvalue equation for energy: its acceptable solutions ψ(x) are the stationary states, and the E values at which they exist are the allowed energies. Use the supplied topic model Y_out = sqrt(0.0381/(3-en)), where Y_out is DECAY LENGTH 1/κ OUTSIDE (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. en is Trial energy E, measured in eV. Keep w (Well width L) = 1 nm fixed. For this exercise use only 0.326625 eV ≤ en ≤ 2.67338 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 = 0.84291 eV and 1.17145 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.
DECAY LENGTH 1/κ OUTSIDE against en
Predictions from the supplied The time-independent Schrödinger equation 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): DECAY LENGTH 1/κ OUTSIDE / nm
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 | DECAY LENGTH 1/κ OUTSIDE / nm |
|---|---|
| 0.32662500000000005 | 0.11938026720318211 |
| 0.9133125000000001 | 0.13512440063928688 |
| 1.5 | 0.15937377450509227 |
| 2.0866875 | 0.20424563648860555 |
| 2.673375 | 0.34153701347424226 |
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 = 0.84291 eV, substitute into Y_out = sqrt(0.0381/(3-en)): Y₀ = 0.132901 nm.
- At en = 1.17145 eV, the same fixed-parameter model gives Y₁ = 0.144348 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0114466 nm; the input change is Δen = 0.328545 eV.
- Average sensitivity is ΔY_out/Δen = 0.0348403 nm 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 = 0.132901 nm, Y₁ = 0.144348 nm; ΔY_out = 0.0114466 nm; average sensitivity = 0.0348403 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.
The time-independent Schrödinger equation: infer an input with uncertainty
The time-independent Schrödinger equation, −(ħ²/2m)ψ″ + U(x)ψ = Eψ, is the eigenvalue equation for energy: its acceptable solutions ψ(x) are the stationary states, and the E values at which they exist are the allowed energies. Use the supplied topic model Y_out = sqrt(0.0381/(3-en)), where Y_out is DECAY LENGTH 1/κ OUTSIDE (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. en is Trial energy E, measured in eV. Keep w (Well width L) = 1 nm fixed. For this exercise use only 0.326625 eV ≤ en ≤ 2.67338 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 = 0.17515664 nm, with a bounded reading uncertainty ±0.0083665 nm. 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 0.000234675 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.
DECAY LENGTH 1/κ OUTSIDE against en
Predictions from the supplied The time-independent Schrödinger equation 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): DECAY LENGTH 1/κ OUTSIDE / nm
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 | DECAY LENGTH 1/κ OUTSIDE / nm |
|---|---|
| 0.32662500000000005 | 0.11938026720318211 |
| 0.9133125000000001 | 0.13512440063928688 |
| 1.5 | 0.15937377450509227 |
| 2.0866875 | 0.20424563648860555 |
| 2.673375 | 0.34153701347424226 |
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(0.326625) = 0.11938 nm and Y_out(2.67338) = 0.341537 nm. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives en_line = 0.326625 + (0.175157 − 0.11938)(2.67338 − 0.326625)/(0.341537 − 0.11938) = 0.915818 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 = 1.75814 eV; substituting back gives Y_out = 0.17515664 nm.
- The interpolation error in the input is |en_line − en| = 0.842324 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.16679014 nm and 0.18352314 nm. Sort the roots even if the response decreases: 1.63043 eV ≤ en ≤ 1.86879 eV.
- The lower and upper input deviations are 0.127712 eV and 0.110647 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 ≈ 1.75814 eV; compatible interval [1.63043, 1.86879] eV. Endpoint interpolation gives 0.915818 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.