University Quantum Mechanics I · topic UQMI-02.04
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″ + V(x)ψ = Eψ, asks for the pairs of a number E and a function ψ for which the Hamiltonian returns ψ multiplied by E; the admissible pairs are the energy eigenvalues and eigenfunctions. Use the supplied topic model Y_out = 2*pi*sqrt(0.0381/en), where Y_out is LEFT WAVELENGTH (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. en is Electron energy E, measured in eV. Keep phi (Phase φ of the left-hand wave) = 45 ° fixed. For this exercise use only 0.29125 eV ≤ en ≤ 1.70875 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 = 0.6031 eV and 0.80155 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.
LEFT WAVELENGTH 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): LEFT WAVELENGTH / nm
Horizontal axis (x): en: Electron energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Electron energy E / eV | LEFT WAVELENGTH / nm |
|---|---|
| 0.29125 | 2.272528765410293 |
| 0.6456249999999999 | 1.5263431668161582 |
| 0.9999999999999999 | 1.226428844542568 |
| 1.3543749999999999 | 1.0538356638692952 |
| 1.7087499999999998 | 0.9382165561340096 |
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 = 0.6031 eV, substitute into Y_out = 2*pi*sqrt(0.0381/en): Y₀ = 1.57924 nm.
- At en = 0.80155 eV, the same fixed-parameter model gives Y₁ = 1.36986 nm.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.209376 nm; the input change is Δen = 0.19845 eV.
- Average sensitivity is ΔY_out/Δen = -1.05506 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 = 1.57924 nm, Y₁ = 1.36986 nm; ΔY_out = -0.209376 nm; average sensitivity = -1.05506 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″ + V(x)ψ = Eψ, asks for the pairs of a number E and a function ψ for which the Hamiltonian returns ψ multiplied by E; the admissible pairs are the energy eigenvalues and eigenfunctions. Use the supplied topic model Y_out = 2*pi*sqrt(0.0381/en), where Y_out is LEFT WAVELENGTH (nm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. en is Electron energy E, measured in eV. Keep phi (Phase φ of the left-hand wave) = 45 ° fixed. For this exercise use only 0.29125 eV ≤ en ≤ 1.70875 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 = 1.140716 nm, with a bounded reading uncertainty ±0.030375 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 0.00014175 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.
LEFT WAVELENGTH 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): LEFT WAVELENGTH / nm
Horizontal axis (x): en: Electron energy E / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| en: Electron energy E / eV | LEFT WAVELENGTH / nm |
|---|---|
| 0.29125 | 2.272528765410293 |
| 0.6456249999999999 | 1.5263431668161582 |
| 0.9999999999999999 | 1.226428844542568 |
| 1.3543749999999999 | 1.0538356638692952 |
| 1.7087499999999998 | 0.9382165561340096 |
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(0.29125) = 2.27253 nm and Y_out(1.70875) = 0.938217 nm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives en_line = 0.29125 + (1.14072 − 2.27253)(1.70875 − 0.29125)/(0.938217 − 2.27253) = 1.49363 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 = 1.15592 eV; substituting back gives Y_out = 1.140716 nm.
- The interpolation error in the input is |en_line − en| = 0.337701 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.110341 nm and 1.171091 nm. Sort the roots even if the response decreases: 1.09674 eV ≤ en ≤ 1.22003 eV.
- The lower and upper input deviations are 0.0591856 eV and 0.0641091 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 ≈ 1.15592 eV; compatible interval [1.09674, 1.22003] eV. Endpoint interpolation gives 1.49363 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.