University Quantum Mechanics I · topic UQMI-02.01
The time-dependent 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-dependent Schrödinger equation: predict a finite change
The time-dependent Schrödinger equation, iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + VΨ, is the postulated law of motion for a wavefunction: given V and Ψ at one instant, it fixes Ψ at every later instant. Use the supplied topic model Y_out = 1.994711/(2*(1+exp(-d^2/0.08)))*4*exp(-d^2/(0.08*(1+(1.447095*t)^2)))/sqrt((1+(1.447095*t)^2)), where Y_out is DENSITY AT x = 0 (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ. t is Time since release t, measured in fs. Keep d (Starting offset of each packet d) = 1 nm fixed. For this exercise use only 1.29688 fs ≤ t ≤ 3.26563 fs 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 t = 1.73 fs and 2.00563 fs. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δt across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DENSITY AT x = 0 against t
Predictions from the supplied The time-dependent 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): DENSITY AT x = 0 / nm⁻¹
Horizontal axis (x): t: Time since release t / fs
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since release t / fs | DENSITY AT x = 0 / nm⁻¹ |
|---|---|
| 1.296875 | 0.11823345080310765 |
| 1.7890625 | 0.2836651117021689 |
| 2.28125 | 0.4044822834139529 |
| 2.7734375 | 0.46450257606941586 |
| 3.265625 | 0.483349519507742 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of t separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At t = 1.73 fs, substitute into Y_out = 1.994711/(2*(1+exp(-d^2/0.08)))*4*exp(-d^2/(0.08*(1+(1.447095*t)^2)))/sqrt((1+(1.447095*t)^2)): Y₀ = 0.264988 nm⁻¹.
- At t = 2.00563 fs, the same fixed-parameter model gives Y₁ = 0.344924 nm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.079936 nm⁻¹; the input change is Δt = 0.275625 fs.
- Average sensitivity is ΔY_out/Δt = 0.290017 nm⁻¹ per fs. 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 = 0.264988 nm⁻¹, Y₁ = 0.344924 nm⁻¹; ΔY_out = 0.079936 nm⁻¹; average sensitivity = 0.290017 nm⁻¹ per fs.
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-dependent Schrödinger equation: infer an input with uncertainty
The time-dependent Schrödinger equation, iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + VΨ, is the postulated law of motion for a wavefunction: given V and Ψ at one instant, it fixes Ψ at every later instant. Use the supplied topic model Y_out = 1.994711/(2*(1+exp(-d^2/0.08)))*4*exp(-d^2/(0.08*(1+(1.447095*t)^2)))/sqrt((1+(1.447095*t)^2)), where Y_out is DENSITY AT x = 0 (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ. t is Time since release t, measured in fs. Keep d (Starting offset of each packet d) = 1 nm fixed. For this exercise use only 1.29688 fs ≤ t ≤ 3.26563 fs 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 = 0.43739029 nm⁻¹, with a bounded reading uncertainty ±0.0068939 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 t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000196875 fs, 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.
DENSITY AT x = 0 against t
Predictions from the supplied The time-dependent 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): DENSITY AT x = 0 / nm⁻¹
Horizontal axis (x): t: Time since release t / fs
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since release t / fs | DENSITY AT x = 0 / nm⁻¹ |
|---|---|
| 1.296875 | 0.11823345080310765 |
| 1.7890625 | 0.2836651117021689 |
| 2.28125 | 0.4044822834139529 |
| 2.7734375 | 0.46450257606941586 |
| 3.265625 | 0.483349519507742 |
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(t) = 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(1.29688) = 0.118233 nm⁻¹ and Y_out(3.26563) = 0.48335 nm⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 1.29688 + (0.43739 − 0.118233)(3.26563 − 1.29688)/(0.48335 − 0.118233) = 3.01781 fs. 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 t = 2.49781 fs; substituting back gives Y_out = 0.43739029 nm⁻¹.
- The interpolation error in the input is |t_line − t| = 0.519995 fs. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.43049639 nm⁻¹ and 0.44428419 nm⁻¹. Sort the roots even if the response decreases: 2.44571 fs ≤ t ≤ 2.55516 fs.
- The lower and upper input deviations are 0.0520975 fs and 0.0573523 fs. 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 t ≈ 2.49781 fs; compatible interval [2.44571, 2.55516] fs. Endpoint interpolation gives 3.01781 fs.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.