University Quantum Mechanics II · topic UQMII-12.03
First-order amplitudes and a perturbation switched on at t = 0
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.
First-order amplitudes and a perturbation turned on at t = 0: predict a finite change
The first-order amplitude cₘ(t) is the amplitude for reaching |m⟩ when the perturbation may act only once on the initial state |n⟩, at any moment t′ between 0 and t, each moment weighted by the phase e^(iωₘₙt′). Use the supplied topic model Y_out = 1000/(t-ramp/2), where Y_out is FIRST ZEROS ±1/(t − T/2) (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. t is Time since switch-on t, measured in ns. Keep ramp (Linear ramp duration T) = 0 ns fixed. For this exercise use only 46.375 ns ≤ t ≤ 93.625 ns 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 = 56.77 ns and 63.385 ns. 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.
FIRST ZEROS ±1/(t − T/2) against t
Predictions from the supplied First-order amplitudes and a perturbation turned on at t = 0 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): FIRST ZEROS ±1/(t − T/2) / MHz
Horizontal axis (x): t: Time since switch-on t / ns
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since switch-on t / ns | FIRST ZEROS ±1/(t − T/2) / MHz |
|---|---|
| 46.375 | 21.5633423180593 |
| 58.1875 | 17.185821697099893 |
| 70 | 14.285714285714286 |
| 81.8125 | 12.223071046600458 |
| 93.625 | 10.68090787716956 |
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 = 56.77 ns, substitute into Y_out = 1000/(t-ramp/2): Y₀ = 17.6149 MHz.
- At t = 63.385 ns, the same fixed-parameter model gives Y₁ = 15.7766 MHz.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.83833 MHz; the input change is Δt = 6.615 ns.
- Average sensitivity is ΔY_out/Δt = -0.277904 MHz per ns. 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 = 17.6149 MHz, Y₁ = 15.7766 MHz; ΔY_out = -1.83833 MHz; average sensitivity = -0.277904 MHz per ns.
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.
First-order amplitudes and a perturbation turned on at t = 0: infer an input with uncertainty
The first-order amplitude cₘ(t) is the amplitude for reaching |m⟩ when the perturbation may act only once on the initial state |n⟩, at any moment t′ between 0 and t, each moment weighted by the phase e^(iωₘₙt′). Use the supplied topic model Y_out = 1000/(t-ramp/2), where Y_out is FIRST ZEROS ±1/(t − T/2) (MHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. t is Time since switch-on t, measured in ns. Keep ramp (Linear ramp duration T) = 0 ns fixed. For this exercise use only 46.375 ns ≤ t ≤ 93.625 ns 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 = 13.298314 MHz, with a bounded reading uncertainty ±0.39261 MHz. 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 0.004725 ns, 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.
FIRST ZEROS ±1/(t − T/2) against t
Predictions from the supplied First-order amplitudes and a perturbation turned on at t = 0 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): FIRST ZEROS ±1/(t − T/2) / MHz
Horizontal axis (x): t: Time since switch-on t / ns
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Time since switch-on t / ns | FIRST ZEROS ±1/(t − T/2) / MHz |
|---|---|
| 46.375 | 21.5633423180593 |
| 58.1875 | 17.185821697099893 |
| 70 | 14.285714285714286 |
| 81.8125 | 12.223071046600458 |
| 93.625 | 10.68090787716956 |
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(46.375) = 21.5633 MHz and Y_out(93.625) = 10.6809 MHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives t_line = 46.375 + (13.2983 − 21.5633)(93.625 − 46.375)/(10.6809 − 21.5633) = 82.2606 ns. 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 = 75.1975 ns; substituting back gives Y_out = 13.298314 MHz.
- The interpolation error in the input is |t_line − t| = 7.06309 ns. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 12.905704 MHz and 13.690924 MHz. Sort the roots even if the response decreases: 73.0411 ns ≤ t ≤ 77.4851 ns.
- The lower and upper input deviations are 2.15641 ns and 2.28762 ns. 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 ≈ 75.1975 ns; compatible interval [73.0411, 77.4851] ns. Endpoint interpolation gives 82.2606 ns.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.