University Advanced Quantum Mechanics · topic UAQM-01.05
Time reversal for a spinless particle: Theta = K and real wavefunctions
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.
Spinless time reversal: Θ = K and real wavefunctions: predict a finite change
For a particle without spin, time reversal is complex conjugation of the position-space wavefunction combined with running time backwards: Θψ(x, t) = ψ*(x, −t). Use the supplied topic model Y_out = (v0/2)/hypot(0.05788*b, v0/2), where Y_out is OVERLAP |⟨Θψ|ψ⟩| (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²). v0 is T-even well depth V₀, measured in meV. Keep b (T-odd field B along the axis) = 2 T fixed. For this exercise use only 0.150938 meV ≤ v0 ≤ 0.899062 meV 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 v0 = 0.315525 meV and 0.420262 meV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δv0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
OVERLAP |⟨Θψ|ψ⟩| against v0
Predictions from the supplied Spinless time reversal: Θ = K and real wavefunctions 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): OVERLAP |⟨Θψ|ψ⟩|
Horizontal axis (x): v0: T-even well depth V₀ / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v0: T-even well depth V₀ / meV | OVERLAP |⟨Θψ|ψ⟩| |
|---|---|
| 0.1509375 | 0.5461311887505322 |
| 0.33796875 | 0.8249906214527497 |
| 0.5249999999999999 | 0.9149806380200805 |
| 0.7120312499999999 | 0.9509909345658355 |
| 0.8990624999999999 | 0.9684064916237454 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of v0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At v0 = 0.315525 meV, substitute into Y_out = (v0/2)/hypot(0.05788*b, v0/2): Y₀ = 0.80624.
- At v0 = 0.420262 meV, the same fixed-parameter model gives Y₁ = 0.875885.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0696447; the input change is Δv0 = 0.104737 meV.
- Average sensitivity is ΔY_out/Δv0 = 0.664945 1 per meV. 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 = 0.80624, Y₁ = 0.875885; ΔY_out = 0.0696447; average sensitivity = 0.664945 1 per meV.
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.
Spinless time reversal: Θ = K and real wavefunctions: infer an input with uncertainty
For a particle without spin, time reversal is complex conjugation of the position-space wavefunction combined with running time backwards: Θψ(x, t) = ψ*(x, −t). Use the supplied topic model Y_out = (v0/2)/hypot(0.05788*b, v0/2), where Y_out is OVERLAP |⟨Θψ|ψ⟩| (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²). v0 is T-even well depth V₀, measured in meV. Keep b (T-odd field B along the axis) = 2 T fixed. For this exercise use only 0.150938 meV ≤ v0 ≤ 0.899062 meV 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 = 0.93440069, with a bounded reading uncertainty ±0.0051009. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000748125 meV, 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.
OVERLAP |⟨Θψ|ψ⟩| against v0
Predictions from the supplied Spinless time reversal: Θ = K and real wavefunctions 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): OVERLAP |⟨Θψ|ψ⟩|
Horizontal axis (x): v0: T-even well depth V₀ / meV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v0: T-even well depth V₀ / meV | OVERLAP |⟨Θψ|ψ⟩| |
|---|---|
| 0.1509375 | 0.5461311887505322 |
| 0.33796875 | 0.8249906214527497 |
| 0.5249999999999999 | 0.9149806380200805 |
| 0.7120312499999999 | 0.9509909345658355 |
| 0.8990624999999999 | 0.9684064916237454 |
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(v0) = 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(0.150938) = 0.546131 and Y_out(0.899062) = 0.968406. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives v0_line = 0.150938 + (0.934401 − 0.546131)(0.899062 − 0.150938)/(0.968406 − 0.546131) = 0.838816 meV. 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 v0 = 0.607294 meV; substituting back gives Y_out = 0.93440069.
- The interpolation error in the input is |v0_line − v0| = 0.231522 meV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.92929979 and 0.93950159. Sort the roots even if the response decreases: 0.582551 meV ≤ v0 ≤ 0.634993 meV.
- The lower and upper input deviations are 0.0247425 meV and 0.0276993 meV. 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 v0 ≈ 0.607294 meV; compatible interval [0.582551, 0.634993] meV. Endpoint interpolation gives 0.838816 meV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.