University Quantum Mechanics I · topic UQMI-08.09
Midterm consolidation: eight units on one paper
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.
Midterm consolidation: eight units on one paper: predict a finite change
Consolidating for the midterm means re-deriving each core result of Units 1–8 from its starting point — the Schrödinger equation, the Born rule, an operator, a commutator — and testing it by normalisation, dimensions, a limiting case and classical correspondence. Use the supplied topic model Y_out = 0.0010798*exp(2*((pi-asin(sqrt(u)))*sqrt(1/u-1))*b)/((pi-asin(sqrt(u)))*u*(1-u)), where Y_out is LIFETIME τ = 2L/(vT) (ps). 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; unmarked trigonometric functions use radians. b is Barrier width b, measured in nm. Keep u (Level position E/V₀) = 0.259 fixed. For this exercise use only 0.574375 nm ≤ b ≤ 1.12562 nm 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 b = 0.69565 nm and 0.772825 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δb across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LIFETIME τ = 2L/(vT) against b
Predictions from the supplied Midterm consolidation: eight units on one paper 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): LIFETIME τ = 2L/(vT) / ps
Horizontal axis (x): b: Barrier width b / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Barrier width b / nm | LIFETIME τ = 2L/(vT) / ps |
|---|---|
| 0.574375 | 0.34234911727811457 |
| 0.7121875 | 1.154634161133024 |
| 0.8499999999999999 | 3.8942120156609703 |
| 0.9878124999999999 | 13.133932576563671 |
| 1.1256249999999999 | 44.296557103720446 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of b separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At b = 0.69565 nm, substitute into Y_out = 0.0010798*exp(2*((pi-asin(sqrt(u)))*sqrt(1/u-1))*b)/((pi-asin(sqrt(u)))*u*(1-u)): Y₀ = 0.997901 ps.
- At b = 0.772825 nm, the same fixed-parameter model gives Y₁ = 1.9713 ps.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.973403 ps; the input change is Δb = 0.077175 nm.
- Average sensitivity is ΔY_out/Δb = 12.6129 ps per nm. 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 = 0.997901 ps, Y₁ = 1.9713 ps; ΔY_out = 0.973403 ps; average sensitivity = 12.6129 ps per nm.
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.
Midterm consolidation: eight units on one paper: infer an input with uncertainty
Consolidating for the midterm means re-deriving each core result of Units 1–8 from its starting point — the Schrödinger equation, the Born rule, an operator, a commutator — and testing it by normalisation, dimensions, a limiting case and classical correspondence. Use the supplied topic model Y_out = 0.0010798*exp(2*((pi-asin(sqrt(u)))*sqrt(1/u-1))*b)/((pi-asin(sqrt(u)))*u*(1-u)), where Y_out is LIFETIME τ = 2L/(vT) (ps). 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; unmarked trigonometric functions use radians. b is Barrier width b, measured in nm. Keep u (Level position E/V₀) = 0.259 fixed. For this exercise use only 0.574375 nm ≤ b ≤ 1.12562 nm 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 = 6.6485766 ps, with a bounded reading uncertainty ±0.94593 ps. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000055125 nm, 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.
LIFETIME τ = 2L/(vT) against b
Predictions from the supplied Midterm consolidation: eight units on one paper 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): LIFETIME τ = 2L/(vT) / ps
Horizontal axis (x): b: Barrier width b / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Barrier width b / nm | LIFETIME τ = 2L/(vT) / ps |
|---|---|
| 0.574375 | 0.34234911727811457 |
| 0.7121875 | 1.154634161133024 |
| 0.8499999999999999 | 3.8942120156609703 |
| 0.9878124999999999 | 13.133932576563671 |
| 1.1256249999999999 | 44.296557103720446 |
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(b) = 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(0.574375) = 0.342349 ps and Y_out(1.12562) = 44.2966 ps. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 0.574375 + (6.64858 − 0.342349)(1.12562 − 0.574375)/(44.2966 − 0.342349) = 0.653464 nm. 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 b = 0.910638 nm; substituting back gives Y_out = 6.6485766 ps.
- The interpolation error in the input is |b_line − b| = 0.257173 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 5.7026466 ps and 7.5945066 ps. Sort the roots even if the response decreases: 0.89324 nm ≤ b ≤ 0.925717 nm.
- The lower and upper input deviations are 0.0173976 nm and 0.0150794 nm. 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 b ≈ 0.910638 nm; compatible interval [0.89324, 0.925717] nm. Endpoint interpolation gives 0.653464 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.