University Quantum Mechanics I · topic UQMI-11.09
Where the matrix comes from: ammonia, double wells, dimers
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.
Where the matrix comes from: ammonia, double wells, dimers: predict a finite change
A physical two-state model writes a real system's Hamiltonian in two basis states the experiment can tell apart, with each state's own energy on the diagonal and a coupling A, supplied from outside the model, off it. Use the supplied topic model Y_out = 13.98*exp(-0.5928*b), where Y_out is COUPLING A (meV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. b is Barrier width b, measured in nm. Keep d (Site asymmetry δ) = 0 meV fixed. For this exercise use only 2.19062 nm ≤ b ≤ 7.30937 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 = 3.31675 nm and 4.03337 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.
COUPLING A against b
Predictions from the supplied Where the matrix comes from: ammonia, double wells, dimers 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): COUPLING A / meV
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 | COUPLING A / meV |
|---|---|
| 2.190625 | 3.81532265608782 |
| 3.4703124999999995 | 1.7868003150304461 |
| 4.75 | 0.836798261530679 |
| 6.0296875 | 0.3918912060908358 |
| 7.309374999999999 | 0.18353135334005402 |
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 = 3.31675 nm, substitute into Y_out = 13.98*exp(-0.5928*b): Y₀ = 1.95709 meV.
- At b = 4.03337 nm, the same fixed-parameter model gives Y₁ = 1.27972 meV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.677367 meV; the input change is Δb = 0.716625 nm.
- Average sensitivity is ΔY_out/Δb = -0.945218 meV per nm. 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 = 1.95709 meV, Y₁ = 1.27972 meV; ΔY_out = -0.677367 meV; average sensitivity = -0.945218 meV 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.
Where the matrix comes from: ammonia, double wells, dimers: infer an input with uncertainty
A physical two-state model writes a real system's Hamiltonian in two basis states the experiment can tell apart, with each state's own energy on the diagonal and a coupling A, supplied from outside the model, off it. Use the supplied topic model Y_out = 13.98*exp(-0.5928*b), where Y_out is COUPLING A (meV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. b is Barrier width b, measured in nm. Keep d (Site asymmetry δ) = 0 meV fixed. For this exercise use only 2.19062 nm ≤ b ≤ 7.30937 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 = 0.59932233 meV, with a bounded reading uncertainty ±0.062369 meV. 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 0.000511875 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.
COUPLING A against b
Predictions from the supplied Where the matrix comes from: ammonia, double wells, dimers 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): COUPLING A / meV
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 | COUPLING A / meV |
|---|---|
| 2.190625 | 3.81532265608782 |
| 3.4703124999999995 | 1.7868003150304461 |
| 4.75 | 0.836798261530679 |
| 6.0296875 | 0.3918912060908358 |
| 7.309374999999999 | 0.18353135334005402 |
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(2.19062) = 3.81532 meV and Y_out(7.30937) = 0.183531 meV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 2.19062 + (0.599322 − 3.81532)(7.30937 − 2.19062)/(0.183531 − 3.81532) = 6.72335 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 = 5.31306 nm; substituting back gives Y_out = 0.59932233 meV.
- The interpolation error in the input is |b_line − b| = 1.41028 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.53695333 meV and 0.66169133 meV. Sort the roots even if the response decreases: 5.14606 nm ≤ b ≤ 5.49843 nm.
- The lower and upper input deviations are 0.167003 nm and 0.185372 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 ≈ 5.31306 nm; compatible interval [5.14606, 5.49843] nm. Endpoint interpolation gives 6.72335 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.