University Quantum Mechanics I · topic UQMI-14.02
Additive Cartesian potentials and the three-dimensional box
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.
Additive Cartesian potentials and the three-dimensional box: predict a finite change
A potential is additive when it splits as V = Vₓ(x) + V_y(y) + V_z(z); its stationary states can then be taken as products X(x)Y(y)Z(z) of one-dimensional solutions, and each energy is the sum Eₓ + E_y + E_z of three one-dimensional eigenvalues. Use the supplied topic model Y_out = 0.376*(1+1/b^2+1/c^2), where Y_out is E (1,1,1) (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Edge b (a fixed at 1.00 nm), measured in nm. Keep c (Edge c) = 1 nm fixed. For this exercise use only 0.86375 nm ≤ b ≤ 1.33625 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 = 0.9677 nm and 1.03385 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.
E (1,1,1) against b
Predictions from the supplied Additive Cartesian potentials and the three-dimensional box 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): E (1,1,1) / eV
Horizontal axis (x): b: Edge b (a fixed at 1.00 nm) / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Edge b (a fixed at 1.00 nm) / nm | E (1,1,1) / eV |
|---|---|
| 0.86375 | 1.2559781687648304 |
| 0.981875 | 1.1420097283635076 |
| 1.1 | 1.0627438016528925 |
| 1.218125 | 1.0053985538360044 |
| 1.33625 | 0.9625777148502618 |
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 = 0.9677 nm, substitute into Y_out = 0.376*(1+1/b^2+1/c^2): Y₀ = 1.15352 eV.
- At b = 1.03385 nm, the same fixed-parameter model gives Y₁ = 1.10378 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0497379 eV; the input change is Δb = 0.06615 nm.
- Average sensitivity is ΔY_out/Δb = -0.751896 eV 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 = 1.15352 eV, Y₁ = 1.10378 eV; ΔY_out = -0.0497379 eV; average sensitivity = -0.751896 eV 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.
Additive Cartesian potentials and the three-dimensional box: infer an input with uncertainty
A potential is additive when it splits as V = Vₓ(x) + V_y(y) + V_z(z); its stationary states can then be taken as products X(x)Y(y)Z(z) of one-dimensional solutions, and each energy is the sum Eₓ + E_y + E_z of three one-dimensional eigenvalues. Use the supplied topic model Y_out = 0.376*(1+1/b^2+1/c^2), where Y_out is E (1,1,1) (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Edge b (a fixed at 1.00 nm), measured in nm. Keep c (Edge c) = 1 nm fixed. For this exercise use only 0.86375 nm ≤ b ≤ 1.33625 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 = 1.035336 eV, with a bounded reading uncertainty ±0.010914 eV. 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 0.00004725 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.
E (1,1,1) against b
Predictions from the supplied Additive Cartesian potentials and the three-dimensional box 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): E (1,1,1) / eV
Horizontal axis (x): b: Edge b (a fixed at 1.00 nm) / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Edge b (a fixed at 1.00 nm) / nm | E (1,1,1) / eV |
|---|---|
| 0.86375 | 1.2559781687648304 |
| 0.981875 | 1.1420097283635076 |
| 1.1 | 1.0627438016528925 |
| 1.218125 | 1.0053985538360044 |
| 1.33625 | 0.9625777148502618 |
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(0.86375) = 1.25598 eV and Y_out(1.33625) = 0.962578 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives b_line = 0.86375 + (1.03534 − 1.25598)(1.33625 − 0.86375)/(0.962578 − 1.25598) = 1.21908 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 = 1.15197 nm; substituting back gives Y_out = 1.035336 eV.
- The interpolation error in the input is |b_line − b| = 0.0671031 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.024422 eV and 1.04625 eV. Sort the roots even if the response decreases: 1.13041 nm ≤ b ≤ 1.17482 nm.
- The lower and upper input deviations are 0.0215658 nm and 0.0228491 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 ≈ 1.15197 nm; compatible interval [1.13041, 1.17482] nm. Endpoint interpolation gives 1.21908 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.