University Physics V · topic UPV-03.09
Discretised states and operators as matrices
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.
Discretised states and matrix operators: predict a finite change
Discretisation replaces the state |ψ⟩ by its samples on a uniform grid, turning the Hilbert space into C^N with inner product Δx ψ†φ, so every observable becomes an N×N Hermitian matrix and every expectation value a vector-matrix-vector product. Use the supplied topic model Y_out = pi/max(h,0.01), where Y_out is NYQUIST k = π/Δx (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. h is Grid spacing Δx, measured in nm. Keep k (Wavenumber k) = 4 nm⁻¹ fixed. For this exercise use only 0.12975 nm ≤ h ≤ 0.35025 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 h = 0.17826 nm and 0.20913 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δh across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
NYQUIST k = π/Δx against h
Predictions from the supplied Discretised states and matrix operators 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): NYQUIST k = π/Δx / nm⁻¹
Horizontal axis (x): h: Grid spacing Δx / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| h: Grid spacing Δx / nm | NYQUIST k = π/Δx / nm⁻¹ |
|---|---|
| 0.12975 | 24.212660143273936 |
| 0.184875 | 16.993063711100977 |
| 0.24000000000000005 | 13.089969389957469 |
| 0.2951250000000001 | 10.644956047741777 |
| 0.35025000000000006 | 8.969572172990128 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of h separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At h = 0.17826 nm, substitute into Y_out = pi/max(h,0.01): Y₀ = 17.6237 nm⁻¹.
- At h = 0.20913 nm, the same fixed-parameter model gives Y₁ = 15.0222 nm⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -2.60145 nm⁻¹; the input change is Δh = 0.03087 nm.
- Average sensitivity is ΔY_out/Δh = -84.2713 nm⁻¹ 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 = 17.6237 nm⁻¹, Y₁ = 15.0222 nm⁻¹; ΔY_out = -2.60145 nm⁻¹; average sensitivity = -84.2713 nm⁻¹ 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.
Discretised states and matrix operators: infer an input with uncertainty
Discretisation replaces the state |ψ⟩ by its samples on a uniform grid, turning the Hilbert space into C^N with inner product Δx ψ†φ, so every observable becomes an N×N Hermitian matrix and every expectation value a vector-matrix-vector product. Use the supplied topic model Y_out = pi/max(h,0.01), where Y_out is NYQUIST k = π/Δx (nm⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. h is Grid spacing Δx, measured in nm. Keep k (Wavenumber k) = 4 nm⁻¹ fixed. For this exercise use only 0.12975 nm ≤ h ≤ 0.35025 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 = 11.888489 nm⁻¹, with a bounded reading uncertainty ±0.43784 nm⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown h by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00002205 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.
NYQUIST k = π/Δx against h
Predictions from the supplied Discretised states and matrix operators 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): NYQUIST k = π/Δx / nm⁻¹
Horizontal axis (x): h: Grid spacing Δx / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| h: Grid spacing Δx / nm | NYQUIST k = π/Δx / nm⁻¹ |
|---|---|
| 0.12975 | 24.212660143273936 |
| 0.184875 | 16.993063711100977 |
| 0.24000000000000005 | 13.089969389957469 |
| 0.2951250000000001 | 10.644956047741777 |
| 0.35025000000000006 | 8.969572172990128 |
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(h) = 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(0.12975) = 24.2127 nm⁻¹ and Y_out(0.35025) = 8.96957 nm⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives h_line = 0.12975 + (11.8885 − 24.2127)(0.35025 − 0.12975)/(8.96957 − 24.2127) = 0.308026 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 h = 0.264255 nm; substituting back gives Y_out = 11.888489 nm⁻¹.
- The interpolation error in the input is |h_line − h| = 0.0437712 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 11.450649 nm⁻¹ and 12.326329 nm⁻¹. Sort the roots even if the response decreases: 0.254868 nm ≤ h ≤ 0.274359 nm.
- The lower and upper input deviations are 0.00938653 nm and 0.0101044 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 h ≈ 0.264255 nm; compatible interval [0.254868, 0.274359] nm. Endpoint interpolation gives 0.308026 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.