University Quantum Mechanics I · topic UQMI-04.01
Observables, Hermitian operators, and real eigenvalues
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.
Observables, Hermitian operators, and real eigenvalues: predict a finite change
An observable is represented by a Hermitian operator Q̂: one for which ∫ ψ*(Q̂ψ) dx is real for every allowed state ψ, which is the same as ∫ f*(Q̂g) dx = ∫ (Q̂f)* g dx for every allowed pair f and g. Use the supplied topic model Y_out = sqrt(max(-1-b*c,0)), where Y_out is IM PART OF EIGENVALUES, ± (q₀). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. c is Reverse coupling Q₂₁, measured in q₀. Keep b (Coupling Q₁₂) = 1 q₀; phi (Relative phase φ) = 90 ° fixed. For this exercise use only -1.575 q₀ ≤ c ≤ -1.05 q₀ 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 c = -1.4595 q₀ and -1.386 q₀. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δc across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
IM PART OF EIGENVALUES, ± against c
Predictions from the supplied Observables, Hermitian operators, and real eigenvalues 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): IM PART OF EIGENVALUES, ± / q₀
Horizontal axis (x): c: Reverse coupling Q₂₁ / q₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| c: Reverse coupling Q₂₁ / q₀ | IM PART OF EIGENVALUES, ± / q₀ |
|---|---|
| -1.575 | 0.758287544405155 |
| -1.44375 | 0.6661456297237115 |
| -1.3125 | 0.5590169943749475 |
| -1.18125 | 0.42573465914815994 |
| -1.05 | 0.22360679774997907 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of c separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At c = -1.4595 q₀, substitute into Y_out = sqrt(max(-1-b*c,0)): Y₀ = 0.677864 q₀.
- At c = -1.386 q₀, the same fixed-parameter model gives Y₁ = 0.621289 q₀.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0565753 q₀; the input change is Δc = 0.0735 q₀.
- Average sensitivity is ΔY_out/Δc = -0.769732 q₀ per q₀. 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 = 0.677864 q₀, Y₁ = 0.621289 q₀; ΔY_out = -0.0565753 q₀; average sensitivity = -0.769732 q₀ per q₀.
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.
Observables, Hermitian operators, and real eigenvalues: infer an input with uncertainty
An observable is represented by a Hermitian operator Q̂: one for which ∫ ψ*(Q̂ψ) dx is real for every allowed state ψ, which is the same as ∫ f*(Q̂g) dx = ∫ (Q̂f)* g dx for every allowed pair f and g. Use the supplied topic model Y_out = sqrt(max(-1-b*c,0)), where Y_out is IM PART OF EIGENVALUES, ± (q₀). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value. c is Reverse coupling Q₂₁, measured in q₀. Keep b (Coupling Q₁₂) = 1 q₀; phi (Relative phase φ) = 90 ° fixed. For this exercise use only -1.575 q₀ ≤ c ≤ -1.05 q₀ 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 = 0.50472765 q₀, with a bounded reading uncertainty ±0.038034 q₀. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown c by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000525 q₀, 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.
IM PART OF EIGENVALUES, ± against c
Predictions from the supplied Observables, Hermitian operators, and real eigenvalues 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): IM PART OF EIGENVALUES, ± / q₀
Horizontal axis (x): c: Reverse coupling Q₂₁ / q₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| c: Reverse coupling Q₂₁ / q₀ | IM PART OF EIGENVALUES, ± / q₀ |
|---|---|
| -1.575 | 0.758287544405155 |
| -1.44375 | 0.6661456297237115 |
| -1.3125 | 0.5590169943749475 |
| -1.18125 | 0.42573465914815994 |
| -1.05 | 0.22360679774997907 |
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(c) = 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(-1.575) = 0.758288 q₀ and Y_out(-1.05) = 0.223607 q₀. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives c_line = -1.575 + (0.504728 − 0.758288)(-1.05 − -1.575)/(0.223607 − 0.758288) = -1.32603 q₀. 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 c = -1.25475 q₀; substituting back gives Y_out = 0.50472765 q₀.
- The interpolation error in the input is |c_line − c| = 0.071281 q₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.46669365 q₀ and 0.54276165 q₀. Sort the roots even if the response decreases: -1.29459 q₀ ≤ c ≤ -1.2178 q₀.
- The lower and upper input deviations are 0.0398402 q₀ and 0.036947 q₀. 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 c ≈ -1.25475 q₀; compatible interval [-1.29459, -1.2178] q₀. Endpoint interpolation gives -1.32603 q₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.