University Physics IV · topic UPIV-08.05
Ladder operators and the algebraic solution
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.
Ladder operators and the algebraic solution: predict a finite change
The lowering operator a and the raising operator a† are the non-Hermitian combinations of position and momentum that obey [a, a†] = 1 and move an oscillator eigenstate one rung of ħω down or up the energy ladder. Use the supplied topic model Y_out = (n+0.5)*w, where Y_out is ENERGY (n + ½)ħω (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Quantum ħω, measured in eV. Keep n (Rung n) = 2 fixed. For this exercise use only 0.17125 eV ≤ w ≤ 0.32875 eV 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 w = 0.2059 eV and 0.22795 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δw across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ENERGY (n + ½)ħω against w
Predictions from the supplied Ladder operators and the algebraic solution 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): ENERGY (n + ½)ħω / eV
Horizontal axis (x): w: Quantum ħω / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Quantum ħω / eV | ENERGY (n + ½)ħω / eV |
|---|---|
| 0.17124999999999999 | 0.428125 |
| 0.21062499999999998 | 0.5265624999999999 |
| 0.25 | 0.625 |
| 0.289375 | 0.7234375 |
| 0.32875 | 0.8218749999999999 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of w separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At w = 0.2059 eV, substitute into Y_out = (n+0.5)*w: Y₀ = 0.51475 eV.
- At w = 0.22795 eV, the same fixed-parameter model gives Y₁ = 0.569875 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.055125 eV; the input change is Δw = 0.02205 eV.
- Average sensitivity is ΔY_out/Δw = 2.5 eV per eV. 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 = 0.51475 eV, Y₁ = 0.569875 eV; ΔY_out = 0.055125 eV; average sensitivity = 2.5 eV per eV.
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.
Ladder operators and the algebraic solution: infer an input with uncertainty
The lowering operator a and the raising operator a† are the non-Hermitian combinations of position and momentum that obey [a, a†] = 1 and move an oscillator eigenstate one rung of ħω down or up the energy ladder. Use the supplied topic model Y_out = (n+0.5)*w, where Y_out is ENERGY (n + ½)ħω (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. w is Quantum ħω, measured in eV. Keep n (Rung n) = 2 fixed. For this exercise use only 0.17125 eV ≤ w ≤ 0.32875 eV 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 = 0.6683125 eV, with a bounded reading uncertainty ±0.023034 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown w by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00001575 eV, 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.
ENERGY (n + ½)ħω against w
Predictions from the supplied Ladder operators and the algebraic solution 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): ENERGY (n + ½)ħω / eV
Horizontal axis (x): w: Quantum ħω / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| w: Quantum ħω / eV | ENERGY (n + ½)ħω / eV |
|---|---|
| 0.17124999999999999 | 0.428125 |
| 0.21062499999999998 | 0.5265624999999999 |
| 0.25 | 0.625 |
| 0.289375 | 0.7234375 |
| 0.32875 | 0.8218749999999999 |
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(w) = 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(0.17125) = 0.428125 eV and Y_out(0.32875) = 0.821875 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives w_line = 0.17125 + (0.668312 − 0.428125)(0.32875 − 0.17125)/(0.821875 − 0.428125) = 0.267325 eV. 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 w = 0.267325 eV; substituting back gives Y_out = 0.6683125 eV.
- The interpolation error in the input is |w_line − w| = 1.11022e-16 eV. This is within the requested numerical tolerance.
- Invert both limits Y_out = 0.6452785 eV and 0.6913465 eV. Sort the roots even if the response decreases: 0.258111 eV ≤ w ≤ 0.276539 eV.
- The lower and upper input deviations are 0.0092136 eV and 0.0092136 eV. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer w ≈ 0.267325 eV; compatible interval [0.258111, 0.276539] eV. Endpoint interpolation gives 0.267325 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.