University Physics IV · topic UPIV-01.09
Old quantum theory and the correspondence principle
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.
Old quantum theory and correspondence: predict a finite change
Old quantum theory keeps Newtonian orbits but allows only those whose action integral ∮p dq around one period equals a whole number of Planck constants, and it settles what that rule leaves open by demanding agreement with classical physics at large quantum numbers. Use the supplied topic model Y_out = 3289842*k*(2*n-k)/(max(n-k,1)^2*n^2), where Y_out is QUANTUM ν(n → n−k) (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. n is Upper level n, dimensionless. Keep k (Jump size k in n → n−k) = 1 fixed. For this exercise use only 20 ≤ n ≤ 100, at the specified test settings separated by 1. 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 n = 38 and 39. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δn across this interval. State whether the response increases or decreases; explain why a finite difference between these settings does not require interpolation between them.
QUANTUM ν(n → n−k) against n
Predictions from the supplied Old quantum theory and correspondence 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): QUANTUM ν(n → n−k) / GHz
Horizontal axis (x): n: Upper level n
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Upper level n | QUANTUM ν(n → n−k) / GHz |
|---|---|
| 20 | 888.5307340720221 |
| 40 | 106.79549556213017 |
| 60 | 31.240320310255672 |
| 80 | 13.095980161432463 |
| 100 | 6.6797118457300275 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of n separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At n = 38, substitute into Y_out = 3289842*k*(2*n-k)/(max(n-k,1)^2*n^2): Y₀ = 124.815 GHz.
- At n = 39, the same fixed-parameter model gives Y₁ = 115.337 GHz.
- Signed change: ΔY_out = Y₁ − Y₀ = -9.47749 GHz; the input change is Δn = 1.
- Average sensitivity is ΔY_out/Δn = -9.47749 GHz per unit input. Its negative sign means that this response decreases over the specified interval.
- This finite difference compares two specified test settings without interpolating between them. A point derivative is not required. Any quantum numbers or particle counts must also obey their physical admissibility conditions.
Answer Y₀ = 124.815 GHz, Y₁ = 115.337 GHz; ΔY_out = -9.47749 GHz; average sensitivity = -9.47749 GHz per unit input.
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.
Old quantum theory and correspondence: infer an input with uncertainty
Old quantum theory keeps Newtonian orbits but allows only those whose action integral ∮p dq around one period equals a whole number of Planck constants, and it settles what that rule leaves open by demanding agreement with classical physics at large quantum numbers. Use the supplied topic model Y_out = 3289842*k*(2*n-k)/(max(n-k,1)^2*n^2), where Y_out is QUANTUM ν(n → n−k) (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. n is Upper level n, dimensionless. Keep k (Jump size k in n → n−k) = 1 fixed. For this exercise use only 20 ≤ n ≤ 100, at the specified test settings separated by 1. 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 = 20.472911 GHz, with a bounded reading uncertainty ±70.548 GHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of n consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.
QUANTUM ν(n → n−k) against n
Predictions from the supplied Old quantum theory and correspondence 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): QUANTUM ν(n → n−k) / GHz
Horizontal axis (x): n: Upper level n
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: Upper level n | QUANTUM ν(n → n−k) / GHz |
|---|---|
| 20 | 888.5307340720221 |
| 40 | 106.79549556213017 |
| 60 | 31.240320310255672 |
| 80 | 13.095980161432463 |
| 100 | 6.6797118457300275 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.
Reveal the worked solution
- Endpoint predictions are Y_out(20) = 888.531 GHz and Y_out(100) = 6.67971 GHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives n_line = 20 + (20.4729 − 888.531)(100 − 20)/(6.67971 − 888.531) = 98.7487. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Evaluate the model at each permitted setting. The central reading corresponds to n = 69; only the allowed settings 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100 lie inside the stated reading bound.
- The interpolation error in the input is |n_line − n| = 29.7487.
- The compatible allowed values span 43 to 100. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
- The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.
Answer Compatible n: 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100. Endpoint interpolation predicts 98.7487; check admissibility against the discrete settings.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.