University Advanced Quantum Mechanics · topic UAQM-03.07
Exact Klein-Gordon Coulomb levels: wrong for hydrogen, right for pionic atoms
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.
Exact Klein-Gordon Coulomb levels and pionic atoms: predict a finite change
The Klein-Gordon Coulomb spectrum is the set of exact bound energies of a spinless charged particle around a point nucleus, E = mc²[1 + (Zα)²/N²]^(−1/2), where N = n − (l + ½) + √((l + ½)² − (Zα)²) depends on l as well as on n. Use the supplied topic model Y_out = 350.38/n^3*2*(lam-1)/3, where Y_out is GAP ns − np₁/₂ (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lam is Spin terms switched on, λ, dimensionless. Keep n (Principal quantum number n) = 2 fixed. For this exercise use only 0.10625 ≤ lam ≤ 0.89375 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 lam = 0.2795 and 0.38975. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlam across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
GAP ns − np₁/₂ against lam
Predictions from the supplied Exact Klein-Gordon Coulomb levels and pionic atoms 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): GAP ns − np₁/₂ / GHz
Horizontal axis (x): lam: Spin terms switched on, λ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: Spin terms switched on, λ | GAP ns − np₁/₂ / GHz |
|---|---|
| 0.10625 | -26.09601041666667 |
| 0.303125 | -20.347588541666667 |
| 0.5 | -14.599166666666667 |
| 0.6968749999999999 | -8.850744791666669 |
| 0.8937499999999999 | -3.102322916666669 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lam separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lam = 0.2795, substitute into Y_out = 350.38/n^3*2*(lam-1)/3: Y₀ = -21.0374 GHz.
- At lam = 0.38975, the same fixed-parameter model gives Y₁ = -17.8183 GHz.
- Signed change: ΔY_out = Y₁ − Y₀ = 3.21912 GHz; the input change is Δlam = 0.11025.
- Average sensitivity is ΔY_out/Δlam = 29.1983 GHz per unit input. 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 = -21.0374 GHz, Y₁ = -17.8183 GHz; ΔY_out = 3.21912 GHz; average sensitivity = 29.1983 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.
Exact Klein-Gordon Coulomb levels and pionic atoms: infer an input with uncertainty
The Klein-Gordon Coulomb spectrum is the set of exact bound energies of a spinless charged particle around a point nucleus, E = mc²[1 + (Zα)²/N²]^(−1/2), where N = n − (l + ½) + √((l + ½)² − (Zα)²) depends on l as well as on n. Use the supplied topic model Y_out = 350.38/n^3*2*(lam-1)/3, where Y_out is GAP ns − np₁/₂ (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. lam is Spin terms switched on, λ, dimensionless. Keep n (Principal quantum number n) = 2 fixed. For this exercise use only 0.10625 ≤ lam ≤ 0.89375 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 = -12.069861 GHz, with a bounded reading uncertainty ±1.3451 GHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lam by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, 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.
GAP ns − np₁/₂ against lam
Predictions from the supplied Exact Klein-Gordon Coulomb levels and pionic atoms 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): GAP ns − np₁/₂ / GHz
Horizontal axis (x): lam: Spin terms switched on, λ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: Spin terms switched on, λ | GAP ns − np₁/₂ / GHz |
|---|---|
| 0.10625 | -26.09601041666667 |
| 0.303125 | -20.347588541666667 |
| 0.5 | -14.599166666666667 |
| 0.6968749999999999 | -8.850744791666669 |
| 0.8937499999999999 | -3.102322916666669 |
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(lam) = 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(0.10625) = -26.096 GHz and Y_out(0.89375) = -3.10232 GHz. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lam_line = 0.10625 + (-12.0699 − -26.096)(0.89375 − 0.10625)/(-3.10232 − -26.096) = 0.586625. 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 lam = 0.586625; substituting back gives Y_out = -12.069861 GHz.
- The interpolation error in the input is |lam_line − lam| = 1.11022e-16. This is within the requested numerical tolerance.
- Invert both limits Y_out = -13.414961 GHz and -10.724761 GHz. Sort the roots even if the response decreases: 0.540557 ≤ lam ≤ 0.632693.
- The lower and upper input deviations are 0.0460677 and 0.0460677. 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 lam ≈ 0.586625; compatible interval [0.540557, 0.632693]. Endpoint interpolation gives 0.586625.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.