University Advanced Quantum Mechanics · topic UAQM-12.09
Scattering of light: the Kramers-Heisenberg formula, Rayleigh and Thomson
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.
Scattering of light: Kramers-Heisenberg, Rayleigh, Thomson: predict a finite change
The Kramers-Heisenberg formula is the cross-section for an atom to absorb one photon and emit another, built from the A² coupling at first order and the A⋅p coupling at second order through intermediate states in both time orders. Use the supplied topic model Y_out = (100*(log((0.5*((1-10^lw)/((1-10^lw)^2+10^(2*lg)/4)+1/(1+10^lw)))^2+(0.25*10^lg/((1-10^lw)^2+10^(2*lg)/4))^2)+4*lw))/100, where Y_out is ALL TERMS, log₁₀ σ/σ_T (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. lg is Line width log₁₀(Γ/ω₀), dimensionless. Keep lw (Photon frequency log₁₀(ω/ω₀)) = -0.7 fixed. For this exercise use only -1.84062 ≤ lg ≤ -0.659375 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 lg = -1.58075 and -1.41538. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlg across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ALL TERMS, log₁₀ σ/σ_T (before display rounding) against lg
Predictions from the supplied Scattering of light: Kramers-Heisenberg, Rayleigh, Thomson 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): ALL TERMS, log₁₀ σ/σ_T (before display rounding)
Horizontal axis (x): lg: Line width log₁₀(Γ/ω₀)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lg: Line width log₁₀(Γ/ω₀) | ALL TERMS, log₁₀ σ/σ_T (before display rounding) |
|---|---|
| -1.840625 | -2.764743352932341 |
| -1.5453125 | -2.764829188607747 |
| -1.25 | -2.765163385672076 |
| -0.9546875 | -2.7664620556106434 |
| -0.659375 | -2.7714710733039065 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lg separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lg = -1.58075, substitute into Y_out = (100*(log((0.5*((1-10^lw)/((1-10^lw)^2+10^(2*lg)/4)+1/(1+10^lw)))^2+(0.25*10^lg/((1-10^lw)^2+10^(2*lg)/4))^2)+4*lw))/100: Y₀ = -2.76481.
- At lg = -1.41538, the same fixed-parameter model gives Y₁ = -2.76492.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.000111953; the input change is Δlg = 0.165375.
- Average sensitivity is ΔY_out/Δlg = -0.000676962 1 per unit input. 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 = -2.76481, Y₁ = -2.76492; ΔY_out = -0.000111953; average sensitivity = -0.000676962 1 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.
Scattering of light: Kramers-Heisenberg, Rayleigh, Thomson: infer an input with uncertainty
The Kramers-Heisenberg formula is the cross-section for an atom to absorb one photon and emit another, built from the A² coupling at first order and the A⋅p coupling at second order through intermediate states in both time orders. Use the supplied topic model Y_out = (100*(log((0.5*((1-10^lw)/((1-10^lw)^2+10^(2*lg)/4)+1/(1+10^lw)))^2+(0.25*10^lg/((1-10^lw)^2+10^(2*lg)/4))^2)+4*lw))/100, where Y_out is ALL TERMS, log₁₀ σ/σ_T (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. lg is Line width log₁₀(Γ/ω₀), dimensionless. Keep lw (Photon frequency log₁₀(ω/ω₀)) = -0.7 fixed. For this exercise use only -1.84062 ≤ lg ≤ -0.659375 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 = -2.7655313, with a bounded reading uncertainty ±0.00011819. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lg by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000118125, 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.
ALL TERMS, log₁₀ σ/σ_T (before display rounding) against lg
Predictions from the supplied Scattering of light: Kramers-Heisenberg, Rayleigh, Thomson 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): ALL TERMS, log₁₀ σ/σ_T (before display rounding)
Horizontal axis (x): lg: Line width log₁₀(Γ/ω₀)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lg: Line width log₁₀(Γ/ω₀) | ALL TERMS, log₁₀ σ/σ_T (before display rounding) |
|---|---|
| -1.840625 | -2.764743352932341 |
| -1.5453125 | -2.764829188607747 |
| -1.25 | -2.765163385672076 |
| -0.9546875 | -2.7664620556106434 |
| -0.659375 | -2.7714710733039065 |
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(lg) = 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(-1.84062) = -2.76474 and Y_out(-0.659375) = -2.77147. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives lg_line = -1.84062 + (-2.76553 − -2.76474)(-0.659375 − -1.84062)/(-2.77147 − -2.76474) = -1.70228. 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 lg = -1.12006; substituting back gives Y_out = -2.7655313.
- The interpolation error in the input is |lg_line − lg| = 0.582223. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -2.7656495 and -2.7654131. Sort the roots even if the response decreases: -1.154 ≤ lg ≤ -1.0907.
- The lower and upper input deviations are 0.0339458 and 0.0293599. 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 lg ≈ -1.12006; compatible interval [-1.154, -1.0907]. Endpoint interpolation gives -1.70228.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.