University Physics V · topic UPV-13.06
Spin-orbit coupling in many-electron atoms: LS against jj
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.
Spin–orbit coupling: LS against jj: predict a finite change
Inside a single LS term, the many-electron spin–orbit sum Σᵢ ξ(rᵢ) lᵢ⋅sᵢ reduces to one constant times L⋅S, so the entire fine structure of that term is the single number A — provided the term has not yet mixed with its neighbours. Use the supplied topic model Y_out = (0.5*z-((d-0.5*z)-sqrt((d+0.5*z)^2+2*z^2))/2)/max(((d-0.5*z)-sqrt((d+0.5*z)^2+2*z^2))/2+z,0.01), where Y_out is (E₂−E₁)/(E₁−E₀), LS: 2 (dimensionless). 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. d is Electrostatic gap Δ, measured in 10³ cm⁻¹. Keep z (Spin–orbit constant ζ) = 2 10³ cm⁻¹ fixed. For this exercise use only 1.97812 10³ cm⁻¹ ≤ d ≤ 5.52187 10³ cm⁻¹ 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 d = 2.75775 10³ cm⁻¹ and 3.25387 10³ cm⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
(E₂−E₁)/(E₁−E₀), LS: 2 against d
Predictions from the supplied Spin–orbit coupling: LS against jj 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): (E₂−E₁)/(E₁−E₀), LS: 2
Horizontal axis (x): d: Electrostatic gap Δ / 10³ cm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Electrostatic gap Δ / 10³ cm⁻¹ | (E₂−E₁)/(E₁−E₀), LS: 2 |
|---|---|
| 1.978125 | 5.889358588440983 |
| 2.8640624999999997 | 4.579151649269328 |
| 3.75 | 3.911334438749599 |
| 4.6359375 | 3.510994135170142 |
| 5.521875 | 3.2460202549680925 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = 2.75775 10³ cm⁻¹, substitute into Y_out = (0.5*z-((d-0.5*z)-sqrt((d+0.5*z)^2+2*z^2))/2)/max(((d-0.5*z)-sqrt((d+0.5*z)^2+2*z^2))/2+z,0.01): Y₀ = 4.68997.
- At d = 3.25387 10³ cm⁻¹, the same fixed-parameter model gives Y₁ = 4.23787.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.452104; the input change is Δd = 0.496125 10³ cm⁻¹.
- Average sensitivity is ΔY_out/Δd = -0.911269 1 per 10³ cm⁻¹. 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 = 4.68997, Y₁ = 4.23787; ΔY_out = -0.452104; average sensitivity = -0.911269 1 per 10³ cm⁻¹.
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.
Spin–orbit coupling: LS against jj: infer an input with uncertainty
Inside a single LS term, the many-electron spin–orbit sum Σᵢ ξ(rᵢ) lᵢ⋅sᵢ reduces to one constant times L⋅S, so the entire fine structure of that term is the single number A — provided the term has not yet mixed with its neighbours. Use the supplied topic model Y_out = (0.5*z-((d-0.5*z)-sqrt((d+0.5*z)^2+2*z^2))/2)/max(((d-0.5*z)-sqrt((d+0.5*z)^2+2*z^2))/2+z,0.01), where Y_out is (E₂−E₁)/(E₁−E₀), LS: 2 (dimensionless). 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. d is Electrostatic gap Δ, measured in 10³ cm⁻¹. Keep z (Spin–orbit constant ζ) = 2 10³ cm⁻¹ fixed. For this exercise use only 1.97812 10³ cm⁻¹ ≤ d ≤ 5.52187 10³ cm⁻¹ 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 = 3.7127026, with a bounded reading uncertainty ±0.070002. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000354375 10³ cm⁻¹, 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.
(E₂−E₁)/(E₁−E₀), LS: 2 against d
Predictions from the supplied Spin–orbit coupling: LS against jj 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): (E₂−E₁)/(E₁−E₀), LS: 2
Horizontal axis (x): d: Electrostatic gap Δ / 10³ cm⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Electrostatic gap Δ / 10³ cm⁻¹ | (E₂−E₁)/(E₁−E₀), LS: 2 |
|---|---|
| 1.978125 | 5.889358588440983 |
| 2.8640624999999997 | 4.579151649269328 |
| 3.75 | 3.911334438749599 |
| 4.6359375 | 3.510994135170142 |
| 5.521875 | 3.2460202549680925 |
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(d) = 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(1.97812) = 5.88936 and Y_out(5.52187) = 3.24602. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 1.97812 + (3.7127 − 5.88936)(5.52187 − 1.97812)/(3.24602 − 5.88936) = 4.89622 10³ cm⁻¹. 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 d = 4.13981 10³ cm⁻¹; substituting back gives Y_out = 3.7127026.
- The interpolation error in the input is |d_line − d| = 0.756412 10³ cm⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.6427006 and 3.7827046. Sort the roots even if the response decreases: 3.99293 10³ cm⁻¹ ≤ d ≤ 4.2987 10³ cm⁻¹.
- The lower and upper input deviations are 0.14688 10³ cm⁻¹ and 0.158885 10³ cm⁻¹. 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 d ≈ 4.13981 10³ cm⁻¹; compatible interval [3.99293, 4.2987] 10³ cm⁻¹. Endpoint interpolation gives 4.89622 10³ cm⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.