University Quantum Mechanics II · topic UQMII-07.06
Excited helium: parahelium, orthohelium and the exchange integral
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.
Parahelium, orthohelium and the exchange integral: predict a finite change
The exchange integral K is the matrix element of the electron–electron repulsion between ψ₁ₛ(r₁)ψ₂ₛ(r₂) and the same product with the electrons swapped; relative to E⁽⁰⁾ + J it raises the spatially symmetric 1s2s state by K and lowers the antisymmetric one by K. Use the supplied topic model Y_out = 100*sin(atan(0.353553*(1-r1)*exp(r1)))^2, where Y_out is ELECTRON 2 IN 1s, GIVEN r₁ (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; unmarked trigonometric functions use radians. r1 is Radius of electron 1, r₁, measured in a₀. Keep j (Direct integral J) = 11.42 eV; k (Exchange integral K) = 1.19 eV fixed. For this exercise use only 1.09325 a₀ ≤ r1 ≤ 2.69188 a₀ 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 r1 = 1.44495 a₀ and 1.66875 a₀. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δr1 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ELECTRON 2 IN 1s, GIVEN r₁ against r1
Predictions from the supplied Parahelium, orthohelium and the exchange integral 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): ELECTRON 2 IN 1s, GIVEN r₁ / %
Horizontal axis (x): r1: Radius of electron 1, r₁ / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r1: Radius of electron 1, r₁ / a₀ | ELECTRON 2 IN 1s, GIVEN r₁ / % |
|---|---|
| 1.09325 | 0.9585365289037387 |
| 1.49290625 | 37.55443300697992 |
| 1.8925625 | 81.43253105021144 |
| 2.29221875 | 95.33680522713055 |
| 2.691875 | 98.73326726934124 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of r1 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At r1 = 1.44495 a₀, substitute into Y_out = 100*sin(atan(0.353553*(1-r1)*exp(r1)))^2: Y₀ = 30.8072 %.
- At r1 = 1.66875 a₀, the same fixed-parameter model gives Y₁ = 61.1445 %.
- Signed change: ΔY_out = Y₁ − Y₀ = 30.3373 %; the input change is Δr1 = 0.223807 a₀.
- Average sensitivity is ΔY_out/Δr1 = 135.551 % per a₀. 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 = 30.8072 %, Y₁ = 61.1445 %; ΔY_out = 30.3373 %; average sensitivity = 135.551 % per a₀.
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.
Parahelium, orthohelium and the exchange integral: infer an input with uncertainty
The exchange integral K is the matrix element of the electron–electron repulsion between ψ₁ₛ(r₁)ψ₂ₛ(r₂) and the same product with the electrons swapped; relative to E⁽⁰⁾ + J it raises the spatially symmetric 1s2s state by K and lowers the antisymmetric one by K. Use the supplied topic model Y_out = 100*sin(atan(0.353553*(1-r1)*exp(r1)))^2, where Y_out is ELECTRON 2 IN 1s, GIVEN r₁ (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; unmarked trigonometric functions use radians. r1 is Radius of electron 1, r₁, measured in a₀. Keep j (Direct integral J) = 11.42 eV; k (Exchange integral K) = 1.19 eV fixed. For this exercise use only 1.09325 a₀ ≤ r1 ≤ 2.69188 a₀ 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 = 89.932294 %, with a bounded reading uncertainty ±1.3201 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r1 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000159862 a₀, 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.
ELECTRON 2 IN 1s, GIVEN r₁ against r1
Predictions from the supplied Parahelium, orthohelium and the exchange integral 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): ELECTRON 2 IN 1s, GIVEN r₁ / %
Horizontal axis (x): r1: Radius of electron 1, r₁ / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r1: Radius of electron 1, r₁ / a₀ | ELECTRON 2 IN 1s, GIVEN r₁ / % |
|---|---|
| 1.09325 | 0.9585365289037387 |
| 1.49290625 | 37.55443300697992 |
| 1.8925625 | 81.43253105021144 |
| 2.29221875 | 95.33680522713055 |
| 2.691875 | 98.73326726934124 |
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(r1) = 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(1.09325) = 0.958537 % and Y_out(2.69188) = 98.7333 %. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives r1_line = 1.09325 + (89.9323 − 0.958537)(2.69188 − 1.09325)/(98.7333 − 0.958537) = 2.54798 a₀. 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 r1 = 2.06841 a₀; substituting back gives Y_out = 89.932294 %.
- The interpolation error in the input is |r1_line − r1| = 0.479567 a₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 88.612194 % and 91.252394 %. Sort the roots even if the response decreases: 2.03306 a₀ ≤ r1 ≤ 2.10884 a₀.
- The lower and upper input deviations are 0.0353514 a₀ and 0.0404246 a₀. 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 r1 ≈ 2.06841 a₀; compatible interval [2.03306, 2.10884] a₀. Endpoint interpolation gives 2.54798 a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.