University Quantum Mechanics II · topic UQMII-10.06
Better trial functions for helium, and what an energy certifies
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.
Better helium trial functions, and what an energy certifies: predict a finite change
The correlation energy is the part of an atom's energy that no single product of orbitals can reach: the exact nonrelativistic, fixed-nucleus energy minus the Hartree-Fock limit, the lowest energy any such product achieves. Use the supplied topic model Y_out = -27.2114*(3.375+13*gam+15.8125*gam^2)^2*(1-(dl/100)^2)/(4*(1+3.125*gam+4*gam^2)*(1+4.375*gam+6*gam^2)), where Y_out is ⟨H⟩ OF THIS TRIAL (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. gam is Correlation factor γ = c/ζ, dimensionless. Keep dl (Scale error δ in ζ) = 0 % fixed. For this exercise use only 0.053125 ≤ gam ≤ 0.184375 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 gam = 0.082 and 0.100375. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δgam across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
⟨H⟩ OF THIS TRIAL against gam
Predictions from the supplied Better helium trial functions, and what an energy certifies 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): ⟨H⟩ OF THIS TRIAL / eV
Horizontal axis (x): gam: Correlation factor γ = c/ζ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gam: Correlation factor γ = c/ζ | ⟨H⟩ OF THIS TRIAL / eV |
|---|---|
| 0.053125 | -78.13631297172546 |
| 0.0859375 | -78.38174058250843 |
| 0.11875 | -78.54002536573445 |
| 0.15156250000000002 | -78.63048766697328 |
| 0.184375 | -78.66829836976521 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of gam separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At gam = 0.082, substitute into Y_out = -27.2114*(3.375+13*gam+15.8125*gam^2)^2*(1-(dl/100)^2)/(4*(1+3.125*gam+4*gam^2)*(1+4.375*gam+6*gam^2)): Y₀ = -78.3573 eV.
- At gam = 0.100375, the same fixed-parameter model gives Y₁ = -78.4609 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.10355 eV; the input change is Δgam = 0.018375.
- Average sensitivity is ΔY_out/Δgam = -5.63539 eV 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 = -78.3573 eV, Y₁ = -78.4609 eV; ΔY_out = -0.10355 eV; average sensitivity = -5.63539 eV 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.
Better helium trial functions, and what an energy certifies: infer an input with uncertainty
The correlation energy is the part of an atom's energy that no single product of orbitals can reach: the exact nonrelativistic, fixed-nucleus energy minus the Hartree-Fock limit, the lowest energy any such product achieves. Use the supplied topic model Y_out = -27.2114*(3.375+13*gam+15.8125*gam^2)^2*(1-(dl/100)^2)/(4*(1+3.125*gam+4*gam^2)*(1+4.375*gam+6*gam^2)), where Y_out is ⟨H⟩ OF THIS TRIAL (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. gam is Correlation factor γ = c/ζ, dimensionless. Keep dl (Scale error δ in ζ) = 0 % fixed. For this exercise use only 0.053125 ≤ gam ≤ 0.184375 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 = -78.587203 eV, with a bounded reading uncertainty ±0.012164 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown gam by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000013125, 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.
⟨H⟩ OF THIS TRIAL against gam
Predictions from the supplied Better helium trial functions, and what an energy certifies 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): ⟨H⟩ OF THIS TRIAL / eV
Horizontal axis (x): gam: Correlation factor γ = c/ζ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gam: Correlation factor γ = c/ζ | ⟨H⟩ OF THIS TRIAL / eV |
|---|---|
| 0.053125 | -78.13631297172546 |
| 0.0859375 | -78.38174058250843 |
| 0.11875 | -78.54002536573445 |
| 0.15156250000000002 | -78.63048766697328 |
| 0.184375 | -78.66829836976521 |
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(gam) = 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(0.053125) = -78.1363 eV and Y_out(0.184375) = -78.6683 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives gam_line = 0.053125 + (-78.5872 − -78.1363)(0.184375 − 0.053125)/(-78.6683 − -78.1363) = 0.164367. 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 gam = 0.133188; substituting back gives Y_out = -78.587203 eV.
- The interpolation error in the input is |gam_line − gam| = 0.0311798. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -78.599367 eV and -78.575039 eV. Sort the roots even if the response decreases: 0.129083 ≤ gam ≤ 0.137654.
- The lower and upper input deviations are 0.00410432 and 0.0044662. 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 gam ≈ 0.133188; compatible interval [0.129083, 0.137654]. Endpoint interpolation gives 0.164367.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.