University Advanced Quantum Mechanics · topic UAQM-10.08
Closed-shell helium in restricted Hartree-Fock, solved numerically
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.
Restricted Hartree-Fock helium, solved numerically: predict a finite change
Restricted Hartree-Fock helium puts both electrons in one spatial orbital with opposite spins and finds the orbital that is the lowest eigenfunction of the mean field it creates: the nucleus's attraction plus the Hartree potential of the partner electron. Use the supplied topic model Y_out = 27.2114*(z*z/2-2*z+0.625*n*z), where Y_out is MEAN-FIELD ⟨h + nV_H⟩ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. z is Trial orbital exponent ζ, measured in a₀⁻¹. Keep n (Electrons counted in the mean field n) = 1 fixed. For this exercise use only 1.40875 a₀⁻¹ ≤ z ≤ 2.14375 a₀⁻¹ 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 z = 1.57045 a₀⁻¹ and 1.67335 a₀⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δz across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
MEAN-FIELD ⟨h + nV_H⟩ against z
Predictions from the supplied Restricted Hartree-Fock helium, solved numerically 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): MEAN-FIELD ⟨h + nV_H⟩ / eV
Horizontal axis (x): z: Trial orbital exponent ζ / a₀⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| z: Trial orbital exponent ζ / a₀⁻¹ | MEAN-FIELD ⟨h + nV_H⟩ / eV |
|---|---|
| 1.40875 | -25.70777881984375 |
| 1.5924999999999998 | -25.079641916875005 |
| 1.7762499999999999 | -23.53273760359375 |
| 1.96 | -21.06706588 |
| 2.14375 | -17.682626746093753 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of z separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At z = 1.57045 a₀⁻¹, substitute into Y_out = 27.2114*(z*z/2-2*z+0.625*n*z): Y₀ = -25.2035 eV.
- At z = 1.67335 a₀⁻¹, the same fixed-parameter model gives Y₁ = -24.5122 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.691333 eV; the input change is Δz = 0.1029 a₀⁻¹.
- Average sensitivity is ΔY_out/Δz = 6.71849 eV per a₀⁻¹. 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 = -25.2035 eV, Y₁ = -24.5122 eV; ΔY_out = 0.691333 eV; average sensitivity = 6.71849 eV per a₀⁻¹.
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.
Restricted Hartree-Fock helium, solved numerically: infer an input with uncertainty
Restricted Hartree-Fock helium puts both electrons in one spatial orbital with opposite spins and finds the orbital that is the lowest eigenfunction of the mean field it creates: the nucleus's attraction plus the Hartree potential of the partner electron. Use the supplied topic model Y_out = 27.2114*(z*z/2-2*z+0.625*n*z), where Y_out is MEAN-FIELD ⟨h + nV_H⟩ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. z is Trial orbital exponent ζ, measured in a₀⁻¹. Keep n (Electrons counted in the mean field n) = 1 fixed. For this exercise use only 1.40875 a₀⁻¹ ≤ z ≤ 2.14375 a₀⁻¹ 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 = -22.561034 eV, with a bounded reading uncertainty ±0.47201 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown z by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000735 a₀⁻¹, 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.
MEAN-FIELD ⟨h + nV_H⟩ against z
Predictions from the supplied Restricted Hartree-Fock helium, solved numerically 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): MEAN-FIELD ⟨h + nV_H⟩ / eV
Horizontal axis (x): z: Trial orbital exponent ζ / a₀⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| z: Trial orbital exponent ζ / a₀⁻¹ | MEAN-FIELD ⟨h + nV_H⟩ / eV |
|---|---|
| 1.40875 | -25.70777881984375 |
| 1.5924999999999998 | -25.079641916875005 |
| 1.7762499999999999 | -23.53273760359375 |
| 1.96 | -21.06706588 |
| 2.14375 | -17.682626746093753 |
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(z) = 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(1.40875) = -25.7078 eV and Y_out(2.14375) = -17.6826 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives z_line = 1.40875 + (-22.561 − -25.7078)(2.14375 − 1.40875)/(-17.6826 − -25.7078) = 1.69695 a₀⁻¹. 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 z = 1.8571 a₀⁻¹; substituting back gives Y_out = -22.561034 eV.
- The interpolation error in the input is |z_line − z| = 0.160149 a₀⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -23.033044 eV and -22.089024 eV. Sort the roots even if the response decreases: 1.81967 a₀⁻¹ ≤ z ≤ 1.89183 a₀⁻¹.
- The lower and upper input deviations are 0.0374335 a₀⁻¹ and 0.0347293 a₀⁻¹. 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 z ≈ 1.8571 a₀⁻¹; compatible interval [1.81967, 1.89183] a₀⁻¹. Endpoint interpolation gives 1.69695 a₀⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.