University Advanced Quantum Mechanics · topic UAQM-10.07
Koopmans' and Brillouin's theorems: what orbital energies mean
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.
What orbital energies mean: Koopmans and Brillouin: predict a finite change
Koopmans' theorem: with every other orbital frozen, removing an electron from canonical Hartree-Fock orbital k costs −εₖ. Brillouin's theorem: H has no matrix element between the converged determinant and any singly excited one. Use the supplied topic model Y_out = v*v/gap, where Y_out is SECOND ORDER v²/Δ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. gap is Rotation gap Δ (both panels), measured in eV. Keep b (Neutral coupling b = ⟨Φᵢᵃ|H|Φ₀⟩) = 0 eV; v (Hole's pull v on the frozen ion) = 8.4 eV fixed. For this exercise use only 19.6125 eV ≤ gap ≤ 46.3875 eV 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 gap = 25.503 eV and 29.2515 eV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δgap across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SECOND ORDER v²/Δ against gap
Predictions from the supplied What orbital energies mean: Koopmans and Brillouin 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): SECOND ORDER v²/Δ / eV
Horizontal axis (x): gap: Rotation gap Δ (both panels) / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gap: Rotation gap Δ (both panels) / eV | SECOND ORDER v²/Δ / eV |
|---|---|
| 19.6125 | 3.5977055449330786 |
| 26.306250000000002 | 2.6822523164647185 |
| 33 | 2.138181818181818 |
| 39.69375 | 1.7776098252243742 |
| 46.3875 | 1.5210994341147939 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of gap separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At gap = 25.503 eV, substitute into Y_out = v*v/gap: Y₀ = 2.76673 eV.
- At gap = 29.2515 eV, the same fixed-parameter model gives Y₁ = 2.41218 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.354549 eV; the input change is Δgap = 3.7485 eV.
- Average sensitivity is ΔY_out/Δgap = -0.0945843 eV per eV. 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 = 2.76673 eV, Y₁ = 2.41218 eV; ΔY_out = -0.354549 eV; average sensitivity = -0.0945843 eV per eV.
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.
What orbital energies mean: Koopmans and Brillouin: infer an input with uncertainty
Koopmans' theorem: with every other orbital frozen, removing an electron from canonical Hartree-Fock orbital k costs −εₖ. Brillouin's theorem: H has no matrix element between the converged determinant and any singly excited one. Use the supplied topic model Y_out = v*v/gap, where Y_out is SECOND ORDER v²/Δ (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. gap is Rotation gap Δ (both panels), measured in eV. Keep b (Neutral coupling b = ⟨Φᵢᵃ|H|Φ₀⟩) = 0 eV; v (Hole's pull v on the frozen ion) = 8.4 eV fixed. For this exercise use only 19.6125 eV ≤ gap ≤ 46.3875 eV 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 = 1.9629854 eV, with a bounded reading uncertainty ±0.066283 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown gap by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0026775 eV, 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.
SECOND ORDER v²/Δ against gap
Predictions from the supplied What orbital energies mean: Koopmans and Brillouin 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): SECOND ORDER v²/Δ / eV
Horizontal axis (x): gap: Rotation gap Δ (both panels) / eV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gap: Rotation gap Δ (both panels) / eV | SECOND ORDER v²/Δ / eV |
|---|---|
| 19.6125 | 3.5977055449330786 |
| 26.306250000000002 | 2.6822523164647185 |
| 33 | 2.138181818181818 |
| 39.69375 | 1.7776098252243742 |
| 46.3875 | 1.5210994341147939 |
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(gap) = 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(19.6125) = 3.59771 eV and Y_out(46.3875) = 1.5211 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives gap_line = 19.6125 + (1.96299 − 3.59771)(46.3875 − 19.6125)/(1.5211 − 3.59771) = 40.69 eV. 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 gap = 35.9452 eV; substituting back gives Y_out = 1.9629854 eV.
- The interpolation error in the input is |gap_line − gap| = 4.74473 eV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.8967024 eV and 2.0292684 eV. Sort the roots even if the response decreases: 34.7712 eV ≤ gap ≤ 37.2014 eV.
- The lower and upper input deviations are 1.1741 eV and 1.25616 eV. 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 gap ≈ 35.9452 eV; compatible interval [34.7712, 37.2014] eV. Endpoint interpolation gives 40.69 eV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.