University Physics V · topic UPV-12.05
Hartree-Fock and the self-consistent field
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.
Hartree-Fock and the self-consistent field: predict a finite change
Hartree-Fock is the best single Slater determinant: the one whose orbitals minimise the energy, each orbital an eigenfunction of a Fock operator built from the averaged Coulomb and exchange field of all the others. Use the supplied topic model Y_out = max(abs((1-a)+a*k),0.0001)^n, where Y_out is DEVIATION SIZE AT n (x d0). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; max(a,b) selects the larger value. n is SCF cycle n, measured in cycles. Keep k (Feedback coefficient k) = -1.5; a (Damping / mixing fraction a) = 0.5 fixed. For this exercise use only 2.125 cycles ≤ n ≤ 17.875 cycles 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 n = 5.59 cycles and 7.795 cycles. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δn across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DEVIATION SIZE AT n against n
Predictions from the supplied Hartree-Fock and the self-consistent field 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): DEVIATION SIZE AT n / x d0
Horizontal axis (x): n: SCF cycle n / cycles
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: SCF cycle n / cycles | DEVIATION SIZE AT n / x d0 |
|---|---|
| 2.125 | 0.052556025953357156 |
| 6.0625 | 0.00022387794023551543 |
| 10 | 9.5367431640625e-7 |
| 13.9375 | 4.062457876895583e-9 |
| 17.875 | 1.7305241126490306e-11 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of n separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At n = 5.59 cycles, substitute into Y_out = max(abs((1-a)+a*k),0.0001)^n: Y₀ = 0.000431007 x d0.
- At n = 7.795 cycles, the same fixed-parameter model gives Y₁ = 0.0000202741 x d0.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.000410733 x d0; the input change is Δn = 2.205 cycles.
- Average sensitivity is ΔY_out/Δn = -0.000186274 x d0 per cycles. 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 = 0.000431007 x d0, Y₁ = 0.0000202741 x d0; ΔY_out = -0.000410733 x d0; average sensitivity = -0.000186274 x d0 per cycles.
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.
Hartree-Fock and the self-consistent field: infer an input with uncertainty
Hartree-Fock is the best single Slater determinant: the one whose orbitals minimise the energy, each orbital an eigenfunction of a Fock operator built from the averaged Coulomb and exchange field of all the others. Use the supplied topic model Y_out = max(abs((1-a)+a*k),0.0001)^n, where Y_out is DEVIATION SIZE AT n (x d0). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|; max(a,b) selects the larger value. n is SCF cycle n, measured in cycles. Keep k (Feedback coefficient k) = -1.5; a (Damping / mixing fraction a) = 0.5 fixed. For this exercise use only 2.125 cycles ≤ n ≤ 17.875 cycles 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 = 8.6363682e-8 x d0, with a bounded reading uncertainty ±1.2952e-8 x d0. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown n by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.001575 cycles, 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.
DEVIATION SIZE AT n against n
Predictions from the supplied Hartree-Fock and the self-consistent field 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): DEVIATION SIZE AT n / x d0
Horizontal axis (x): n: SCF cycle n / cycles
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| n: SCF cycle n / cycles | DEVIATION SIZE AT n / x d0 |
|---|---|
| 2.125 | 0.052556025953357156 |
| 6.0625 | 0.00022387794023551543 |
| 10 | 9.5367431640625e-7 |
| 13.9375 | 4.062457876895583e-9 |
| 17.875 | 1.7305241126490306e-11 |
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(n) = 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(2.125) = 0.052556 x d0 and Y_out(17.875) = 1.73052e-11 x d0. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives n_line = 2.125 + (8.63637e-8 − 0.052556)(17.875 − 2.125)/(1.73052e-11 − 0.052556) = 17.875 cycles. 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 n = 11.7325 cycles; substituting back gives Y_out = 8.6363682e-8 x d0.
- The interpolation error in the input is |n_line − n| = 6.14247 cycles. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 7.3411682e-8 x d0 and 9.9315682e-8 x d0. Sort the roots even if the response decreases: 11.6317 cycles ≤ n ≤ 11.8497 cycles.
- The lower and upper input deviations are 0.100798 cycles and 0.117208 cycles. 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 n ≈ 11.7325 cycles; compatible interval [11.6317, 11.8497] cycles. Endpoint interpolation gives 17.875 cycles.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.