University Advanced Quantum Mechanics · topic UAQM-15.05
The Hartree-Fock spectrum and its logarithmic singularity at k_F
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.
The Hartree-Fock spectrum and its singularity at k_F: predict a finite change
The Hartree-Fock spectrum of the electron gas is the energy εₖ = ħ²k²/2m + Σₓ(k) of a plane-wave orbital in the self-consistent field of the filled Fermi sea, the exchange term Σₓ(k) being the electron's attraction to its own exchange hole. Use the supplied topic model Y_out = (3.683169/rs^2+2.443548/rs*(0.5-ks*atan(1/ks)-ks^2/8*ln(1+4/ks^2)+ks/2*atan(2/ks)))*13.60569, where Y_out is SCREENED BANDWIDTH (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; unmarked trigonometric functions use radians. ks is Screening wavenumber kₛ/k_F, dimensionless. Keep rs (Density parameter rₛ) = 3.93 fixed. For this exercise use only 0.310312 ≤ ks ≤ 2.23969 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 ks = 0.734775 and 1.00489. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δks across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
SCREENED BANDWIDTH against ks
Predictions from the supplied The Hartree-Fock spectrum and its singularity at k_F 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): SCREENED BANDWIDTH / eV
Horizontal axis (x): ks: Screening wavenumber kₛ/k_F
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ks: Screening wavenumber kₛ/k_F | SCREENED BANDWIDTH / eV |
|---|---|
| 0.3103125 | 5.618542505545603 |
| 0.7926562499999998 | 4.110333678598147 |
| 1.275 | 3.5770911580355347 |
| 1.7573437499999998 | 3.387496935890547 |
| 2.2396874999999996 | 3.313164526129436 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of ks separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At ks = 0.734775, substitute into Y_out = (3.683169/rs^2+2.443548/rs*(0.5-ks*atan(1/ks)-ks^2/8*ln(1+4/ks^2)+ks/2*atan(2/ks)))*13.60569: Y₀ = 4.22146 eV.
- At ks = 1.00489, the same fixed-parameter model gives Y₁ = 3.80581 eV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.415643 eV; the input change is Δks = 0.270112.
- Average sensitivity is ΔY_out/Δks = -1.53878 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 = 4.22146 eV, Y₁ = 3.80581 eV; ΔY_out = -0.415643 eV; average sensitivity = -1.53878 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.
The Hartree-Fock spectrum and its singularity at k_F: infer an input with uncertainty
The Hartree-Fock spectrum of the electron gas is the energy εₖ = ħ²k²/2m + Σₓ(k) of a plane-wave orbital in the self-consistent field of the filled Fermi sea, the exchange term Σₓ(k) being the electron's attraction to its own exchange hole. Use the supplied topic model Y_out = (3.683169/rs^2+2.443548/rs*(0.5-ks*atan(1/ks)-ks^2/8*ln(1+4/ks^2)+ks/2*atan(2/ks)))*13.60569, where Y_out is SCREENED BANDWIDTH (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm; unmarked trigonometric functions use radians. ks is Screening wavenumber kₛ/k_F, dimensionless. Keep rs (Density parameter rₛ) = 3.93 fixed. For this exercise use only 0.310312 ≤ ks ≤ 2.23969 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 = 3.4706298 eV, with a bounded reading uncertainty ±0.02362 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown ks by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000192937, 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.
SCREENED BANDWIDTH against ks
Predictions from the supplied The Hartree-Fock spectrum and its singularity at k_F 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): SCREENED BANDWIDTH / eV
Horizontal axis (x): ks: Screening wavenumber kₛ/k_F
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ks: Screening wavenumber kₛ/k_F | SCREENED BANDWIDTH / eV |
|---|---|
| 0.3103125 | 5.618542505545603 |
| 0.7926562499999998 | 4.110333678598147 |
| 1.275 | 3.5770911580355347 |
| 1.7573437499999998 | 3.387496935890547 |
| 2.2396874999999996 | 3.313164526129436 |
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(ks) = 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(0.310312) = 5.61854 eV and Y_out(2.23969) = 3.31316 eV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives ks_line = 0.310312 + (3.47063 − 5.61854)(2.23969 − 0.310312)/(3.31316 − 5.61854) = 2.1079. 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 ks = 1.48723; substituting back gives Y_out = 3.4706298 eV.
- The interpolation error in the input is |ks_line − ks| = 0.620673. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.4470098 eV and 3.4942498 eV. Sort the roots even if the response decreases: 1.43136 ≤ ks ≤ 1.55034.
- The lower and upper input deviations are 0.0558728 and 0.0631099. 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 ks ≈ 1.48723; compatible interval [1.43136, 1.55034]. Endpoint interpolation gives 2.1079.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.