University Advanced Quantum Mechanics · topic UAQM-15.02
The Fermi sea: k_F and a kinetic energy of 2.21/r_s^2 Ry
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 Fermi sea: k_F and a kinetic energy of 2.21/rₛ² Ry: predict a finite change
The Fermi sea is the ground state of free electrons in a box: every plane wave of either spin with |k| ≤ k_F holds one electron and every other is empty, so the density is n = k_F³/(3π²). Use the supplied topic model Y_out = 5173.93/rs^5*((1+z)^(5/3)+max(1-z,0)^(5/3))/2, where Y_out is DEGENERACY PRESSURE P (GPa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. rs is Density parameter rₛ, dimensionless. Keep z (Spin polarisation ζ) = 0 fixed. For this exercise use only 1.97812 ≤ rs ≤ 5.52187 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 rs = 2.75775 and 3.25387. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δrs across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DEGENERACY PRESSURE P against rs
Predictions from the supplied The Fermi sea: k_F and a kinetic energy of 2.21/rₛ² Ry 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): DEGENERACY PRESSURE P / GPa
Horizontal axis (x): rs: Density parameter rₛ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rs: Density parameter rₛ | DEGENERACY PRESSURE P / GPa |
|---|---|
| 1.978125 | 170.825181254169 |
| 2.8640624999999997 | 26.84776228229173 |
| 3.75 | 6.976927499588478 |
| 4.6359375 | 2.41619892643721 |
| 5.521875 | 1.0078309536173289 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of rs separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At rs = 2.75775, substitute into Y_out = 5173.93/rs^5*((1+z)^(5/3)+max(1-z,0)^(5/3))/2: Y₀ = 32.4374 GPa.
- At rs = 3.25387, the same fixed-parameter model gives Y₁ = 14.1846 GPa.
- Signed change: ΔY_out = Y₁ − Y₀ = -18.2528 GPa; the input change is Δrs = 0.496125.
- Average sensitivity is ΔY_out/Δrs = -36.7908 GPa 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 = 32.4374 GPa, Y₁ = 14.1846 GPa; ΔY_out = -18.2528 GPa; average sensitivity = -36.7908 GPa 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 Fermi sea: k_F and a kinetic energy of 2.21/rₛ² Ry: infer an input with uncertainty
The Fermi sea is the ground state of free electrons in a box: every plane wave of either spin with |k| ≤ k_F holds one electron and every other is empty, so the density is n = k_F³/(3π²). Use the supplied topic model Y_out = 5173.93/rs^5*((1+z)^(5/3)+max(1-z,0)^(5/3))/2, where Y_out is DEGENERACY PRESSURE P (GPa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. rs is Density parameter rₛ, dimensionless. Keep z (Spin polarisation ζ) = 0 fixed. For this exercise use only 1.97812 ≤ rs ≤ 5.52187 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 = 4.2551727 GPa, with a bounded reading uncertainty ±0.4871 GPa. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown rs by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000354375, 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.
DEGENERACY PRESSURE P against rs
Predictions from the supplied The Fermi sea: k_F and a kinetic energy of 2.21/rₛ² Ry 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): DEGENERACY PRESSURE P / GPa
Horizontal axis (x): rs: Density parameter rₛ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| rs: Density parameter rₛ | DEGENERACY PRESSURE P / GPa |
|---|---|
| 1.978125 | 170.825181254169 |
| 2.8640624999999997 | 26.84776228229173 |
| 3.75 | 6.976927499588478 |
| 4.6359375 | 2.41619892643721 |
| 5.521875 | 1.0078309536173289 |
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(rs) = 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(1.97812) = 170.825 GPa and Y_out(5.52187) = 1.00783 GPa. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rs_line = 1.97812 + (4.25517 − 170.825)(5.52187 − 1.97812)/(1.00783 − 170.825) = 5.45411. 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 rs = 4.13981; substituting back gives Y_out = 4.2551727 GPa.
- The interpolation error in the input is |rs_line − rs| = 1.3143. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.7680727 GPa and 4.7422727 GPa. Sort the roots even if the response decreases: 4.05104 ≤ rs ≤ 4.2417.
- The lower and upper input deviations are 0.0887699 and 0.10189. 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 rs ≈ 4.13981; compatible interval [4.05104, 4.2417]. Endpoint interpolation gives 5.45411.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.