University Advanced Quantum Mechanics · unit UAQM-15
Many-Body Applications: the Electron Gas and the Bose Gas
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
The jellium Hamiltonian in second quantisation: infer an input with uncertainty
Jellium is N electrons in a box whose ions are smeared into a uniform positive background; on plane waves its Hamiltonian is Σεₖc†c plus (1/2V)Σ′v_q c†c†cc with v_q = e²/(ε₀q²) and q = 0 left out, and its only parameter is rₛ, the radius per electron in Bohr radii. Use the supplied topic model Y_out = 20.380+2*cut, where Y_out is ONE TERM, LOG₁₀(X/eV) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. cut is Cutoff μ⁻¹ as log₁₀(μ⁻¹/m), dimensionless. Keep nexp (Electrons N as log₁₀ N) = 23 fixed. For this exercise use only -7.83125 ≤ cut ≤ 0.83125 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 = 15.28575, with a bounded reading uncertainty ±1.0135. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown cut by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00086625, 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.
ONE TERM, LOG₁₀(X/eV) against cut
Predictions from the supplied The jellium Hamiltonian in second quantisation 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): ONE TERM, LOG₁₀(X/eV)
Horizontal axis (x): cut: Cutoff μ⁻¹ as log₁₀(μ⁻¹/m)
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| cut: Cutoff μ⁻¹ as log₁₀(μ⁻¹/m) | ONE TERM, LOG₁₀(X/eV) |
|---|---|
| -7.83125 | 4.717499999999999 |
| -5.665625 | 9.048749999999998 |
| -3.500000000000001 | 13.379999999999997 |
| -1.3343750000000014 | 17.711249999999996 |
| 0.831249999999998 | 22.042499999999997 |
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(cut) = 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(-7.83125) = 4.7175 and Y_out(0.83125) = 22.0425. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives cut_line = -7.83125 + (15.2858 − 4.7175)(0.83125 − -7.83125)/(22.0425 − 4.7175) = -2.54712. 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 cut = -2.54713; substituting back gives Y_out = 15.28575.
- The interpolation error in the input is |cut_line − cut| = 8.88178e-16. This is within the requested numerical tolerance.
- Invert both limits Y_out = 14.27225 and 16.29925. Sort the roots even if the response decreases: -3.05388 ≤ cut ≤ -2.04038.
- The lower and upper input deviations are 0.50675 and 0.50675. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer cut ≈ -2.54713; compatible interval [-3.05388, -2.04038]. Endpoint interpolation gives -2.54712.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
The exchange energy of the electron gas: −0.916/r_s Ry: infer an input with uncertainty
The exchange energy of the electron gas is the Coulomb energy per electron of the Hartree-Fock Fermi sea once the background has cancelled the direct term: it is negative, because antisymmetry keeps electrons of the same spin apart. Use the supplied topic model Y_out = 100*(8/3)*(min(q,1)-0.75*min(q,1)^2+0.125*min(q,1)^4), where Y_out is SHARE FROM TRANSFERS BELOW q (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value. q is Momentum transfer q/2k_F, dimensionless. Keep rs (Density parameter r_s) = 3.93 fixed. For this exercise use only 0.172187 ≤ q ≤ 0.98725 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 = 95.579458 %, with a bounded reading uncertainty ±0.66304 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000815063, 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.
SHARE FROM TRANSFERS BELOW q against q
Predictions from the supplied The exchange energy of the electron gas: −0.916/r_s 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): SHARE FROM TRANSFERS BELOW q / %
Horizontal axis (x): q: Momentum transfer q/2k_F
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Momentum transfer q/2k_F | SHARE FROM TRANSFERS BELOW q / % |
|---|---|
| 0.1721875 | 40.01626082331371 |
| 0.375953125 | 72.6519231682698 |
| 0.5797187500000001 | 91.14175479992036 |
| 0.783484375 | 98.71991614825637 |
| 0.98725 | 99.99972452463552 |
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(q) = 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(0.172187) = 40.0163 % and Y_out(0.98725) = 99.9997 %. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives q_line = 0.172187 + (95.5795 − 40.0163)(0.98725 − 0.172187)/(99.9997 − 40.0163) = 0.927187. 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 q = 0.669376; substituting back gives Y_out = 95.579458 %.
- The interpolation error in the input is |q_line − q| = 0.257811. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 94.916418 % and 96.242498 %. Sort the roots even if the response decreases: 0.653095 ≤ q ≤ 0.687318.
- The lower and upper input deviations are 0.0162802 and 0.0179421. 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 q ≈ 0.669376; compatible interval [0.653095, 0.687318]. Endpoint interpolation gives 0.927187.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Hartree-Fock binding of a metal, and what correlation adds: infer an input with uncertainty
Hartree-Fock binding is the minimum of the energy per electron of jellium held in one Slater determinant, the density at which the slopes of the kinetic energy, 2.21/rₛ² Ry, and the exchange energy, −0.916/rₛ Ry, cancel. Use the supplied topic model Y_out = 1170.6*(4.42/rs^3-0.916/rs^2-lam*0.2846*(0.52645/sqrt(rs)+0.3334)/(1+1.0529*sqrt(rs)+0.3334*rs)^2)/rs^2, where Y_out is PRESSURE, UNPOLARISED GAS (GPa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. rs is Density parameter rₛ, dimensionless. Keep lam (Correlation energy added, fraction λ) = 0 fixed. For this exercise use only 3.6375 ≤ rs ≤ 5.8425 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 = -0.054898482 GPa, with a bounded reading uncertainty ±0.015798 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 0.0002205, 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.
PRESSURE, UNPOLARISED GAS against rs
Predictions from the supplied Hartree-Fock binding of a metal, and what correlation adds 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): PRESSURE, UNPOLARISED GAS / 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 | PRESSURE, UNPOLARISED GAS / GPa |
|---|---|
| 3.6375 | 2.000056954949522 |
| 4.188750000000001 | 0.5293380107662309 |
| 4.74 | 0.03823854529675329 |
| 5.29125 | -0.12045539592844048 |
| 5.8425 | -0.16021567124405342 |
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(3.6375) = 2.00006 GPa and Y_out(5.8425) = -0.160216 GPa. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives rs_line = 3.6375 + (-0.0548985 − 2.00006)(5.8425 − 3.6375)/(-0.160216 − 2.00006) = 5.735. 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 = 4.98255; substituting back gives Y_out = -0.054898482 GPa.
- The interpolation error in the input is |rs_line − rs| = 0.752452. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -0.070696482 GPa and -0.039100482 GPa. Sort the roots even if the response decreases: 4.93171 ≤ rs ≤ 5.03967.
- The lower and upper input deviations are 0.0508412 and 0.0571175. 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 ≈ 4.98255; compatible interval [4.93171, 5.03967]. Endpoint interpolation gives 5.735.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.