University Physics V · unit UPV-12
Multi-Electron Atoms
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.
Identical particles and the exchange operator: infer an input with uncertainty
The exchange operator P̂ᵢⱼ swaps the labels of two identical particles in a state, and because it is Hermitian and squares to the identity its only eigenvalues are +1 for a symmetric state and −1 for an antisymmetric one. Use the supplied topic model Y_out = 2*(1+eta)*sin(pi*x0)*sin(nb*pi*x0)/max(abs(2*sin(nb*pi*x0))+abs(2*eta*sin(pi*x0)),0.35), where Y_out is Ψ AT x₁ = x₂ (SCALED) (dimensionless). 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; unmarked trigonometric functions use radians. eta is Exchange mixing η, dimensionless. Keep x0 (Position x₂ of particle 2) = 0.23 L; nb (Second orbital n_b) = 2 fixed. For this exercise use only -0.7875 ≤ eta ≤ 0.7875 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 = 0.69555898, with a bounded reading uncertainty ±0.011944. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown eta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575, 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.
Ψ AT x₁ = x₂ (SCALED) against eta
Predictions from the supplied Identical particles and the exchange operator 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): Ψ AT x₁ = x₂ (SCALED)
Horizontal axis (x): eta: Exchange mixing η
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| eta: Exchange mixing η | Ψ AT x₁ = x₂ (SCALED) |
|---|---|
| -0.7875 | 0.09215471162832578 |
| -0.39375000000000004 | 0.31757042524744444 |
| -1.1102230246251565e-16 | 0.6613118653236517 |
| 0.39374999999999993 | 0.7300845858781455 |
| 0.7874999999999998 | 0.7751837507559168 |
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(eta) = 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(-0.7875) = 0.0921547 and Y_out(0.7875) = 0.775184. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives eta_line = -0.7875 + (0.695559 − 0.0921547)(0.7875 − -0.7875)/(0.775184 − 0.0921547) = 0.603893. 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 eta = 0.17325; substituting back gives Y_out = 0.69555898.
- The interpolation error in the input is |eta_line − eta| = 0.430643. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.68361498 and 0.70750298. Sort the roots even if the response decreases: 0.108459 ≤ eta ≤ 0.243479.
- The lower and upper input deviations are 0.0647908 and 0.0702288. 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 eta ≈ 0.17325; compatible interval [0.108459, 0.243479]. Endpoint interpolation gives 0.603893.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Electron spin as a two-state space: infer an input with uncertainty
Electron spin is a two-dimensional complex vector space that every electron carries alongside its wavefunction, acted on by S = (ħ/2)σ, whose measurement along any axis returns only +ħ/2 or −ħ/2. Use the supplied topic model Y_out = 0.5*cosd(b-a), where Y_out is ⟨S⋅n⟩ IN STATE m (ħ). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. a is State direction θm from +z, measured in °. Keep b (Analyser direction θn from +z) = 120 ° fixed. For this exercise use only 123.075 ° ≤ a ≤ 160.875 ° 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 = 0.44888702 ħ, with a bounded reading uncertainty ±0.007559 ħ. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00378 °, 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.
⟨S⋅n⟩ IN STATE m against a
Predictions from the supplied Electron spin as a two-state space 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): ⟨S⋅n⟩ IN STATE m / ħ
Horizontal axis (x): a: State direction θm from +z / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: State direction θm from +z / ° | ⟨S⋅n⟩ IN STATE m / ħ |
|---|---|
| 123.07499999999999 | 0.4992800858026867 |
| 132.52499999999998 | 0.4881007372945333 |
| 141.975 | 0.4636736096308058 |
| 151.425 | 0.4266616913463868 |
| 160.875 | 0.37806954089016137 |
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(a) = 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(123.075) = 0.49928 ħ and Y_out(160.875) = 0.37807 ħ. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a_line = 123.075 + (0.448887 − 0.49928)(160.875 − 123.075)/(0.37807 − 0.49928) = 138.79 °. 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 a = 146.133 °; substituting back gives Y_out = 0.44888702 ħ.
- The interpolation error in the input is |a_line − a| = 7.34272 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.44132802 ħ and 0.45644602 ħ. Sort the roots even if the response decreases: 144.092 ° ≤ a ≤ 148.036 °.
- The lower and upper input deviations are 2.04112 ° and 1.90256 °. 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 a ≈ 146.133 °; compatible interval [144.092, 148.036] °. Endpoint interpolation gives 138.79 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Slater determinants and occupation numbers: infer an input with uncertainty
A Slater determinant is the antisymmetrised product of N orthonormal spin-orbitals, Ψ = (1/√N!) det[φ_i(x_j)] — the simplest N-fermion state, and the one whose occupation numbers are all exactly 0 or 1. Use the supplied topic model Y_out = 2*sin(pi*p)^2*sin(2*pi*p)^2/max(sin(pi*p)^2+sin(2*pi*p)^2,0.05), where Y_out is DIRECT TERM at x₂ = x₁ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; unmarked trigonometric functions use radians. p is Reference electron x₁ / L, measured in L. Keep s (Exchange weight s (−1 = fermion)) = -1 fixed. For this exercise use only 0.323 L ≤ p ≤ 0.4175 L 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 = 0.60469016, with a bounded reading uncertainty ±0.023362. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00000945 L, 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.
DIRECT TERM at x₂ = x₁ against p
Predictions from the supplied Slater determinants and occupation numbers 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): DIRECT TERM at x₂ = x₁
Horizontal axis (x): p: Reference electron x₁ / L / L
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Reference electron x₁ / L / L | DIRECT TERM at x₂ = x₁ |
|---|---|
| 0.323 | 0.7604364228224935 |
| 0.346625 | 0.7256916010268496 |
| 0.37024999999999997 | 0.6506468924163751 |
| 0.393875 | 0.5355857334365328 |
| 0.4175 | 0.38880510785453665 |
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(p) = 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(0.323) = 0.760436 and Y_out(0.4175) = 0.388805. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives p_line = 0.323 + (0.60469 − 0.760436)(0.4175 − 0.323)/(0.388805 − 0.760436) = 0.362604 L. 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 p = 0.380645 L; substituting back gives Y_out = 0.60469016.
- The interpolation error in the input is |p_line − p| = 0.0180412 L. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.58132816 and 0.62805216. Sort the roots even if the response decreases: 0.375578 L ≤ p ≤ 0.385367 L.
- The lower and upper input deviations are 0.00506701 L and 0.00472202 L. 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 p ≈ 0.380645 L; compatible interval [0.375578, 0.385367] L. Endpoint interpolation gives 0.362604 L.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Helium: exchange and the variational bound: infer an input with uncertainty
The variational bound on helium is the lowest expectation value ⟨ψ|H|ψ⟩ reachable from a chosen family of trial states; it is guaranteed to lie at or above the true ground-state energy, never below it. Use the supplied topic model Y_out = 27.2114*zp*(zp-4+0.625*lam), where Y_out is TRIAL ENERGY E(Z′) (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. zp is Trial nuclear charge Z′, dimensionless. Keep lam (Repulsion switched on, λ) = 1 fixed. For this exercise use only 1.71063 ≤ zp ≤ 2.34063 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 = -72.971753 eV, with a bounded reading uncertainty ±0.67536 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown zp by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000063, 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.
TRIAL ENERGY E(Z′) against zp
Predictions from the supplied Helium: exchange and the variational bound 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): TRIAL ENERGY E(Z′) / eV
Horizontal axis (x): zp: Trial nuclear charge Z′
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| zp: Trial nuclear charge Z′ | TRIAL ENERGY E(Z′) / eV |
|---|---|
| 1.710625 | -77.47416155992187 |
| 1.868125 | -76.60093072679688 |
| 2.0256250000000002 | -74.37767431117187 |
| 2.183125 | -70.80439231304688 |
| 2.340625 | -65.88108473242187 |
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(zp) = 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(1.71063) = -77.4742 eV and Y_out(2.34063) = -65.8811 eV. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives zp_line = 1.71063 + (-72.9718 − -77.4742)(2.34063 − 1.71063)/(-65.8811 − -77.4742) = 1.9553. 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 zp = 2.09493; substituting back gives Y_out = -72.971753 eV.
- The interpolation error in the input is |zp_line − zp| = 0.139627. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -73.647113 eV and -72.296393 eV. Sort the roots even if the response decreases: 2.06323 ≤ zp ≤ 2.12432.
- The lower and upper input deviations are 0.0316909 and 0.0293978. 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 zp ≈ 2.09493; compatible interval [2.06323, 2.12432]. Endpoint interpolation gives 1.9553.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.