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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.

Extra challenging · 01

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.

Explore the question data

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)

Ψ AT x₁ = x₂ (SCALED) against etaPredictions 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. Horizontal axis: eta: Exchange mixing η. Vertical axis: Ψ AT x₁ = x₂ (SCALED). Exact values are available in the data table below.0.064830.24930.43370.61810.8025-0.8505-0.4253-1.11e-160.42520.8505xy

Horizontal axis (x): eta: Exchange mixing η

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1eta: Exchange mixing η: -0.7875Ψ AT x₁ = x₂ (SCALED): 0.09215
Read the exact data as a table
Ψ AT x₁ = x₂ (SCALED) against eta — supplied values
eta: Exchange mixing ηΨ AT x₁ = x₂ (SCALED)
-0.78750.09215471162832578
-0.393750000000000040.31757042524744444
-1.1102230246251565e-160.6613118653236517
0.393749999999999930.7300845858781455
0.78749999999999980.7751837507559168

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |eta_line − eta| = 0.430643. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.68361498 and 0.70750298. Sort the roots even if the response decreases: 0.108459 ≤ eta ≤ 0.243479.
  6. 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.

Extra challenging · 02

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.

Explore the question data

⟨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

⟨S⋅n⟩ IN STATE m against aPredictions 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. Horizontal axis: a: State direction θm from +z / °. Vertical axis: ⟨S⋅n⟩ IN STATE m / ħ. Exact values are available in the data table below.0.37320.40590.43870.47140.5041121.6131.8142152.2162.4xy

Horizontal axis (x): a: State direction θm from +z / °

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1a: State direction θm from +z / °: 123.1⟨S⋅n⟩ IN STATE m / ħ: 0.4993
Read the exact data as a table
⟨S⋅n⟩ IN STATE m against a — supplied values
a: State direction θm from +z / °⟨S⋅n⟩ IN STATE m / ħ
123.074999999999990.4992800858026867
132.524999999999980.4881007372945333
141.9750.4636736096308058
151.4250.4266616913463868
160.8750.37806954089016137

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |a_line − a| = 7.34272 °. This is larger than the requested numerical tolerance.
  5. 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 °.
  6. 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.

Extra challenging · 03

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.

Explore the question data

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

DIRECT TERM at x₂ = x₁ against pPredictions 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. Horizontal axis: p: Reference electron x₁ / L / L. Vertical axis: DIRECT TERM at x₂ = x₁. Exact values are available in the data table below.0.37390.47430.57460.6750.77530.31920.34470.37020.39580.4213xy

Horizontal axis (x): p: Reference electron x₁ / L / L

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1p: Reference electron x₁ / L / L: 0.323DIRECT TERM at x₂ = x₁: 0.7604
Read the exact data as a table
DIRECT TERM at x₂ = x₁ against p — supplied values
p: Reference electron x₁ / L / LDIRECT TERM at x₂ = x₁
0.3230.7604364228224935
0.3466250.7256916010268496
0.370249999999999970.6506468924163751
0.3938750.5355857334365328
0.41750.38880510785453665

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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |p_line − p| = 0.0180412 L. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

Extra challenging · 04

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.

Explore the question data

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

TRIAL ENERGY E(Z′) against zpPredictions 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. Horizontal axis: zp: Trial nuclear charge Z′. Vertical axis: TRIAL ENERGY E(Z′) / eV. Exact values are available in the data table below.-77.94-74.81-71.68-68.55-65.421.6851.8562.0262.1962.366xy

Horizontal axis (x): zp: Trial nuclear charge Z′

Axes show the supplied calibration interval; the origin may lie outside the plot.

Reading 1zp: Trial nuclear charge Z′: 1.711TRIAL ENERGY E(Z′) / eV: -77.47
Read the exact data as a table
TRIAL ENERGY E(Z′) against zp — supplied values
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
  1. 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.
  2. 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.
  3. 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.
  4. The interpolation error in the input is |zp_line − zp| = 0.139627. This is larger than the requested numerical tolerance.
  5. 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.
  6. 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.

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