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University Advanced Quantum Mechanics · unit UAQM-10

Second Quantisation and the Hartree-Fock Method

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

From the Pauli field to occupation numbers on any basis: infer an input with uncertainty

An occupation-number state lists, for an ordered set of orthonormal orbitals, how many identical particles occupy each; a creation operator raises one occupation, with a factor √(n + 1) for bosons and, for fermions, a −1 for each occupied orbital earlier in the order. Use the supplied topic model Y_out = (1-abs(op-floor(s/2^(k-1))%2))*(-1)^((s%2^(k-1))-floor((s%2^(k-1))/2)-floor((s%2^(k-1))/4)-floor((s%2^(k-1))/8)), where Y_out is FERMION COEFFICIENT (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; abs(x) means |x|. k is Orbital acted on, k, dimensionless. Keep s (Pattern s = n₁ + 2n₂ + 4n₃ + 8n₄ + 16n₅) = 5; op (Operator: 0 = c†ₖ, 1 = cₖ) = 0 fixed. For this exercise use only 2 ≤ k ≤ 4, at the specified test settings separated by 1. 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, with a bounded reading uncertainty ±0.16. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of k consistent with that bound. Compare with a straight-line interpolation between the two endpoint predictions, and explain why a fractional setting may be inadmissible. Do not assume that the supplied calibration points prove monotonicity outside this branch.

Explore the question data

FERMION COEFFICIENT against k

Predictions from the supplied From the Pauli field to occupation numbers on any basis model, with all other inputs fixed as stated in the question. 3 calibration points support a trend comparison; they are not measured data or a straight-line law.

Vertical axis (y): FERMION COEFFICIENT

FERMION COEFFICIENT against kPredictions from the supplied From the Pauli field to occupation numbers on any basis model, with all other inputs fixed as stated in the question. 3 calibration points support a trend comparison; they are not measured data or a straight-line law. Horizontal axis: k: Orbital acted on, k. Vertical axis: FERMION COEFFICIENT. Exact values are available in the data table below.-1.08-0.5400.541.081.922.4633.544.08xy

Horizontal axis (x): k: Orbital acted on, k

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

Reading 1k: Orbital acted on, k: 2FERMION COEFFICIENT: -1
Read the exact data as a table
FERMION COEFFICIENT against k — supplied values
k: Orbital acted on, kFERMION COEFFICIENT
2-1
30
41

These are the supplied values, not a worked answer. No fitted trend is drawn.

Show a hint

Enumerate the permitted settings in the stated range, evaluate Y_out at each, and retain every setting satisfying |Y_out − Y_measured| ≤ u. The plotted predictions are supplied calibration evidence.

Reveal the worked solution
  1. Endpoint predictions are Y_out(2) = -1 and Y_out(4) = 1. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives k_line = 2 + (0 − -1)(4 − 2)/(1 − -1) = 3. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
  3. Evaluate the model at each permitted setting. The central reading corresponds to k = 3; only the allowed settings 3 lie inside the stated reading bound.
  4. The interpolation error in the input is |k_line − k| = 0.
  5. The compatible allowed values span 3 to 3. Values between specified test settings are outside this exercise's input domain even if an interpolation passes through them.
  6. The quoted output bound restricts which specified settings remain compatible. It does not change physical restrictions on quantum numbers or counts, and the chosen grid alone does not establish physical quantisation.

Answer Compatible k: 3. Endpoint interpolation predicts 3; check admissibility against the discrete settings.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 02

Field operators, change of basis and the wavefunction: infer an input with uncertainty

The field operator ψ(x) = Σᵢ φᵢ(x)cᵢ removes a particle at position r with spin σ; it is the same operator whichever complete orthonormal orbitals expand it, and its matrix elements between the vacuum and an N-particle state are that state's wavefunction. Use the supplied topic model Y_out = sqrt(2)*(sin(pi/4)*(cosd(mix)*sin(2*pi*xp) + sind(mix)*sin(3*pi*xp)) - (cosd(mix) + sin(pi/4)*sind(mix))*sin(pi*xp)), where Y_out is Ψ(L/4, x₂) × L (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; sind/cosd/tand use degrees; unmarked trigonometric functions use radians. mix is Mixing with empty φ₃, β, measured in °. Keep th (Occupied rotation θ) = 0 °; xp (Second electron x₂) = 0.3 L fixed. For this exercise use only 9.5625 ° ≤ mix ≤ 66.2625 ° 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 = -0.48679633, with a bounded reading uncertainty ±0.0072932. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown mix by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00567 °, 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(L/4, x₂) × L against mix

Predictions from the supplied Field operators, change of basis and the wavefunction 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(L/4, x₂) × L

Ψ(L/4, x₂) × L against mixPredictions from the supplied Field operators, change of basis and the wavefunction 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: mix: Mixing with empty φ₃, β / °. Vertical axis: Ψ(L/4, x₂) × L. Exact values are available in the data table below.-0.5459-0.4752-0.4044-0.3337-0.2637.29422.637.9153.2268.53xy

Horizontal axis (x): mix: Mixing with empty φ₃, β / °

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

Reading 1mix: Mixing with empty φ₃, β / °: 9.563Ψ(L/4, x₂) × L: -0.2734
Read the exact data as a table
Ψ(L/4, x₂) × L against mix — supplied values
mix: Mixing with empty φ₃, β / °Ψ(L/4, x₂) × L
9.5625-0.2734453176442383
23.7375-0.37800624570671876
37.9125-0.4595483270530312
52.087500000000006-0.5131060229050558
66.2625-0.5354179152166867

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(mix) = 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(9.5625) = -0.273445 and Y_out(66.2625) = -0.535418. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives mix_line = 9.5625 + (-0.486796 − -0.273445)(66.2625 − 9.5625)/(-0.535418 − -0.273445) = 55.7391 °. 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 mix = 44.1495 °; substituting back gives Y_out = -0.48679633.
  4. The interpolation error in the input is |mix_line − mix| = 11.5896 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = -0.49408953 and -0.47950313. Sort the roots even if the response decreases: 42.3476 ° ≤ mix ≤ 46.0838 °.
  6. The lower and upper input deviations are 1.80192 ° and 1.93435 °. 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 mix ≈ 44.1495 °; compatible interval [42.3476, 46.0838] °. Endpoint interpolation gives 55.7391 °.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 03

One-body operators: the sum of ⟨i|f|j⟩ c†ᵢ cⱼ: infer an input with uncertainty

A one-body operator is the sum of one single-particle operator f over all the particles; in second quantisation it is F = Σᵢⱼ ⟨i|f|j⟩ c†ᵢ cⱼ, each term emptying orbital j and filling orbital i with amplitude ⟨i|f|j⟩. Use the supplied topic model Y_out = cosd(theta), where Y_out is OVERLAP ⟨u|v⟩ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. theta is Mixing angle θ, measured in °. Keep k (Admixed box level k) = 2 fixed. For this exercise use only 9.5625 ° ≤ theta ≤ 80.4375 ° 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 = 0.60465125, with a bounded reading uncertainty ±0.057218. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0070875 °, 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

OVERLAP ⟨u|v⟩ against theta

Predictions from the supplied One-body operators: the sum of ⟨i|f|j⟩ c†ᵢ cⱼ 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): OVERLAP ⟨u|v⟩

OVERLAP ⟨u|v⟩ against thetaPredictions from the supplied One-body operators: the sum of ⟨i|f|j⟩ c†ᵢ cⱼ 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: theta: Mixing angle θ / °. Vertical axis: OVERLAP ⟨u|v⟩. Exact values are available in the data table below.0.13330.35470.57610.79751.0196.72825.864564.1483.27xy

Horizontal axis (x): theta: Mixing angle θ / °

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

Reading 1theta: Mixing angle θ / °: 9.563OVERLAP ⟨u|v⟩: 0.9861
Read the exact data as a table
OVERLAP ⟨u|v⟩ against theta — supplied values
theta: Mixing angle θ / °OVERLAP ⟨u|v⟩
9.56250.9861049757675088
27.281250.8887672777860675
450.7071067811865476
62.718750.45835873062127136
80.43750.16612337814576558

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(theta) = 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(9.5625) = 0.986105 and Y_out(80.4375) = 0.166123. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives theta_line = 9.5625 + (0.604651 − 0.986105)(80.4375 − 9.5625)/(0.166123 − 0.986105) = 42.5334 °. 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 theta = 52.7962 °; substituting back gives Y_out = 0.60465125.
  4. The interpolation error in the input is |theta_line − theta| = 10.2628 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.54743325 and 0.66186925. Sort the roots even if the response decreases: 48.5574 ° ≤ theta ≤ 56.8089 °.
  6. The lower and upper input deviations are 4.23884 ° and 4.01265 °. 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 theta ≈ 52.7962 °; compatible interval [48.5574, 56.8089] °. Endpoint interpolation gives 42.5334 °.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

Extra challenging · 04

Two-body operators and the many-electron Hamiltonian: infer an input with uncertainty

A two-body operator in second quantisation removes a pair of particles from any two orbitals and puts them into any two, the same or different, weighted by the interaction's matrix element between the pairs: V̂ = ½Σ⟨ij|v|kl⟩c†ᵢc†ⱼcₗcₖ, with the annihilators in reverse order. Use the supplied topic model Y_out = ord*(11.422-1.1945*sind(2*th)), where Y_out is ⟨V̂⟩ IN ψ(θ) (eV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. th is Mixing angle θ, measured in °. Keep ord (Annihilator order: +1 cₗcₖ, −1 cₖcₗ) = 1 fixed. For this exercise use only 51.975 ° ≤ th ≤ 70.875 ° 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 = 10.46813 eV, with a bounded reading uncertainty ±0.03081 eV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00189 °, 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

⟨V̂⟩ IN ψ(θ) against th

Predictions from the supplied Two-body operators and the many-electron Hamiltonian 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): ⟨V̂⟩ IN ψ(θ) / eV

⟨V̂⟩ IN ψ(θ) against thPredictions from the supplied Two-body operators and the many-electron Hamiltonian 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: th: Mixing angle θ / °. Vertical axis: ⟨V̂⟩ IN ψ(θ) / eV. Exact values are available in the data table below.10.24610.35910.47310.58610.69951.2256.3261.4266.5371.63xy

Horizontal axis (x): th: Mixing angle θ / °

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

Reading 1th: Mixing angle θ / °: 51.97⟨V̂⟩ IN ψ(θ) / eV: 10.26
Read the exact data as a table
⟨V̂⟩ IN ψ(θ) against th — supplied values
th: Mixing angle θ / °⟨V̂⟩ IN ψ(θ) / eV
51.97499999999999410.262730017441827
56.69999999999999610.325742099620484
61.42510.418508249867715
66.1510.538510657048052
70.87510.682492277549404

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(th) = 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(51.975) = 10.2627 eV and Y_out(70.875) = 10.6825 eV. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives th_line = 51.975 + (10.4681 − 10.2627)(70.875 − 51.975)/(10.6825 − 10.2627) = 61.2232 °. 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 th = 63.504 °; substituting back gives Y_out = 10.46813 eV.
  4. The interpolation error in the input is |th_line − th| = 2.28076 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 10.43732 eV and 10.49894 eV. Sort the roots even if the response decreases: 62.2389 ° ≤ th ≤ 64.6989 °.
  6. The lower and upper input deviations are 1.26506 ° and 1.19489 °. 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 th ≈ 63.504 °; compatible interval [62.2389, 64.6989] °. Endpoint interpolation gives 61.2232 °.

Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.

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