Skip to main content
Course and syllabus guidance

University Quantum Mechanics II · unit UQMII-07

Identical Particles and Many-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

The exchange operator: why [P, H] = 0 forces no symmetry: infer an input with uncertainty

The exchange operator P swaps the labels of two particles, PΨ(r₁, r₂) = Ψ(r₂, r₁); because P² = 1 its only eigenvalues are +1, for symmetric states, and −1, for antisymmetric ones. Use the supplied topic model Y_out = (1000*sind(2*th))/1000, where Y_out is EXCHANGE MEAN ⟨P⟩ (before display rounding) (dimensionless). 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 c (Particle detected at c) = 0.25 L; n (Upper level n) = 2 fixed. For this exercise use only 141.975 ° ≤ th ≤ 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 = -0.79855147, with a bounded reading uncertainty ±0.025793. 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

EXCHANGE MEAN ⟨P⟩ (before display rounding) against th

Predictions from the supplied The exchange operator: why [P, H] = 0 forces no symmetry 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): EXCHANGE MEAN ⟨P⟩ (before display rounding)

EXCHANGE MEAN ⟨P⟩ (before display rounding) against thPredictions from the supplied The exchange operator: why [P, H] = 0 forces no symmetry 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: EXCHANGE MEAN ⟨P⟩ (before display rounding). Exact values are available in the data table below.-0.9846-0.8897-0.7948-0.6999-0.605141.2146.3151.4156.5161.6xy

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 / °: 142EXCHANGE MEAN ⟨P⟩ (before display rounding): -0.9705
Read the exact data as a table
EXCHANGE MEAN ⟨P⟩ (before display rounding) against th — supplied values
th: Mixing angle θ / °EXCHANGE MEAN ⟨P⟩ (before display rounding)
141.975-0.9705064734685425
146.7-0.9177546256839815
151.425-0.8400935538989416
156.15-0.7396310949786097
160.875-0.6190939493098343

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(141.975) = -0.970506 and Y_out(160.875) = -0.619094. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives th_line = 141.975 + (-0.798551 − -0.970506)(160.875 − 141.975)/(-0.619094 − -0.970506) = 151.223 °. 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 = 153.504 °; substituting back gives Y_out = -0.79855147.
  4. The interpolation error in the input is |th_line − th| = 2.28075 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = -0.82434447 and -0.77275847. Sort the roots even if the response decreases: 152.239 ° ≤ th ≤ 154.699 °.
  6. The lower and upper input deviations are 1.26504 ° and 1.19488 °. 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 ≈ 153.504 °; compatible interval [152.239, 154.699] °. Endpoint interpolation gives 151.223 °.

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

Extra challenging · 02

The symmetrisation postulate and spin-statistics: infer an input with uncertainty

The symmetrisation postulate requires a state of identical particles to be unchanged by every pairwise exchange (bosons) or to change sign under every one (fermions); the spin-statistics connection, quoted from field theory, makes integer-spin particles bosons and half-integer-spin ones fermions. Use the supplied topic model Y_out = s*(1-m) + m*cosd(2*th-120), where Y_out is ⟨P₂₃⟩ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. th is Direction θ in the mixed plane, measured in °. Keep m (Weight m in the mixed plane) = 1; s (Allowed sector s: +1 sym, −1 antisym) = 1 fixed. For this exercise use only 19.125 ° ≤ th ≤ 56.925 ° 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 = 0.81275518, with a bounded reading uncertainty ±0.027223. 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 0.00378 °, 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

⟨P₂₃⟩ against th

Predictions from the supplied The symmetrisation postulate and spin-statistics 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): ⟨P₂₃⟩

⟨P₂₃⟩ against thPredictions from the supplied The symmetrisation postulate and spin-statistics 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: Direction θ in the mixed plane / °. Vertical axis: ⟨P₂₃⟩. Exact values are available in the data table below.0.10950.33920.56890.79861.02817.6127.8238.0248.2358.44xy

Horizontal axis (x): th: Direction θ in the mixed plane / °

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

Reading 1th: Direction θ in the mixed plane / °: 19.13⟨P₂₃⟩: 0.1435
Read the exact data as a table
⟨P₂₃⟩ against th — supplied values
th: Direction θ in the mixed plane / °⟨P₂₃⟩
19.1250.1434926219911793
28.5750.4563215909004759
38.0250.7199457301444867
47.4749999999999940.9059386379797358
56.9250.9942448326331054

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(19.125) = 0.143493 and Y_out(56.925) = 0.994245. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives th_line = 19.125 + (0.812755 − 0.143493)(56.925 − 19.125)/(0.994245 − 0.143493) = 48.8612 °. 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 = 42.183 °; substituting back gives Y_out = 0.81275518.
  4. The interpolation error in the input is |th_line − th| = 6.67818 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.78553218 and 0.83997818. Sort the roots even if the response decreases: 40.885 ° ≤ th ≤ 43.5689 °.
  6. The lower and upper input deviations are 1.29804 ° and 1.38591 °. 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 ≈ 42.183 °; compatible interval [40.885, 43.5689] °. Endpoint interpolation gives 48.8612 °.

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

Extra challenging · 03

Pauli exclusion and the Slater determinant: infer an input with uncertainty

A Slater determinant is the normalised antisymmetric state of N fermions in N orthonormal spin-orbitals, (1/√(N!)) det[ψ_aⱼ(xᵢ)]; it vanishes whenever two of the spin-orbitals coincide, and that vanishing is the Pauli exclusion principle. Use the supplied topic model Y_out = (1000*0.5642*exp(-xi^2)*(cosd(theta) + sqrt(2)*sind(theta)*xi))/1000, where Y_out is BARE PRODUCT AT ξ₁ = ξ₂ (before display rounding) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; exp(x) means eˣ; sind/cosd/tand use degrees. theta is Angle θ between orbitals a and b, measured in °. Keep xi (Particle 2 at ξ₂ = x₂/ℓ) = 0.5; eta (Exchange sign η (−1 same-spin fermions, +1 bosons)) = -1 fixed. For this exercise use only 37.9125 ° ≤ 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 = 0.47254076, with a bounded reading uncertainty ±0.0097555. 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 0.0042525 °, 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

BARE PRODUCT AT ξ₁ = ξ₂ (before display rounding) against theta

Predictions from the supplied Pauli exclusion and the Slater determinant 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): BARE PRODUCT AT ξ₁ = ξ₂ (before display rounding)

BARE PRODUCT AT ξ₁ = ξ₂ (before display rounding) against thetaPredictions from the supplied Pauli exclusion and the Slater determinant 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: Angle θ between orbitals a and b / °. Vertical axis: BARE PRODUCT AT ξ₁ = ξ₂ (before display rounding). Exact values are available in the data table below.0.37310.41580.45850.50120.543936.2147.6959.1770.6682.14xy

Horizontal axis (x): theta: Angle θ between orbitals a and b / °

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

Reading 1theta: Angle θ between orbitals a and b / °: 37.91BARE PRODUCT AT ξ₁ = ξ₂ (before display rounding): 0.5376
Read the exact data as a table
BARE PRODUCT AT ξ₁ = ξ₂ (before display rounding) against theta — supplied values
theta: Angle θ between orbitals a and b / °BARE PRODUCT AT ξ₁ = ξ₂ (before display rounding)
37.91250.5375774857321434
48.543750.5237628793859499
59.1750.49196736041319206
69.806250.4432824770527133
80.43750.3793795937107995

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(37.9125) = 0.537577 and Y_out(80.4375) = 0.37938. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives theta_line = 37.9125 + (0.472541 − 0.537577)(80.4375 − 37.9125)/(0.37938 − 0.537577) = 55.395 °. 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 = 63.8528 °; substituting back gives Y_out = 0.47254076.
  4. The interpolation error in the input is |theta_line − theta| = 8.4578 °. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.46278526 and 0.48229626. Sort the roots even if the response decreases: 61.6004 ° ≤ theta ≤ 65.9531 °.
  6. The lower and upper input deviations are 2.25237 ° and 2.1004 °. 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 ≈ 63.8528 °; compatible interval [61.6004, 65.9531] °. Endpoint interpolation gives 55.395 °.

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

Extra challenging · 04

Exchange effects and mean separation in a well: infer an input with uncertainty

An exchange effect is a difference between identical and distinguishable particles that comes from the symmetry of the state alone, with no interaction in the Hamiltonian; for positions it appears as a correlation between where the two particles are found. Use the supplied topic model Y_out = sqrt((1/6-(1+1/m^2)/(2*pi^2))-64*m^2*(1+cos(pi*m))/(pi^4*(m^2-1)^4)), where Y_out is RMS x₁ − x₂, BOSONS (a). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. m is Second level m (first is n = 1), dimensionless. Keep p (Particle 1 found at x₁) = 0.3 a fixed. For this exercise use only 2 ≤ m ≤ 5, 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 = 0.33529761 a, with a bounded reading uncertainty ±0.011322 a. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of m 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

RMS x₁ − x₂, BOSONS against m

Predictions from the supplied Exchange effects and mean separation in a well model, with all other inputs fixed as stated in the question. 4 calibration points support a trend comparison; they are not measured data or a straight-line law.

Vertical axis (y): RMS x₁ − x₂, BOSONS / a

RMS x₁ − x₂, BOSONS against mPredictions from the supplied Exchange effects and mean separation in a well model, with all other inputs fixed as stated in the question. 4 calibration points support a trend comparison; they are not measured data or a straight-line law. Horizontal axis: m: Second level m (first is n = 1). Vertical axis: RMS x₁ − x₂, BOSONS / a. Exact values are available in the data table below.0.19040.22860.26680.30510.34331.882.693.54.315.12xy

Horizontal axis (x): m: Second level m (first is n = 1)

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

Reading 1m: Second level m (first is n = 1): 2RMS x₁ − x₂, BOSONS / a: 0.1961
Read the exact data as a table
RMS x₁ − x₂, BOSONS against m — supplied values
m: Second level m (first is n = 1)RMS x₁ − x₂, BOSONS / a
20.19608615402130644
30.33223052267770525
40.335297605889589
50.33760872496523403

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) = 0.196086 a and Y_out(5) = 0.337609 a. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives m_line = 2 + (0.335298 − 0.196086)(5 − 2)/(0.337609 − 0.196086) = 4.95101. 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 m = 4; only the allowed settings 3, 4, 5 lie inside the stated reading bound.
  4. The interpolation error in the input is |m_line − m| = 0.951009.
  5. The compatible allowed values span 3 to 5. 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 m: 3, 4, 5. Endpoint interpolation predicts 4.95101; check admissibility against the discrete settings.

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

Choose a topic