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University Quantum Mechanics II · unit UQMII-01

Deriving the Spherical Harmonics

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 associated Legendre equation and its poles: infer an input with uncertainty

The associated Legendre equation is the polar part of the angular equation written in x = cos θ; its only singular points on the sphere are the poles x = ±1, and at each pole exactly one of its two solutions is regular. Use the supplied topic model Y_out = 57.29578*10^(-n), where Y_out is CAP HALF-ANGLE θ₀ (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. n is Cap size n, with θ₀ = 10^(−n) rad, dimensionless. Keep m (|m|, azimuthal quantum number) = 0 fixed. For this exercise use only 0.978125 ≤ n ≤ 4.52187 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 = 0.041525047 °, with a bounded reading uncertainty ±0.0059703 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown n 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.

Explore the question data

CAP HALF-ANGLE θ₀ against n

Predictions from the supplied The associated Legendre equation and its poles 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): CAP HALF-ANGLE θ₀ / °

CAP HALF-ANGLE θ₀ against nPredictions from the supplied The associated Legendre equation and its poles 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: n: Cap size n, with θ₀ = 10^(−n) rad. Vertical axis: CAP HALF-ANGLE θ₀ / °. Exact values are available in the data table below.-0.23921.3873.0144.646.2670.83641.7932.753.7074.664xy

Horizontal axis (x): n: Cap size n, with θ₀ = 10^(−n) rad

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

Reading 1n: Cap size n, with θ₀ = 10^(−n) rad: 0.9781CAP HALF-ANGLE θ₀ / °: 6.026
Read the exact data as a table
CAP HALF-ANGLE θ₀ against n — supplied values
n: Cap size n, with θ₀ = 10^(−n) radCAP HALF-ANGLE θ₀ / °
0.97812499999999996.025563063026317
1.86406249999999970.7835381305946377
2.750.10188790585611991
3.63593750.013249062112480949
4.5218750.0017228506699142692

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(n) = 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(0.978125) = 6.02556 ° and Y_out(4.52187) = 0.00172285 °. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives n_line = 0.978125 + (0.041525 − 6.02556)(4.52187 − 0.978125)/(0.00172285 − 6.02556) = 4.49846. 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 n = 3.13981; substituting back gives Y_out = 0.041525047 °.
  4. The interpolation error in the input is |n_line − n| = 1.35865. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.035554747 ° and 0.047495347 °. Sort the roots even if the response decreases: 3.08147 ≤ n ≤ 3.20723.
  6. The lower and upper input deviations are 0.0583409 and 0.0674125. 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 n ≈ 3.13981; compatible interval [3.08147, 3.20723]. Endpoint interpolation gives 4.49846.

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

Extra challenging · 02

Legendre's equation by series, and why it must terminate: infer an input with uncertainty

Legendre's equation has a convergent series solution for every separation constant λ, but a solution finite at both poles exists only when that series terminates, which happens exactly when λ = l(l + 1) with l a non-negative integer. Use the supplied topic model Y_out = (sqrt(0.25+lam)-0.5), where Y_out is ν = √(λ + ¼) − ½ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. lam is Separation constant λ, dimensionless. Keep no other adjustable input fixed. For this exercise use only 1.4875 ≤ lam ≤ 12.5125 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 = 2.4090806, with a bounded reading uncertainty ±0.099508. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lam by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0011025, 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

ν = √(λ + ¼) − ½ against lam

Predictions from the supplied Legendre's equation by series, and why it must terminate 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): ν = √(λ + ¼) − ½

ν = √(λ + ¼) − ½ against lamPredictions from the supplied Legendre's equation by series, and why it must terminate 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: lam: Separation constant λ. Vertical axis: ν = √(λ + ¼) − ½. Exact values are available in the data table below.0.7281.3371.9452.5543.1631.0464.02379.97712.95xy

Horizontal axis (x): lam: Separation constant λ

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

Reading 1lam: Separation constant λ: 1.488ν = √(λ + ¼) − ½: 0.8181
Read the exact data as a table
ν = √(λ + ¼) − ½ against lam — supplied values
lam: Separation constant λν = √(λ + ¼) − ½
1.48750.8181426326464067
4.24374999999999951.6198466925700075
6.9999999999999992.192582403567252
9.756252.663265717577327
12.51253.07246413557925

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(lam) = 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(1.4875) = 0.818143 and Y_out(12.5125) = 3.07246. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives lam_line = 1.4875 + (2.40908 − 0.818143)(12.5125 − 1.4875)/(3.07246 − 0.818143) = 9.26815. 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 lam = 8.21275; substituting back gives Y_out = 2.4090806.
  4. The interpolation error in the input is |lam_line − lam| = 1.0554. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 2.3095726 and 2.5085886. Sort the roots even if the response decreases: 7.6437 ≤ lam ≤ 8.80161.
  6. The lower and upper input deviations are 0.569052 and 0.588855. 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 lam ≈ 8.21275; compatible interval [7.6437, 8.80161]. Endpoint interpolation gives 9.26815.

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

Extra challenging · 03

Legendre polynomials: Rodrigues, orthogonality and norm: infer an input with uncertainty

The Legendre polynomial Pₗ(x) is the degree-l solution of Legendre's equation that stays finite at x = ±1, scaled so that Pₗ(1) = 1; the set is orthogonal on −1 ≤ x ≤ 1, with ∫Pₗ² dx = 2/(2l + 1). Use the supplied topic model Y_out = 2/(2*l+1), where Y_out is NORM OF Pₗ 2/(2l + 1) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. l is Degree l of the solid curve, dimensionless. Keep k (Degree k of the dashed curve) = 0 fixed. For this exercise use only 0 ≤ l ≤ 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 = 0.28571429, with a bounded reading uncertainty ±0.14545. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. Determine every allowed value of l 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

NORM OF Pₗ 2/(2l + 1) against l

Predictions from the supplied Legendre polynomials: Rodrigues, orthogonality and norm 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): NORM OF Pₗ 2/(2l + 1)

NORM OF Pₗ 2/(2l + 1) against lPredictions from the supplied Legendre polynomials: Rodrigues, orthogonality and norm 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: l: Degree l of the solid curve. Vertical axis: NORM OF Pₗ 2/(2l + 1). Exact values are available in the data table below.0.10910.61.0911.5822.073-0.21.152.53.855.2xy

Horizontal axis (x): l: Degree l of the solid curve

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

Reading 1l: Degree l of the solid curve: 0NORM OF Pₗ 2/(2l + 1): 2
Read the exact data as a table
NORM OF Pₗ 2/(2l + 1) against l — supplied values
l: Degree l of the solid curveNORM OF Pₗ 2/(2l + 1)
02
10.6666666666666666
30.2857142857142857
40.2222222222222222
50.18181818181818182

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(0) = 2 and Y_out(5) = 0.181818. The target lies between them. On the selected branch the response is strictly decreasing.
  2. Endpoint interpolation gives l_line = 0 + (0.285714 − 2)(5 − 0)/(0.181818 − 2) = 4.71429. 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 l = 3; only the allowed settings 2, 3, 4, 5 lie inside the stated reading bound.
  4. The interpolation error in the input is |l_line − l| = 1.71429.
  5. The compatible allowed values span 2 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 l: 2, 3, 4, 5. Endpoint interpolation predicts 4.71429; check admissibility against the discrete settings.

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

Extra challenging · 04

Generating function and multipole expansion of 1/|r − r′|: infer an input with uncertainty

The Legendre generating function G(x, t) = (1 − 2xt + t²)^(−1/2) is the function whose coefficient of tˡ is Pₗ(x); with x = cos γ and t = r_</r_> it equals r_> times 1/|r − r′|, which makes its series the multipole expansion of the Coulomb interaction. Use the supplied topic model Y_out = t^(lmax+1)/(1-t), where Y_out is WORST ERROR, AT γ = 0 (× 1/r_>). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. t is Radius ratio t = smaller/larger, dimensionless. Keep lmax (Highest multipole kept, L) = 2; gam (Angle γ between r and r′) = 60 ° fixed. For this exercise use only 0.174375 ≤ t ≤ 0.725625 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 = 0.27208677 × 1/r_>, with a bounded reading uncertainty ±0.03985 × 1/r_>. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000055125, 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

WORST ERROR, AT γ = 0 against t

Predictions from the supplied Generating function and multipole expansion of 1/|r − r′| 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): WORST ERROR, AT γ = 0 / × 1/r_>

WORST ERROR, AT γ = 0 against tPredictions from the supplied Generating function and multipole expansion of 1/|r − r′| 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: t: Radius ratio t = smaller/larger. Vertical axis: WORST ERROR, AT γ = 0 / × 1/r_>. Exact values are available in the data table below.-0.049020.32520.69951.0741.4480.15230.30120.450.59880.7477xy

Horizontal axis (x): t: Radius ratio t = smaller/larger

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

Reading 1t: Radius ratio t = smaller/larger: 0.1744WORST ERROR, AT γ = 0 / × 1/r_>: 0.006422
Read the exact data as a table
WORST ERROR, AT γ = 0 against t — supplied values
t: Radius ratio t = smaller/largerWORST ERROR, AT γ = 0 / × 1/r_>
0.1743750.006421992985900833
0.31218750.0442360627537909
0.450.16568181818181818
0.58781250000000010.4927443287558049
0.7256251.3924902842041569

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(t) = 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(0.174375) = 0.00642199 × 1/r_> and Y_out(0.725625) = 1.39249 × 1/r_>. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives t_line = 0.174375 + (0.272087 − 0.00642199)(0.725625 − 0.174375)/(1.39249 − 0.00642199) = 0.280032. 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 t = 0.510637; substituting back gives Y_out = 0.27208677 × 1/r_>.
  4. The interpolation error in the input is |t_line − t| = 0.230606. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.23223677 × 1/r_> and 0.31193677 × 1/r_>. Sort the roots even if the response decreases: 0.49083 ≤ t ≤ 0.528032.
  6. The lower and upper input deviations are 0.0198074 and 0.017394. 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 t ≈ 0.510637; compatible interval [0.49083, 0.528032]. Endpoint interpolation gives 0.280032.

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

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