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University Advanced Quantum Mechanics · unit UAQM-06

The Non-Relativistic Limit: g = 2, Spin-Orbit Coupling and the Darwin Term

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

Large and small components: eliminating χ to order v/c: infer an input with uncertainty

Eliminating the small component means solving the lower two rows of the Dirac equation for χ in terms of the upper pair φ, χ ≈ (σ⋅π/2mc)φ for a slow electron, so that φ obeys a closed two-component equation of its own. Use the supplied topic model Y_out = sqrt(t*(t+1021.998))/(t+1021.998), where Y_out is FREE EXACT |χ|/|φ| (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. t is Free electron's kinetic energy T, measured in keV. Keep z (Nuclear charge Z of the 1s ion) = 92 fixed. For this exercise use only 25.7188 keV ≤ t ≤ 179.281 keV 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 = 0.32342261, with a bounded reading uncertainty ±0.0094344. 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 0.0153563 keV, 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

FREE EXACT |χ|/|φ| against t

Predictions from the supplied Large and small components: eliminating χ to order v/c 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): FREE EXACT |χ|/|φ|

FREE EXACT |χ|/|φ| against tPredictions from the supplied Large and small components: eliminating χ to order v/c 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: Free electron's kinetic energy T / keV. Vertical axis: FREE EXACT |χ|/|φ|. Exact values are available in the data table below.0.14750.20950.27150.33350.395519.5861.04102.5144185.4xy

Horizontal axis (x): t: Free electron's kinetic energy T / keV

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

Reading 1t: Free electron's kinetic energy T / keV: 25.72FREE EXACT |χ|/|φ|: 0.1567
Read the exact data as a table
FREE EXACT |χ|/|φ| against t — supplied values
t: Free electron's kinetic energy T / keVFREE EXACT |χ|/|φ|
25.718750.1566761838398359
64.1093750.2429541882182584
102.50.3019135390524935
140.8906250.3480743226194173
179.281250.38631844835115303

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(25.7188) = 0.156676 and Y_out(179.281) = 0.386318. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives t_line = 25.7188 + (0.323423 − 0.156676)(179.281 − 25.7188)/(0.386318 − 0.156676) = 137.223 keV. 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 = 119.392 keV; substituting back gives Y_out = 0.32342261.
  4. The interpolation error in the input is |t_line − t| = 17.8307 keV. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 0.31398821 and 0.33285701. Sort the roots even if the response decreases: 111.777 keV ≤ t ≤ 127.339 keV.
  6. The lower and upper input deviations are 7.61456 keV and 7.94758 keV. 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 ≈ 119.392 keV; compatible interval [111.777, 127.339] keV. Endpoint interpolation gives 137.223 keV.

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

Extra challenging · 02

The Pauli equation and g = 2 from minimal coupling: infer an input with uncertainty

The Pauli equation, iħ∂φ/∂t = [π²/2m − (qħ/2m)σ⋅B + V]φ, is the two-component equation the upper Dirac components obey at leading order in v/c; its σ⋅B term is the energy of a spin moment (q/m)S, which is g = 2. Use the supplied topic model Y_out = (1/3+2*gs/3)*13.996*b, where Y_out is D2 ½ → −½ SHIFT (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. b is Magnetic field B, measured in T. Keep gs (Spin g-factor gₛ) = 2 fixed. For this exercise use only 0.3275 T ≤ b ≤ 1.2725 T 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 = 21.08614 GHz, with a bounded reading uncertainty ±1.2896 GHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000945 T, 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

D2 ½ → −½ SHIFT against b

Predictions from the supplied The Pauli equation and g = 2 from minimal coupling 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): D2 ½ → −½ SHIFT / GHz

D2 ½ → −½ SHIFT against bPredictions from the supplied The Pauli equation and g = 2 from minimal coupling 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: b: Magnetic field B / T. Vertical axis: D2 ½ → −½ SHIFT / GHz. Exact values are available in the data table below.6.75812.7118.6624.6130.560.28970.54480.81.0551.31xy

Horizontal axis (x): b: Magnetic field B / T

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

Reading 1b: Magnetic field B / T: 0.3275D2 ½ → −½ SHIFT / GHz: 7.639
Read the exact data as a table
D2 ½ → −½ SHIFT against b — supplied values
b: Magnetic field B / TD2 ½ → −½ SHIFT / GHz
0.32757.639483333333333
0.5637513.150408333333331
0.799999999999999818.661333333333328
1.0362524.17225833333333
1.272499999999999729.683183333333325

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(b) = 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(0.3275) = 7.63948 GHz and Y_out(1.2725) = 29.6832 GHz. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives b_line = 0.3275 + (21.0861 − 7.63948)(1.2725 − 0.3275)/(29.6832 − 7.63948) = 0.90395 T. 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 b = 0.90395 T; substituting back gives Y_out = 21.08614 GHz.
  4. The interpolation error in the input is |b_line − b| = 0 T. This is within the requested numerical tolerance.
  5. Invert both limits Y_out = 19.79654 GHz and 22.37574 GHz. Sort the roots even if the response decreases: 0.848666 T ≤ b ≤ 0.959234 T.
  6. The lower and upper input deviations are 0.0552844 T and 0.0552844 T. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.

Answer b ≈ 0.90395 T; compatible interval [0.848666, 0.959234] T. Endpoint interpolation gives 0.90395 T.

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

Extra challenging · 03

Dirac Landau levels: g = 2 as an exact degeneracy: infer an input with uncertainty

The Dirac Landau levels are the exact energies E = √(m²c⁴ + p_z²c² + 2nħ|q|Bc²) of a spin-½ Dirac particle in a uniform magnetic field, where each level n ≥ 1 holds orbit n with its spin moment along B and orbit n − 1 with its spin moment against B. Use the supplied topic model Y_out = 510.999*(sqrt(1+2*b)-1), where Y_out is n = 1 PAIR AT g = 2 (keV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. b is Field B, measured in B_c. Keep gs (Spin g-factor g (range exaggerated)) = 2 fixed. For this exercise use only 0.257188 B_c ≤ b ≤ 1.79281 B_c 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 = 429.5498 keV, with a bounded reading uncertainty ±23.056 keV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown b by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000153562 B_c, 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

n = 1 PAIR AT g = 2 against b

Predictions from the supplied Dirac Landau levels: g = 2 as an exact degeneracy 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): n = 1 PAIR AT g = 2 / keV

n = 1 PAIR AT g = 2 against bPredictions from the supplied Dirac Landau levels: g = 2 as an exact degeneracy 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: b: Field B / B_c. Vertical axis: n = 1 PAIR AT g = 2 / keV. Exact values are available in the data table below.99.22224.9350.5476.2601.90.19580.61041.0251.441.854xy

Horizontal axis (x): b: Field B / B_c

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

Reading 1b: Field B / B_c: 0.2572n = 1 PAIR AT g = 2 / keV: 117.8
Read the exact data as a table
n = 1 PAIR AT g = 2 against b — supplied values
b: Field B / B_cn = 1 PAIR AT g = 2 / keV
0.2571875117.83608703416385
0.64109375260.96262456054586
1.025381.42238751995967
1.4089062500000002487.45336041182406
1.7928125000000001583.2579621075463

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(b) = 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(0.257188) = 117.836 keV and Y_out(1.79281) = 583.258 keV. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives b_line = 0.257188 + (429.55 − 117.836)(1.79281 − 0.257188)/(583.258 − 117.836) = 1.28566 B_c. 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 b = 1.19392 B_c; substituting back gives Y_out = 429.5498 keV.
  4. The interpolation error in the input is |b_line − b| = 0.0917451 B_c. This is larger than the requested numerical tolerance.
  5. Invert both limits Y_out = 406.4938 keV and 452.6058 keV. Sort the roots even if the response decreases: 1.11189 B_c ≤ b ≤ 1.27798 B_c.
  6. The lower and upper input deviations are 0.0820294 B_c and 0.0840651 B_c. 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 b ≈ 1.19392 B_c; compatible interval [1.11189, 1.27798] B_c. Endpoint interpolation gives 1.28566 B_c.

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

Extra challenging · 04

Anomalous moments: Pauli's term and g = 2(1 + a): infer an input with uncertainty

An anomalous magnetic moment is the part of a spin-½ particle's moment beyond the g = 2 that minimal coupling gives; writing g = 2(1 + a), it enters the Dirac equation through Pauli's term, proportional to σ^(μν)F_(μν), with coefficient μ′ = aqħ/2m. Use the supplied topic model Y_out = 2*(1+a), where Y_out is g = 2(1 + a) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. a is Anomaly a = (g − 2)/2, dimensionless. Keep n (Principal quantum number n) = 2 fixed. For this exercise use only 0.0002125 ≤ a ≤ 0.0017875 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 = 2.0023465, with a bounded reading uncertainty ±0.00018428. 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 1.575e-7, 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

g = 2(1 + a) against a

Predictions from the supplied Anomalous moments: Pauli's term and g = 2(1 + a) 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): g = 2(1 + a)

g = 2(1 + a) against aPredictions from the supplied Anomalous moments: Pauli's term and g = 2(1 + a) 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: Anomaly a = (g − 2)/2. Vertical axis: g = 2(1 + a). Exact values are available in the data table below.2.00032.001152.0022.002852.00371.50e-45.75e-40.0010.0014250.00185xy

Horizontal axis (x): a: Anomaly a = (g − 2)/2

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

Reading 1a: Anomaly a = (g − 2)/2: 2.12e-4g = 2(1 + a): 2
Read the exact data as a table
g = 2(1 + a) against a — supplied values
a: Anomaly a = (g − 2)/2g = 2(1 + a)
0.00021252.000425
0.00060624999999999992.0012125
0.00099999999999999982.002
0.00139374999999999982.0027875
0.00178749999999999982.003575

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(0.0002125) = 2.00042 and Y_out(0.0017875) = 2.00358. The target lies between them. On the selected branch the response is strictly increasing.
  2. Endpoint interpolation gives a_line = 0.0002125 + (2.00235 − 2.00042)(0.0017875 − 0.0002125)/(2.00358 − 2.00042) = 0.00117325. 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 = 0.00117325; substituting back gives Y_out = 2.0023465.
  4. The interpolation error in the input is |a_line − a| = 9.84456e-17. This is within the requested numerical tolerance.
  5. Invert both limits Y_out = 2.0021622 and 2.0025308. Sort the roots even if the response decreases: 0.00108111 ≤ a ≤ 0.00126539.
  6. The lower and upper input deviations are 0.00009214 and 0.00009214. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.

Answer a ≈ 0.00117325; compatible interval [0.00108111, 0.00126539]. Endpoint interpolation gives 0.00117325.

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

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