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.
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.
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|
Horizontal axis (x): t: Free electron's kinetic energy T / keV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Free electron's kinetic energy T / keV | FREE EXACT |χ|/|φ| |
|---|---|
| 25.71875 | 0.1566761838398359 |
| 64.109375 | 0.2429541882182584 |
| 102.5 | 0.3019135390524935 |
| 140.890625 | 0.3480743226194173 |
| 179.28125 | 0.38631844835115303 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |t_line − t| = 17.8307 keV. This is larger than the requested numerical tolerance.
- 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.
- 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.
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.
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
Horizontal axis (x): b: Magnetic field B / T
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Magnetic field B / T | D2 ½ → −½ SHIFT / GHz |
|---|---|
| 0.3275 | 7.639483333333333 |
| 0.56375 | 13.150408333333331 |
| 0.7999999999999998 | 18.661333333333328 |
| 1.03625 | 24.17225833333333 |
| 1.2724999999999997 | 29.683183333333325 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |b_line − b| = 0 T. This is within the requested numerical tolerance.
- 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.
- 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.
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.
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
Horizontal axis (x): b: Field B / B_c
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Field B / B_c | n = 1 PAIR AT g = 2 / keV |
|---|---|
| 0.2571875 | 117.83608703416385 |
| 0.64109375 | 260.96262456054586 |
| 1.025 | 381.42238751995967 |
| 1.4089062500000002 | 487.45336041182406 |
| 1.7928125000000001 | 583.2579621075463 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |b_line − b| = 0.0917451 B_c. This is larger than the requested numerical tolerance.
- 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.
- 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.
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.
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)
Horizontal axis (x): a: Anomaly a = (g − 2)/2
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Anomaly a = (g − 2)/2 | g = 2(1 + a) |
|---|---|
| 0.0002125 | 2.000425 |
| 0.0006062499999999999 | 2.0012125 |
| 0.0009999999999999998 | 2.002 |
| 0.0013937499999999998 | 2.0027875 |
| 0.0017874999999999998 | 2.003575 |
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
- 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.
- 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.
- 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.
- The interpolation error in the input is |a_line − a| = 9.84456e-17. This is within the requested numerical tolerance.
- Invert both limits Y_out = 2.0021622 and 2.0025308. Sort the roots even if the response decreases: 0.00108111 ≤ a ≤ 0.00126539.
- 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.