University Advanced Quantum Mechanics · topic UAQM-06.03
Exact Landau levels of a Dirac electron: g = 2 as an exact degeneracy
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.
Further practice
Work through a fresh case before tackling the extension.
Dirac Landau levels: g = 2 as an exact degeneracy: predict a finite change
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. Calculate Y_out at b = 0.595025 B_c and 0.810013 B_c. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δb across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of b separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At b = 0.595025 B_c, substitute into Y_out = 510.999*(sqrt(1+2*b)-1): Y₀ = 245.219 keV.
- At b = 0.810013 B_c, the same fixed-parameter model gives Y₁ = 316.129 keV.
- Signed change: ΔY_out = Y₁ − Y₀ = 70.91 keV; the input change is Δb = 0.214987 B_c.
- Average sensitivity is ΔY_out/Δb = 329.833 keV per B_c. Its positive sign means that this response increases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 245.219 keV, Y₁ = 316.129 keV; ΔY_out = 70.91 keV; average sensitivity = 329.833 keV per B_c.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
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.