University Advanced Quantum Mechanics · topic UAQM-03.05
A spinless particle in a uniform magnetic field: relativistic Landau levels
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.
Relativistic Landau levels of a spinless particle: predict a finite change
Relativistic Landau levels are the exact energies E = ±√(m²c⁴ + p_z²c² + (2n + 1)ħ|q|Bc²) of a spinless charge in a uniform magnetic field, with |q|B/h states per unit area in every level. Use the supplied topic model Y_out = 1+(n+0.5)*b+pz^2/2, where Y_out is NON-RELATIVISTIC mc² + ε (mc²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. pz is Momentum along B, p_z, measured in mc. Keep b (Field strength B/B_c) = 0.4; n (Highlighted level n) = 1 fixed. For this exercise use only -0.945 mc ≤ pz ≤ -0.063 mc 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 pz = -0.75096 mc and -0.62748 mc. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δpz across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
NON-RELATIVISTIC mc² + ε against pz
Predictions from the supplied Relativistic Landau levels of a spinless particle 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): NON-RELATIVISTIC mc² + ε / mc²
Horizontal axis (x): pz: Momentum along B, p_z / mc
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| pz: Momentum along B, p_z / mc | NON-RELATIVISTIC mc² + ε / mc² |
|---|---|
| -0.945 | 2.0465125 |
| -0.7244999999999999 | 1.862450125 |
| -0.504 | 1.727008 |
| -0.2835 | 1.640186125 |
| -0.06299999999999994 | 1.6019845000000001 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of pz separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At pz = -0.75096 mc, substitute into Y_out = 1+(n+0.5)*b+pz^2/2: Y₀ = 1.88197 mc².
- At pz = -0.62748 mc, the same fixed-parameter model gives Y₁ = 1.79687 mc².
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0851049 mc²; the input change is Δpz = 0.12348 mc.
- Average sensitivity is ΔY_out/Δpz = -0.68922 mc² per mc. Its negative sign means that this response decreases 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 = 1.88197 mc², Y₁ = 1.79687 mc²; ΔY_out = -0.0851049 mc²; average sensitivity = -0.68922 mc² per mc.
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.
Relativistic Landau levels of a spinless particle: infer an input with uncertainty
Relativistic Landau levels are the exact energies E = ±√(m²c⁴ + p_z²c² + (2n + 1)ħ|q|Bc²) of a spinless charge in a uniform magnetic field, with |q|B/h states per unit area in every level. Use the supplied topic model Y_out = 1+(n+0.5)*b+pz^2/2, where Y_out is NON-RELATIVISTIC mc² + ε (mc²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. pz is Momentum along B, p_z, measured in mc. Keep b (Field strength B/B_c) = 0.4; n (Highlighted level n) = 1 fixed. For this exercise use only -0.945 mc ≤ pz ≤ -0.063 mc 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 = 1.6828164 mc², with a bounded reading uncertainty ±0.012125 mc². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown pz by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000882 mc, 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.
NON-RELATIVISTIC mc² + ε against pz
Predictions from the supplied Relativistic Landau levels of a spinless particle 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): NON-RELATIVISTIC mc² + ε / mc²
Horizontal axis (x): pz: Momentum along B, p_z / mc
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| pz: Momentum along B, p_z / mc | NON-RELATIVISTIC mc² + ε / mc² |
|---|---|
| -0.945 | 2.0465125 |
| -0.7244999999999999 | 1.862450125 |
| -0.504 | 1.727008 |
| -0.2835 | 1.640186125 |
| -0.06299999999999994 | 1.6019845000000001 |
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(pz) = 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(-0.945) = 2.04651 mc² and Y_out(-0.063) = 1.60198 mc². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives pz_line = -0.945 + (1.68282 − 2.04651)(-0.063 − -0.945)/(1.60198 − 2.04651) = -0.223381 mc. 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 pz = -0.40698 mc; substituting back gives Y_out = 1.6828164 mc².
- The interpolation error in the input is |pz_line − pz| = 0.183599 mc. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.6706914 mc² and 1.6949414 mc². Sort the roots even if the response decreases: -0.435755 mc ≤ pz ≤ -0.376009 mc.
- The lower and upper input deviations are 0.0287753 mc and 0.0309711 mc. 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 pz ≈ -0.40698 mc; compatible interval [-0.435755, -0.376009] mc. Endpoint interpolation gives -0.223381 mc.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.