University Quantum Mechanics II · topic UQMII-13.08
Beyond the dipole approximation: magnetic-dipole and quadrupole lines
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.
Beyond the dipole: magnetic-dipole and quadrupole lines: predict a finite change
A magnetic-dipole (M1) or electric-quadrupole (E2) transition is driven by the first correction to the dipole approximation: the wave's magnetic field acting on (μ_B/ħ)(L + 2S), or its field gradient acting on products of the electron's coordinates. Both connect states of equal parity. Use the supplied topic model Y_out = 9.30667-3*(lw+7)+2*log(a), where Y_out is E1 RATE, log₁₀(A / s⁻¹) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. a is Orbital size a, measured in a₀. Keep lw (Cursor wavelength, log₁₀(λ / 1 m)) = -6.3 fixed. For this exercise use only 1.31875 a₀ ≤ a ≤ 3.68125 a₀ 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 a = 1.8385 a₀ and 2.16925 a₀. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
E1 RATE, log₁₀(A / s⁻¹) against a
Predictions from the supplied Beyond the dipole: magnetic-dipole and quadrupole lines 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): E1 RATE, log₁₀(A / s⁻¹)
Horizontal axis (x): a: Orbital size a / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Orbital size a / a₀ | E1 RATE, log₁₀(A / s⁻¹) |
|---|---|
| 1.31875 | 7.446994945283536 |
| 1.909375 | 7.768452463845296 |
| 2.5 | 8.002550017344076 |
| 3.090625 | 8.186762626554547 |
| 3.68125 | 8.338660624262353 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of a separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a = 1.8385 a₀, substitute into Y_out = 9.30667-3*(lw+7)+2*log(a): Y₀ = 7.7356.
- At a = 2.16925 a₀, the same fixed-parameter model gives Y₁ = 7.87929.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.143692; the input change is Δa = 0.33075 a₀.
- Average sensitivity is ΔY_out/Δa = 0.434443 1 per a₀. 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 = 7.7356, Y₁ = 7.87929; ΔY_out = 0.143692; average sensitivity = 0.434443 1 per a₀.
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.
Beyond the dipole: magnetic-dipole and quadrupole lines: infer an input with uncertainty
A magnetic-dipole (M1) or electric-quadrupole (E2) transition is driven by the first correction to the dipole approximation: the wave's magnetic field acting on (μ_B/ħ)(L + 2S), or its field gradient acting on products of the electron's coordinates. Both connect states of equal parity. Use the supplied topic model Y_out = 9.30667-3*(lw+7)+2*log(a), where Y_out is E1 RATE, log₁₀(A / s⁻¹) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; log(x) means log₁₀x. a is Orbital size a, measured in a₀. Keep lw (Cursor wavelength, log₁₀(λ / 1 m)) = -6.3 fixed. For this exercise use only 1.31875 a₀ ≤ a ≤ 3.68125 a₀ 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 = 8.0884488, with a bounded reading uncertainty ±0.037532. 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 0.00023625 a₀, 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.
E1 RATE, log₁₀(A / s⁻¹) against a
Predictions from the supplied Beyond the dipole: magnetic-dipole and quadrupole lines 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): E1 RATE, log₁₀(A / s⁻¹)
Horizontal axis (x): a: Orbital size a / a₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Orbital size a / a₀ | E1 RATE, log₁₀(A / s⁻¹) |
|---|---|
| 1.31875 | 7.446994945283536 |
| 1.909375 | 7.768452463845296 |
| 2.5 | 8.002550017344076 |
| 3.090625 | 8.186762626554547 |
| 3.68125 | 8.338660624262353 |
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(1.31875) = 7.44699 and Y_out(3.68125) = 8.33866. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives a_line = 1.31875 + (8.08845 − 7.44699)(3.68125 − 1.31875)/(8.33866 − 7.44699) = 3.0183 a₀. 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 = 2.75987 a₀; substituting back gives Y_out = 8.0884488.
- The interpolation error in the input is |a_line − a| = 0.25843 a₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.0509168 and 8.1259808. Sort the roots even if the response decreases: 2.64316 a₀ ≤ a ≤ 2.88174 a₀.
- The lower and upper input deviations are 0.116715 a₀ and 0.121869 a₀. 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 a ≈ 2.75987 a₀; compatible interval [2.64316, 2.88174] a₀. Endpoint interpolation gives 3.0183 a₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.