University Quantum Mechanics II · topic UQMII-09.09
Hyperfine levels in a magnetic field, and the hydrogen maser
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.
Hyperfine levels in a magnetic field, and the hydrogen maser: predict a finite change
The Breit–Rabi levels of hydrogen are the exact energies of the 1s electron and proton spins in a uniform field: two levels linear in B for m_F = ±1, and a repelling pair −A/4 ± (A/2)√(1 + x²) for m_F = 0. Use the supplied topic model Y_out = 1420.406*(b/50.607)/(50.607*sqrt(1+(b/50.607)^2)), where Y_out is SLOPE df₀₀/dB (MHz mT⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. b is Magnetic field B, measured in mT. Keep no other adjustable input fixed. For this exercise use only 15.9375 mT ≤ b ≤ 134.063 mT 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 = 41.925 mT and 58.4625 mT. 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.
SLOPE df₀₀/dB against b
Predictions from the supplied Hyperfine levels in a magnetic field, and the hydrogen maser 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): SLOPE df₀₀/dB / MHz mT⁻¹
Horizontal axis (x): b: Magnetic field B / mT
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Magnetic field B / mT | SLOPE df₀₀/dB / MHz mT⁻¹ |
|---|---|
| 15.9375 | 8.430965691805858 |
| 45.46875 | 18.7584058286315 |
| 75 | 23.266201376277497 |
| 104.53125 | 25.262520234607404 |
| 134.0625 | 26.258766348291832 |
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 = 41.925 mT, substitute into Y_out = 1420.406*(b/50.607)/(50.607*sqrt(1+(b/50.607)^2)): Y₀ = 17.9058 MHz mT⁻¹.
- At b = 58.4625 mT, the same fixed-parameter model gives Y₁ = 21.2211 MHz mT⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 3.31524 MHz mT⁻¹; the input change is Δb = 16.5375 mT.
- Average sensitivity is ΔY_out/Δb = 0.200468 MHz mT⁻¹ per mT. 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 = 17.9058 MHz mT⁻¹, Y₁ = 21.2211 MHz mT⁻¹; ΔY_out = 3.31524 MHz mT⁻¹; average sensitivity = 0.200468 MHz mT⁻¹ per mT.
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.
Hyperfine levels in a magnetic field, and the hydrogen maser: infer an input with uncertainty
The Breit–Rabi levels of hydrogen are the exact energies of the 1s electron and proton spins in a uniform field: two levels linear in B for m_F = ±1, and a repelling pair −A/4 ± (A/2)√(1 + x²) for m_F = 0. Use the supplied topic model Y_out = 1420.406*(b/50.607)/(50.607*sqrt(1+(b/50.607)^2)), where Y_out is SLOPE df₀₀/dB (MHz mT⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. b is Magnetic field B, measured in mT. Keep no other adjustable input fixed. For this exercise use only 15.9375 mT ≤ b ≤ 134.063 mT 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 = 24.330524 MHz mT⁻¹, with a bounded reading uncertainty ±0.28924 MHz mT⁻¹. 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 0.0118125 mT, 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.
SLOPE df₀₀/dB against b
Predictions from the supplied Hyperfine levels in a magnetic field, and the hydrogen maser 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): SLOPE df₀₀/dB / MHz mT⁻¹
Horizontal axis (x): b: Magnetic field B / mT
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| b: Magnetic field B / mT | SLOPE df₀₀/dB / MHz mT⁻¹ |
|---|---|
| 15.9375 | 8.430965691805858 |
| 45.46875 | 18.7584058286315 |
| 75 | 23.266201376277497 |
| 104.53125 | 25.262520234607404 |
| 134.0625 | 26.258766348291832 |
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(15.9375) = 8.43097 MHz mT⁻¹ and Y_out(134.063) = 26.2588 MHz mT⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives b_line = 15.9375 + (24.3305 − 8.43097)(134.063 − 15.9375)/(26.2588 − 8.43097) = 121.286 mT. 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 = 87.9937 mT; substituting back gives Y_out = 24.330524 MHz mT⁻¹.
- The interpolation error in the input is |b_line − b| = 33.2924 mT. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 24.041284 MHz mT⁻¹ and 24.619764 MHz mT⁻¹. Sort the roots even if the response decreases: 83.9984 mT ≤ b ≤ 92.4448 mT.
- The lower and upper input deviations are 3.99533 mT and 4.45107 mT. 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 ≈ 87.9937 mT; compatible interval [83.9984, 92.4448] mT. Endpoint interpolation gives 121.286 mT.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.