University Quantum Mechanics II · topic UQMII-09.04
The intermediate field: diagonalising the n = 2 matrix
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.
The intermediate field: diagonalising the n = 2 matrix: predict a finite change
In the intermediate field μ_B B is comparable with the fine-structure splitting Δ, so the n = 2 energies come from diagonalising the fine structure and the Zeeman term together, exactly, inside each block of fixed l and mⱼ. Use the supplied topic model Y_out = (delta-(13.996*bmax))/2+sqrt(delta^2-2*delta*(13.996*bmax)/3+(13.996*bmax)^2)/2, where Y_out is UPPER mⱼ = −1/2, RIGHT EDGE (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. bmax is Field range, 0 to B, measured in T. Keep delta (Fine-structure splitting Δ/h) = 10.95 GHz; lamb (Lamb shift L/h of 2S₁/₂) = 0 GHz fixed. For this exercise use only 0.575938 T ≤ bmax ≤ 4.47406 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. Calculate Y_out at bmax = 1.43352 T and 1.97926 T. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δbmax across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
UPPER mⱼ = −1/2, RIGHT EDGE against bmax
Predictions from the supplied The intermediate field: diagonalising the n = 2 matrix 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): UPPER mⱼ = −1/2, RIGHT EDGE / GHz
Horizontal axis (x): bmax: Field range, 0 to B / T
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| bmax: Field range, 0 to B / T | UPPER mⱼ = −1/2, RIGHT EDGE / GHz |
|---|---|
| 0.5759375000000001 | 7.057862575625717 |
| 1.5504687499999998 | 5.021880795002332 |
| 2.525 | 4.4696064148134464 |
| 3.4995312499999995 | 4.230376967772273 |
| 4.4740625 | 4.098437540096157 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of bmax separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At bmax = 1.43352 T, substitute into Y_out = (delta-(13.996*bmax))/2+sqrt(delta^2-2*delta*(13.996*bmax)/3+(13.996*bmax)^2)/2: Y₀ = 5.13838 GHz.
- At bmax = 1.97926 T, the same fixed-parameter model gives Y₁ = 4.71101 GHz.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.427368 GHz; the input change is Δbmax = 0.545737 T.
- Average sensitivity is ΔY_out/Δbmax = -0.783101 GHz per T. 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 = 5.13838 GHz, Y₁ = 4.71101 GHz; ΔY_out = -0.427368 GHz; average sensitivity = -0.783101 GHz per T.
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.
The intermediate field: diagonalising the n = 2 matrix: infer an input with uncertainty
In the intermediate field μ_B B is comparable with the fine-structure splitting Δ, so the n = 2 energies come from diagonalising the fine structure and the Zeeman term together, exactly, inside each block of fixed l and mⱼ. Use the supplied topic model Y_out = (delta-(13.996*bmax))/2+sqrt(delta^2-2*delta*(13.996*bmax)/3+(13.996*bmax)^2)/2, where Y_out is UPPER mⱼ = −1/2, RIGHT EDGE (GHz). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. bmax is Field range, 0 to B, measured in T. Keep delta (Fine-structure splitting Δ/h) = 10.95 GHz; lamb (Lamb shift L/h of 2S₁/₂) = 0 GHz fixed. For this exercise use only 0.575938 T ≤ bmax ≤ 4.47406 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 = 4.3441434 GHz, with a bounded reading uncertainty ±0.036856 GHz. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown bmax by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000389812 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.
UPPER mⱼ = −1/2, RIGHT EDGE against bmax
Predictions from the supplied The intermediate field: diagonalising the n = 2 matrix 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): UPPER mⱼ = −1/2, RIGHT EDGE / GHz
Horizontal axis (x): bmax: Field range, 0 to B / T
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| bmax: Field range, 0 to B / T | UPPER mⱼ = −1/2, RIGHT EDGE / GHz |
|---|---|
| 0.5759375000000001 | 7.057862575625717 |
| 1.5504687499999998 | 5.021880795002332 |
| 2.525 | 4.4696064148134464 |
| 3.4995312499999995 | 4.230376967772273 |
| 4.4740625 | 4.098437540096157 |
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(bmax) = 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(0.575938) = 7.05786 GHz and Y_out(4.47406) = 4.09844 GHz. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives bmax_line = 0.575938 + (4.34414 − 7.05786)(4.47406 − 0.575938)/(4.09844 − 7.05786) = 4.15042 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 bmax = 2.95379 T; substituting back gives Y_out = 4.3441434 GHz.
- The interpolation error in the input is |bmax_line − bmax| = 1.19663 T. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.3072874 GHz and 4.3809994 GHz. Sort the roots even if the response decreases: 2.81288 T ≤ bmax ≤ 3.11021 T.
- The lower and upper input deviations are 0.140912 T and 0.156419 T. 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 bmax ≈ 2.95379 T; compatible interval [2.81288, 3.11021] T. Endpoint interpolation gives 4.15042 T.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.